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Zeta Function Regularization Nicolas M Robles
Department of Theoretical Physics Imperial College London
Date
Supervior: Dr A Ra jantie
Submitted in partial ful llment of the requirements for the degree of Master of Science in Quantum Fields and Fundamental Forces of Imperial College London
Abstract We review (...)
1
Acknowledgements I should very much like to thank (...)
2
Table of contents
3
Introduction Special functions have arisen constantly in mathematical and theoretical physics throughout the XIX and XX centuries. Indeed, the study of theoretical physics is plagued with special functions: gamma, Dirac, Legendre, Bessel functions - to name a few - are abundant in the indices of most modern treatises on physics. The most common special functions have been the ever-present gamma function and those which are solutions to dierential equations that model physical systems. The Riemann zeta ζ function is de ned for a complex variable
ζ(s) :=
s as
∞ Y X 1 1 = s n 1 − p−s p n=1
for Re(s) > 1 and analytically continued to the whole complex plane except at s = 1 where it has a simple pole with residue 1. Note that the product is taken over all primes p. It is
not a solution to any dierential equation [1] which sets it apart from other special functions which have a more transparent physical meaning. Traditionally, the Riemann zeta function has had its implications in analytic number theory and especially in the distribution of prime numbers. As such it has been regarded mostly as a function that fell completely within the realm of pure mathematics and it was temporarily excluded. This changed in the last half of the XX century when papers from S. Hawking [2], S. Elizalde, S. Odintsov and A. Romeo [3] explained how the zeta regularization assigns nite values to super cially divergent sums. At a more academic level however, zeta regularization is hardly ever mentioned in undergraduate quantum mechanics books, nor is it mentioned either in Peskin and Schroeder or in Weinberg which are some of the standard books on quantum eld theory. It is precisely in QFT where the zeta function becomes apparent. In this MSc dissertation we will explain how the zeta function comes into play in both quantum mechanics and quantum eld theory. In order to achieve this, the dissertation will be divided in four chapters as follows. The rst chapter will be devoted to the study of the gamma and zeta functions. The lack of a course on special functions at Imperial College gives us a welcome opportunity to discuss thoroughly these objects. The treatment will be rigorous de nition-theorem-proof style and the only essential pre-requisites are those of complex analysis: convergence, analytic continuation and residue calculus. The functional equation is of particular importance
ζ(s) = 2(2π)s−1 Γ(1 − s) sin as well as the formulas
1 ζ(0) = − , 2
sπ 2
ζ(1 − s)
1 ζ 0 (0) = − log(2π) 2
which will be used time and again. As we have just stated, zeta regularization consists in assigning nite values to super cially divergent sums and this is precisely what the two above formulas do, since when written formally we have the odd looking sums
1 1 + 1 + 1 + ··· = − , 2
log 1 + log 2 + log 3 + · · · =
1 log(2π). 2
4 Of course, this needs a note of clari cation. The ζ function can be written as
ζ(s) =
∞ ∞ X X 1 (−1)n+1 1 = . ns 1 − 21−s n=1 ns n=1
These two sums agree on, however, when Re(s) > 1, however, the RHS is the step to analytic continuation and it goes − 21 to as s → 0 hence ζ(0) = − 12 in this sense. Indeed, this shows that in a certain sense zeta regularization can be thought of as a technique of complex analysis, namely analytic continuation. Because of the equivalence with other regularizations, this principle can be extended to other renormalizations. We have limited the discussion, however, to the essential aspects of the zeta functions that we will need for the rest of the dissertation and thus avoided the beautiful connection between number theory and the zeta function. The determinant of an operator
A can be written as the in nite product of its eigenvalues
log det A = log
Y
an = tr log A =
n
X
log an
n
When we take the operator to be A := −d2 /dt2 + · · · , the ζ function arises naturally by using
ζA (s) := trA
−s
X 1 = , asn n
X d ζA (s) =− log an , ds s=0 n
and the functional determinant of the operator is 0 det A = exp(−ζA (0)).
Quantum mechanical partition functions will then be computed via zeta regularization, by taking operator to be
AQM = −
d2 + ω2 . dτ 2
Indeed, we shall see that the partition function of a quantum mechanical harmonic oscillator is
ˆ = Z(β) = tr exp(−β H)
Z
ˆ dx < x| exp(−β H)|x >=
1 . 2 sinh(βω/2)
We will explain how the Riemann zeta function is used in the bosonic case, whereas a slightly more exotic zeta function (Hurwitz) is needed for fermions. The partition function of the harmonic oscillator will be our main example. Furthermore, we will make use of formulas proved in chapter 1 showing the necessity of having discussed it at length. Since fermionic theories demand anti-commutation relations we will need Grassmann numbers and therefore we will explain these in detail but only in so much as is needed to compute partition functions for fermions. Finally, equipped with all the tools we have developed in the preceding three chapters we will see how special functions can be used in eld theory. Because of the way special functions arise in eld theory we have devoted a whole chapter to explain the path integral approach to quantum eld theory. This approach is complementary to the QFT/AQFT courses from the MSc.
5 The only eld theory we shall consider is ϕ4 however it is important to note that zeta regularization can be applied to more complex eld theories and even string theory [3]. We shall keep this section brief and avoid topics such as Feynman diagrams whenever possible. We will consider quantum eld theories of a scalar eld in the presence of an external source J. It will be shown how the generating functional [4]
ZE [J] = e−iE[J] = < Ω|e−iHT |Ω > =
Z Dφ exp i d4 x(L[φ] + Jφ)
Z
can be written as
ZE [J] = NE p
[−∂µ
∂µ
exp(SEuc [φ0 ], J) + m2 + V 00 [φ0 ]δ(x1 − x2 )]
and can be computed as done in chapter 2 via zeta function regularization as the determinant of the operator is already visible on the denominator of the LHS. By taking more complex operators, which account for eld theoretical Lagrangians, such as
AFT = −∂ 2 + m2 +
λ 2 φ (x), 2 0
we shall see how zeta regularization is also used in eld theory. We will show that by taking into account rst order quantum corrections, the potential in the bare Lagrangian of the eld theory is renormalized as [5]
V (φcl ) =
λ 4 λ2 φ4cl φcl + 4! 256π 2
log
25 φ2cl − M2 6
where φcl is the classical eld de ned in terms of the source
J and vacuum state Ω as
φcl = < Ω|φ(x)|Ω >J . We will show this rstly by constructing a special zeta function associated with the operator
A. The key steps, for both the quantum mechanical and eld cases, are the connection between the zeta function ζA of the operator and its eigenvalues,
ζ−d2 /dt2 (s) =
−2s ∞ X nπ n=1
β
2s β = ζ(2s), π
∂ as well as the link between the heat function G(x, y, t) - solution of Ax G(x, y, t) = − ∂t G(x, y, t)
- and the zeta function
Z ζA (s)Γ(s) =
∞
dtts−1
Z
d4 xG(x, x, t).
0
We will then discuss how coupling constants evolve in terms of scale dependence and by exploring the analogy between eld theory and statistical mechanics we will compute the partition function of a QFT system. This will be the culmination of the formulas we have proved in chapter 1, the techniques developed in chapter 2 and the theory explained in chapter 3. Furthermore, this encapsulates the spirit of the technique we will use repeatedly. Finally, as a means of a check to make sure the technique yields the same results as other
6 regularization techniques we will derive the same results by using more standard methods such as the one-loop expansion approach. In terms of the knowledge of physics, the only pre-requisite comes from eld theory up to the notes of QFT/AQFT from the MSc in QFFF. The appendix contains some formulas that did not t in the presentation and makes the whole dissertation almost self-contained. References have been provided in each individual chapter, including pages where the main ideas have been explored. Overall, the literature on zeta regularization is sparse. The main sources for this dissertation have been Grosche and Steiner, Kleinert, Ramond and the superb treatise by Elizalde, Odintsov, and Romeo. It has been an objective to try to put results from dierent sources in a new light, by clarifying proofs and creating a coherent sent of examples and applications which are related to each other and which are of increasing complexity. Finally, a few remarks which I have not been able to nd in the literature have been made concerning a potential relationship between zeta functions and families of elementary particles. Dierent zeta functions come into play in quantum physics by computing partition functions. As we have said above, the Riemann zeta function is used in bosonic partition functions, whereas a slightly dierent zeta function is needed for the fermionic partition function. Riemann zeta function ⇔ bosons
Hurwitz zeta function ⇔ fermions
It would be an interesting subject of research to investigate how the zeta function behaves with respect to its corresponding particle, and what kind of knowledge we can extract about the particle given its special zeta function. For instance, a photon has an associated zeta function, and its partition function can be computed by using known facts of this zeta function, on the other hand, known facts about the boson might bring clari cations to properties of this zeta function. The celebrated Riemann hypothesis claims that all the nontrivial zeros of the Riemann zeta function are of the form ζ( 12 + it) with
t real.
Hence, another example would be to consider an operator which brings the zeta function to the form ζ( 21 + it) and investigate the behaviour of zeta function with respect to this operator. This would mean a 'translation' of the Riemann hypothesis and a study of its physical interpretations. REFENCES
The Riemann Hypothesis Stephen Hawking Zeta function regularization of path integrals in curved spacetime Elizalde, Odintsov, and Romeo Zeta Regularization Techniques with Applications Peskin, Michael E. and Schroeder, Daniel V. An Introduction to Quantum Field Theory Ramond, Pierre Field Theory: A Modern Premier
.1. J. Brian Conrey .2. .3. .4. .5.
7
Introduction to Special Functions Often in mathematics, it is more natural to de ne a function in terms of an integral depending on a parameter rather than through power series. The gamma function is one such case. Traditionally, it can be approached as a Weierstrass product or as a parameter-dependent integral. The approach chosen to introduce the gamma function follows from a course in complex analysis such as the one delivered by Eberhard Freitag and Rolf Busam [2]. We adopt the notation s = σ + it which was introduced by Riemann in 1859 and which has become the standard in the literature of the zetafunction. Let us de ne the Euler Gamma function.
De nition 1 The integral (1.1) ∞
Z
dtts−1 e−t
Γ(s) := 0
is well de ned and de nes a holomorphic function in the right half complex plane, where
Re(s) = σ > 0. The rst lemma generalises the factorial function as follows
Lemma 1 For any n ∈ N we have Γ(n) = (n − 1)!. R Proof. Note that Γ(1) := 0∞ dte−t and by integration by parts we have (1.2) Z Γ(s + 1) =
∞ s −t
dtt e
=
−ts e−t |∞ 0
Z +s
0
for any
∞
dtts−1 e−t = sΓ(s)
0
s in the right half-plane. Now, for any positive integer n, we have (1.3) Γ(n) = (n − 1)Γ(n − 1) = (n − 1)!Γ(1) = (n − 1)!
so the result is proved. In order to have a complete view of the gamma function we need to extend it to a meromorphic function in the whole complex plane.
Lemma 2 P∞
n=0
Let cn where n ∈ Z + be a sequence of complex numbers such that the sum
|cn | converges. Furthermore, let S = {−n|n ∈ Z + andcn 6= 0}. Then (1.4) f (s) =
∞ X cn s +n n=0
converges absolutely for s ∈ C −S and uniformly on bounded subsets of C −S . The function
f is a meromorphic function on C with simple poles at the points in S and the residues are by given ress=−n f (s) = cn by for any −n ∈ S .
Proof.
Let us start by nding upper bounds. If |s| < R, then |s + n| ≥ |n − R| for all
n ≥ R. Therefore, we have |s + n|−1 ≤ (n − R)−1 for |s| < R and n ≥ R. From this we can
8 deduce that for n0 > R we have
∞ ∞ ∞ ∞ X X X X cn |cn | 1 |cn | ≤ ≤ |cn |. ≤ n=n s + n n=n |s + n| n=n n − R n − R n=n 0
As such the series
0
0
0
converges absolutely and uniformly on the disk |s| < R P∞ n=0 cn /(s + n) is a meromorphic
P
n>R cn /(s + n)
and de nes there a holomorphic function. It follows that
function on that disk with simple poles at the points of S on the disk |s| < R. Thus, P∞ n=0 cn /(s + n) is a meromorphic function with simple poles at the points in S and for
any −n ∈ S we can write
f (s) =
cn + s+n
X −k∈S−{n}
ck cn = + g(s), s+k s+n
where g is holomorphic at −n. From this we see that residues are indeed ress=−n f (s) = cn . This concludes the proof. Equipped with this lemma we are in a position to extend the Γ function as we wanted.
Theorem 1
The Γ function extends to a meromorphic function on the complex plane.
It has simple poles at 0, −1, −2, −3, · · · . The residues of Γ at are given by (1.7)
res Γ(s) =
s=−k
(−1)k , k!
for any k ∈ Z + .
Proof.
Let us split the gamma function as (1.8)
Z
1 s−1 −t
Γ(s) =
dtt
e
∞
Z
dtts−1 e−t ,
+
0
1
the second integral converges for any complex
s and it is an entire function. Let us expand
the exponential function in the rst integral
Z
1 s−1 −t
dtt 0
e
Z =
1 s−1
dtt 0
∞ X (−1)k
k!
k=0
k
t =
Z ∞ X (−1)k k=0
k!
1
dttk+s−1 =
0
∞ X (−1)k k=0
k!
1 , s+k
these operations are valid for s ∈ C as the exponential function is entire and converges uniformly on compact sets of the complex plane. The gamma function can now be written in a form where Lemma 2 can be used, i.e. (1.9)
Z Γ(s) = 1
for any
∞
dtts−1 e−t +
∞ X (−1)k k=0
k!
1 , s+k
s in the right half-plane. By Lemma 2, the RHS de nes a meromorphic function on
the complex plane with simple poles at 0, −1, −2, −3, · · · . The residues are given a direct application of the lemma.
9
Theorem 2 For s ∈ C any we have Γ(s + 1) = sΓ(s). Proof. This follows directly from Lemma 1 and Theorem 1. We three-dimensional representation of the gamma function looks like (Wikipedia) Another important function related to the Γ function, and also discovered by Euler, is the Beta function which we proceed to develop as follows. Let Re(p), Re(q) > 0 and in the integral that de nes the gamma function, make the change of variable t = u2 to obtain ∞
Z
dttp−1 e−t = 2
Γ(p) =
Z
0
∞
2
duu2p−1 e−u .
0
In an analogous form we have ∞
Z
dvv 2q−1 e−v
Γ(q) = 2
2
0
Multiplying these two together we have
Z
∞
∞
Z
0
2
dudve−(u
Γ(p)Γ(q) = 4
+v 2 ) 2p−1 2q−1
u
v
,
0
and switching to polar coordinates u = r cos θ, v = r sin θ, dudv = rdrdθ
Z∞ Zπ/2 2 Γ(p)Γ(q) = 4 drdθ e−r r2(p+q)−1 cos2p−1 θ sin2q−1 θ 0
Z∞
= 2
0
π/2 Z Zπ/2 2q−1 −r 2 2(p+q)−1 2p−1 dr e r dθ cos θ sin θ = 2Γ(p+q) dθ cos2p−1 θ sin2q−1 θ. 2
0
0
0
The integral can be simpli ed by setting z = sin2 θ
Zπ/2 Z1 2q−1 2p−1 2 dθ cos θ sin θ = dz z q−1 (1 − z)p−1 . 0
0
Next we de ne
Z1 B(p, q) :=
dz z p−1 (1 − z)q−1
0
for Re(p), Re(q) > 0, and this gives the identity
B(p, q) = B(q, p) = We denote by
Γ(p)Γ(q) . Γ(p + q)
B the Beta function. Moreover, if 0 < x < 1 we have
Γ(x)Γ(1 − x) Γ(x)Γ(1 − x) = = B(x, 1 − x) = Γ(1)
Z1 0
dz z x−1 (1 − z)−x .
10 We will evaluate this integral with an appropriate contour, but rst we need to make one last change z = u/(u + 1) which yields
Z1 dz z
x−1
−x
(1 − z)
0
Z∞ =
ux−1 du (u + 1)2 (u + 1)x−1
u 1− u+1
Z∞
−x =
du
ux−1 . 1+u
0
0
Lemma 3 For 0 < y < 1 we have (1.10) Z∞ du
π u−y = . 1+u sin πy
0
Proof.
Let us use a keyhole contour, which is accomplished by cutting the complex plane
along the positive real axis. On this region we de ne the function
s−y 1+s
f (s) =
with argument of s−y equal to 0 on the upper side of the cut. Furthermore, the function has a rst order pole at s = −1 with residue e
−iyπ
f
. See Figure 1 below.
Figure 1 We are now to integrate this function along the path described in Figure 1: the path goes along the upper side of the cut from > 0 to R, then along the circle CR of radius R centred at the origin, then along the side of the cut from
R to and at the end around the origin
via the circle C of radius also centred at the origin. An application of the Cauchy residue theorem gives
2πie−πiy =
ZR du ε
u−y + 1+u
I dz
z −y − e−2πiy 1+z
ZR du ε
CR
u−y − 1+u
I dz
z −y . 1+z
Cε
We can get rid of the integrals around the arcs by appropriate estimates. Note that for we have the following
|z −y | = |e−y log z | = e−yRe(log z) = e−y log |z| = |z|−y , −y −y z |z|−y ≤ |z| ≤ , 1 + z |1 + z| |1 − |z|| and the integrals can be estimated
I 1−y −y z ≤ 2π R dz → 0, 1 + z R − 1 R→∞ CR
I 1−y −y z dz ≤ 2π ε → 0; 1 + z 1 − ε ε→0 Cε
11 so that we are left with
(1 − e
−2πiy
Z∞ )
du
u−y = 2πie−πiy . 1+u
0
We may re-write this to obtain the nal result (1.11)
(e
πiy
−e
−πiy
Z∞ )
Z∞
u−y du = 2πi 1+u
⇒
du
0
u−y π = . 1+u sin πy
0
This proves the claim of the lemma. Let us now make the concluding remarks.
Theorem 3 (Euler Re ection Formula) For all s ∈ C Γ(s)Γ(1 − s) =
Proof.
one has
π . sin πs
The lemma we have just proved can be written as
Γ(x)Γ(1 − x) =
π π = , sin π(1 − x) sin πx
for 0 < x < 1. However, both sides of the equation above are meromorphic, hence we have proved the theorem.
Corollary 1 One has Γ( 12 ) = Proof.
R∞
dt t−1/2 e−t =
R∞
2
dt e−t =
√
π.
−∞
0
This follows by substituting s = 1/2 in the Euler re ection formula.
Theorem 4 The Γ function has no zeroes. Proof. Since s → sin(πs) is an entire function,
the RHS of Theorem 3 has no zeroes,
therefore Γ(s) = 0 only happens where s → Γ(1 − s) has poles. However, as we have argued before, the poles of Γ are at 0, −1, −2, −3, · · · so it follows that must have poles at
1, 2, 3, · · · . By the factorial formula, Γ(n + 1) = n! 6= 0 and so Γ has no zeroes. Intrinsically connected to the Γ function is the Euler γ constant. Let us rst de ne it and prove its existence.
Lemma 4
If then sn = 1 +
1 2
+ ··· +
1 n
− log n, exists. This limit is called the Euler
γ constant.
Proof.
Consider tn = 1 +
1 2
+···+
1 n−1
− log n geometrically, it represents the area of the Rn dx x−1 . Therefore 1
n − 1 regions between the upper Riemann sum and the exact value of tn increases with n. We can write tn =
n−1 X k=1
1 k+1 − log k k
lim n→∞ tn
=
∞ X 1 k=1
1 − log 1 + . k k
12 The series on the right converges to a positive constant since
0<
1 1 1 1 1 1 − log 1 + = 2 − 3 + 4 − ··· ≤ 2, k k 2k 3k 4k 2k
next, the following hold
sn+1 − sn =
1 1 − log 1 + , n+1 n
tn+1 − tn =
1 1 − log 1 + n n
which means that
1 1 1 < log 1 + < ⇒ sn+1 − sn < 0 < tn+1 − tn . n+1 n n Convergence now follows because sn decreases monotonically whereas tn increase monotonically and the dierences are negative. Hence sn is monotonically decreasing and bounded below thus convergent.
Lemma 5
Weierstrass product of the Γ function (1.12) n Y 1 s −s/k sγ =lim se 1 + e . Γ(s) n→∞ k k=1
Proof.
Using the fact that (1 − t/n)n → e−t as n → ∞ it can be shown that
Γ(s)
=lim n→∞
Zn dt t
s−1
t 1− n
n
=lim n→∞
1 nn
Zn
dt ts−1 (n − t)n
0
0
and integrating by parts yields
Γ(s)
=lim n→∞
1 n nn s
Zn
s
n−1
dt t (n − t)
=lim n→∞
n(n − 1) · · · 1 1 nn s(s + 1) · · · (s + n − 1)
0
Zn
dt ts+n−1
0
=lim n→∞
ns s
1 s+1
2 s+2
···
n s+n
.
Inverting both sides n Y 1 s s lim s −s −s =lim sn (s + 1) 1 + · · · 1 + = sn 1+ n→∞ n→∞ Γ(s) 2 n k k=1
In order to be limit we need to insert the convergence factor e−s/k to obtain n Y 1 s −s/k −s s(1+1/2+···+1/n) =lim sn e 1 + e Γ(s) n→∞ k k=1
" =lim n→∞
e
s(1+1/2+···+1/n−log n)
s
n Y k=1
# s −s/k 1+ e . k
13 However by the use of Lemma 4, we know that the sum converges γ to so that we have shown the result (1.12). The Hurwitz zeta function ζ(s, a) is initially de ned for σ > 1 by the series (1.13)
ζ(s, a) :=
∞ X
1 . (n + a)s n=0
This is provided that n + a 6= 0. The reason why we work with a generalized zeta function, rather than with the zeta function itself, is because fermions require this special kind of zeta function for their regularization. Note however that bosons require the Riemann zeta function (a = 1). For the special case of the Riemann zeta function, the 3-d plot looks like (mathworld.com). The discussion presented here of the properties of the Riemann zeta function has its foundations in Titchmarsh [3]. Although the roots of the functional equation go back to Riemann, the development of the Hurwitz zeta can be traced back to Apostol [1] which in turn is taken from Ingham. Figure Let us now examine the properties of the Hurwitz zeta function.
Proposition 1
The series ζ(s, a) for converges absolutely for σ > 1. The convergence
is uniform in every half-plane σ ≥ 1 + δ with δ > 0 so ζ(s, a) is analytic function of
s in the
half-plane σ > 1.
Proof.
From the inequalities ∞ ∞ ∞ X X X (n + a)−s = (n + a)−σ ≤ (n + a)−(1+δ) . n=1
n=1
n=1
all the statements follow and this proves the claim. The analytic continuation of the zeta function to a meromorphic function in the complex plane is more complicated that in the case of the gamma function.
Proposition 2 For σ > 1 we have the integral representation (1.14) Z∞ Γ(s)ζ(s, a) =
dx
xs−1 e−ax . 1 − e−x
0
In the case of the Riemann zeta function, that is when a = 1 we have (1.15)
Z∞ Γ(s)ζ(s) =
dx
xs−1 e−x . 1 − e−x
0
Proof.
First we consider the case when
s is real and s > 1, then extend the result to
14 complex
s by analytic continuation. In the integral for the gamma function we make the
change of variable x = (n + a)t where n ≥ 0 and this yields s
Z∞
Γ(s) = (n + a)
dte−(n+a)s ts−1
0
which can be re-arranged to
(n + a)
−s
Z∞ Γ(s) =
dte−nt e−at ts−1 .
0
Next, we sum over all n ≥ 0 and this gives ∞
ζ(s, a)Γ(s) =
∞ Z X
dte−nt e−at ts−1 ,
n=0 0
where the series on the right is convergent if Re(s) > 1. To nish the proof we need to interchange the sum and the integral signs. This interchange is valid by the theory of Lebesgue integration; however we do not proceed to prove this more rigorously because it would take us too far from the subject at matter. Therefore, we may write ∞
ζ(s, a)Γ(s) =
∞ Z X
dte
−nt −at s−1
e
t
Z∞ =
n=0 0
dt
∞ X
e−nt e−at ts−1 .
n=0
0
However, if Im(s) = t > 0 we have 0 < e−t < 1 and therefore we may sum ∞ X
e−nt =
n=0
1 , 1 − e−t
by geometric summation. Therefore the integrand becomes
Z∞ ζ(s, a)Γ(s) =
dt 0
∞ X
−nt −at s−1
e
e
t
n=0
Z∞ =
dt
e−at ts−1 . 1 − e−t
0
Now we have the rst part of the argument and we need to extend this to all complex
s
with Re(s) > 1. To this end, note that both members are analytic for Re(s) > 1. In order to show that the right member is analytic we assume 1 + δ ≤ σ ≤ c where c > 1 and δ > 0. We then have
Z∞ 0
1 −at s−1 Z∞ Z Z∞ −at σ−1 e t e−at tσ−1 ≤ dt e t dt = dt + dt . −t −t 1−e 1−e 1 − e−t 0
0
1
Notice the analogy of splitting the integral as in the proof of Theorem 1. If 0 ≤ t ≤ 1 we have tσ−1 ≤ tδ and if t ≥ 1 we have tσ−1 ≤ tc−1 . Also since et − 1 ≥ t for
15
t ≥ 0 we then have Z1
and
e−at tσ−1 ≤ dt 1 − e−t
Z1
0
0
Z∞
Z∞
e−at tσ−1 ≤ dt 1 − e−t
e(1−a)t tδ dt t ≤ e(1−a) e −1
dt tδ−1 =
e1−a , δ
0
e−at tc−1 dt ≤ 1 − e−t
1
1
Z1
Z∞ dt
e−at tc−1 = ζ(c, a)Γ(c). 1 − e−t
0
This proves that the integral in the statement of the theorem converges uniformly in every strip 1 + δ ≤ σ ≤ c, where δ > 0, and therefore represents an analytic function in every such strip, hence also in the half-plane σ = Re(s) > 1. Therefore, by analytic continuation, (1.14) holds for all
s with Re(s) > 1.
Consider the keyhole contour
C : a loop around the negative real axis as show in Figure
2. The loop is made of three parts C1 , C2 , and C3 . The C2 part is a positively oriented circle of radius ε < 2π above the origin, and C1 and C3 are the lower and upper edges of a cut in the z -plane along the negative real axis. Figure 2 This can be translated into the following parametrizations: z = re−πi on C1 and z = reπi on C3 where
r varies from ε to ∞.
Proposition 3 If 0 < a ≤ 1 the function de ned by the contour integral (1.16) 1 I(s, a) = 2πi
Z dz
z s−1 eaz 1 − ez
C
is an entire function of s. Moreover, we have (1.17)
ζ(s, a) = Γ(1 − s)I(s, a) if Re(s) = σ > 1.
Proof.
Write z s = rs e−πis on C1 and z s = rs eπis on C3 . Let us consider an arbitrary
compact disk |s| ≤ M and we proceed to prove that the integrals along C1 and C3 converge
uniformly on every such disk. Since the integrand is an entire function of s, this will prove that the integral I(s, a) is entire. Along C1 we have for r ≥ 1,
|z s−1 | = rσ−1 |e−πi(σ−1+it) | = rσ−1 eπt ≤ rM −1 eπM since |s| ≤ M . The same on C3 gives
|z s−1 | = rσ−1 |eπi(σ−1+it) | = rσ−1 e−πt ≤ rM −1 eπM also for r ≥ 1. Therefore, independently on which side of the cut we place ourselves, we
16 have that for r ≥ 1,
s−1 az z e rM −1 eπM e−ar rM −1 eπM e(1−a)r ≤ = . 1 − ez 1 − e−r er − 1 However, er − 1 > er /2 when r > log 2 so the integrand is bounded by ArM −1 e−ar where A R∞ is a constant depending on M but not on r. The integral ε dr rM −1 e−ar converges if ε > 0 so this proves that the integrals along C1 and C3 converge uniformly on every compact disk
|s| ≤ M and hence I(s, a) is indeed an entire function of s.
To prove the equation of the theorem, we have to split up the integral as
Z
Z
2πiI(s, a) =
dz +
C1
Z dz +
C2
dz z s−1 g(z)
C3
where g(z) = eaz /(1 − ez ). According to the parametrizations we have on C1 and C3 that
g(z) = g(−r) but on the circle C2 we write z = εeiθ , where −π ≤ θ ≤ π . This gives us Zε 2πiI(s, a) =
dr r
s−1 −iπs
e
Zπ g(−r) + i
∞
s−1 (s−1)iθ
dθ ε
e
iθ
iθ
Z∞
εe g(εe ) +
−π
dr rs−1 eiπs g(−r).
ε
Divide by 2i and names the integrals I1 and I2
πI(s, a) = sin(πs)I1 (s, ε) + I2 (s, ε). If we let ε → 0 we see that lim ε→0 I1 (s, ε)
Z∞ =
dr
rs−1 e−ar = Γ(s)ζ(s, a), 1 − e−r
0
as long as σ > 1. In |z| < 2π the function
g is analytic except for a rst order pole at
z = 0. Therefore zg(z) is analytic everywhere inside |z| < 2π and hence is bounded there, say |g(z)| ≤ Ω/|z|, where |z| = ε < 2π and Ω is a constant. We can then write
εσ |I2 (s, ε)| ≤ 2
Zπ
dθ
e−tθ
Ω ε
≤ Ωeπ|t| εσ−1 .
−π
When we let ε → 0 and provided that σ > 1 we nd that I2 (s, ε) → 0 hence we have
πI(s, a) = sin(πs)Γ(s)ζ(s, a). Finally, by the use of the formula that we proved earlier Γ(s)Γ(1 − s) = π/ sin(πs) we have a proof of (1.17). Now we have to extend this result for complex numbers such that σ ≤ 1. In the statement that we have just proved the functions I(s, a) and Γ(1 − s) make sense for every complex s, and thus we can use this equation to
de ne ζ(s, a) for σ ≤ 1.
17
De nition 2 If σ ≤ 1 we de ne ζ(s, a) by the equation (1.18) ζ(s, a) = I(s, a)Γ(1 − s). This provides the analytic continuation of ζ(s, a) in the entire
Theorem 5
s plane.
The function ζ(s, a) de ned above is analytic for all
s except for a simple
pole at s = 1 with residue 1.
Proof.
The function I(s, a) is entire so the only possible singularities of ζ(s, a) must be
the poles of Γ(1 − s), and we have shown those to be the points s = 1, 2, 3, · · · . However Theorem 1 shows that ζ(s, a) is analytic at s = 2, 3, · · · so s = 1 is the only possible pole of
ζ(s, a).
If s in an integer s = n the integrand in the contour integral for I(s, a) takes the same value on both C1 and C3 and hence the integrals along C1 and C3 cancel, yielding
1 I(n, a) = 2πi
Z dz
z n−1 eaz z n−1 eaz = res . z=0 1 − ez 1 − ez
C2
In this case we have s = 1 and so
eaz zeaz z −1 lim = =lim =lim = −1. z→0 z→0 z→0 z z z z=0 1 − e 1−e 1−e ez
I(1, a) = res
Finally, the residue of ζ(s, a) at s = 1 is computed as lim s→1 (s
lim − 1)ζ(s, a) = −lim s→1 (1 − s)Γ(1 − s)I(s, a) = −I(1, a)s→1 Γ(2 − s) = Γ(1) = 1,
now the claim is complete: ζ(s, a) has a simple pole at s = 1 with residue 1. Let us remark that since ζ(s, a) is analytic at s = 2, 3, · · · and Γ(1 − s) has poles at these points, then (2) implies that I(s, a) vanishes at these points. Also we have proved that the Riemann zeta function ζ(s) is analytic everywhere except for a simple pole at s = 1 with residue 1.
Lemma 6
Let S(r) designate the region that remains when we remove from the
z plane
all open circular disks of radius r, 0 < r < π with centres at z = 2nπi, n = 0, ±1, ±2, · · · . Then if 0 < a ≤ 1 the function
g(s) :=
eas 1 − es
is bounded in S(r).
Proof.
With our usual notation: s = σ + it we consider the rectangle H(r) with the circle
at n = 0 this rectangle has an indentation as follows
H(r) = {s : |σ| ≤ 1, |t| ≤ π, |s| ≥ r} , as shown in Figure 3 below.
18 Figure 3
H so de ned is compact so g is bounded on H. Also, because of the periodicity |g(s + 2πi)| = |g(s)|, g is bounded in the perforated in nite strip The set
{s : |σ| ≤ 1, |s − 2nπi| ≥ r, n = 0, ±1, ±2, · · ·} . Let us suppose that |σ| ≥ 1 and consider
as aσ e eaσ = e ≤ . |g(s)| = s s 1−e |1 − e | |1 − eσ | We can examine the numerator and denominator for σ ≥ 1 giving |1 − eσ | = eσ − 1 and
eaσ ≤ eσ so |g(s)| ≤
eσ 1 e 1 ≤ = = . eσ − 1 1 − e−σ 1 − e−1 e−1
A similar argument when σ ≤ −1 gives |1 − eσ | = 1 − eσ and so
|g(s)| ≤
1 1 e eσ ≤ ≤ = . 1 − eσ 1 − eσ 1 − e−1 e−1
And therefore, as we claimed |g(s)| ≤ e/(e − 1) for |σ| ≤ 1.
De nition 3 The periodic zeta function is de ned as F (x, s) := where
∞ X e2πinx , ns n=1
x is real and σ > 1.
Let us remark the following properties of the periodic zeta function. It is indeed a periodic function of
Theorem 6
x with period 1 and F (1, s) = ζ(s).
The series converges absolutely if σ > 1. If
also converges (conditionally) for σ > 0.
Proof.
This is because for each xed non-integral
x is not an integer the series
x the coe cients have bounded partial
sums.
Theorem 7 (Hurwitz's formula) If 0 < a ≤ 1 and σ > 1 we have (1.19) ζ(1 − s) =
Γ(s) −πis/2 (e F (a, s) + eπis/2 F (−a, s)). (2π)s
If a 6= 1 this is also valid for σ > 0.
Proof.
Consider the function de ned by the contour integral (1.20)
IN (s, a) :=
1 2πi
I dz C(N )
z s−1 eaz , 1 − ez
19 where is the contour show in Figure 4 and
N is an integer. It is the same keyhole contour
as that of the gamma function, only that it has been rotated for convenience. The poles are located on the y -axis, symmetric to the origin, at multiplies of 2iπ Figure 4 Let us rst prove that if σ < 0 then
lim N →∞ IN (s, a)
= I(s, a). The method to prove this
is to show that the integral along the outer circle tends to 0 as N → ∞. On the outer circle we have z = R eiθ , −π ≤ θ ≤ π , hence
|z s−1 | = |Rs−1 eiθ(s−1) | = Rσ−1 e−tθ ≤ Rσ−1 eπ|t| . The out circle is inside the domain S(r) described in the lemma, the integrand is bounded
A is the bound for |g(s)| implied by the lemma, hence the whole integral is bounded by 2πeπ|t| Rσ which tends to 0 as R tends to in nity as long as σ < 0. Now when we replace s by 1 − s this yields
by Aeπ|t| Rσ−1 where
lim N →∞ IN (1
− s, a) =lim N →∞
1 2πi
I dz
z −s eaz = I(1 − s, a) 1 − ez
C(N )
for σ > 1. We are left with the problem of computing IN (1 − s, a) which we proceed to do by the use of the Cauchy residue theorem. Formally, n=N X
IN (1 − s, a) = −
res f (z) = −
n=−N,n6=0
z=n
n=N X n=1
res f (z) + res f (z)
z=n
z=−n
where f (z) is the integrand of IN (1 − s, a) and residues are calculated as follows
res f (z) =
z=n
res
z=2nπi
z −s eaz 1 − ez
=lim z→2nπi (z−2nπi)
e2nπia lim e2nπia z −s eaz z − 2nπi = = − , z→2nπi 1 − ez (2nπi)s 1 − ez (2nπi)s
which in turn gives (1.20)
IN (1 − s, a) =
N N X e2nπia X e−2nπia + . (2nπi)s n=1 (2nπi)s n=1
Now we make the following replacements i−s = e−πis/2 and (−i)−s = eπis/2 which allow us to write
IN (1 − s, a) =
N N e−πis/2 X e2nπia eπis/2 X e−2nπia + (2π)s n=1 ns (2π)s n=1 (2nπi)s
and we let N → ∞
I(1 − s, a) =
e−πis/2 eπis/2 F (a, s) + F (−a, s). s (2π) (2π)s
20 We have thus arrived at the following result (1.21)
ζ(1 − s, a) = Γ(s)I(1 − s, a) =
o Γ(s) n −πis/2 e F (a, s) + eπis/2 F (−a, s) . s (2π)
This proves the claim. The simplest particular case (and the most important one) is when we take a = 1 this gives us the functional equation of the Riemann zeta function (1.22)
ζ(1 − s) =
o Γ(s) n −πis/2 Γ(s) πs πis/2 e ζ(s) + e ζ(s) = 2 cos ζ(s). (2π)s (2π)s 2
This is valid for σ > 1 but it also holds for all formulation can be obtained by switching
s by analytic continuation. Another useful
s with 1 − s (1.23)
ζ(s) = 2(2π)s−1 Γ(1 − s) sin
πs ζ(1 − s). 2
Let us now see the consequences of this equation. Taking s = 2n + 1 in (1.22) when
n in
an integer the cosine factor vanishes and we nd the trivial zeroes of ζ(s) (1.24)
n = 1, 2, 3, · · · .
ζ(−2n) = 0
Later we will need to use certain other values of the Riemann zeta function which we can now compute. In particular the value of ζ(−n, a) can be calculated if
n is a non-negative
integer. Taking s = −n in the formula ζ(−s, a) = Γ(1 − s)I(s, a) we nd that
ζ(−n, a) = Γ(1 + n)I(−n, a) = n!I(−n, a), where
I(−n, a) = res
z=0
z −n−1 eaz 1 − ez
,
the evaluation of this residue requires special functions of its own (special type of polynomials, rather) which are known as the Bernoulli polynomials.
De nition 4 For any complex s
we de ne the functions Bn (s) as ∞ X Bn (s) n zesz = z ez − 1 n=0 n!
provided that |z| < 2π . A particular case of the polynomials are the Bernoulli number Bn = Bn (0) i.e. ∞ X z Bn (0) n = z . ez − 1 n=0 n!
Lemma 7 One has the following equations Bn (s) =
Pn
n n−k . k=0 (k ) Bk s
In particular when
21
s = 1 this yields Bn =
Proof.
Pn
n k=0 (k ) Bk .
Using a Taylor expansion and comparing coe cients on both sides we have ∞ X Bn (s) n z z = z esz = n! e − 1 n=0
∞ X Bn n z n! n=0
!
∞ X sn n z n! n=0
! ,
n
Bn (s) X Bk sn−k = n! k! (n − k)! k=0
and by passing n! to the RHS we obtain the lemma. Now we can write the values of the zeta function in terms of Bernoulli numbers.
Lemma 8 For every integer n ≥ 0 we have (1.25) ζ(−n, a) = −
Proof.
Bn+1 (a) . n+1
This follows from the previous observation that ζ(−n, a) = n!I(−n, a), so we just
have to evaluate the integral
I(−n, a) = res
z=0
z −n−1 eaz 1 − ez
I by the Cauchy residue theorem.
= − res
z=0
z −n−2 zeaz ez − 1
= − res
z=0
z
−n−2
∞ X Bk (a) k=0
k!
! z
k
=−
Bn+1 (a) , (n + 1)!
and dividing by n! we have the end of the proof.
Lemma 9
The recursion Bn (s + 1) − Bn (s) = nsn−1 is valid for Bernoulli polynomials
if n ≥ 1 and in particular for Bernoulli numbers when n > 1: Bn (0) = Bn (1).
Proof.
From the identity
z it follows that
esz e(s+1)z −z z = zesz z e −1 e −1
∞ ∞ X Bn (s + 1) − Bn (s) n X sn n+1 z = z , n! n! n=0 n=0
and as we did before, we equate coe cients of z n to obtain the rst statement and then set
s = 0 to obtain the second statement. Using the de nition, the rst Bernoulli number is B0 = 1 and the rest can be computed by recursion. We obtain the following values Building on from the Bernoulli numbers we can construct the polynomials by the use of the lemmas, the rst ones as s Bernoulli polynomial Bernoulli number Table
22 Note that for n ≥ 0 we have by setting a = 1 in (1.25) (1.26)
ζ(−n) = −
Bn+1 n+1
and because of the trivial zeroes of the Riemann zeta function ζ(−2n) = 0 we con rm our observation that the odd Bernoulli numbers are zero, i.e. B2n+1 = 0 . Also note (1.27)
1 ζ(0) = − . 2 Finally we can write a compact formula for the even values of the zeta function in terms of Bernoulli numbers.
Theorem 8 Suppose n
is a positive integer, then (1.28)
ζ(2n) = (−1)n+1
Proof.
(2π)2n B2n . 2(2n)!
This follows from the functional equation by setting s = 2n
ζ(1 − 2n) = 2(2π)−2n Γ(2n) cos(πn)ζ(2n) re-arranging we obtain
−
B2n = 2(2π)−2n (2n − 1)!(−1)n ζ(2n), 2n
from which the result follows. Using the tabulated values given above we have the well-know Euler formulas
ζ(2) =
π2 , 6
ζ(4) =
π4 . 90
Let us remark that no such formula for odd values is known and in fact the value of ζ(3) remains one of the most elusive mysteries of modern mathematics. A signi cant advance was achieved by Roger Apery who was able to show in 1979 that is an irrational number.
Theorem 9 One has ζ 0 (0) = − 21 log(2π) Proof. We may write (1.16) and (1.17) as ζ(s) 1 = Γ(1 − s) 2πi
Z C
where
(−z)s−1 1 dz z = e −1 2πi
Z
dz (−z)s z ez − 1
C
C is the same contour as that of Figure 2, except shifted to the positive in nity
instead of negative in nity and to account for this we have (−z)s = exp[s log(−z)]. Let us
23
s and then set s = 1, we obtain
dierentiate with respect to
Z
1 2πi
dz (−z) log(−z) . z ez − 1
C
The integral on the RHS can be split as
1 2πi
Zε
dz (−z)(log z − iπ) 1 + z z e −1 2πi
+∞
dz (−z)(log ε + iθ − iπ) 1 + z z e −1 2πi
I
Z∞
dz (−z)(log z + iπ) z ez − 1
ε
|z|=ε
and writing z = εei(φ+π) in the middle integral we have
Z∞ −
dz log ε − z e −1 2πi
ε
I
dz z 1 − z z e − 1 2πi
Zπ dφ · φ
ez
z , −1
−π
|z|=ε
at this point we need to evaluate all three integrals. The rst one can be expanded as
Z∞ −
dz =− ez − 1
Z∞ dz ε
ε
∞ X
−nz
e
n=1
z=∞ ∞ ∞ X X e−nz (e−ε )n =− = − −n n n=1 n=1 z=ε
ε3 ε ε2 + − · · · = log ε + log 1 − + · · · . = log(1 − e−ε ) = log ε − 2 6 2 The second integral is solved by Cauchy's integral theorem by noting that at z = 0 we have
z(ez − 1)−1 → 1, and therefore we have log ε − 2πi
I
dz z = − log ε. z ez − 1
|z|=ε
Finally, the third integral goes to zero as ε → 0. Putting all these facts together we have shown that
1 2πi
Z
dz (−z) log(−z) = log ε − log ε + O(εα ) = 0. z ez − 1
C
Re-arranging the functional equation (1.23)
ζ(s) πs = 2(2π)s−1 ζ(1 − s) sin Γ(1 − s) 2 and because the derivative at s = 1 is zero, as we have just shown, its logarithmic derivative
log(2π) −
ζ 0 (1 − s) π cos(πs/2) + ζ(1 − s) 2 sin(πs/2)
must also be 0 at s = 1 and consequently (1.29)
log(2π) =
ζ 0 (0) , ζ(0)
nally yielding the result of Theorem 9 by use of (1.27).
24
Theorem 10 One has ζ(0, 12 ) = 0 and ζ 0 (0, 21 ) = − 12 log 2. Proof. We have the following identity ζ(s, 1/2) + ζ(s) = 2s
∞ X n=1
1 1 + = 2s ζ(s), (2n − 1)s (2n)s
from which it follows that (1.30)
ζ(s, 1/2) = (2s − 1)ζ(s) and hence the rst formula is shown. By dierentiating (1.30) with respect to
s we have
the second formula of the theorem. REFERENCES There exist many sources for the development of the Riemann zeta function. The most comprehensive one is by E.C. Titchmarsh (and revised by D.R. Heath-Brown). Here we have followed the course on analytic number theory by Tom Apostol [2] for the presentation of the zeta function and its analytic continuation. The part of the gamma function follows from the course on Complex Analysis by Freitag and Busam and Whitaker [1] and Watson's Modern Analysis [4].
25
Zeta Regularization in Quantum Mechanics Zeta Regularization in Quantum Mechanics One of the rst instances where the Riemann zeta function occurs in quantum physics is in the path integral development of the harmonic oscillator; speci cally it takes place when computing the partition function of the spectrum of the harmonic oscillator. We will follow Exercise 9.2 p 312 from Peskin and Schroeder and the path integral development from Chapter 2 of Kleinert. The action of the one-dimensional harmonic oscillator is given by (2.1)
Ztf S=
dtL, ti
where the Lagrangian is (2.2)
L=
1 1 mx˙ 2 − mω 2 x2 . 2 2
As we know from the functional approach to quantum mechanics, the transition amplitude is the functional integral (2.3)
Z < xf , tf | xi , ti > =
DxeiS[x(t)] .
The extremum of S, xc (t), satis es (2.4)
δS[x] = 0. δx x=xc (t) We now proceed to expand the action around xc (t). This indicates that xc (t) is the classical trajectory connecting both space-time points of the amplitude and therefore it satis es the Euler-Lagrange equation (2.5)
x ¨c + ω 2 xc = 0. The solution to the equation above with conditions xc (ti ) = xi and xc (tf ) = xf is (2.6)
xc (t) = (sin ωT )−1 [xf sin ω(t − ti ) + xi sin ω(tf − t)], where T = tf − ti . We next plug this solution into the action
Sc := S[xc ] =
S (2.7)
mω [(x2 + x2i ) cos ωT − 2xf xi ]. 2 sin ωT f
As we intended originally, we now expand S[x] around x = xc to obtain (2.8)
Z S[xc + z] = S[xc ] +
Z δS[x] δ 2 S[x] 1 dtz(t) dt1 dt2 z(t1 )z(t2 ) + δx(t) x=xc 2! δx(t1 )δx(t2 ) x=xc
26 where z(t) satis es the boundary condition (2.9)
z(ti ) = z(tf ) = 0. The expansion ends at second order because the action is in second order in x. Noting that
δS[x]/δx = 0 at x = xc we are left with the rst and last terms only. Because the expansion is nite (2.10)
1 S[xc + z] = S[xc ] + 2!
Z
δ 2 S[x] dt1 dt2 z(t1 )z(t2 ) δx(t1 )δx(t2 ) x=xc
the problem can be solved analytically. Let us now compute the second order functional derivative in the integrand; this can be accomplished as follows (2.11)
δ δx(t1 )
Ztf
1 1 ˙ 2 − mω 2 x(t)2 dt mx(t) 2 2
= −m
ti
d2 2 + ω x(t1 ). dt21
Next, using the rule (2.12)
δx(t1 ) = δ(t1 − t2 ) δx(t2 ) we obtain the following expression for the second order functional derivative (2.13)
δ 2 S[x] = −m δx(t1 )δx(t2 )
d2 2 + ω δ(t1 − t2 ). dt21
Plugging this back into the equation for the expansion, we can use the delta function to get rid of the
t variables (2.14)
S[xc +z] = S[xc ]−
m 2!
Z dt1 dt2 z(t1 )z(t2 )δ(t1 − t2 )
d2 + ω2 dt21
= S[xc ]+
m 2!
Z
dt(z˙ 2 − ω 2 z 2 ),
where we have simpli ed the expression by using (2.9). A crucial point to be made is that because Dx is invariant, we may replace it by Dz and this will give us (2.15)
< xf , tf | xi , ti > = eiS[xc ]
Z
m Dz exp i 2
Ztf
dt(z˙ 2 − ω 2 z 2 ).
ti
z(ti )=z(tf )=0
The uctuation part (integral at z(0) = z(T ) = 0) (2.16)
Z If := z(0)=z(T )=0
m Dz exp i 2
ZT
dt(z˙ 2 − ω 2 z 2 )
0
is computed as follows. First, let us shift the variables so that time start at 0 and ends at
27
T. We Fourier expand z(t) as (2.17) z(t) =
∞ X
nπt . T
an sin
n=1
This choice satis es the (2.9). The integral in the exponential gives (2.18)
ZT 0
∞ T X 2 nπ 2 2 a −ω . dt(z˙ − ω z ) = 2 n=1 n T 2
2 2
In order to have a well-de ned transformation we must check that the number of variables is the same before and after the transformation. Indeed, the Fourier transformation from
y(t) to an may be thought of as a change of variables in the integration. To check this, take the number of the time slice to be N + 1, including both t = 0 and t = T , for which there exist N − 1 independent zk variables. Therefore, we must set an = 0 for all n > N − 1. Next, we compute the corresponding Jacobian. Denote by tk the k th time slice when the interval [0, T ] is split into
N in nitesimal parts, then (2.19) JN
∂zk nπtk = det . = det sin ∂an T
We evaluate the Jacobian for the easiest possible case, which is that of the free particle. Therefore, let us make a digression to evaluate the above mentioned probability amplitude. For a free particle, the Lagrangian is L =
1 ˙ 2. 2 mx
Carefully derived solutions to the free
particle can be found in Chapter 2 of Kleinert as well as in Chapter 3 of Grosche and Steiner. The amplitude is computed by rst noting that the Hamiltonian is given (2.20)
H = px˙ − L =
p2 , 2m
so that (2.21) ˆ
< xf , tf | xi , ti > = < xf |e−iHT | xi > = Z =
dp ip(xf −xi ) −iT p2 /(2m) e e = 2π
r
Z
ˆ
dp < xf |e−iHT | p >< p| xi >
m exp 2πiT
im(xf − xi )2 2T
,
where T = tf − ti as noted before and ε here denotes the discretization of time ε = T /N . Similarly to the theory of functional integration we have amplitude (2.22)
< xf , tf | xi , ti > =
lim n→∞
" n # m n/2 Z X m xk − xk−1 2 dx1 · · · dxn−1 exp iε . 2πiε 2 ε k=1
We now change the coordinates to zk = (m/(2ε))1/2 xk so that the amplitude becomes (2.23)
< xf , tf | xi , ti > =
lim n→∞
" n # m n/2 2ε (n−1)/2 Z X 2 dz1 · · · dzn−1 exp i (zk − zk−1 ) . 2πiε m k=1
28 In the appendix, we prove by induction that (2.24)
"
Z
dz1 · · · dzn−1 exp i
n X
# (zk − zk−1 )
2
=
k=1
(iπ)n−1 n
1/2
2
ei(zn −z0 )
/n
.
The expression reduces to (2.25)
hxf , tf | xi , ti i =
lim n→∞
r m n/2 2πiε (n−1)/2 1 m im im(xf −x0 )2 /(2nε) √ e = exp (xf − xi )2 . 2πiε m 2πiT 2T n
Taking (2.25) into account we arrive at the (2.26)
< xf , T | xi , 0 > =
1/2
1 2πiT
1/2 im 1 2 exp eiS[xc ] . (xf − xi ) = 2T 2πiT
When we write this in terms of a path integral we obtain (2.27)
Z
eiS[xc ]
m Dz exp i 2
Ztf
dtz˙ 2 .
ti
z(0)=z(T )=0
Now, from (2.18) we have (2.28)
m 2
ZT
dt z˙ 2 → m
0
N X a2n n2 π 2 4T n=1
and when we compare both path integral expressions one has the equality (2.29)
1 2πiT
1/2
Z
m Dz exp i 2
=
Ztf
dtz˙
2
=
lim N →∞ JN
1 2πiε
1/2 Z da1 . . . daN −1 exp im
n=1
ti
z(0)=z(T )=0
N −1 X
a2n n2 π 2 4T
Now comes the process of evaluating the Gaussian integrals (2.30)
1 2πiT
1/2 =
lim N →∞ JN
1 2πiε
N/2 NY −1 n=1
1 n
4πiT π2
1/2 =
lim N →∞ JN
1 2πiε
N/2
1 (N − 1)!
4πiT π2
(N −1)/2
and from this we obtain a formula for the Jacobian (2.31)
JN = 2−(N −1)/2 N −N/2 π N −1 (N − 1)! → ∞. N →∞
The Jacobian is divergent, but this divergence is not relevant because JN is combined with other divergent factors. Let us now return to the original problem of the probability amplitude of a harmonic oscillator. The amplitude was (2.32)
< xf , T | xi , 0 > =
lim N →∞ JN
1 2πiT
N/2 e
iS[xc ]
Z
# N −1 mT X 2 nπ 2 2 −ω . da1 . . . daN −1 exp i a 4 n=1 n T "
! .
29 As we did with the free particle, we carry out the computation of the Gaussian integrals using the formula (2.33)
Z
1/2 " 2 #−1/2 ωT imT 2 nπ 2 4iT 1− dan exp − ω2 . a = 4 n T πn2 nπ
This breaks the amplitude (2.32) into smaller parts (2.34)
hxf , tf | xi , ti i =
lim N →∞ JN
=
lim N →∞
N 2πiT
N/2 e
iS[xc ]
N −1 Y k=1
1 2πiT
1/2 e
iS[xc ]
" " 1/2 # NY 2 #−1/2 −1 1 4iT ωT 1− k π nπ n=1
N −1 Y
"
1−
n=1
ωT nπ
2 #−1/2 .
It can be shown that his product is equal to (2.35) lim N →∞
N Y
"
1−
n=1
ωT nπ
2 # =
sin ωT . ωT
The divergence of Jn cancels the divergence of the other terms and therefore we are left with a nite value con rming what we stated above concerning the irrelevance of J∞ . When we insert the value of the product we arrive at the nal result (2.36)
1/2 1/2 2 ω ω iω iS[xc ] 2 (xf + xi ) cos ωT − 2xi xf . < xf , tf | xi , ti > = e = exp 2πi sin ωT 2πi sin ωT 2 sin ωT
ˆ whose spectrum is bounded from below then, by adding a If we have a Hamiltonian H ˆ positive de nite, i.e. (2.37) positive constant to the Hamiltonian, we can make H ˆ = {0 < E0 ≤ E1 ≤ · · · ≤ En ≤ · · ·} . specH Also we assume that the ground state is not degenerate. The spectral decomposition of ˆ
e−iHt is (2.38) ˆ
e−iHt =
X
e−iEn t |n >< n|
n
ˆ > = and this decomposition is analytic in the lower half-plane of t, where we have H|n En |n >. As we have done when evaluating the Gaussian integrals we introduce the Wick rotation ˆ
ˆ
t = −iτ where τ is real and positive, this gives us x˙ = idx/dτ and e−iHt = e−Hτ so that (2.39)
Ztf i ti
# # 2 2 Zτf " Zτf " 1 1 dx 1 dx dt mx˙ 2 − V (x) = i(−i) dt − m − V (x) = − dt m + V (x) 2 2 dτ 2 dτ
τi
τi
30 Consequently, the path integral becomes (2.40)
ˆ
< xf , tf | xi , ti > = < xf |e−H(τf −τi ) | xi > =
Z
¯ exp − Dx
Zτf
"
1 m dτ 2
dx dτ
#
2
+ V (x) ,
τi
¯ is the integration measure in the imaginary time τ . Equation (2.40) shows the where Dx connection between the functional approach and statistical mechanics.
ˆ as Z(β) = Let us now de ne partition function [see Kleinert, p.77] of a Hamiltonian H ˆ
Tre−β H , where β is a positive constant and the trace is over the Hilbert space associated ˆ . This partition can be written in terms of eigenstates of energy {|En >} with (2.41) with H ˆ = En |En >, H
< Em | En >= δmn .
In this case (2.42)
Z(β) =
X
ˆ
< En |e−β H | En > =
X
< En |e−βEn | En > =
n
n
X
e−βEn ,
n
or in terms of the eigenvector |x > of the position operator x ˆ, (2.43)
Z
ˆ
dx < x|e−β H | x > .
Z(β) =
Initially we had an arbitrary β but if we set it to be β = iT we nd that (2.44) ˆ
ˆ
< xf |e−iHT | xi > = < xf |e−β H | xi >, and from this we have the path integral expression of the partition function, (2.45)
Z Z(β) =
Z dz
¯ exp − Dx
Zβ
dτ
1 mx˙ 2 + V (x) = 2
0
x(0)=x(β)=z
Z periodic
¯ exp − Dx
Zβ
dτ
1 mx˙ 2 + V (x) , 2
0
where the periodic integral indicates that the integral is over all paths which are periodic in the interval [0, β]. When we apply this to the harmonic oscillator, the partition function is simply (2.46) ˆ −β H
Tre
=
∞ X
e−β(n+1/2)ω .
n=0
Although there are many ways of evaluating the partition function, here we choose one where the use of the zeta regularization is illustrated. Proceed as follows. Set imaginary time τ = iT to obtain the path integral (2.47)
Z z(0)=z(T )=0
Z i d2 2 Dz exp dt z − 2 − ω z → 2 dt
Z
Z 1 d2 2 ¯ Dz exp − dτ z − 2 + ω z , 2 dt
z(0)=z(β)=0
¯ indicates the path integration measure with imaginary time. Suppose we in this case Dz
31 have an n × n Hermitian matrix
M with positive-de nite eigenvalues λk where 1 ≤ k ≤ n
then we show in the appendix that (see Rajantie) (2.48) n Y
Z∞
k=1
"
1 dxk exp − 2
# X
−∞
xp Apq xq = π n/2
p,q
n Y
π n/2 1 √ = . det M λk k=1
This is a matrix generalization of the scalar Gaussian integral
Z∞
1 dx exp − λx2 2
r
=
2π , λ
λ > 0.
−∞
The next task is to de ne the determinant of an operator O by the in nite product of its Q eigenvalues λk . This is accomplished by setting DetO = k λk . Note that Det with capital d denotes the determinant of an operator, whereas with a small d it denotes the determinant of a matrix, same applies for Tr and tr. Using this we can we can write the integral over imaginary time as (2.49)
Z
−1/2 Z 2 2 ¯ exp − 1 dτ z − d + ω 2 z = DetD − d + ω 2 Dz 2 dt2 dτ 2
z(0)=z(β)=0
here the
D denotes the Dirichlet boundary condition z(0) = z(β) = 0.
Similarly to what we did with (2.18) we see that the general solution (2.50) ∞ 1 X nπτ z(τ ) = √ zn sin . β β n=1
We are restricted to having the coe cients zn real as
z is a real function. We are now in
a position to write formal expressions. Knowing that the eigenvalues of the eigenfuction
sin(nπτ /β) are λn = (nπ/β)2 + ω 2 we may write the determinant of the operator as (2.51) DetD
d2 − 2 + ω2 dτ
" " # 2 Y 2 # ∞ ∞ ∞ 2 Y Y nπ nπ βω 2 1+ λn = +ω = . = β β mπ n=1 n=1 m=1 n=1 ∞ Y
It is now time to identify the rst in nite product with the functional determinant, i.e. (2.52)
2 ∞ Y d2 nπ DetD − 2 ↔ . dτ β n=1
Here is where the ζ function comes into play. Suppose O is an operator with positive-de ne eigenvalues λn . Following Grosche and Steiner, in this case, we take the log (2.53)
log DetO = log
Y
λn = Tr log O =
n
n
Now we de ne the spectral ζ function as (2.54)
ζO (s) :=
X
X n
λ−s n .
log λn .
32 The sum converges for su ciently large Re(s ) and ζO (s) is analytic in Additionally, it can be analytically continued to the whole
s in this region.
s plane except at a possible
nite number of points. The derivative of the spectral zeta-function is linked to functional determinant by (2.55)
X dζO (s) = − log λn . ds s=0 n
And therefore the expression for DetO is (2.56)
dζO (s) DetO = exp − . ds s=0 The operator we are interested in is O = −d2 /dτ 2 so this yields (2.57)
ζ−d2 /dτ 2 (s) =
−2s ∞ X nπ β
n=1
2s β ζ(2s). π
=
As we proved in Chapter 1, the zeta function is analytic over the whole complex
s plane
except at the simple pole at s = 1. The values
ζ(0) = −
1 2
1 ζ 0 (0) = − log(2π) 2
were also calculated, and we can use them now to obtain (2.58) 0 ζ−d 2 /dτ 2 (0)
β = 2 log ζ(0) + 2ζ 0 (0) = − log(2β). π
Putting this into the expression for the determinant with Dirichlet conditions we have (2.59)
d2 DetD − 2 = elog(2β) = 2β dτ and nally (2.60)
DetD
d2 − 2 + ω2 dτ
= 2β
∞ Y
"
1+
p=1
βω pπ
2 # .
Note how the in nite product now becomes convergent due to ζ(0) and ζ 0 (0)! Let us go back to the partition function (2.61)
Tre
ˆ −β H
=
∞ X
" e
−β(n+1/2)ω
n=0
=
2β
= 2β
∞ Y p=1
π sinh βω βω
−1/2
( 1+
βω pπ
π ω tanh(βω/2)
2 )#−1/2
1/2 =
π ω tanh(βω/2)
1/2
1 . 2 sinh(βω/2)
It is important to note that there is a more direct way of computing this partition function and which is more satisfying for solvable cases such as the harmonic oscillator but which fails under more complicated Lagrangians.
33 This can be done by computing (2.43)
Z
ˆ
Z(β) = Tre−β H =
ˆ
dx < x|e−β H | x >.
Recall that
hxf , tf | xi , ti i =
1/2 2 ω iω exp (xf + x2i ) cos ωT − 2xi xf , 2πi sin ωT 2 sin ωT
so that
Z(β) =
ω 2πi(−i sinh βω) =
ω 2π sinh βω
1/2 Z
1/2
ω 2 2 dx exp i (2x cosh βω − 2x ) −2i sinh βω
π ω tanh βω/2
1/2 =
1 . 2 sinh(βω/2)
The quantisation of bosonic particles is done by using commutation relations, however, the quantisation of fermionic particles require a more dierent approach, namely that of anticommutation relations. This in turn requires anti-commuting numbers which are called Grassmann numbers. In analogy with the bosonic harmonic oscillator which was described by the Hamiltonian, (2.62)
H = 21 (a† a + aa† ), where
a and a† satisfy the commutation relations (2.63) [a, a† ] = 1
[a, a] = [a† , a† ] = 0.
The Hamiltonian has eigenvalues (n+ 21 )ω where n is an integer with eigenvector |n > (2.64)
H|n >= (n + 12 )ω|n > . From now we drop the hat notation in the operators whenever there is no risk for confusion with the eigenvalue. The prescription for the fermionic Hamiltonian is to set (2.65)
H = 12 (c† c − cc† )ω. This may be thought of as a Fourier component of the Dirac Hamiltonian, which describes relativistic fermions. However, it is evident that if
c and c† were to satisfy commutation
relations then the Hamiltonian would be a constant, and therefore it is more appropriate to consider anti-commutation relations (2.65)
c, c† = 1
{c, c} = c† , c† = 0.
In this case the Hamiltonian becomes (2.66)
H=
1 2
† c c − (1 − c† c) ω = N − 21 ω,
34
N are either 0 or 1 since
where N = c† c. The eigenvalues of
N 2 = c† cc† c = N ⇔ N (N − 1) = 0. Next we need a description of the Hilbert space of the Hamiltonian. To this end, let |n > be an eigenvector of
H with eigenvalue n (necessarily n = 0, 1). Then the follow equations
hold (2.67)
H |0i = −
ω |0i 2
H |1i =
ω |1i 2
For the sake of convenience we introduce the spin-notation (2.68)
0
|0i = c† |0i = |1i
1
! ,
|1i =
1
! .
0
c† |1i = 0 c |0i = 0 c |1i = |0i .
When the basis vectors of the space have this form then the operators take the matrix representations (2.69)
c=
0
0
1
0
! †
c =
0
1
0
0
! N=
1
0
0
0
!
ω H= 2
1
0
0
−1
! .
Instead of having the bosonic commutation relation [x, p] = i we now have [x, p] = 0. The anti-commutation relation {c, c† } = 1 is replaced by {θ, θ∗ } = 0 where θ and θ∗ are anticommuting numbers, i.e. Grassmann numbers which we proceed to develop in further detail in the appendix. The Hamiltonian of the fermionic harmonic oscillator is H = (c† c − 1/2)ω, with eigenvalues
±ω/2. From our previous discussion of the partition function and Grassmann numbers we know that (2.105)
Z(β) = Tre−βH =
1 X
< n|e−βH | n > = eβω/2 + e−βω/2 = 2 cosh(βω/2).
n=0
As with the bosonic case, we can evaluate Z(β) in two dierent ways using the path integral formalism. Let us start with some preliminary propositions. Let
H be the Hamiltonian of
a fermionic harmonic oscillator, its partition function is written as (2.106)
Tre
−βH
Z =
∗
dθ∗ dθ < −θ|e−βH | θ > e−θ θ .
We can show this by the inserting (2.104) into the partition function (2.105) this yields (2.107)
Z(β) =
X n=0,1
=
XZ n
< n|e−βH | n > =
XZ n
dθ∗ dθ|θ >< θ|e−θ
∗
θ
< n|e−βH | n > =
XZ
dθ∗ dθe−θ
n
dθ∗ dθ(1 − θ∗ θ) (< n | 0 > + < n | 1 > θ) < 0|e−βH | n > +θ∗ < 1|e−βH | n >
∗
θ
< n | θ >< θ|e−βH | n
35
=
XZ
dθ∗ dθ(1 − θ∗ θ)
n
× < 0|e−βH | n >< n | 0 > −θ∗ θ < 1|e−βH | n >< n | 1 > +θ < 0|e−βH | n >< n | 1 > +θ∗ < 1|e−βH | n >< n | 0 > . Note now that the last term of the integrand does not contribute to the integral and therefore we may substitute θ∗ to −θ∗ which implies that (2.108)
Z(β) =
XZ
dθ∗ dθ(1 − θ∗ θ)
n
× < 0|e−βH | n >< n | 0 > −θ∗ θ < 1|e−βH | n >< n | 1 > +θ < 0|e−βH | n >< n | 1 > −θ∗ < 1|e−βH | n >< n | 0 > Z ∗ = dθ∗ dθe−θ θ < −θ|e−βH | θ > . Unlike the bosonic case, we have to impose an anti-periodic boundary over [0, β] in the trace since the Grassmann variable is θ when τ = 0 and it is −θ when τ = β . By invoking the expression (2.109)
e−βH = lim (1 − βH/N )N N →∞
and inserting the completeness relation (2.104) at each step one has the following expression for the partition function (2.110)
Z Z(β) = lim
N →∞
Z = lim
N →∞
= lim
N →∞
dθ∗ dθ
N −1 Y
" dθk∗ dθk exp − "
dθk∗ dθk
N −1 X
∗
θ
−θ (1 − βH/N )N θ
# θn∗ θn ×h−θ | (1 − εH)| θN −1 i hθN −1 | · · · | θ1 i hθ1 | (1 − εH)| θi
n=1
k=1
Z Y N
dθ∗ dθe−θ
exp −
N X
# θn∗ θn
×hθN | (1 − εH)| θN −1 i hθN −1 | · · · | θ1 i hθ1 | (1 − εH)| − θN i ,
n=1
k=1
where we have the usual conventions and we have been using all along (2.111)
ε = β/N
and θ = −θN = θ0 ,
∗
∗
θ∗ = −θN = −θ0 .
Matrix elements are evaluated (up to rst order) as (2.112)
< θk |H| θk−1 > < θk |(1 − εH)| θk−1 > = < θk | θk−1 > 1 − ε < θk | θk−1 > ∗
∗
∗
= hθk | θk−1 i exp(−ε < θk |H| θk−1 > / < θk |θk−1 >) = eθk θk−1 e−εω(θk θk−1 −1/2) = eεω/2 e(1−εω)θk θk−1 . In terms of the path integral the partition function becomes (2.113)
Z(β) = lim e N →∞
βω/2
N Z Y k=1
" dθk∗ dθk
exp −
N X n=1
# θn∗ θn
" exp (1 − εω)
N X n=1
# θn∗ θn
36
=e
βω/2
N Z Y
lim
N →∞
" dθk∗ dθk
exp −
N X
# {θn∗ (θn
− θn−1 ) +
εωθn∗ θn−1 }
n=1
k=1
= eβω/2 lim
N →∞
N Z Y
†
dθk∗ dθk exp(−θ · B · θ),
k=1
where we have the following vector and matrix elements (2.114)
θ1 . . θ= . , θN
†
θ =
θ1∗
···
∗ θN
,
BN
1
0
···
0
=
y
1
0
···
0 .. .
y .. .
1 .. .
··· .. .
0
0
···
y
−y
0 0 , .. . 1
where y = −1 + εω . The computation is ended by recalling that from the de nition of the Grassmann Gaussian integral one had (2.115)
βω 1 + (1 − βω/N )N = eβω/2 (1+eβω ) = 2 cosh . N →∞ 2
Z(β) = eβω/2 lim det BN = eβω/2 lim N →∞
As with the bosonic partition function, we can arrive to the same result using the ζ function (a generalization of it), and this will prove useful later. Recall that we showed that (2.115)
Z(β) = eβω/2 lim
N →∞
N Z Y
†
dθk∗ dθk exp(−θ · B · θ) = eβω/2
Z
Dθk∗ Dθk exp −
k=1
Zβ
dτ θ∗
d + ω θ (1 − εω) dτ
0
=e
βω/2
d Detθ(β)=−θ(0) (1 − εω) +ω . dτ
The subscript θ(β) = −θ(0) indicates that the eigenvalue should be evaluated for the solutions of the anti-periodic boundary condition θ(β) = −θ(0). First, we expand the orbit θ(τ ) in the Fourier modes. The eigenmodes and the corresponding eigenvalues are (2.116)
exp where
πi(2n + 1)τ β
,
(1 − εω)
πi(2n + 1) + ω, β
n runs as n = 0, ±1, ±2, · · · . The number of degrees of freedom is N (= β/ε) so the
coherent states are (over)complete. The presence ε in operator will account for the fact that the in nite contribution of the eigenvalues is nite. Since one complex variable has two real degrees of freedom, we need to truncate the product at −N/4 ≤ k ≤ N/4. Following this prescription, one has (2.117) N/4
Z(β) = e
βω/2
lim
N →∞
Y k=−N/4
" # 2 ∞ Y π(2n − 1) 2π(n − 1/2) βω/2 −βω/2 2 i(1 − εω) +ω =e e +ω β β k=1
37
=
2 ∞ ∞ Y π(2k − 1) Y β
k=1
"
1+
n=1
βω π(2n − 1)
2 # .
The trouble comes from the rst in nite product, Ξ, which is divergent and as such it is in need of ζ values to become nite. This can be accomplished as follows (2.118)
log Ξ = 2
∞ X k=1
2π(k − 1/2) , log β
and we de ne the corresponding ζ function by (which is the Hurwitz ζ function) (2.119)
−s ∞ X 2π(k − 1/2)
ζfermion (s) =
β
k=1
=
β 2π
s ζ(s, 1/2),
where (see Chapter 1, Eq 1.13) (2.120)
ζ(s, a) =
∞ X
(k + a)−s ,
k=0
where 0 < a < 1. This gives (2.121) 0 Ξ = exp(−2ζfermion (0)).
So now we are left with the issue of dierentiating the ζfermion function at s = 0 which is done as follows (2.122) 0 ζfermion (0) = log
β 2π
1 ζ(0, 1/2) + ζ 0 (0, 1/2) = − log 2, 2
since we showed in Chapter 1 (Theorem 10) that (2.123)
ζ(0, 12 ) = 0,
ζ 0 (0, 12 ) = − 12 log 2.
Putting this together, we obtain the surprising result (2.124) 0 Ξ = exp(−2ζfermion (0)) = elog 2 = 2.
This result indicates that Ξ is independent of β once the regularization is performed. Finally, the partition function is evaluated using all these facts (2.125) ∞ Y
Z(β) = 2
"
1+
n=1
βω π(2n − 1)
2 # = 2 cosh
βω , 2
by the virtue of the formula (2.126) ∞ Y n=1
" 1+
x π(2n − 1)
2 #
x = cosh . 2
This is in agreement with partition function we computed earlier (2.35) and (2.61).
38 Richard Feynman was an advocate of using solutions of known problems in unknown problems, quoting him 'The
same equations have the same solutions '. The rationale behind this
statement is once we solve a mathematical problem, we can re-use the solution in another physical situation. Feynman was skilled in transforming a problem into one that he could solve. This is precisely what we have done in this case. We can prove (2.126) using path integrals.
REFERENCES The discussion of the role of the zeta function in quantum mechanics and zeta function regularization as explained in the above chapter can be traced back to the Handbook of Feynman Path Integrals C. Grosche and F. Steiner [pages 37 to 44, 55 to 59 for the fermionic case], Peskin and Schroeder [pages 299 to 301] and the Advanced Quantum Field Theory course by Dr Rajantie. It is also complemented by the treatise on path integrals in quantum mechanics from Hagen Kleinert [pages 81 to 83 and 161 to 163].
39
Path Integrals in Field Theory From the quantum mechanical case we can build a generalization with several degrees of freedom, a eld theory. We will exclusively deal with φ4 , where φ(x) is a real scalar eld. Let us summarize standard results from eld theory from Peskin and Schroder as well as Rajantie. The action is built from the Lagrangian (3.1)
Z S= where it is understood that
dx L(φ(x), ∂µ φ(x)),
L is the Lagrangian density. The equations of motion (EOM)
are given by the Euler-Lagrange equation (3.2)
∂ ∂L ∂L =− . ∂xµ ∂(∂µ φ(x)) ∂φ(x) From the free scalar eld Lagrangian (3.3)
1 L0 (φ(x), ∂µ φ(x)) = − (∂µ ∂ µ φ + m2 φ2 ) 2 we can derive the Klein-Gordon equation (3.4)
(∂µ ∂ µ − m2 )φ = 0. When there is a source J present the vacuum amplitude has functional representation (3.5)
Z < 0, ∞|0, −∞ >J = Z[J] = N
Z i Dφ exp i dx L0 + Jφ + εφ2 , 2
with the arti cial iε is added to make sure the integral converges. Integrating by parts we obtain (3.6)
Z Z[J] =
Z Z Z i 1 Dφ exp i dx L0 + Jφ + εφ2 = Dφ exp i dx φ(∂µ ∂ µ φ − m2 )φ + iεφ2 + Jφ . 2 2
In this case, the Klein-Gordon becomes the slightly more generalized equation (3.7)
(∂µ ∂ µ − m2 + iε)φc = −J. Working in
d dimensions and de ning the Feynman propagator as (3.8) ∆(x − y) =
−1 (2π)d
Z
dd k
k2
eik(x−y) + m2 − iε
the solution to the generalized Klein-Gordon equation becomes (3.9)
Z φc (x) = −
dy∆(x − y)J(y).
40 The Feynman propagator obeys (3.10)
(∂µ ∂ µ − m2 + iε)∆(x − y) = δ d (x − y). Hence the vacuum amplitude can be written in terms of the source
J as (3.11)
Z i dxdyJ(x)∆(x − y)J(y) < 0, ∞|0, −∞ >J = N exp − 2 or (3.12)
Z i Z0 [J] = Z0 [0] exp − dxdyJ(x)∆(x − y)J(y) . 2 by setting < 0, ∞|0, −∞ >J := Z0 [J]. The (Feynman) propagator is also computed by the functional derivative of Z0 [J] (3.13)
∆(x − y) =
i δ 2 Z0 [J] . Z0 [0] δJ(x)δJ(y) J=0
In order to evaluate Z0 [0] (which is the vacuum to vacuum amplitude when there is no source) we need to introduce imaginary time x4 = t = ix0 and operator ∂¯µ ∂¯µ = ∂ 2 + ∇2 so τ
that (3.14)
Z Z0 [0] =
Z 1 ¯ exp 1 dxφ(∂¯µ ∂¯µ − m2 )φ = q Dφ , 2 Det(∂¯µ ∂¯µ − m2 )
with capital d, the determinant is the product of eigenvalues with corresponding boundary condition. With term sources, the Lagrangian of the free complex scalar eld takes the form (3.15) 2
L0 = −∂µ φ∗ ∂ µ φ − m2 |φ| + Jφ∗ + J ∗ φ, and consequently the generating functional becomes (3.16) ∗
Z0 [J, J ] =
Z
Z Z Z 2 ∗ ∗ µ 2 ∗ ∗ DφDφ exp i dx(L0 − iε |φ| ) = DφDφ exp i dxφ (∂µ ∂ − m − iε)φ + Jφ + J φ , ∗
dierentiating we obtain the propagator (3.17)
∆(x − y) =
δ 2 Z0 [J, J ∗ ] i . Z0 [0, 0] δJ ∗ (x)δJ(y) J=J ∗ =0
We may split the function by virtue of the Klein-Gordon equations (∂µ ∂ µ − m2 )φ = −J and (∂µ ∂ µ − m2 )φ∗ = −J ∗ (3.18)
Z ∗ Z0 [J, J ] = Z0 [0, 0] exp −i dxdyJ (x)∆(x − y)J(y) ∗
and by using another Wick rotation we have (3.19)
Z Z0 [0, 0] =
Z ∗ µ 2 DφDφ exp −i dxφ (∂µ ∂ − m − iε)φ = ∗
1 . ¯ ¯ Det(∂µ ∂ µ − m2 )
41 The presence of a potential in the Lagrangian (3.20)
L(φ, ∂µ φ) = L0 (φ, ∂µ φ) − V (φ) comes at a double price: the form of the potential is limited by symmetry and renormalization of the theory but this theory needs to be handled perturbatively. The potential is usually of the form V (φ) =
α n n! φ
where α is a real number that sets the strength of the
interaction and n > 2 is an integer. As with the free theory, the generating functional is (3.21)
Z 1 Dφ exp i dx φ(∂µ ∂ µ − m2 )φ − V (φ) + Jφ 2 Z Z Z = Dφ exp −i dxV (φ) exp i dx (L0 (φ, ∂µ φ) + Jφ) Z
Z[J] =
Z = exp −i dxV −i
=
∞ Z X k=0
Z dx1 · · ·
δ δJ(x)
Z
(−i)k dxk V k!
Z Dφ exp i dx (L0 (φ, ∂µ φ) + Jφ)
δ −i δJ(x1 )
· · · V −i
δ Z0 [J]. δJ(xk )
The Green function (which is the vacuum expectation of the order time product of eld operators) (3.22)
Gn (x1 , · · · , xn ) := h0 | T [φ(x1 ) · · · φ(xn )]| 0i =
(−i)n δ n Z[J] δJ(x1 ) · · · δJ(xn ) J=0
is generated by the generating functional Z[J]. However, we can see that this is the
n th functional derivative of Z[J] around J = 0 and
therefore we may plug it into the Taylor expansion of the exponential above and we obtain (3.23)
" n Z # Z ∞ X 1 Y dxi J(xi ) h0 | T [φ(x1 ) · · · φ(xn )]| 0i = 0 T exp dxJ(x)φ(x) 0 . Z[J] = k! i=1 k=1
Connected n -point functions are generated by (3.24)
Z[J] = exp(−W [J]), and the eective action is de ned by a Legendre transformation as follows (3.25)
Z Γ[φcl ] := W [J] −
dτ dxi Jφcl
where (3.26)
φcl := hφiJ =
δW [J] . δJ
We will also see that Γ[φcl ] generates 1-particle irreducible diagrams (see Ramond and Bailin and Love). It is convenient now to derive the above discussion in a formal manner and with a closer
42 analogy to statistical mechanics. The generating functional of correlation functions for a eld theory with Lagrangian
L is given by (3.5) (3.27)
Z Dφ exp i d4 x(L + Jφ) ,
Z Z[J] =
where the time variable is contained between -T and
T, with T → ∞(1 − iε). Furthermore
we have the following (3.28)
R h0 | T φ(x1 )φ(x2 )| 0i =
lim
T →∞(1−iε)
= Z[J]−1 −i
δ δJ(x1 )
h R i T Dφφ(x1 )φ(x2 ) exp i −T d4 xL h R i R T Dφ exp i −T d4 xL
−i
δ δJ(x2 )
Z[J]
. J=0
Let us do some manipulations on the time variable; when we derived the path integral formulation of quantum mechanics (see AQFT) it is shown that the time integration was tilted into the complex plane in the direction that would allow the contour of integration to be rotated clockwise onto the imaginary axis. We assume that the original in nitesimal rotation gives the correct imaginary in nitesimal to produce the Feynman propagator. Now, the wick rotation of the time coordinate t → −ix0 yields a Euclidean 4-vector product 2
2
2
x2 = t2 − |x| → −(x0 )2 − |x| = − |xE | , and similarly we assume that the analytic continuation of the time variables in any Green's function of a quantum eld theory produces a correlation function
invariant under the
rotational symmetry of four-dimensional Euclidean space. Let us now apply this to the φ4 theory. As we know the action in this case is (3.29)
Z S=
Z
4
d x(L + Jφ) =
4
d x
1 1 2 2 λ 4 2 (∂µ φ) − m φ − φ + Jφ , 2 2 4!
and performing the Wick rotation
Z i
Z
4
d xE (LE − Jφ) = i
4
d xE
1 1 2 2 λ 4 2 (∂Eµ φ) + m φ − φ − Jφ , 2 2 4!
which in turn gives the Wick-rotated generating functional (3.30)
Z Z[J] =
Z 4 Dφ exp − d xE (LE − Jφ) .
The functional LE [φ] is bounded from below and when the eld φ has large amplitude or large gradient the functional becomes large. These two facts would imply that LE [φ] has the form of an energy and consequently it a is a possible candidate for a statistical weight for the uctuations of φ. Within this light, the Wick rotated functional Z[J] is the partition function describing the statistical mechanics of a macroscopic system when approximating the uctuating variable as a eld.
43 Finally, let us push this analogy between eld theory and statistical mechanics further by presenting the Green's function of φ(xE ) (3.31)
Z < φ(xE1 )φ(xE2 ) > =
d4 kE eikE ·(xE1 −xE2 ) , 2 + m2 (2π)4 kE
which is in fact the Feynman propagator evaluated in the spacelike region and this falls o
exp(−m|xE1 − xE2 |). This correspondence between quantum eld theory and statistical mechanics will play an important part in understanding ultraviolet divergences. Recalling the generating functional of correlation functions, we de ne an energy functional
E[J] by (3.32) Z Z[J] = exp(−iE[J]) =
Z 4 Dφ exp i d x(L + Jφ) = < Ω|e−iHT |Ω >,
with the constraints on time explained above. Note the −i factor in the exponential in contrast to (3.24). The functional E[J] is, as we have said before, the vacuum energy as
a function of the external source J. Let us perform the functional derivative of E[J] with respect to J(x)
δ δ E[J] = i log Z = − δJ(x) δJ(x)
Z
Z −1 Z Z Dφ exp i d4 x(L + Jφ) Dφφ(x) exp i d4 x(L + Jφ)
and set (3.33)
δ E[J] = − < Ω|φ(x)|Ω >J δJ(x) the vacuum expectation value in the presence of a source J. Next, we de ne - the classical eld as (3.34)
φcl (x) = < Ω|φ(x)|Ω >J , a weighted average over all possible uctuations, and dependent on the source J. - the eective action as the Legendre transformation of E[J] i.e. as in (3.25) (3.35)
Z Γ[φcl ] := −E[J] −
d4 x0 J(x0 )φcl (x0 ).
By virtue of (3.33) we have the following
δ δ Γ[φcl ] = − E[J] − δφcl (x) δφcl (x) Z =−
δJ(x0 ) δE[J] d x − δφcl (x) δJ(x0 ) 4 0
Z
d4 x0
Z
d4 x0
δJ(x0 ) φcl (x0 ) − J(x) δφcl (x)
δJ(x0 ) φcl (x0 ) − J(x) = −J(x) δφcl (x)
which means that when the source is set to zero, the equation (3.36)
δ Γ[φcl ] = 0 δφcl (x) is satis ed by the eective action. This equation has solutions which are the values of
44
hφ(x)i in the stable states of the theory. It will be assumed that the possible vacuum states are invariant under translation and Lorentz transformations. This implies a substantial simpli cation of (3.36) as for each possible vacuum state the corresponding solution φcl will
be independent of x, and hence it is just solving an ODE of one variable.
Thermodynamically, Γ is proportional to the volume of the spacetime region over which the functional integral is taken, and therefore, it can be a large quantity. Consequently, in terms of volume
V and associated T of the region we may write (3.37) Γ[φcl ] = −(V T )Veff (φcl ),
where Veff is the eective
potential. In order that Γ[φcl ] have an extremum we need the
following to hold (3.38)
δ Veff (φcl ) = 0. δφcl Each solution of (3.38) is a translation invariant state without source J = 0. Therefore, the eective action (3.35) is −E in this case (Γ = −E) and consequently Veff (φcl ) evaluated at the solution of (3.38) is the energy density of the corresponding state. The eective potential de ned by (3.37) and (3.38) yields a function whose minimization de nes the exact vacuum sate of the eld theory including all eects of quantum corrections. The evaluation of Veff (φcl ) will follow from the path integral formulation. In order to accomplish this we will follow Peskin and Schroeder's method which in turn follows from R. Jackiw and dates back to 1974. The idea is to compute the eective action Γ directly from its path integral de nition and then obtain Veff by focusing on constant values of φcl . Because we are using renormalized perturbation theory, the Lagrangian
L=
1 λ0 1 (∂µ φ)2 − m20 φ2 − φ4 2 2 4!
ought to be split as
L = L1 + δL, which is analogous to the split of the Lagrangian in renormalized φ4 theory done in AQFT (see Rajantie and Bailin and Love 7.4), i.e. rescaling the eld φ = Z 1/2 φr where residue in the LSZ reduction formula Z d4 x hΩ | T φ(x)φ(0)| Ωi eip·x = with
Z is the
iZ + (terms regular at p2 = m2 ), p2 − m2
m being the physical mass. This rescale changes the Lagrangian into L=
1 1 λ0 1 1 δλ (∂µ φr )2 − m20 φ2r − φ4r + δZ (∂µ φr )2 − δm φ2r − φ4r , 2 2 4! 2 2 4!
by the use of δZ = Z − 1, δm = m20 Z − m2 and δλ = λ2 Z − λ where
m and λ are
physically measured.. The last three terms are known as the counter terms and they take into account the in nite and unobservable shifts between the bare parameters and the physical parameters. At the lowest order in perturbation theory the relationship between the source and the
45 classical eld is (3.39)
δL + J(x) = 0. δφ φ=φcl
Because the functional Z[J] depends on φcl through its dependence on
J and our goal is to
compute Γ as a function of φcl . This will be the starting point. Next, we de ne J1 to be the function that exactly satis es the classical eld equation above for higher orders, i.e. when L = L1 (3.40)
δL1 + J1 (x) = 0, δφ φ=φcl and the dierence between both sources
J and J1 will be written as (see Peskin and
Schroeder 11.4) (3.41)
J(x) = J1 (x) + δJ(x), where δJ has to be determined, order by order in perturbation theory by use of (3.34), that is by using the equation hφ(x)iJ = φcl (x). We may now write (3.32) as (3.42)
e
−iE[J]
Z Z 4 4 Dφ exp i d x(L1 [φ] + J1 φ) exp i d x(δL[φ] + δJφ) ,
Z =
where all the counter terms are in the second exponential. Let us concentrate on the rst exponential rst. Expanding the exponential about φ(x) = φcl (x) + η(x) yields (3.43)
Z
Z
4
d x(L1 [φ] + J1 φ) = 1 + 3!
Z
Z Z δL1 δ 2 L1 1 4 4 4 d x(L1 [φcl ] + J1 φcl )+ d xη(x) + J1 + d xd yη(x)η(y) δφ 2! δφ(x)δφ(y) 4
4
4
4
d xd yd zη(x)η(y)η(z)
δ 3 L1 δφ(x)δφ(y)δφ(z)
+ ···
and it is understood that the functional derivatives of L1 are evaluated at φcl . The second integral on the RHS vanishes by (3.40) and therefore the integral over η is a Gaussian integral, where the perturbative corrections are given by the cubic and higher terms. Let us assume that the coe cients of (3.43) (i.e. the successive functional derivatives of L1 ) give well-de ned operators. If we keep only terms up to quadratic order in η and we only focus of the rst integral of (3.42) we nd that there is a pure Gaussian integral which can be evaluated in terms of a functional determinant as we have computed in the Appendix (3.44)
Z 1 δ 2 L1 4 4 4 d xd yη η Dη exp i d x(L1 [φcl ] + J1 φcl + 2 δφ(x)δφ(y) Z −1/2 δ 2 L1 4 = exp i d x(L1 [φcl ] + J1 φcl ) det − , δφ(x)δφ(y)
Z
the lowest-order quantum correction to the eective action is given by the determinant. If we now consider the second integral of (3.42) which consists of the counter terms of the Lagrangian and expanding as we have done before we have (3.45)
(δL[φcl ] + δJφcl ) + (δL[φcl + η] − δL[φcl ] + δJη).
46 The cubic and higher order terms in η in (3.43) produce Feynman diagram expansion of the functional integral in (3.42) in which the propagator is the operator invert (3.46)
−i
δ 2 L1 δφ(x)δφ(y)
−1 ,
and hence when the second term in (3.45) is expanded as a Taylor series in η the successive terms give counter term vertices which can be included in the above mentioned Feynman diagrams. The rst term is a constant with respect to the integral over thus it gives additional terms in the exponent of (3.44). Taking (3.44) with the contributions from higher order vertices and counter terms together we obtain an expression for the functional integral (3.42). Feynman diagrams representing the higher order terms can be arranged in such a way that they yield the exponential of the sum of the connected diagrams, obtaining the expression for E[J] (3.47)
Z −iE[J] = i
d4 x(L1 [φcl ] + J1 φcl ) − Z +i
1 δ 2 L1 log det − + (connected diagrams) 2 δφ(x)δφ(y)
d4 x(δL[φcl ] + δJφcl ),
and nally by virtue of (3.41) and (3.35) we nally have (3.48)
Z Γ[φcl ] =
d4 xL1 [φcl ] +
Z δ 2 L1 i log det − − i(connected diagrams) + d4 x(δL[φcl ]). 2 δφ(x)δφ(y)
This indeed the expression we were seeking since Γ is a function of φcl , taking away the
J dependence. The Feynman diagrams in the expression for Γ have no external lines and they all contain at least two loops. The last term of (3.48) gives the counter terms that are needed for the renormalization conditions on Γ and cancel the divergences that appear in the evaluation of the determinant and the diagrams. We shall ignore any one-particle irreducible one-point diagram (these diagrams are cancelled by the adjustment of δJ ). We now go back to the Euclidean space de nition of the generating functional (3.49)
Z ZE [J] = NE
Z 1 1 Dφ exp − d4 x ¯ ∂¯µ φ∂¯µ φ + m2 φ2 + V (φ) − Jφ . 2 2
This will be evaluated by expanding the action. In order to do this, set φ0 to be a eld con guration, then (3.50)
Z SEuc [φ, J] :=
d4 x ¯
1 ¯ ¯µ 1 ∂µ φ∂ φ + m2 φ2 + V (φ) − Jφ 2 2 Z
SEuc [φ, J] = SEuc [φ0 , J] + 1 + 2
Z
4
4
d x ¯1 d x ¯2
δSEuc d x ¯ (φ − φ0 ) δφ 4
δ 2 SEuc (φ(¯ x1 ) − φ0 (¯ x1 ))(φ(¯ x2 ) − φ0 (¯ x2 )) + · · · δφ(¯ x1 )δφ(¯ x2 )
It is understood that the functional derivatives are evaluated at φ0 . We know from AQFT
47 that the classical limit can be recovered, by taking SEuc to be stationary at φ0 and this implies that φ0 satis es the classical EOM with the source term (3.51)
δSEuc = −∂¯µ ∂¯µ φ(x0 ) + m2 φ(x0 ) + V 0 [φ(x0 )] − J = 0. δφ x0 Integrating by parts gives (3.52)
1 SEuc [φ(x0 ), J] = 2
Z
d x ¯ 2 − φ(x0 ) 4
d [−Jφ(x0 ) + V 0 [φ(x0 )]], dφ(x0 )
whereas the second derivative is an operator (3.53)
δ2 SEuc = δ(¯ x1 − x ¯2 ) −∂¯µ ∂¯µ + m2 + V 00 [φ(x1 )] . δφ(x1 )δφ(x2 ) We need to make regression now on how to evaluate integrals of the sort (3.54)
Z I :=
dx exp(−α(x))
This can be evaluated by expanding the exponential around a point x0 where α is stationary, i.e. (3.55)
The
1 α(x) = α(x0 ) + (x − x0 )200 α00 (x0 ) + O(x3 ). 2
I integral is then approximated (3.56) Z I = exp(−α(x0 ))
1 dx exp − (x − x0 )2 α00 (x0 ) , 2
and we can recognize this as a Gaussian integral (when the higher derivatives are ignored)! The degree of this approximation depends obviously on α. When α is smallest then the integrand is largest and the points away from the minimum do not add a substantial contribution. Equipped with this method we can apply it to the functional (see Ramond 3.4) we have just arrived at a crucially important result (3.57)
Z ZE [J] = NE exp {−SEuc [φ(x0 ), J]}
= NE0 q
Z δ 2 SEuc 1 d4 x ¯ 1 d4 x ¯2 φ(¯ x1 ) φ(¯ x2 ) Dφ exp − 2 δφ(¯ x1 )δφ(¯ x2 )
exp {−SEuc [φ(x0 ), J]} , det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
where we have ignored the higher order terms. The precise derivation is in the Appendix. Let us do some re-write to make this expression easier to handle, rst the determinant, which we will call
M can be taken care of by using the identity (3.58) det M = exp(tr log M),
48 (3.59)
ZE [J] =
NE0
1 µ 2 00 ¯ ¯ exp −SEuc [φ(x0 ), J] − tr log(−∂µ ∂ + m + V [φ(x0 )])δ(x1 − x2 ) , 2
the delta term accounts for the quantum perturbations (or corrections) to Z[J], whereas the rst term accounts for the classical contribution. Also note that we set by convention the determinant of an operator to be the product of its eigenvalues and that because φ0 satis es δSEuc /δφ|x0 = 0 then it is a functional of J.
The concluding remark is to nd the equivalence of these results in terms of the classical eld φcl and explore the corresponding eective action. When we work in Euclidean space, the classical eld was de ned as (3.60)
φcl (¯ x) = −
δZE δSE ≈− + O(~), δJ(¯ x) δJ(¯ x)
and by using (3.57) and (3.58) we can have φcl as a function (functional, rather) of J, however at the cost of doing it order by order in λ. The term O(~) stands for quantum corrections. This relationship can be inverted and we can nd J(¯ x) as a function of φcl . This inversion can be carried out to give (see Ramond) (3.61)
λ x). J(¯ x) = (∂¯2 − m2 )φcl (¯ x) − φ3cl (¯ 3! An attractive (and indispensable) feature is that there are no higher terms in λ, comparison with
0=
δSEuc = −∂¯µ ∂¯µ φ(x0 ) + m2 φ(x0 ) + V 0 [φ(x0 )] − J δφ x0
gives
φcl (¯ x) = φ0 (¯ x) + O(~). By integrating J(x) = −δΓ[φcl ]/δφcl we see that to this order the eective action is (3.62)
Z Γeff [φcl ] = −
1 λ 4 2 2 ¯ d x ¯ φcl (∂ − m )φcl − φcl (¯ x) , 2 4! 4
which is a eective action.
REFERENCES The above summary on quantum eld theory has its roots several sources, overall it follows the presentation given by Peskin and Schroeder [275 to 294, 306 to 308 and 365 to 372] as well as Ramond and section 4.4 of Bailin and Love. At times it is complemented by the courses of QED and AQFT from the MSc QFFF. However, it is broad enough to be explained in most QFT books.
49
Zeta Regularization in Field Theory At this point we can put together some of the concepts acquired in the second and third chapters. Firstly we will further develop the theory of quantum corrections by evaluating the determinant of the M matrix and then see how this is related to the use of the zeta function
in quantum theory as we explained in Chapter 2. In what follows, we will extrapolate the results found in the quantum mechanical section to eld theory, this section borrows some its contents from 3.5 and 3.6 from Ramond as well as 22.1 to 22.3 of Har eld. As we mentioned earlier, the determinant in the expression (3.57) (4.1)
ZE [J] = q
NE0 exp {−SEuc [φ(x0 ), J]} det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
must be interpreted as the product of the eigenvalues of the operator. In order to discretize these eigenvalues we truncate the space (by use of a box). We then multiply the resulting eigenvalues and then let the size of the box increase to in nity. First we state some preliminary results that will become useful later on. As it is shown in a course on partial dierential equations, the heat function
G(¯ x, y¯, t) :=
X
exp(−λn t)ψn (¯ x)ψn∗ (¯ y)
n
satis es the heat equation
Ax¯ G(¯ x, y¯, t) = −
∂ G(¯ x, y¯, t). ∂t
Also, similar to our expression (1.15) we have the following result (4.2)
Z∞ ζA (s)Γ(s) =
dt ts−1
Z
d4 x ¯G(¯ x, x ¯, t),
0
this in fact follows from the orthogonality of eigenfunctions ∗ hψn , ψm i=
Z
dxψn∗ (x)ψm (x) = δnm ,
X
ψn∗ (x)ψn (x0 ) = δ(x − x0 )
n
By this same orthogonality we have (4.3)
G(¯ x, y¯, t = 0) = δ(¯ x − y¯). This is the analytic representation of the zeta function we are seeking. We can explain this when we take the trace of
G
Z Tr(G) =
dxG(x, x, t) =
X
e−λn t
n
(remember Tr(G) is a function of t ) multiply it by e−λt and integrate it with respect to
t
50 and then with respect to λ to obtain
Z∞
Z dλ
dtTr(G)e
−λt
dλ
n
0
The determinant of
=
XZ
X 1 = log(λn + λ). λn + λ n
A then is written as (*) Z Z∞ det A = exp dλ dt Tr(G)e−λt λ=0 0
ignoring the factor that would show from the λ integration. Swapping the integrals yields
Z∞
det A = exp −
Tr(G) dt . t
0
The relationship between the trace of the heat kernel TrG and the ζ function is given by a Mellin transformation
1 ζA (s) = Γ(s)
Z∞
dtts−1 Tr(G),
0
which justi es (4.2). A solution of the heat equation (4.4)
−∂¯x2 G0 (¯ x, y¯, t) = −
∂ G0 , ∂t
with the boundary condition G0 (¯ x, y¯, t = 0) = δ(¯ x − y¯) is (4.5)
1 1 2 G0 (¯ x, y¯, t) = exp − (¯ x − y¯) , 16π 2 r2 4r also a classic result from partial dierential equations.
A i runs from 1 to n and its eigenfunctions
Let us generalize some of the techniques we used in Chapter 2. Consider an operator with positive real discrete eigenvalues ai where
are fn (x) i.e. Afi (x) = ai fi (x). From here we set (4.6)
ζA (s) =
X
a−s n ,
n
which we call the zeta function associated to the operator eigenvalues and
A. The sum is over all the
A is a real variable. If the operator A were the one-dimensional harmonic
oscillator Hamiltonian, then ζA would be the Riemann zeta function (excluding the singular zero-point energy). The rst observation is that (4.7)
X Y d ζA (s) =− log asn exp(−s log asn )|s=0 = − log an , ds s=0 n n
51 which gives the determinant (4.8)
det A =
Y
0 an = exp(−ζA (0)).
n
As we know from our discussion of chapter 2. The zeta-function ζA is not always singular at s = 0 for physically interesting operators and hence the convenience of writing this representation for det A. Algorithmically we have a procedure to compute the determinant of
A and it can be done
as follows. First we need to nd the solution to the heat equation subject to the deltafunction initial condition (4.3). Second, once this is done we can insert the solution into the
ζ representation above and we have ζA . Finally evaluate at s = 0 and compute exp(−ζA0 (0)). In our case, our operator is (4.9)
λ A = −∂¯2 + m2 + φ20 (¯ x), 2 where φ0 (¯ x) is a solution of the classical equations with source J as we discussed in Chapter 3. Also note that in (4.1) we have a V 00 factor in the determinant which takes the form
V 00 = (λ/2)φ20 in the
A operator.
As we have pointed out a solution of the heat equation with the boundary condition
G0 (¯ x, y¯, t = 0) = δ(¯ x − y¯) is (4.10) 1 1 2 exp − (¯ x − y¯) . G0 (¯ x, y¯, t) = 16π 2 r2 4r However, this is only a part of the operator as we want to nd G0 (¯ x, y¯, t) subject to the initial condition (4.3) which obeys the whole
A operator (4.11)
∂ λ 2 2 2 ¯ x) G(¯ x, y¯, t) = − G(¯ x, y¯, t). −∂ + m + φ0 (¯ 2 ∂t For arbitrary elds φ0 we need to expand the eective action as (0)
(1)
ΓE [φcl ] = ΓE [φcl ] + ~ΓE [φcl ] + · · · , the ~ indicates quantum terms. Given the fact that the eective potential at 1 loop is of the form (Ramond) (1)
ΓE [φcl ]
Z
d4 x =
1 log det(−∂ 2 + m20 + 2
λ0 2 2 φcl )
−
1 log det(−∂ 2 + m20 ), 2
as we also argued in Chapter 3 Eq 3.48, allows us to write (4.12)
1 (1) ΓE [φcl ] = − ζ 0 ¯2 2 λ 2 (0), 2 [−∂ +m + 2 φ0 (¯x)] by use of (4.1) and (4.8) combined with (4.9). Note that replacing φ0 by φcl is not a problem as there are no new quantum errors, that is errors up to O(~). By the expansion of the
52 eective potential we can swap m and m0 . Also the eective action can be written as (4.13)
Z ΓE [φcl ] =
d4 x ¯ V (φcl (¯ x)) + F (φcl )∂¯µ φcl (¯ x)∂¯µ φcl (¯ x) + · · ·
constant v is a constant independent of x¯.
To compute the quantum O(~) contribution to V (φcl (¯ x)) we need to consider a eld con guration: i.e. suppose we take φcl (¯ x) = v where Then (4.14)
Z
d4 x ¯V (v),
ΓE [φcl ] =
and it is proportional to the in nite volume element
d4 x ¯, since the Euclidean space R4 is unbounded. This can be temporarily solved by taking the space to be the sphere S4 then R
the volume is just that of the 5-dimensional sphere and hence nite. While the radius is nite we need not worry about the infrared divergence. We then let the radius of the sphere tend to in nity. Taking
V out of the integral we have [see Hart eld] (4.15) Z V (v)
1 d4 x ¯ = − ζ 0 ¯2 2 λ 2 (0). 2 [−∂ +m + 2 v ]
We can proceed to integrate (4.11) when
G(¯ x, y¯, t) =
v is constant this yields (4.16)
µ4 µ2 (¯ x − y¯)2 λ 2 t 2 v exp − exp −m + , 16π 2 t2 4t 2 µ2
where the µ factor needs to be explained: it has dimensions of mass so that t is dimensionless. Using the Mellin transform (4.2) we obtain (4.17)
1 ζ(s) = Γ(s)
Z∞ dt t
s−1
Z
µ4 d x ¯ exp 16π 2 t2 4
λ −m + v 2 2 2
m2 + λ2 v 2 µ2
t µ4 = 2 µ 16π 2 t2
!2−s
Γ(s − 2) Γ(s)
Z
d4 x ¯,
0
the volume element
R
d4 x ¯ is present because it is in (4.15) and here
t has been rescaled
since the integration over t is valid when s > 2 however the ζ function is de ned everywhere by analytic continuation as we know from Chapter 1. Comparing these two equations we have (4.18)
µ4 d 1 V (v) = − 2 32π ds (s − 2)(s − 1)
m2 + λ2 v 2 µ2
Equipped with this functional form of
!2−s
1 = 64π 2
λ m + v2 2 2
2
m2 + λ2 v 2 3 log − 2 µ 2
s=0
V the eective potential can be written as (4.19)
1 λ ~ V (φcl ) = m2 φ2cl (¯ x) + φ4cl (¯ x) + 2 4! 64π 2
2 λ m + φ2cl (¯ x) 2 2
m2 + λ2 v 2 3 log − µ2 2
! ,
ignoring terms of order ~2 . We make a pause now to examine this result. Super cially the rst striking observation is that there is a strong dependence on the unknown and arbitrary scale µ2 . This would seem to imply that the potential is therefore arbitrary. However
V
depends on the parameters m and λ which are unde ned, except for the fact that they are 2
! .
53 included in the classical Lagrangian. Let us take the special massless case. This yields the following (4.20)
d2 V = 0. dφ2 φ=0 Now we
de ne the mass squared as the coe cient of the terms φ2 in the Lagrangian eval-
uated at φ = 0. To rst quantum corrections the coe cient is zero, if it is classically zero. The λ term is de ned to be the coe cient of the fourth derivative of
V evaluated at some
constant point φ = M, i.e. (4.21)
d4 V λ := . dφ4 φ=M We cannot take φ = 0 as with the mass squared factor because of the infrared divergence coming from the logarithm. When we dierentiate (4.19), set m2 = 0 and use (4.21) we see that the above condition requires (4.22)
log
λM 2 8 =− . 2µ2 3
In this case, we may use M 2 instead of 2µ2 /λ and write the result as (4.23)
λ λ2 φ4cl V (φcl ) = φ4cl (¯ x) + 4! 256π 2
φ2cl 25 log 2 − . M 6
This was proved by Coleman and Weinberg in 1973. The main result that can be extracted is that we need to be careful with how we de ne the input parameters in the Lagrangian if we are to take into account quantum corrections. Again, super cially it seems that (4.23) depends on another arbitrary scale M 2 , but in fact it does not. Given the normalization condition, if we change the scale from M 2 to M 02 we simultaneously have to change at the same time λ to λ0 by use of (4.21) (4.24)
λ0 = λ +
3λ2 M0 log . 16π 2 M
Therefore the potential (see Ramond) (4.25)
V (φcl ) =
λ0 4 λ02 φ4cl φcl (¯ x) + 4! 256π 2
log
φ2cl 25 − M 02 6
is indeed invariant under the representation V (λ0 , M 0 ) = V (λ, M ). This proves that the physics behind remains unchanged but our way of interpreting the coe cients changes. Let us now look more closely to the scaling of determinants and the coupling constants. The ζ function technique just used allows us derive scaling properties for determinants.
(0). λ [−∂¯2 + 2 φ2cl ]
First, we will need the computation of ζ function ζ
This can be accomplished
54 by taking the asymptotic expansion of G(¯ x, y¯, t) at µ2 = 1, (4.26)
G(¯ x, y¯, t) = e−εt
2 ∞ e−(¯x−¯y) /(4t) X an (¯ x, y¯)tn , 16π 2 t2 n=0
with ε > 0 as a convergence factor. The boundary condition (4.3) sets the condition (4.27)
a0 (¯ x, x ¯) = 1. Additionally, when we insert (4.26) into the PDE (4.11) we nd recursion relations for the
an coe cients (4.28) (¯ x − y¯)µ
∂ a0 (¯ x, y¯) = 0 ∂x ¯µ
and for n = 0, 1, 2, · · · (4.29)
∂ λ (n + 1) + (¯ x − y¯)µ an+1 (¯ x, y¯) = ∂¯x2 − φ2cl + ε an (¯ x, y¯). ∂x ¯µ 2 When we compute the rst terms we have (4.30)
λ a1 (¯ x, x ¯) = − φ2cl + ε 2 a2 (¯ x, x ¯) =
λ2 4 λ ε ε2 φcl (¯ x) − ∂¯x2 φcl (¯ x) − λφ2cl (¯ x) + . 8 4 2 2
Let us now use these results. We work under a scale change A → A0 = ead A, where the natural dimension of
A. By the de nition of the zeta function we have (4.31)
d is
ζA0 (s) = e−ads ζA (s) which implies that (4.32)
det(ead A) = eadζA (0) det A. Let us illustrate this with an example. Under the transformation (4.33)
xµ → x0µ = ea xµ
φcl → φ0cl = e−a φcl
the massless classical action (4.34)
Z SE [φcl ] = −
d4 x ¯
λ 1 ¯2 φcl ∂ φcl − φ4cl 2 4!
is unchanged. However, the path integral corresponding to this action is not scale invariant since in the steepest descent approximation the change in the eective action to quantum order is as follows (4.35) eff eff 0 eff SE [φcl ] → SE [φcl ] = SE [φcl ] − ~aζ
(0). λ [−∂¯2 + 2 φ2cl ]
Plugging (4.26) into (4.2) and with the assistance of the rst two terms a1 and a2 and
55 integrating out the ∂¯2 with the divergence theorem yields (4.36)
1 ζ(0) = 16π 2
Z
d4 x ¯
λ2 4 φ (¯ x). 8 cl
A small digression is now required to further explain this. In 4 dimensions and in the presence of mass the heat kernel is
2 e−(¯x−¯y) /(4t) λ0 φ2cl 2 G(x, x, t) = exp − m0 + t 16π 2 t2 2 and like in (4.17) and (4.18)
ζ(s) =
1 16π 2
m20 +
λ0 φ2cl 2
2−s
Γ(s − 2) Γ(s)
Z
d4 x ¯.
Furthermore, note that in the more restricted case B = −∂ 2 + ω 2 we have kernel
(x − x0 )2 1 exp − exp(−ω 2 t) G(x, x t) = √ 4t 4πt 0
and the determinant of
B is
1 log det B = − √ 4π
Z∞
dtt−3/2 exp(−ω 2 t)
Z
ω d4 x ¯ = −√ 4π
Z
d4 x ¯
0
Z∞
dte−t t−3/2
0
ω −√ 4π
Z
∞ h i d4 x ¯ t−1/2 e−t − 2π 1/2 0
integration by parts was used in the last line. Ignoring the divergent factor we have the simpli cation
Z det B = exp ω d4 x ¯ .
This can nd justi cation the integral in log det B is expressed in terms of Γ functions in this divergence zone by analytic continuation. Finally, the divergence of Γ(s − 2) at s = 0 cancels that the divergence of Γ(0), i.e.
1 Γ(s − 2) = Γ(s) (s − 2)(s − 1) and this process makes ζ(0) regular. In terms of the eective action (4.35) (4.37) eff 0 SE [φcl ]
=
eff SE [φcl ]
λ2 − ~a 128π 2
Z
d4 x ¯φ4cl (¯ x).
Consequently the eect of the transformation to quantum order has been to change the coupling constant λ by the following (4.38)
λ λ0 λ λ2 3λ2 → = − ~a ⇔ λ0 = λ − ~a. 2 4! 4! 4! 128π 16π 2
56 What this is means is that
the dimensionless coupling constant λ evolves as a result of
quantum eects. This evolution is in term of scale dependence. At large scales the coupling constant decreases indicating that the non-interaction theory is a good approximation for the asymptotic states. On the other hand, if the scale decreases, the coupling increases. Independently of how small λ was at the beginning this increment might throw away the results obtained in the perturbation of λ. Moreover, this scaling law is like the one we found earlier, and they are both correct to quantum orders.
t a quantum mechanical system with one degree of freedom, q and canonically conjugate momentum p, is described in terms of the spectrum We recall that for any given time
of its Hamiltonian H(p, q). From the path integral formulation we know that if the system at an initial time ti is measured to be in the state q i , then the probability that the system will be found in the state q f , at a nal time tf is exactly (4.39)
E E D D qtff qtii = q f e−i(tf −ti )H q i , and it can be written in terms of path integrals as (4.40)
tf Z Z E Z D q f e−i(tf −ti )H q i = Dq Dp exp i dt[pq˙ − H(p, q)], ti
the factor Dq denotes integration between the initial and nal con gurations q i and q f ; the dot over
q denotes the derivative of q with respect to time.
Quantum eld theory and statistical mechanics share certain common elements and precisely this analogy will allow us to apply path integrals to the description of dynamics systems at nite temperature (see Peskin and Schroeder, and Kleinert). The rst step, as we did in Chapter 2, is to compute the partition function (4.41)
Z = Tr[e−βH ], where the constant is (4.42)
β = (kT )−1 , taking into account that the trace is taken to be the sum over all the possible con gurations the system is allowed to take. Note that the time is singled out. Now, the probability for the system to be in state of energy
E is identi ed with (4.43) P = Z −1 e−βH .
The value of any function for the dynamical variable f (q, p) is given by (4.44)
hf i = Tr(f P ) = Z −1 Tr(f e−βH ). This does show a special similarity with zero temperature quantum mechanics and QFT but the degree of this similarity is not fully understood. However, we can push the analogy to calculate partition functions, specially this one. Let us start with a system which can be
57 regarded as a eld theory in zero space dimensions. We can compare this expression to the partition function for the same system at temperature
β −1 (4.45) Z = Tr[e−βH ] =
X q e−βH q . q
Let us draw comparisons between (4.40) and (4.45). If we set i(tf − ti ) = β or alternatively set ti = 0 and then itf = β since the origin of time is arbitrary. Next, set q f = q i which means that the initial and nal con gurations are the same, and since the dierence is a β factor, the only requirement is that the relevant con guration is periodic in the functional integrals (4.46)
q(β) = q(0). Thus the functional integration Dq is over the space of sum over
periodic functions. In this case, the
q in (4.40) is implicit. When we do the comparison we can write (4.47) Z = Tr(e−βH ) =
Z
Z Dq
β Z dq Dp exp dτ ip − H , dτ 0
bearing in mind again that Dq is over periodic functions. If we take a well-behaved potential V (q) we could scale the temperature dependence purely into the
q integral. To do this, we make the following transformations (4.48) τ¯ = τ β −1 ,
q¯ = qβ −1/2 ,
p¯ = pβ 1/2 ,
then the exponent of the integrand becomes (4.49)
Z1
p¯2 d¯ q − − βV (β 1/2 q¯) . d¯ τ i¯ p d¯ τ 2
0
Furthermore we can drop all the bars because the path integral measure is invariant under the changes (4.48) thus we can write the partition function as (4.50)
Z Z=
Z Dq
1 Z 1 Dp exp dτ ipq˙ − p2 − βV (qβ 1/2 ) . 2 0
Next, set p0 = p − iq˙, so that the measures are equal (4.51)
Dp0 = Dp, which in turn, by competing the square in the exponent, allows us to write (4.52)
Z Z=
1 Dp0 exp − 2
Z1 0
dτ p02
Z
Z1
Dq exp −
dτ
1 2 q˙ + βV (qβ 1/2 ) . 2
0
As we have done repeatedly in AQFT we can ignore the p0 integral (even though it is
58 in nite) because it is independent of β . We call this integral be ignored is because
N, and the reason why it can
N is usually always present in the numerator and denominators of
correlation functions. The only example we could tackle is that of an integral that can be evaluated, i.e. a Gaussian integral. The integral becomes Gaussian when we take potential V (q) = 12 ω 2 q 2 , the partition function is (4.53)
Z1
Z Dq exp −
Z=N
dτ
1 2 1 2 2 2 q˙ + β ω q . 2 2
0
It is one of the very few types that can actually be integrated. By virtue of (4.46) we have (4.54)
Z1
1 2
dτ
dq dτ
2
1 =− 2
0
Z1 dτ q
d2 q, dτ 2
0
since the extra surface term is eliminated due to (4.46) and therefore (4.55)
Z
1 Dq exp − 2
Z=N
Z1
d2 dτ q − 2 + ω 2 β 2 q . dτ
0
If we proceed by analogy with the discrete case we have (4.56)
Z Dq exp −
1 2
Z1
dτ q − 0
where N 0 is a constant, and
N0 d + ω2 β 2 q = √ , 2 dτ det A 2
A is the operator (4.57) A=−
d2 + ω2 β 2 dτ 2
with positive de nite eigenvalues (it must not contain zero eigenvalues as these would create in nities which have to be removed). In order to prove (4.56) we need to express q(τ ) in terms of its Fourier components (QFT course) then transform into the normal modes of
A
and integrate each one using (4.58)
Z∞
r 1 2π dqn exp − an qn2 = . 2 an
0
A operates on periodic functions with unit period, which can all be expanded in terms of the complete Fourier set {e2πinτ }. The eigenvalues of A are (4.59) The operator
(4π 2 n2 + ω 2 β 2 );
n ∈ Z.
Note the analogy with the eigenvalues of the quantum
mechanical operator (2.116).
59 Hence multiplying we have the determinant of the operator
Y
det A =
A, (4.60)
(4π 2 n2 + ω 2 β 2 ).
n∈Z
Setting x2 = ω 2 β 2 yields the following (4.61)
X X 1 1 1 d log det A = = 2 +2 . 2 2 2 2 2 2 dx 4π n + x x 4π n + x2 n∈Z
n≥1
Substituting the formula that was shown in the Appendix (4.62)
1 2x X 1 + 2 πx π x + n2
coth πx =
n≥1
1 d x log det A = coth dx2 2x 2 and from here we can integrate to nd (4.63)
log
det A = C
Z dx coth
x x ωβ = 2 log sinh = 2 log sinh . 2 2 2
Removing the logs we have the formula for the determinant (4.64)
det A = C sinh2
ωβ . 2
When we clean the expression we arrive to (4.65)
F =− where
D 1 1 1 log Z = − + ω + log(1 − e−ωβ ), β β 2 β
D is another constant, the zero-point energy is identi ed at 1 = e−ωβ . This formula
is called the thermodynamic potential. Let us go back to our discussion of the ζ function. The heat equation associated with the
A operator (4.57) is (see Ramond, Elizalde, Odintsov, Romeo, Bytsenk and Zerbini) (4.66) X
G(t, t0 , σ) =
exp 2πin(t − t0 ) − (ω 2 β 2 + 4π 2 n2 )σ ,
n∈Z
and recalling our expression (4.2) for the ζ function (4.67)
1 ζA [s] = Γ(s)
Z∞ dσσ
s−1
0
Z1 dt 0
X
exp −(ω 2 β 2 + 4π 2 n2 )σ .
n∈Z
Scaling σ by ω 2 β 2 + 4π 2 n2 leads nowhere as we simply come back to the expression (4.68)
ζA [s] =
X n∈
1 . (ω 2 β 2 + 4π 2 n2 )s
The technique we need to use is somewhat messier, it involves expanding in powers of ωβ
60 then integrating and re-arranging the sums. With this in mind we have (4.69)
X
e−4π
2
n2 σ
=1+2
n∈Z
X
e−4π
2
n2 σ
,
n≥1
and integrating we have (4.70) ∞
ζA [s] = (ωβ)−2s +
∞ ∞ Z X 2 2 2 X (ωβ)2k (−1)−k dσσ s+k−1 e−4π n σ . Γ(s) k! n=1 k=0
0
Now it is the time when we can rescale σ by 4π 2 n2 and we can also identify the sum over P −2s n with the Riemann ζ function ζ(2s) = ∞ , which gives (4.71) n=1 n ∞
ζA [s] = (ωβ)−2s +
2 2 X (ωβ)2k (−1)−k Γ(s + k)ζ(2s + 2k). ζ(2s) + (4π 2 ) Γ(s) k! (4π 2 )s+k k=1
In order to dierentiate ζA at s = 0 we note that the sum is well behaved at s = 0 and as
s → 0 a non zero term arises from the derivative of Γ−1 (s). Recalling our formulas from Chapter 1, (4.72)
1 ζ(0) = − , 2
1 ζ 0 (0) = − log 2π, 2
(4.73) 0 ζA [0] = −2 log(ωβ) − 2 log 2π + 2 log 2π + 2
∞ X (ωβ)2k (−1)k ζ(2k) k=1
k(4π 2 )k
.
Recalling the formula for the even values of the Riemann ζ function in terms of Bernoulli numbers (4.74)
ζ(2k) =
(−1)k+1 (2π)2k B2k , 2(2k)!
we can continue simplifying (4.73) (4.75) 0 ζA [0] = −2 log(ωβ) −
∞ X (ωβ)2k 1 B2k . (2k)! k
k=1
Using another formula from the appendix (4.76) ∞
coth x =
X (2x)2k−1 1 +2 B2k , x (2k)! k=1
and integrating yields (4.77) ∞
Z dx coth x = log x +
1 X (2x)2k B2k , 2 k(2k)! k=1
and nally by comparing with (4.76) and setting x = 12 ωβ , we obtain (4.78) 0 ζA [0] = −2 log(ωβ) + 2 log
ωβ ωβ − 2 log sinh = −ωβ − 2 log(1 − e−ωβ ). 2 2
61 Solving for
Z yields (4.79) log Z =
1 1 0 ζ [0] = − ωβ − log(1 − e−ωβ ), 2 A 2
which is similar to what we found earlier on the quantum mechanical case. Although this technique is lengthy it has enabled us to show the connection between the ζ function and quantum eld theory. It is desirable to be able to obtain (4.19) using another regularization and see how these two compare. In the Uni cation course, it was shown that spontaneous symmetry breaking requires the development of a vacuum expectation value (VEV) from the scalar eld. Furthermore, as we have shown in this Chapters 3 and 4, the VEV is determined by the minimisation of the eective potential, (4.80)
dV = 0. dϕcl The eective potential is given exclusively by the potential
V of the Lagrangian if no
quantum eects are taken into account. Perturbation theory allows us to place the quantum terms, however this would clash with the non-perturbative nature of spontaneous symmetry breaking. The alternative parameter is the loop expansion which we now describe. Let us re-write the theory around (3.24) as follows. With a generating functional
X of the
connected Green functions in a scalar eld theory, (4.81)
Z[J] = exp(i~−1 X[J]) = N 0
Z
Z Dϕ exp i~−1 d4 x(L + Jϕ)
with normalization constant N 0 chosen so that (as we pointed out in Chapter 3) (4.82)
Z[0] = 1 X[0] = 0. As we have also pointed out, the one-particle-irreducible Green functions Γ(n) are generated by the eective action Γ[ϕcl ] by the use of (4.83)
Γ[ϕcl ] =
Z ∞ n Z X i d4 x1 · · · d4 xn Γ(n) (x1 , · · · , xn )ϕcl (x1 ) · · · ϕcl (xn ). n! n=1
In (4.81) the factor ~−1 multiplies the whole Lagrangian (not just the interaction part) each
of the V vertices in any diagram will carry a ~−1 factor and each of the I internal lines will ˜ (E) has a propagator. carry a ~ factor. Each of the E external lines in Green functions G With this information, there overall factor is (4.84)
~−V +I+E = ~L+1−E , by virtue of L = I − V + 1. Furthermore, there is a factor of ~L−E in any diagram in ˜ (E) have no propagators, expansion of ~−1 X . The one-particle-irreducible Green functions Γ hence the multiplying factor is only ~L . This means that the power of the ~ indicates the
62 number of
loops. When there are no loops (L = 0) the only non-vanishing Γ˜ (E) are (4.85) ˜ (2) (p, −p) = p2 − µ2 Γ
and (4.86)
˜ (4) (p1 , p2 , p3 , p4 ) = −λ, Γ which gives the approximation to the potential (4.87)
V0 (ϕcl ) =
1 1 2 2 µ ϕcl + λϕ4cl , 2 4!
as classically expected without quantum corrections. Note, however, that the role of ~ is not central to the discussion as no assumptions about its size have been made, it is only an expansion parameter. Bailin and Love [201] give same eective action Γ[ϕcl ] with (4.88)
1 Γ[ϕcl ] = −i log N − λ 8
Z
1 − λi∆F (0) 4
Z
dx[i∆F (0)]2 − dxϕ2cl (x)
1 2
1 − λ 24
Z
Z
dxϕcl (x)(∂ µ ∂µ + µ2 )ϕcl (x) dxϕ4cl (x) + O(λ2 )
The terms that contain a Feynman propagator ∆F (0) come from divergent loop integrals and hence they do not contribute in zeroth order, consequently we may write (4.89)
Γ0 [ϕcl ] = −
1 2
Z
dxϕcl (x)(∂ µ ∂µ + µ2 )ϕcl (x) −
1 λ 4!
Z
dxϕ4cl (x)
Again taking into account that N = 1 implying Γ[0] = 0. The rst order accuracy of ~, the loop expansion that is, of the eective potential
V and
eective action Γ can be computed by writing (4.90)
ϕ(x) = ϕ0 (x) + ϕ(x), ˜ With ϕ0 being the zeroth order approximation to ϕcl . Hence this shift in the functional integration variable must satisfy the following EOM (4.91)
(∂ µ ∂µ + µ2 )ϕ0 (x) +
λ 3 ϕ (x) = J(x). 6 0
When we plug this change into the Lagrangian density (4.92)
L(ϕ) =
1 µ 1 1 (∂ ϕ)(∂µ ϕ) − µ2 ϕ2 − λϕ4 2 2 4!
we have the following integral (4.93)
Z
Z +
d4 x(L + Jϕ) =
Z
d4 x(L(ϕ0 (x)) + Jϕ0 )
Z 1 1 1 d4 x (∂ µ ϕ)(∂ ˜ µ ϕ0 ) − µ2 ϕϕ ˜ 0 − λϕϕ ˜ 30 + J ϕ˜ + d4 x L2 (ϕ, ˜ ϕ0 ) − λϕ˜3 ϕ0 − λϕ˜4 6 6 4
63 with L2 accounting for all the quadratic leftovers in ϕ˜ when the shift is performed, i.e. (4.94)
L2 =
1 µ 1 1 (∂ ϕ)(∂ ˜ µ ϕ) ˜ − µ2 ϕ˜2 − λϕ2 ϕ˜2 . 2 2 4
Now, since ϕ˜ minimises the classical action then the linear term in ϕ˜ of (4.93) disappears. Our next step is to re-scale the integration variable as follows (4.95)
ϕ˜ = ~1/2 ϕ And plug this back into (4.81) yielding (4.96)
Z[J] = N 0 exp i~−1
Z Z 1 λ d4 x[L(ϕ0 ) + Jϕ0 ]× Dϕ exp i d4 x L2 (ϕ, ϕ0 ) − ~1/2 ϕ3 ϕ0 − λ~ϕ4 6 24
Z
We can recover a Gaussian integral out of this by noting that we are only interested in the rst-order corrections, hence terms proportional to ~1/2 and ~ may be discarded. This procedure gives (4.97)
Z
1 d xL2 (ϕ, ϕ0 ) = − 2 4
Z
d4 xd4 x0 ϕ(x0 )A(x0 , x, ϕ0 )ϕ(x)
with, as expected from our previous discussions (4.98)
1 2 µ 2 A(x , x, ϕ0 ) = −∂x0 µ ∂x + µ + λϕ0 δ(x0 − x), 2 0
nally obtaining for the value of 0
−1
Z[J] ≈ N exp i~
Z
Z (4.99) 1 0 d x[L(ϕ0 ) + Jϕ0 ] exp − tr log A(x , x, ϕ0 ) . 2 4
By our conditions (4.82) and ϕ0 [0] = 0 we then obtain (4.100)
i Z[0] = 1 ≈ N 0 exp − tr log A(x0 , x, 0) . 2 Comparison with (4.81) shows that by keeping the same de nition of A we then have (4.101)
Z X0 [J] =
d4 x[L(ϕ0 ) + Jϕ0 ]
X1 [J] =
A(x0 , x, ϕ0 ) i tr log . 2 A(x0 , x, 0)
(4.102)
The eective action is computed by expanding (4.102)
Γ[ϕcl ] = Γ0 [ϕcl ] + ~Γ1 [ϕcl ] + O(~2 ).
64 Now, since ϕcl is a functional of J, then (4.103)
ϕcl (x) =
δX[J] , δJ(x)
we have the following expansion (4.104)
Z Γ0 [ϕcl ] = X0 [J]−
Z
4
d xJϕ0 =
Z
4
4
d xL(ϕ0 ) =
d x
1 µ 1 2 2 1 4 (∂ ϕcl )(∂µ ϕcl ) − µ ϕcl − ϕcl , 2 2 4!
which is the same as (4.89). The additional term in the expanded eective action is (4.105)
Z ~Γ1 [ϕcl ] = X0 [J] − Γ0 [ϕcl ] − Z
Z
4
d x[L(ϕ0 ) + Jϕ0 ] −
=
d4 xJϕcl + ~X1 [J]
d4 x[L(ϕcl ) + Jϕcl ] +
A(x0 , x, ϕ0 ) i~ tr log . 2 A(x0 , x, 0)
Because ϕ0 is a solution of (4.91) the dierence of the two integrals in (4.105) is of order
(ϕ0 − ϕcl )2 = O(~2 ), and we can interchange ϕ0 and ϕcl at this level of accuracy. Therefore (4.106)
Γ1 [ϕcl ] =
A(x0 , x, ϕcl ) i tr log . 2 A(x0 , x, 0)
Next, the eective potential V (ϕcl ) can be derived from Γ[ϕcl ] by setting ϕcl constant, in which case (4.107)
Z Γ[ϕcl ] = −
d4 xV (ϕcl ).
Delta functions allow us to diagonalise A(x0 , x, ϕcl ) which is a prerequisite to properly de ne the logarithmic part of (4.106). This is done as follows (4.108)
0
A(x , x, ϕcl ) = Z =
d4 k (2π)4
−∂x0 µ ∂xµ
1 2 + µ + λϕcl δ(x0 − x) 2 2
Z 0 1 d4 k 1 2 2 2 ik(x0 −x) λϕ −∂x0 µ ∂xµ + µ2 + λϕ2cl eik(x −x) = −k + µ + cl e 4 2 (2π) 2
Hence performing log and trace operation (4.109)
log A(x0 , x, ϕcl ) =
Z
d4 kd4 k 0
0 0 1 2 e−ixk eix k 0 2 2 log −k + µ + λϕ cl δ(k − k) 2 (2π) 2 (2π)2
(4.110)
Z tr log A =
4
4 0
0
0
Z
d xd x δ(x − x) log A(x , x, ϕcl ) =
4
d x
Z
d4 k 1 2 2 2 log −k + µ + λϕcl . (2π)4 2
Summarizing we can now state the following: the one-loop order contribution to the eective potential is (4.111)
−i V1 (ϕcl ) = 2
Z
d4 k log (2π)4
−k 2 + µ2 + 21 λϕ2cl −k 2 + µ2
65 and while at this order, the eective potential is approximated by (4.112)
1 1 i V (ϕcl ) ≈ V0 (ϕcl ) + V1 (ϕcl ) = µ2 ϕ2cl + λϕ4cl − ~ 2 4! 2
1 λϕ2cl d4 k log 1 − , (2π)4 2 k 2 − µ2
Z
where, the Greek parameters are bare parameters, i.e. the ones present in the Lagrangian. The integral term in the potential is ultraviolet divergent and it requires dimensional regularization to deal with i.e. (see for instance Bailin and Love, p.80) (4.113)
d2ω k 2 1 (k − µ2B + iε)−1 = (2π)2ω i
Z I(ω, µB ) =
iµ2B = (M 2 )ω−2 16π 2
d2ω 6 k 2 (6 k + µ2B )−1 (2π)2ω
Z
1 µ2B 0 + O(ω − 2) . + Γ (1) + 1 − log 2−ω 4πM 2
The derivative of the gamma function at -1 is Γ0 (1) = −γ, as it can be seen by taking the logarithmic derivative of Weierstrass product (1.12)
− log Γ(s) = log s + γs +
∞ h X k=1
∞ X Γ0 (s) 1 1 s si 1 − ⇒− = +γ+ − log 1 + k k Γ(s) s k+s k
k=1
and hence 0
Γ (1) = −1 − γ −
∞ X k=1
1 1 − = −γ. k+1 k
By the use of the following transformations for the bare parameters (4.114)
ϕB (x) = Z 1/2 ϕ(x) Zµ2B = µ2 + δµ2
Z 2 λB = λ + δλ,
we can transform the above potential (4.112) to (4.115)
V (ϕcl ) =
1 2 2 1 1 1 i µ ϕcl + λϕ4cl + δµ2 ϕ2cl + δλϕ4cl − ~ 2 4! 2 4! 2
Z
1 λϕ2cl d4 k log 1 − , (2π)4 2 k 2 − µ2
with the new Greek parameters now being the renormalized parameters rather than the bare. It is important to note, however, that these parameters have constraints imposed by the MS scheme. In terms of Green functions, these develop singularities in ω − 2 and these are neutralized by the counter terms which are (4.116)
δλ = M
4−2ω
ˆ M/µ, ω) + a0 (λ,
∞ X ˆ M/µ) ak (λ, k=1
2
2
δµ = µ
ˆ M/µ, ω) + b0 (λ,
ˆ M/µ, ω) + δZ = c0 (λ,
(2 − ω)k
∞ X ˆ M/µ) bk (λ, k=1
!
!
(2 − ω)k
∞ X ˆ M/µ) ck (λ, k=1
(2 − ω)k
ˆ = λM 2ω−4 . Because we have only simple poles, with a0 , b0 and c0 regular as ω → 2 and λ the k -integrals only have simple poles at 2 − ω . This means that ∀k > 1 ⇒ ak = bk = 0.
66 The one-point-irreducible Green functions of the renormalized theory behave as (4.117)
ˆ 2 1 µ2 0 ˜ 2 (p, −p) = p2 (1 + δZ1 ) − µ2 − δµ21 + λµ Γ + Γ + O(ω − 2) . (1) + 1 − log 32π 2 2 − ω 4πM This can be shown by using (4.118)
Z I4 (µB ) =
d4 k 2 (k − µ2B + iε)−1 (2π)4
and (4.113) on the Feynman propagator (4.119)
Z ∆F (0) =
2 d4 k 2 2 −1 2ω−4 iµ (k − µ + iε) = M B (2π)4 16π 2
1 µ2 0 + Γ (1) + 1 − log + O(ω − 2) . 2−ω 4πM
This propagator, in turn, is present in the Green function as (4.120)
1 2 2 2 2 ˜ Γ (p, −p) = p (1 + δZ1 ) + µ + λi∆F (0) + δµ1 + O(λ2 ), 2 as it can be seen by writing the Feynman diagrams to leading order. ˜ 4 is lengthier but follows the same lines, eventually, since Γ ˜ 4 is nite The computation of Γ in any renormalisation scheme we have that as ω → 2 the following holds (4.121)
ˆ 1 3λλ − δλ2 → const, 32π 2 2 − ω with (4.122)
δλ =
∞ X
δλk .
k=2
The nite part of δλ2 is arbitrary. Going back to our discussion of (4.116) now we see that the poles in 2 − ω are neutralized by setting (4.123)
a1 =
3 ˆ2 λ , 32π 2
b1 =
1 ˆ λ. 32π 2
Therefore putting this back together in the potential one has (4.124)
λ 3 1 2 2 0 + Γ (1) + log 4π V (ϕcl ) = µ ϕcl 1 + b0 − 2 32π 2 2 1 4 3λ2 3 0 λϕcl 1 + a0 − + Γ (1) + log 4π 4! 32π 2 2 " # 2 µ2 + 12 λϕ2cl 1 1 µ2 2 4 + µ + λϕcl log − µ log 2 . 64π 2 2 M2 M +
In turns out that in the MS scheme we may choose (see Rajantie or Bailin and Love) (4.125) MS
δλ
ˆ 3λλ = 32π 2
1 0 + Γ (1) + log 4π + O(λ3 ) 2−ω
67
aMS = 0
ˆ2 3λ [Γ0 (1) + log 4π] + O(λ3 ), 32π 2
aMS = 1
ˆ2 3λ + O(λ3 ), 32π 2
This will get rid of the Γ0 (1) + log 4π factors. However, and here is the main objective of this comparison with the zeta function regularization, we could also renormalize V (ϕcl ) by writing it as a function of physical mass and coupling constant. This can be accomplished by setting aphys and bphys such that the following hold (4.126) 0 0
d2 V = µ2 , dϕ2cl ϕcl =0
d4 V = λ, dϕ4cl ϕcl =M
as we did in (4.20) and (4.21). In the case that µ2 is small there could be radiative corrections that generate spontaneous symmetry breaking. Performing the second and fourth derivatives (4.126) along with the discussion in (4.20) to (4.22) leads to the same potential (4.25), namely (4.127)
V (ϕcl ) =
1 2 2 λ λ2 ϕ4cl µ ϕcl + ϕ4cl + 2 4! 256π 2
ϕ2 25 log cl2 − . M 6
Eectively we have compared the zeta function regularization and the one-loop order expansion and have seen that both yield the same results in dierent fashions. The zeta function technique makes use of more powerful tools (specially, more sophisticated functions) whereas the one-loop expansion mostly requires Γ function and its derivative.
REFERENCES The above discussion of the role of the zeta function in quantum eld theory follows from Hart eld [611 to 616], Ramond [81 to 93] and the rst three chapters from Elizalde, Odintsov, and Romeo. Additionally, the exposition is the progression from the ideas presented in Chapter 2 (zeta functions in quantum mechanics) combined with Chapter 3 (path integrals in quantum elds). Finally, the discussion of the eective potential in one-loop order follows from Bailin and Love [200 - 207] and other remarks
68
Appendix Claim 1: The following holds
Z
s
Z
dxn exp iλ (x1 − a)2 + (x2 − x1 )2 + · · · + (b − xn )2 =
dx1 · · ·
Proof.
in π n iλ 2 exp (b − a) (n + 1)λn n+1
(by induction)
Assume it is true for n and show it is true for n + 1 Z Z dx1 · · · dxn+1 exp iλ (x1 − a)2 + (x2 − x1 )2 + · · · + (b − xn )2
s = =
in π n (n + 1)λn
in π n (n + 1)λn
Z dxn+1 exp
2 Z
iλ (xn+1 − a)2 exp[iλ(b − xn+1 )2 ] n+1
dxn+1 exp iλ
1 (xn+1 − a)2 + (b − xn+1 )2 n+1
,
the exponential in the integrand can be worked out as follows
1 n+2 2 = (xn+1 − a)2 + (b − xn+1 )2 y − 2y(b − a) + (b − a)2 xn+1 −a=y n + 1 n+1 2 n+1 1 n+2 y− (b − a) + (b − a)2 . = n+1 n+2 n+2 Finally, let λ − ((n + 1)/(n + 2))(b − a) = z so that the integral becomes
s
in π n (n + 1)λn
Z
s n+2 2 iλ iλ in+1 π n+1 2 2 dz exp iλ z + (b − a) = exp (b − a) , n+1 n+2 (n + 1 + 1)λn+1 n+2
and this concludes the proof by induction. Claim 3: If
J are N
M
is a symmetric N × N matrix with real-valued elements Mij and
and
component vectors with components qi and Ji respectively, then
Z
1 d q exp − qT Mq + JT q 2 N
Z(J) =
Proof.
q
The process is to diagonalize the matrix
relations hold
(2π)N/2 =√ exp det M
M
ΛT Λ = 1
det Λ = 1
Z
˜ T where the following as M = ΛMΛ
m ˜1 . ˜ = . M . 0
··· .. .
0 .. .
···
m ˜N
,
the integral becomes
Z(J) =
1 T −1 J M J 2
1 ˜ T q + JT q , dN q exp − qT ΛMΛ 2
69
˜ = ΛT J from which it follows next we de ne the following q ˜ = ΛT q and J dN q = dN q˜ det Λ = dN q˜. " # Z X 1 1 T ˜ T N 2 ˜ ˜ ˜ M˜ ˜ i q˜i + Ji q˜i Z(J) = d q exp − q q+J q ˜ = d q˜i exp − m 2 2 i ∞ " " #! ! # !−1/2 r Z N N X 1 Y X J2 Y Y ˜2 2π J i i N 2 N/2 = d q˜i exp − m ˜ i q˜i + J˜i q˜i = exp = (2π) m ˜i exp . 2 m ˜i 2m ˜i 2m ˜i i i i i=1 i=1 Z
N
−∞
Finally we note that the inverse of the diagonal matrix is
˜ −1 M
m ˜ −1 1 . = .. 0
··· .. .
0 .. .
···
m ˜ −1 N
and therefore the last product is the determinant of the matrix
Y
˜ = det M, m ˜ i = det M
i
and consequently the result follows
(2π)N/2 Z(J) = √ exp det M
1 ˜T −1 ˜ J M J 2
(2π)N/2 =√ exp det M
1 T −1 J M J . 2
Claim: One has the following
Z WE [J] = NE exp {SEuc [φ(x0 ), J]}
= NE0 q
Z 1 δ 2 SEuc d4 x1 d4 x2 φ(x1 ) φ(x2 ) Dφ exp − 2 δφ(x1 )δφ(x2 )
exp {SEuc [φ(x0 ), J]} . det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
Proof. This follows from Claim 3 with the action Z 1 1 SE [φ, J] = d4 x ¯ ∂¯µ φ∂¯µ φ + m2 φ2 + V (φ) − Jφ 2 2 as expanded on (3.54) and using
δ2 SEuc = δ(¯ x1 − x ¯2 ) −∂¯µ ∂¯µ + m2 + V 00 [φ(x1 )] δφ(x1 )δφ(x2 ) Z WE [J] = NE exp {SEuc [φ(x0 ), J]}
Z 1 δ 2 SEuc Dφ exp − d4 x1 d4 x2 φ(x1 ) φ(x2 ) 2 δφ(x1 )δφ(x2 )
70 The formulas ∞
coth x =
X (2x)2k 1 B2k , +2 x (2k)k
coth πx =
1 2x X 1 + πx π x2 + n2 n≥1
k=1
are almost always quoted as Gradshteyn and Ryzhik, p.35. The rst one represents the Laurent series. There is very little added value in reproducing the proofs here. Grassman Numbers Let us introduce some notation rst. For convenience, we will the same equation labelling as in Chapter 2 to ease the read. The following presentation about Grassmann numbers follows the notes of A. Rajantie and Peskin and Schroeder. Ordinary commuting numbers will be denoted c-numbers (these can be real or complex). Now let n generators {θ1 , · · · , θn } satisfy the anti-commutation relations (2.70)
{θi , θj } = 0 ∀i, j. Then the set of the linear combinations of {θi } with the c-number coe cient is called the Grassmann number and the algebra generated by {θi } is called the Grassmann algebra, denoted by Λn . Let us taken an arbitrary element
g(θ) = g0 +
n X i=1
gi θi +
X
gij θi θj + · · · =
i<j
g of this algebra expand it as (2.71) X
0≤k≤n
1 X gi1 ,··· ,ik θi1 · · · θik , k! {i}
where g0 , gi , gij , · · · and gi1 ,··· ,ik are c-numbers that are anti-symmetric under the exchange of two indices. Additionally, we can write
g(θ) =
X
g as (2.72)
g˜k1 ,··· ,kn θ1k1 · · · θnkn .
ki =0,1
It is impossible for the set of Grassmann numbers to be an ordered set because the generator
θk does not have a magnitude. The only number that is both c-number and Grassmann number is zero, moreover, a Grassmann number commutes with a c-number. From the discussion above it follows that (2.73)
θk2 = 0 θk1 θk2 · · · θkn = εk1 k2 ···kn θ1 θ2 · · · θn θk1 θk2 · · · θkm = 0
(m > n).
The tensor εk1 k2 ···kn is the Levi-Civita symbol, de ned as
εk1 k2 ···kn
+1 if{k1 · · · kn }isanevenpermutationof{1 · · · n} = −1 if{k1 · · · kn }isanoddpermutationof{1 · · · n} . 0 otherwise
Functions of Grassmann numbers are de ned in terms of Taylor expansions of the function.
71 If n = 1 we have the simple expression (2.74)
eθ = 1 + θ since terms O(θ2 ) are zero. Our next step is to develop the theory of dierentiation and integration of Grassmann variables, this theory has a few surprising facts, for instance dierentiation is the same process as integration. We assume that the dierential operator acts on a function from the left, let θi and θj be two Grassmann variables then (2.74)
∂θj ∂ = θj = δij . ∂θi ∂θi Similarly, we assume that the dierential operator anti-commutes with θk . The product rule has the slightly dierent form (2.75)
∂ ∂θj ∂θk (θj θk ) = θk − θj = δij θk − δik θj . ∂θi ∂θi ∂θi Moreover, the following properties hold (2.76)
∂ ∂ ∂ ∂ ∂2 + = 0 ⇒ 2 = 0, ∂θi ∂θj ∂θj ∂θi ∂θi the last equation is termed nil-potency and nally (2.77)
∂ ∂ θj + θj = δij . ∂θi ∂θi Let us now move to integration. To this end, we adopt the notation with respect to a Grassmann variable and
D for dierentiation
I for integration. Let us suppose that these
operations satisfy the relations (2.78)
ID = DI = 0, where
D(A) = 0 ⇒ I(BA) = I(B)A,
A and B are arbitrary functions of Grassmann variables. The rst part of the rst
equation implies that the integral of a derivative gives a surface term and it is set to zero, whereas the second part implies that the derivative of an integral vanishes. The last equation implies that if the derivative of the function is zero then it can be taken out the integral. The relations are satis ed when
D is proportional to I and for normalization purposes we
set I = D and write (2.79)
Z dθg(θ) =
∂f (θ) . ∂θ
From the previous de nition it follows that (2.80)
Z dθ =
∂1 = 0, ∂θ
Z dθ θ =
∂θ = 1. ∂θ
72 We can generalize this for
n generators {θi } as (2.81)
Z dθ1 dθ2 · · · dθn g(θ1 θ2 · · · θn ) =
∂ ∂ ∂ ··· g(θ1 θ2 · · · θn ). ∂θ1 ∂θ2 ∂θn
In a theory where dierentiation is equivalent to integration we would expect some strange behaviour. In order to see this behaviour we consider the simpler case when we have only one generator and we change variables θ0 = aθ where (2.82)
Z dθg(θ) =
a is a complex number, then one has
∂g(θ) ∂g(θ0 /a) = =a ∂θ ∂θ0 /a
Z
dθ0 g(θ0 /a)
which implies that dθ0 = (1/a)dθ. The extension to the general case with
n generators
yields θi → θi0 = aij θj and hence (2.83)
Z dθ1 dθ2 · · · dθn g(θ) =
n X ∂θk0 1 ∂θ0 ∂ ∂ ∂ ∂ ∂ ··· g(θ) = · · · kn 0 · · · 0 g(a−1 θ0 ) ∂θ1 ∂θ2 ∂θn ∂θ1 ∂θn ∂θk1 ∂θkn ki =1
=
n X
εk1 ···kn ak1 1 · · · akn n
ki =1
∂ ∂ · · · 0 g(a−1 θ0 ) = det a ∂θk0 1 ∂θkn
Z
dθ10 · · · dθn0 g(a−1 θ0 ).
Consequently the measure has the Jacobian (2.84)
dθ1 dθ2 · · · dθn = (det a)dθ10 · · · dθn0 . In the case of a single variable, the delta function of a Grassmann variable is de ned in a similar fashion as with c-numbers de ned as (2.85)
Z dθ δ(θ − z)g(θ) = g(z). However, in the case of Grassmann variables we can obtain a closed expression for the delta function. If we set g(z) = a + bz in the de nition of the delta function we have (2.86)
Z dθ δ(θ − z)(a + bθ) = a + bz and this means that (2.87)
δ(θ − z) = θ − z. Again, we can extend this to
n generators if we are careful about the order of the variables
(2.88)
δ n (θ − z) = (θn − zn ) · · · (θ2 − z2 )(θ1 − z1 ). We can nd the integral of the delta function by considering complex Grassmann variables which we proceed to develop later. Consider (2.89)
Z
dξ eiξθ =
Z dξ (1 + iξθ) = iθ
73 so that we have (2.90)
Z
dξ eiξθ .
δ(θ) = θ = −i
One of the most crucial developments of the Grassmann variables is the Grassmann Gaussian integral which will be fundamental when developing the path integral formalism of fermions. Let us evaluate the following integral (2.91)
Z I=
dθ1∗ dθ1 · · · dθn∗ dθn exp
X
θi∗ Mij θj
ij
where it is important to stress that {θi } and {θi∗ } are two independent sets of Grassmann variables. Since Grassmann variables θi and θi∗ anti-commute we can take the n×n c-number matrix
M to be anti-symmetric.
The formula for the transformation of the measure solves P the problem of the computation. Set θi0 = j Mij θj this yields (2.92)
Z I = det M
! dθ1∗ dθ10
· · · dθn∗ dθn0
exp −
X
θi∗ θi0
Z = det M
n dθ dθ(1 + θ θ ) = det M. ∗
0 ∗
i
Complex conjugation is de ned as (2.93)
(θi )∗ = θi∗ ,
(θi∗ )∗ = θi .
In the case of Grassmann variables we have (2.94)
(θi θj )∗ = θj∗ θi∗ . The reasoning behind (2.94) is that the real c-number θi θi∗ does not satisfy (θi θi∗ )∗ = θi θi∗ .
Let us recall that the annihilation and creation operators c and c† satisfy the anti-commutation relations {c, c† } = 1 and {c, c} = {c† , c† } = 0 and that the number operator N = c† c has eigenvectors |0 > and |1 >. We are now in a position to study the Hilbert space Ω spanned by these vectors, i.e. (2.95)
Ω = span {|0 >, |1 >} . An arbitrary vector |ω > ∈ Ω can be written in the form (2.96)
|ω > = |0 > ω0 + |1 > ω1 , with ωi ∈ C where i = 1, 2. Next we de ne the coherent states (2.97)
|θi = |0 > +|1 > θ
< θ| = < 0| + θ∗ < 1|
where θ and θ∗ are Grassmann numbers. The coherent states are eigenstates of
c and c† respectively, that is (2.98)
c|θ > = |0 > θ = |θ > θ,
< θ| c† = θ∗ < 0| = θ∗ < θ|.
74 It can be shown fairly easily that the following identities hold (2.99)
hθ0 | θi = 1 + θ0∗ θ = eθ
0∗
θ
,
hθ | gi = g0 + θ∗ g1 ,
θ |c† | g = hθ | 1i g0 = θ∗ g0 = θ∗ hθ | gi , hθ |c| gi = hθ | 0i g1 =
∂ hθ | gi . ∂θ∗
Finally, we show how matrix elements are represented and the completeness relation. Let (2.100)
p(c, c† ) = p00 + p10 c† + p01 c + p11 c† c,
pij ∈ C
c and c† . The complex matrix elements of p can be written in terms of scalar products as (2.101) be an arbitrary function of
h0 |p| 0i = p00 ,
h0 |p| 1i = p01 ,
h1 |p| 0i = p10 ,
h1 |p| 1i = p00 + p11 .
From these scalar products we can form the more general product (2.102)
hθ |p| θ0 i = (p00 + θ∗ p10 + p01 θ0 + θ∗ θ0 p11 )eθ
∗ 0
θ
.
Moreover, one has (2.103)
Z
∗
−θ ∗ θ
dθ dθ |θi hθ| e Z =
Z =
dθ∗ dθ (|0i + |1i θ) (h0| + θ∗ h1|) (1 − θ∗ θ)
dθ∗ dθ (|0i h0| + |1i θ h0| + |0i θ∗ h1| + |1i θθ∗ h1|) (1 − θ∗ θ) = |0i h0| + |1i h1| = I,
and therefore the completeness relation is (2.104)
Z
dθ∗ dθ |θi hθ| e−θ
∗
θ
= I.
75
References Apostol, Tom "Introduction to Analytic Number Theory" Bailin and Love "Introduction to Gauge Field Theory" Elizalde, Odintsov, and Romeo "Zeta Regularization Techniques with Applications" Freitag, Eberhard and Busam, Rolf "Complex Analysis" Grosche C. and Steiner F, "Handbook of Feynman Path Integrals" Ingham, A.E. "The Distribution of Prime Numbers" Kleinert, Hagen "Path Integrals in Quantum Mechanics, Statistic, Polymer Physics and Financial Markets" Hat eld, Brian "Quantum Field Theory of Point Particles and Strings" Peskin, Michael E. and Schroeder, Daniel V. "An Introduction to Quantum Field Theory" Rajantie, Arttu "Advanced quantum eld theory Course Notes" Ramond, Pierre "Field Theory: A Modern Primer" Rivers, Ray J. "Path Integral Methods in Quantum Field Theory" Sondow, Jonathan and Weisstein, Eric W. Riemann Zeta Function." From MathWorldA Wolfram Web Resource Titchmarsh, E. C. and Heath-Brown, Roger "The Theory of the Riemann Zeta-Function" Weinberg, Steven "The Quantum Theory of Fields: Volume I Foundations" Wikipedia on the Gammafunction Whitaker and Watson "A Course on Modern Analysis"
Department of Theoretical Physics Imperial College London
Date
Supervior: Dr A Ra jantie
Submitted in partial ful llment of the requirements for the degree of Master of Science in Quantum Fields and Fundamental Forces of Imperial College London
Abstract We review (...)
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Acknowledgements I should very much like to thank (...)
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Table of contents
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Introduction Special functions have arisen constantly in mathematical and theoretical physics throughout the XIX and XX centuries. Indeed, the study of theoretical physics is plagued with special functions: gamma, Dirac, Legendre, Bessel functions - to name a few - are abundant in the indices of most modern treatises on physics. The most common special functions have been the ever-present gamma function and those which are solutions to dierential equations that model physical systems. The Riemann zeta ζ function is de ned for a complex variable
ζ(s) :=
s as
∞ Y X 1 1 = s n 1 − p−s p n=1
for Re(s) > 1 and analytically continued to the whole complex plane except at s = 1 where it has a simple pole with residue 1. Note that the product is taken over all primes p. It is
not a solution to any dierential equation [1] which sets it apart from other special functions which have a more transparent physical meaning. Traditionally, the Riemann zeta function has had its implications in analytic number theory and especially in the distribution of prime numbers. As such it has been regarded mostly as a function that fell completely within the realm of pure mathematics and it was temporarily excluded. This changed in the last half of the XX century when papers from S. Hawking [2], S. Elizalde, S. Odintsov and A. Romeo [3] explained how the zeta regularization assigns nite values to super cially divergent sums. At a more academic level however, zeta regularization is hardly ever mentioned in undergraduate quantum mechanics books, nor is it mentioned either in Peskin and Schroeder or in Weinberg which are some of the standard books on quantum eld theory. It is precisely in QFT where the zeta function becomes apparent. In this MSc dissertation we will explain how the zeta function comes into play in both quantum mechanics and quantum eld theory. In order to achieve this, the dissertation will be divided in four chapters as follows. The rst chapter will be devoted to the study of the gamma and zeta functions. The lack of a course on special functions at Imperial College gives us a welcome opportunity to discuss thoroughly these objects. The treatment will be rigorous de nition-theorem-proof style and the only essential pre-requisites are those of complex analysis: convergence, analytic continuation and residue calculus. The functional equation is of particular importance
ζ(s) = 2(2π)s−1 Γ(1 − s) sin as well as the formulas
1 ζ(0) = − , 2
sπ 2
ζ(1 − s)
1 ζ 0 (0) = − log(2π) 2
which will be used time and again. As we have just stated, zeta regularization consists in assigning nite values to super cially divergent sums and this is precisely what the two above formulas do, since when written formally we have the odd looking sums
1 1 + 1 + 1 + ··· = − , 2
log 1 + log 2 + log 3 + · · · =
1 log(2π). 2
4 Of course, this needs a note of clari cation. The ζ function can be written as
ζ(s) =
∞ ∞ X X 1 (−1)n+1 1 = . ns 1 − 21−s n=1 ns n=1
These two sums agree on, however, when Re(s) > 1, however, the RHS is the step to analytic continuation and it goes − 21 to as s → 0 hence ζ(0) = − 12 in this sense. Indeed, this shows that in a certain sense zeta regularization can be thought of as a technique of complex analysis, namely analytic continuation. Because of the equivalence with other regularizations, this principle can be extended to other renormalizations. We have limited the discussion, however, to the essential aspects of the zeta functions that we will need for the rest of the dissertation and thus avoided the beautiful connection between number theory and the zeta function. The determinant of an operator
A can be written as the in nite product of its eigenvalues
log det A = log
Y
an = tr log A =
n
X
log an
n
When we take the operator to be A := −d2 /dt2 + · · · , the ζ function arises naturally by using
ζA (s) := trA
−s
X 1 = , asn n
X d ζA (s) =− log an , ds s=0 n
and the functional determinant of the operator is 0 det A = exp(−ζA (0)).
Quantum mechanical partition functions will then be computed via zeta regularization, by taking operator to be
AQM = −
d2 + ω2 . dτ 2
Indeed, we shall see that the partition function of a quantum mechanical harmonic oscillator is
ˆ = Z(β) = tr exp(−β H)
Z
ˆ dx < x| exp(−β H)|x >=
1 . 2 sinh(βω/2)
We will explain how the Riemann zeta function is used in the bosonic case, whereas a slightly more exotic zeta function (Hurwitz) is needed for fermions. The partition function of the harmonic oscillator will be our main example. Furthermore, we will make use of formulas proved in chapter 1 showing the necessity of having discussed it at length. Since fermionic theories demand anti-commutation relations we will need Grassmann numbers and therefore we will explain these in detail but only in so much as is needed to compute partition functions for fermions. Finally, equipped with all the tools we have developed in the preceding three chapters we will see how special functions can be used in eld theory. Because of the way special functions arise in eld theory we have devoted a whole chapter to explain the path integral approach to quantum eld theory. This approach is complementary to the QFT/AQFT courses from the MSc.
5 The only eld theory we shall consider is ϕ4 however it is important to note that zeta regularization can be applied to more complex eld theories and even string theory [3]. We shall keep this section brief and avoid topics such as Feynman diagrams whenever possible. We will consider quantum eld theories of a scalar eld in the presence of an external source J. It will be shown how the generating functional [4]
ZE [J] = e−iE[J] = < Ω|e−iHT |Ω > =
Z Dφ exp i d4 x(L[φ] + Jφ)
Z
can be written as
ZE [J] = NE p
[−∂µ
∂µ
exp(SEuc [φ0 ], J) + m2 + V 00 [φ0 ]δ(x1 − x2 )]
and can be computed as done in chapter 2 via zeta function regularization as the determinant of the operator is already visible on the denominator of the LHS. By taking more complex operators, which account for eld theoretical Lagrangians, such as
AFT = −∂ 2 + m2 +
λ 2 φ (x), 2 0
we shall see how zeta regularization is also used in eld theory. We will show that by taking into account rst order quantum corrections, the potential in the bare Lagrangian of the eld theory is renormalized as [5]
V (φcl ) =
λ 4 λ2 φ4cl φcl + 4! 256π 2
log
25 φ2cl − M2 6
where φcl is the classical eld de ned in terms of the source
J and vacuum state Ω as
φcl = < Ω|φ(x)|Ω >J . We will show this rstly by constructing a special zeta function associated with the operator
A. The key steps, for both the quantum mechanical and eld cases, are the connection between the zeta function ζA of the operator and its eigenvalues,
ζ−d2 /dt2 (s) =
−2s ∞ X nπ n=1
β
2s β = ζ(2s), π
∂ as well as the link between the heat function G(x, y, t) - solution of Ax G(x, y, t) = − ∂t G(x, y, t)
- and the zeta function
Z ζA (s)Γ(s) =
∞
dtts−1
Z
d4 xG(x, x, t).
0
We will then discuss how coupling constants evolve in terms of scale dependence and by exploring the analogy between eld theory and statistical mechanics we will compute the partition function of a QFT system. This will be the culmination of the formulas we have proved in chapter 1, the techniques developed in chapter 2 and the theory explained in chapter 3. Furthermore, this encapsulates the spirit of the technique we will use repeatedly. Finally, as a means of a check to make sure the technique yields the same results as other
6 regularization techniques we will derive the same results by using more standard methods such as the one-loop expansion approach. In terms of the knowledge of physics, the only pre-requisite comes from eld theory up to the notes of QFT/AQFT from the MSc in QFFF. The appendix contains some formulas that did not t in the presentation and makes the whole dissertation almost self-contained. References have been provided in each individual chapter, including pages where the main ideas have been explored. Overall, the literature on zeta regularization is sparse. The main sources for this dissertation have been Grosche and Steiner, Kleinert, Ramond and the superb treatise by Elizalde, Odintsov, and Romeo. It has been an objective to try to put results from dierent sources in a new light, by clarifying proofs and creating a coherent sent of examples and applications which are related to each other and which are of increasing complexity. Finally, a few remarks which I have not been able to nd in the literature have been made concerning a potential relationship between zeta functions and families of elementary particles. Dierent zeta functions come into play in quantum physics by computing partition functions. As we have said above, the Riemann zeta function is used in bosonic partition functions, whereas a slightly dierent zeta function is needed for the fermionic partition function. Riemann zeta function ⇔ bosons
Hurwitz zeta function ⇔ fermions
It would be an interesting subject of research to investigate how the zeta function behaves with respect to its corresponding particle, and what kind of knowledge we can extract about the particle given its special zeta function. For instance, a photon has an associated zeta function, and its partition function can be computed by using known facts of this zeta function, on the other hand, known facts about the boson might bring clari cations to properties of this zeta function. The celebrated Riemann hypothesis claims that all the nontrivial zeros of the Riemann zeta function are of the form ζ( 12 + it) with
t real.
Hence, another example would be to consider an operator which brings the zeta function to the form ζ( 21 + it) and investigate the behaviour of zeta function with respect to this operator. This would mean a 'translation' of the Riemann hypothesis and a study of its physical interpretations. REFENCES
The Riemann Hypothesis Stephen Hawking Zeta function regularization of path integrals in curved spacetime Elizalde, Odintsov, and Romeo Zeta Regularization Techniques with Applications Peskin, Michael E. and Schroeder, Daniel V. An Introduction to Quantum Field Theory Ramond, Pierre Field Theory: A Modern Premier
.1. J. Brian Conrey .2. .3. .4. .5.
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Introduction to Special Functions Often in mathematics, it is more natural to de ne a function in terms of an integral depending on a parameter rather than through power series. The gamma function is one such case. Traditionally, it can be approached as a Weierstrass product or as a parameter-dependent integral. The approach chosen to introduce the gamma function follows from a course in complex analysis such as the one delivered by Eberhard Freitag and Rolf Busam [2]. We adopt the notation s = σ + it which was introduced by Riemann in 1859 and which has become the standard in the literature of the zetafunction. Let us de ne the Euler Gamma function.
De nition 1 The integral (1.1) ∞
Z
dtts−1 e−t
Γ(s) := 0
is well de ned and de nes a holomorphic function in the right half complex plane, where
Re(s) = σ > 0. The rst lemma generalises the factorial function as follows
Lemma 1 For any n ∈ N we have Γ(n) = (n − 1)!. R Proof. Note that Γ(1) := 0∞ dte−t and by integration by parts we have (1.2) Z Γ(s + 1) =
∞ s −t
dtt e
=
−ts e−t |∞ 0
Z +s
0
for any
∞
dtts−1 e−t = sΓ(s)
0
s in the right half-plane. Now, for any positive integer n, we have (1.3) Γ(n) = (n − 1)Γ(n − 1) = (n − 1)!Γ(1) = (n − 1)!
so the result is proved. In order to have a complete view of the gamma function we need to extend it to a meromorphic function in the whole complex plane.
Lemma 2 P∞
n=0
Let cn where n ∈ Z + be a sequence of complex numbers such that the sum
|cn | converges. Furthermore, let S = {−n|n ∈ Z + andcn 6= 0}. Then (1.4) f (s) =
∞ X cn s +n n=0
converges absolutely for s ∈ C −S and uniformly on bounded subsets of C −S . The function
f is a meromorphic function on C with simple poles at the points in S and the residues are by given ress=−n f (s) = cn by for any −n ∈ S .
Proof.
Let us start by nding upper bounds. If |s| < R, then |s + n| ≥ |n − R| for all
n ≥ R. Therefore, we have |s + n|−1 ≤ (n − R)−1 for |s| < R and n ≥ R. From this we can
8 deduce that for n0 > R we have
∞ ∞ ∞ ∞ X X X X cn |cn | 1 |cn | ≤ ≤ |cn |. ≤ n=n s + n n=n |s + n| n=n n − R n − R n=n 0
As such the series
0
0
0
converges absolutely and uniformly on the disk |s| < R P∞ n=0 cn /(s + n) is a meromorphic
P
n>R cn /(s + n)
and de nes there a holomorphic function. It follows that
function on that disk with simple poles at the points of S on the disk |s| < R. Thus, P∞ n=0 cn /(s + n) is a meromorphic function with simple poles at the points in S and for
any −n ∈ S we can write
f (s) =
cn + s+n
X −k∈S−{n}
ck cn = + g(s), s+k s+n
where g is holomorphic at −n. From this we see that residues are indeed ress=−n f (s) = cn . This concludes the proof. Equipped with this lemma we are in a position to extend the Γ function as we wanted.
Theorem 1
The Γ function extends to a meromorphic function on the complex plane.
It has simple poles at 0, −1, −2, −3, · · · . The residues of Γ at are given by (1.7)
res Γ(s) =
s=−k
(−1)k , k!
for any k ∈ Z + .
Proof.
Let us split the gamma function as (1.8)
Z
1 s−1 −t
Γ(s) =
dtt
e
∞
Z
dtts−1 e−t ,
+
0
1
the second integral converges for any complex
s and it is an entire function. Let us expand
the exponential function in the rst integral
Z
1 s−1 −t
dtt 0
e
Z =
1 s−1
dtt 0
∞ X (−1)k
k!
k=0
k
t =
Z ∞ X (−1)k k=0
k!
1
dttk+s−1 =
0
∞ X (−1)k k=0
k!
1 , s+k
these operations are valid for s ∈ C as the exponential function is entire and converges uniformly on compact sets of the complex plane. The gamma function can now be written in a form where Lemma 2 can be used, i.e. (1.9)
Z Γ(s) = 1
for any
∞
dtts−1 e−t +
∞ X (−1)k k=0
k!
1 , s+k
s in the right half-plane. By Lemma 2, the RHS de nes a meromorphic function on
the complex plane with simple poles at 0, −1, −2, −3, · · · . The residues are given a direct application of the lemma.
9
Theorem 2 For s ∈ C any we have Γ(s + 1) = sΓ(s). Proof. This follows directly from Lemma 1 and Theorem 1. We three-dimensional representation of the gamma function looks like (Wikipedia) Another important function related to the Γ function, and also discovered by Euler, is the Beta function which we proceed to develop as follows. Let Re(p), Re(q) > 0 and in the integral that de nes the gamma function, make the change of variable t = u2 to obtain ∞
Z
dttp−1 e−t = 2
Γ(p) =
Z
0
∞
2
duu2p−1 e−u .
0
In an analogous form we have ∞
Z
dvv 2q−1 e−v
Γ(q) = 2
2
0
Multiplying these two together we have
Z
∞
∞
Z
0
2
dudve−(u
Γ(p)Γ(q) = 4
+v 2 ) 2p−1 2q−1
u
v
,
0
and switching to polar coordinates u = r cos θ, v = r sin θ, dudv = rdrdθ
Z∞ Zπ/2 2 Γ(p)Γ(q) = 4 drdθ e−r r2(p+q)−1 cos2p−1 θ sin2q−1 θ 0
Z∞
= 2
0
π/2 Z Zπ/2 2q−1 −r 2 2(p+q)−1 2p−1 dr e r dθ cos θ sin θ = 2Γ(p+q) dθ cos2p−1 θ sin2q−1 θ. 2
0
0
0
The integral can be simpli ed by setting z = sin2 θ
Zπ/2 Z1 2q−1 2p−1 2 dθ cos θ sin θ = dz z q−1 (1 − z)p−1 . 0
0
Next we de ne
Z1 B(p, q) :=
dz z p−1 (1 − z)q−1
0
for Re(p), Re(q) > 0, and this gives the identity
B(p, q) = B(q, p) = We denote by
Γ(p)Γ(q) . Γ(p + q)
B the Beta function. Moreover, if 0 < x < 1 we have
Γ(x)Γ(1 − x) Γ(x)Γ(1 − x) = = B(x, 1 − x) = Γ(1)
Z1 0
dz z x−1 (1 − z)−x .
10 We will evaluate this integral with an appropriate contour, but rst we need to make one last change z = u/(u + 1) which yields
Z1 dz z
x−1
−x
(1 − z)
0
Z∞ =
ux−1 du (u + 1)2 (u + 1)x−1
u 1− u+1
Z∞
−x =
du
ux−1 . 1+u
0
0
Lemma 3 For 0 < y < 1 we have (1.10) Z∞ du
π u−y = . 1+u sin πy
0
Proof.
Let us use a keyhole contour, which is accomplished by cutting the complex plane
along the positive real axis. On this region we de ne the function
s−y 1+s
f (s) =
with argument of s−y equal to 0 on the upper side of the cut. Furthermore, the function has a rst order pole at s = −1 with residue e
−iyπ
f
. See Figure 1 below.
Figure 1 We are now to integrate this function along the path described in Figure 1: the path goes along the upper side of the cut from > 0 to R, then along the circle CR of radius R centred at the origin, then along the side of the cut from
R to and at the end around the origin
via the circle C of radius also centred at the origin. An application of the Cauchy residue theorem gives
2πie−πiy =
ZR du ε
u−y + 1+u
I dz
z −y − e−2πiy 1+z
ZR du ε
CR
u−y − 1+u
I dz
z −y . 1+z
Cε
We can get rid of the integrals around the arcs by appropriate estimates. Note that for we have the following
|z −y | = |e−y log z | = e−yRe(log z) = e−y log |z| = |z|−y , −y −y z |z|−y ≤ |z| ≤ , 1 + z |1 + z| |1 − |z|| and the integrals can be estimated
I 1−y −y z ≤ 2π R dz → 0, 1 + z R − 1 R→∞ CR
I 1−y −y z dz ≤ 2π ε → 0; 1 + z 1 − ε ε→0 Cε
11 so that we are left with
(1 − e
−2πiy
Z∞ )
du
u−y = 2πie−πiy . 1+u
0
We may re-write this to obtain the nal result (1.11)
(e
πiy
−e
−πiy
Z∞ )
Z∞
u−y du = 2πi 1+u
⇒
du
0
u−y π = . 1+u sin πy
0
This proves the claim of the lemma. Let us now make the concluding remarks.
Theorem 3 (Euler Re ection Formula) For all s ∈ C Γ(s)Γ(1 − s) =
Proof.
one has
π . sin πs
The lemma we have just proved can be written as
Γ(x)Γ(1 − x) =
π π = , sin π(1 − x) sin πx
for 0 < x < 1. However, both sides of the equation above are meromorphic, hence we have proved the theorem.
Corollary 1 One has Γ( 12 ) = Proof.
R∞
dt t−1/2 e−t =
R∞
2
dt e−t =
√
π.
−∞
0
This follows by substituting s = 1/2 in the Euler re ection formula.
Theorem 4 The Γ function has no zeroes. Proof. Since s → sin(πs) is an entire function,
the RHS of Theorem 3 has no zeroes,
therefore Γ(s) = 0 only happens where s → Γ(1 − s) has poles. However, as we have argued before, the poles of Γ are at 0, −1, −2, −3, · · · so it follows that must have poles at
1, 2, 3, · · · . By the factorial formula, Γ(n + 1) = n! 6= 0 and so Γ has no zeroes. Intrinsically connected to the Γ function is the Euler γ constant. Let us rst de ne it and prove its existence.
Lemma 4
If then sn = 1 +
1 2
+ ··· +
1 n
− log n, exists. This limit is called the Euler
γ constant.
Proof.
Consider tn = 1 +
1 2
+···+
1 n−1
− log n geometrically, it represents the area of the Rn dx x−1 . Therefore 1
n − 1 regions between the upper Riemann sum and the exact value of tn increases with n. We can write tn =
n−1 X k=1
1 k+1 − log k k
lim n→∞ tn
=
∞ X 1 k=1
1 − log 1 + . k k
12 The series on the right converges to a positive constant since
0<
1 1 1 1 1 1 − log 1 + = 2 − 3 + 4 − ··· ≤ 2, k k 2k 3k 4k 2k
next, the following hold
sn+1 − sn =
1 1 − log 1 + , n+1 n
tn+1 − tn =
1 1 − log 1 + n n
which means that
1 1 1 < log 1 + < ⇒ sn+1 − sn < 0 < tn+1 − tn . n+1 n n Convergence now follows because sn decreases monotonically whereas tn increase monotonically and the dierences are negative. Hence sn is monotonically decreasing and bounded below thus convergent.
Lemma 5
Weierstrass product of the Γ function (1.12) n Y 1 s −s/k sγ =lim se 1 + e . Γ(s) n→∞ k k=1
Proof.
Using the fact that (1 − t/n)n → e−t as n → ∞ it can be shown that
Γ(s)
=lim n→∞
Zn dt t
s−1
t 1− n
n
=lim n→∞
1 nn
Zn
dt ts−1 (n − t)n
0
0
and integrating by parts yields
Γ(s)
=lim n→∞
1 n nn s
Zn
s
n−1
dt t (n − t)
=lim n→∞
n(n − 1) · · · 1 1 nn s(s + 1) · · · (s + n − 1)
0
Zn
dt ts+n−1
0
=lim n→∞
ns s
1 s+1
2 s+2
···
n s+n
.
Inverting both sides n Y 1 s s lim s −s −s =lim sn (s + 1) 1 + · · · 1 + = sn 1+ n→∞ n→∞ Γ(s) 2 n k k=1
In order to be limit we need to insert the convergence factor e−s/k to obtain n Y 1 s −s/k −s s(1+1/2+···+1/n) =lim sn e 1 + e Γ(s) n→∞ k k=1
" =lim n→∞
e
s(1+1/2+···+1/n−log n)
s
n Y k=1
# s −s/k 1+ e . k
13 However by the use of Lemma 4, we know that the sum converges γ to so that we have shown the result (1.12). The Hurwitz zeta function ζ(s, a) is initially de ned for σ > 1 by the series (1.13)
ζ(s, a) :=
∞ X
1 . (n + a)s n=0
This is provided that n + a 6= 0. The reason why we work with a generalized zeta function, rather than with the zeta function itself, is because fermions require this special kind of zeta function for their regularization. Note however that bosons require the Riemann zeta function (a = 1). For the special case of the Riemann zeta function, the 3-d plot looks like (mathworld.com). The discussion presented here of the properties of the Riemann zeta function has its foundations in Titchmarsh [3]. Although the roots of the functional equation go back to Riemann, the development of the Hurwitz zeta can be traced back to Apostol [1] which in turn is taken from Ingham. Figure Let us now examine the properties of the Hurwitz zeta function.
Proposition 1
The series ζ(s, a) for converges absolutely for σ > 1. The convergence
is uniform in every half-plane σ ≥ 1 + δ with δ > 0 so ζ(s, a) is analytic function of
s in the
half-plane σ > 1.
Proof.
From the inequalities ∞ ∞ ∞ X X X (n + a)−s = (n + a)−σ ≤ (n + a)−(1+δ) . n=1
n=1
n=1
all the statements follow and this proves the claim. The analytic continuation of the zeta function to a meromorphic function in the complex plane is more complicated that in the case of the gamma function.
Proposition 2 For σ > 1 we have the integral representation (1.14) Z∞ Γ(s)ζ(s, a) =
dx
xs−1 e−ax . 1 − e−x
0
In the case of the Riemann zeta function, that is when a = 1 we have (1.15)
Z∞ Γ(s)ζ(s) =
dx
xs−1 e−x . 1 − e−x
0
Proof.
First we consider the case when
s is real and s > 1, then extend the result to
14 complex
s by analytic continuation. In the integral for the gamma function we make the
change of variable x = (n + a)t where n ≥ 0 and this yields s
Z∞
Γ(s) = (n + a)
dte−(n+a)s ts−1
0
which can be re-arranged to
(n + a)
−s
Z∞ Γ(s) =
dte−nt e−at ts−1 .
0
Next, we sum over all n ≥ 0 and this gives ∞
ζ(s, a)Γ(s) =
∞ Z X
dte−nt e−at ts−1 ,
n=0 0
where the series on the right is convergent if Re(s) > 1. To nish the proof we need to interchange the sum and the integral signs. This interchange is valid by the theory of Lebesgue integration; however we do not proceed to prove this more rigorously because it would take us too far from the subject at matter. Therefore, we may write ∞
ζ(s, a)Γ(s) =
∞ Z X
dte
−nt −at s−1
e
t
Z∞ =
n=0 0
dt
∞ X
e−nt e−at ts−1 .
n=0
0
However, if Im(s) = t > 0 we have 0 < e−t < 1 and therefore we may sum ∞ X
e−nt =
n=0
1 , 1 − e−t
by geometric summation. Therefore the integrand becomes
Z∞ ζ(s, a)Γ(s) =
dt 0
∞ X
−nt −at s−1
e
e
t
n=0
Z∞ =
dt
e−at ts−1 . 1 − e−t
0
Now we have the rst part of the argument and we need to extend this to all complex
s
with Re(s) > 1. To this end, note that both members are analytic for Re(s) > 1. In order to show that the right member is analytic we assume 1 + δ ≤ σ ≤ c where c > 1 and δ > 0. We then have
Z∞ 0
1 −at s−1 Z∞ Z Z∞ −at σ−1 e t e−at tσ−1 ≤ dt e t dt = dt + dt . −t −t 1−e 1−e 1 − e−t 0
0
1
Notice the analogy of splitting the integral as in the proof of Theorem 1. If 0 ≤ t ≤ 1 we have tσ−1 ≤ tδ and if t ≥ 1 we have tσ−1 ≤ tc−1 . Also since et − 1 ≥ t for
15
t ≥ 0 we then have Z1
and
e−at tσ−1 ≤ dt 1 − e−t
Z1
0
0
Z∞
Z∞
e−at tσ−1 ≤ dt 1 − e−t
e(1−a)t tδ dt t ≤ e(1−a) e −1
dt tδ−1 =
e1−a , δ
0
e−at tc−1 dt ≤ 1 − e−t
1
1
Z1
Z∞ dt
e−at tc−1 = ζ(c, a)Γ(c). 1 − e−t
0
This proves that the integral in the statement of the theorem converges uniformly in every strip 1 + δ ≤ σ ≤ c, where δ > 0, and therefore represents an analytic function in every such strip, hence also in the half-plane σ = Re(s) > 1. Therefore, by analytic continuation, (1.14) holds for all
s with Re(s) > 1.
Consider the keyhole contour
C : a loop around the negative real axis as show in Figure
2. The loop is made of three parts C1 , C2 , and C3 . The C2 part is a positively oriented circle of radius ε < 2π above the origin, and C1 and C3 are the lower and upper edges of a cut in the z -plane along the negative real axis. Figure 2 This can be translated into the following parametrizations: z = re−πi on C1 and z = reπi on C3 where
r varies from ε to ∞.
Proposition 3 If 0 < a ≤ 1 the function de ned by the contour integral (1.16) 1 I(s, a) = 2πi
Z dz
z s−1 eaz 1 − ez
C
is an entire function of s. Moreover, we have (1.17)
ζ(s, a) = Γ(1 − s)I(s, a) if Re(s) = σ > 1.
Proof.
Write z s = rs e−πis on C1 and z s = rs eπis on C3 . Let us consider an arbitrary
compact disk |s| ≤ M and we proceed to prove that the integrals along C1 and C3 converge
uniformly on every such disk. Since the integrand is an entire function of s, this will prove that the integral I(s, a) is entire. Along C1 we have for r ≥ 1,
|z s−1 | = rσ−1 |e−πi(σ−1+it) | = rσ−1 eπt ≤ rM −1 eπM since |s| ≤ M . The same on C3 gives
|z s−1 | = rσ−1 |eπi(σ−1+it) | = rσ−1 e−πt ≤ rM −1 eπM also for r ≥ 1. Therefore, independently on which side of the cut we place ourselves, we
16 have that for r ≥ 1,
s−1 az z e rM −1 eπM e−ar rM −1 eπM e(1−a)r ≤ = . 1 − ez 1 − e−r er − 1 However, er − 1 > er /2 when r > log 2 so the integrand is bounded by ArM −1 e−ar where A R∞ is a constant depending on M but not on r. The integral ε dr rM −1 e−ar converges if ε > 0 so this proves that the integrals along C1 and C3 converge uniformly on every compact disk
|s| ≤ M and hence I(s, a) is indeed an entire function of s.
To prove the equation of the theorem, we have to split up the integral as
Z
Z
2πiI(s, a) =
dz +
C1
Z dz +
C2
dz z s−1 g(z)
C3
where g(z) = eaz /(1 − ez ). According to the parametrizations we have on C1 and C3 that
g(z) = g(−r) but on the circle C2 we write z = εeiθ , where −π ≤ θ ≤ π . This gives us Zε 2πiI(s, a) =
dr r
s−1 −iπs
e
Zπ g(−r) + i
∞
s−1 (s−1)iθ
dθ ε
e
iθ
iθ
Z∞
εe g(εe ) +
−π
dr rs−1 eiπs g(−r).
ε
Divide by 2i and names the integrals I1 and I2
πI(s, a) = sin(πs)I1 (s, ε) + I2 (s, ε). If we let ε → 0 we see that lim ε→0 I1 (s, ε)
Z∞ =
dr
rs−1 e−ar = Γ(s)ζ(s, a), 1 − e−r
0
as long as σ > 1. In |z| < 2π the function
g is analytic except for a rst order pole at
z = 0. Therefore zg(z) is analytic everywhere inside |z| < 2π and hence is bounded there, say |g(z)| ≤ Ω/|z|, where |z| = ε < 2π and Ω is a constant. We can then write
εσ |I2 (s, ε)| ≤ 2
Zπ
dθ
e−tθ
Ω ε
≤ Ωeπ|t| εσ−1 .
−π
When we let ε → 0 and provided that σ > 1 we nd that I2 (s, ε) → 0 hence we have
πI(s, a) = sin(πs)Γ(s)ζ(s, a). Finally, by the use of the formula that we proved earlier Γ(s)Γ(1 − s) = π/ sin(πs) we have a proof of (1.17). Now we have to extend this result for complex numbers such that σ ≤ 1. In the statement that we have just proved the functions I(s, a) and Γ(1 − s) make sense for every complex s, and thus we can use this equation to
de ne ζ(s, a) for σ ≤ 1.
17
De nition 2 If σ ≤ 1 we de ne ζ(s, a) by the equation (1.18) ζ(s, a) = I(s, a)Γ(1 − s). This provides the analytic continuation of ζ(s, a) in the entire
Theorem 5
s plane.
The function ζ(s, a) de ned above is analytic for all
s except for a simple
pole at s = 1 with residue 1.
Proof.
The function I(s, a) is entire so the only possible singularities of ζ(s, a) must be
the poles of Γ(1 − s), and we have shown those to be the points s = 1, 2, 3, · · · . However Theorem 1 shows that ζ(s, a) is analytic at s = 2, 3, · · · so s = 1 is the only possible pole of
ζ(s, a).
If s in an integer s = n the integrand in the contour integral for I(s, a) takes the same value on both C1 and C3 and hence the integrals along C1 and C3 cancel, yielding
1 I(n, a) = 2πi
Z dz
z n−1 eaz z n−1 eaz = res . z=0 1 − ez 1 − ez
C2
In this case we have s = 1 and so
eaz zeaz z −1 lim = =lim =lim = −1. z→0 z→0 z→0 z z z z=0 1 − e 1−e 1−e ez
I(1, a) = res
Finally, the residue of ζ(s, a) at s = 1 is computed as lim s→1 (s
lim − 1)ζ(s, a) = −lim s→1 (1 − s)Γ(1 − s)I(s, a) = −I(1, a)s→1 Γ(2 − s) = Γ(1) = 1,
now the claim is complete: ζ(s, a) has a simple pole at s = 1 with residue 1. Let us remark that since ζ(s, a) is analytic at s = 2, 3, · · · and Γ(1 − s) has poles at these points, then (2) implies that I(s, a) vanishes at these points. Also we have proved that the Riemann zeta function ζ(s) is analytic everywhere except for a simple pole at s = 1 with residue 1.
Lemma 6
Let S(r) designate the region that remains when we remove from the
z plane
all open circular disks of radius r, 0 < r < π with centres at z = 2nπi, n = 0, ±1, ±2, · · · . Then if 0 < a ≤ 1 the function
g(s) :=
eas 1 − es
is bounded in S(r).
Proof.
With our usual notation: s = σ + it we consider the rectangle H(r) with the circle
at n = 0 this rectangle has an indentation as follows
H(r) = {s : |σ| ≤ 1, |t| ≤ π, |s| ≥ r} , as shown in Figure 3 below.
18 Figure 3
H so de ned is compact so g is bounded on H. Also, because of the periodicity |g(s + 2πi)| = |g(s)|, g is bounded in the perforated in nite strip The set
{s : |σ| ≤ 1, |s − 2nπi| ≥ r, n = 0, ±1, ±2, · · ·} . Let us suppose that |σ| ≥ 1 and consider
as aσ e eaσ = e ≤ . |g(s)| = s s 1−e |1 − e | |1 − eσ | We can examine the numerator and denominator for σ ≥ 1 giving |1 − eσ | = eσ − 1 and
eaσ ≤ eσ so |g(s)| ≤
eσ 1 e 1 ≤ = = . eσ − 1 1 − e−σ 1 − e−1 e−1
A similar argument when σ ≤ −1 gives |1 − eσ | = 1 − eσ and so
|g(s)| ≤
1 1 e eσ ≤ ≤ = . 1 − eσ 1 − eσ 1 − e−1 e−1
And therefore, as we claimed |g(s)| ≤ e/(e − 1) for |σ| ≤ 1.
De nition 3 The periodic zeta function is de ned as F (x, s) := where
∞ X e2πinx , ns n=1
x is real and σ > 1.
Let us remark the following properties of the periodic zeta function. It is indeed a periodic function of
Theorem 6
x with period 1 and F (1, s) = ζ(s).
The series converges absolutely if σ > 1. If
also converges (conditionally) for σ > 0.
Proof.
This is because for each xed non-integral
x is not an integer the series
x the coe cients have bounded partial
sums.
Theorem 7 (Hurwitz's formula) If 0 < a ≤ 1 and σ > 1 we have (1.19) ζ(1 − s) =
Γ(s) −πis/2 (e F (a, s) + eπis/2 F (−a, s)). (2π)s
If a 6= 1 this is also valid for σ > 0.
Proof.
Consider the function de ned by the contour integral (1.20)
IN (s, a) :=
1 2πi
I dz C(N )
z s−1 eaz , 1 − ez
19 where is the contour show in Figure 4 and
N is an integer. It is the same keyhole contour
as that of the gamma function, only that it has been rotated for convenience. The poles are located on the y -axis, symmetric to the origin, at multiplies of 2iπ Figure 4 Let us rst prove that if σ < 0 then
lim N →∞ IN (s, a)
= I(s, a). The method to prove this
is to show that the integral along the outer circle tends to 0 as N → ∞. On the outer circle we have z = R eiθ , −π ≤ θ ≤ π , hence
|z s−1 | = |Rs−1 eiθ(s−1) | = Rσ−1 e−tθ ≤ Rσ−1 eπ|t| . The out circle is inside the domain S(r) described in the lemma, the integrand is bounded
A is the bound for |g(s)| implied by the lemma, hence the whole integral is bounded by 2πeπ|t| Rσ which tends to 0 as R tends to in nity as long as σ < 0. Now when we replace s by 1 − s this yields
by Aeπ|t| Rσ−1 where
lim N →∞ IN (1
− s, a) =lim N →∞
1 2πi
I dz
z −s eaz = I(1 − s, a) 1 − ez
C(N )
for σ > 1. We are left with the problem of computing IN (1 − s, a) which we proceed to do by the use of the Cauchy residue theorem. Formally, n=N X
IN (1 − s, a) = −
res f (z) = −
n=−N,n6=0
z=n
n=N X n=1
res f (z) + res f (z)
z=n
z=−n
where f (z) is the integrand of IN (1 − s, a) and residues are calculated as follows
res f (z) =
z=n
res
z=2nπi
z −s eaz 1 − ez
=lim z→2nπi (z−2nπi)
e2nπia lim e2nπia z −s eaz z − 2nπi = = − , z→2nπi 1 − ez (2nπi)s 1 − ez (2nπi)s
which in turn gives (1.20)
IN (1 − s, a) =
N N X e2nπia X e−2nπia + . (2nπi)s n=1 (2nπi)s n=1
Now we make the following replacements i−s = e−πis/2 and (−i)−s = eπis/2 which allow us to write
IN (1 − s, a) =
N N e−πis/2 X e2nπia eπis/2 X e−2nπia + (2π)s n=1 ns (2π)s n=1 (2nπi)s
and we let N → ∞
I(1 − s, a) =
e−πis/2 eπis/2 F (a, s) + F (−a, s). s (2π) (2π)s
20 We have thus arrived at the following result (1.21)
ζ(1 − s, a) = Γ(s)I(1 − s, a) =
o Γ(s) n −πis/2 e F (a, s) + eπis/2 F (−a, s) . s (2π)
This proves the claim. The simplest particular case (and the most important one) is when we take a = 1 this gives us the functional equation of the Riemann zeta function (1.22)
ζ(1 − s) =
o Γ(s) n −πis/2 Γ(s) πs πis/2 e ζ(s) + e ζ(s) = 2 cos ζ(s). (2π)s (2π)s 2
This is valid for σ > 1 but it also holds for all formulation can be obtained by switching
s by analytic continuation. Another useful
s with 1 − s (1.23)
ζ(s) = 2(2π)s−1 Γ(1 − s) sin
πs ζ(1 − s). 2
Let us now see the consequences of this equation. Taking s = 2n + 1 in (1.22) when
n in
an integer the cosine factor vanishes and we nd the trivial zeroes of ζ(s) (1.24)
n = 1, 2, 3, · · · .
ζ(−2n) = 0
Later we will need to use certain other values of the Riemann zeta function which we can now compute. In particular the value of ζ(−n, a) can be calculated if
n is a non-negative
integer. Taking s = −n in the formula ζ(−s, a) = Γ(1 − s)I(s, a) we nd that
ζ(−n, a) = Γ(1 + n)I(−n, a) = n!I(−n, a), where
I(−n, a) = res
z=0
z −n−1 eaz 1 − ez
,
the evaluation of this residue requires special functions of its own (special type of polynomials, rather) which are known as the Bernoulli polynomials.
De nition 4 For any complex s
we de ne the functions Bn (s) as ∞ X Bn (s) n zesz = z ez − 1 n=0 n!
provided that |z| < 2π . A particular case of the polynomials are the Bernoulli number Bn = Bn (0) i.e. ∞ X z Bn (0) n = z . ez − 1 n=0 n!
Lemma 7 One has the following equations Bn (s) =
Pn
n n−k . k=0 (k ) Bk s
In particular when
21
s = 1 this yields Bn =
Proof.
Pn
n k=0 (k ) Bk .
Using a Taylor expansion and comparing coe cients on both sides we have ∞ X Bn (s) n z z = z esz = n! e − 1 n=0
∞ X Bn n z n! n=0
!
∞ X sn n z n! n=0
! ,
n
Bn (s) X Bk sn−k = n! k! (n − k)! k=0
and by passing n! to the RHS we obtain the lemma. Now we can write the values of the zeta function in terms of Bernoulli numbers.
Lemma 8 For every integer n ≥ 0 we have (1.25) ζ(−n, a) = −
Proof.
Bn+1 (a) . n+1
This follows from the previous observation that ζ(−n, a) = n!I(−n, a), so we just
have to evaluate the integral
I(−n, a) = res
z=0
z −n−1 eaz 1 − ez
I by the Cauchy residue theorem.
= − res
z=0
z −n−2 zeaz ez − 1
= − res
z=0
z
−n−2
∞ X Bk (a) k=0
k!
! z
k
=−
Bn+1 (a) , (n + 1)!
and dividing by n! we have the end of the proof.
Lemma 9
The recursion Bn (s + 1) − Bn (s) = nsn−1 is valid for Bernoulli polynomials
if n ≥ 1 and in particular for Bernoulli numbers when n > 1: Bn (0) = Bn (1).
Proof.
From the identity
z it follows that
esz e(s+1)z −z z = zesz z e −1 e −1
∞ ∞ X Bn (s + 1) − Bn (s) n X sn n+1 z = z , n! n! n=0 n=0
and as we did before, we equate coe cients of z n to obtain the rst statement and then set
s = 0 to obtain the second statement. Using the de nition, the rst Bernoulli number is B0 = 1 and the rest can be computed by recursion. We obtain the following values Building on from the Bernoulli numbers we can construct the polynomials by the use of the lemmas, the rst ones as s Bernoulli polynomial Bernoulli number Table
22 Note that for n ≥ 0 we have by setting a = 1 in (1.25) (1.26)
ζ(−n) = −
Bn+1 n+1
and because of the trivial zeroes of the Riemann zeta function ζ(−2n) = 0 we con rm our observation that the odd Bernoulli numbers are zero, i.e. B2n+1 = 0 . Also note (1.27)
1 ζ(0) = − . 2 Finally we can write a compact formula for the even values of the zeta function in terms of Bernoulli numbers.
Theorem 8 Suppose n
is a positive integer, then (1.28)
ζ(2n) = (−1)n+1
Proof.
(2π)2n B2n . 2(2n)!
This follows from the functional equation by setting s = 2n
ζ(1 − 2n) = 2(2π)−2n Γ(2n) cos(πn)ζ(2n) re-arranging we obtain
−
B2n = 2(2π)−2n (2n − 1)!(−1)n ζ(2n), 2n
from which the result follows. Using the tabulated values given above we have the well-know Euler formulas
ζ(2) =
π2 , 6
ζ(4) =
π4 . 90
Let us remark that no such formula for odd values is known and in fact the value of ζ(3) remains one of the most elusive mysteries of modern mathematics. A signi cant advance was achieved by Roger Apery who was able to show in 1979 that is an irrational number.
Theorem 9 One has ζ 0 (0) = − 21 log(2π) Proof. We may write (1.16) and (1.17) as ζ(s) 1 = Γ(1 − s) 2πi
Z C
where
(−z)s−1 1 dz z = e −1 2πi
Z
dz (−z)s z ez − 1
C
C is the same contour as that of Figure 2, except shifted to the positive in nity
instead of negative in nity and to account for this we have (−z)s = exp[s log(−z)]. Let us
23
s and then set s = 1, we obtain
dierentiate with respect to
Z
1 2πi
dz (−z) log(−z) . z ez − 1
C
The integral on the RHS can be split as
1 2πi
Zε
dz (−z)(log z − iπ) 1 + z z e −1 2πi
+∞
dz (−z)(log ε + iθ − iπ) 1 + z z e −1 2πi
I
Z∞
dz (−z)(log z + iπ) z ez − 1
ε
|z|=ε
and writing z = εei(φ+π) in the middle integral we have
Z∞ −
dz log ε − z e −1 2πi
ε
I
dz z 1 − z z e − 1 2πi
Zπ dφ · φ
ez
z , −1
−π
|z|=ε
at this point we need to evaluate all three integrals. The rst one can be expanded as
Z∞ −
dz =− ez − 1
Z∞ dz ε
ε
∞ X
−nz
e
n=1
z=∞ ∞ ∞ X X e−nz (e−ε )n =− = − −n n n=1 n=1 z=ε
ε3 ε ε2 + − · · · = log ε + log 1 − + · · · . = log(1 − e−ε ) = log ε − 2 6 2 The second integral is solved by Cauchy's integral theorem by noting that at z = 0 we have
z(ez − 1)−1 → 1, and therefore we have log ε − 2πi
I
dz z = − log ε. z ez − 1
|z|=ε
Finally, the third integral goes to zero as ε → 0. Putting all these facts together we have shown that
1 2πi
Z
dz (−z) log(−z) = log ε − log ε + O(εα ) = 0. z ez − 1
C
Re-arranging the functional equation (1.23)
ζ(s) πs = 2(2π)s−1 ζ(1 − s) sin Γ(1 − s) 2 and because the derivative at s = 1 is zero, as we have just shown, its logarithmic derivative
log(2π) −
ζ 0 (1 − s) π cos(πs/2) + ζ(1 − s) 2 sin(πs/2)
must also be 0 at s = 1 and consequently (1.29)
log(2π) =
ζ 0 (0) , ζ(0)
nally yielding the result of Theorem 9 by use of (1.27).
24
Theorem 10 One has ζ(0, 12 ) = 0 and ζ 0 (0, 21 ) = − 12 log 2. Proof. We have the following identity ζ(s, 1/2) + ζ(s) = 2s
∞ X n=1
1 1 + = 2s ζ(s), (2n − 1)s (2n)s
from which it follows that (1.30)
ζ(s, 1/2) = (2s − 1)ζ(s) and hence the rst formula is shown. By dierentiating (1.30) with respect to
s we have
the second formula of the theorem. REFERENCES There exist many sources for the development of the Riemann zeta function. The most comprehensive one is by E.C. Titchmarsh (and revised by D.R. Heath-Brown). Here we have followed the course on analytic number theory by Tom Apostol [2] for the presentation of the zeta function and its analytic continuation. The part of the gamma function follows from the course on Complex Analysis by Freitag and Busam and Whitaker [1] and Watson's Modern Analysis [4].
25
Zeta Regularization in Quantum Mechanics Zeta Regularization in Quantum Mechanics One of the rst instances where the Riemann zeta function occurs in quantum physics is in the path integral development of the harmonic oscillator; speci cally it takes place when computing the partition function of the spectrum of the harmonic oscillator. We will follow Exercise 9.2 p 312 from Peskin and Schroeder and the path integral development from Chapter 2 of Kleinert. The action of the one-dimensional harmonic oscillator is given by (2.1)
Ztf S=
dtL, ti
where the Lagrangian is (2.2)
L=
1 1 mx˙ 2 − mω 2 x2 . 2 2
As we know from the functional approach to quantum mechanics, the transition amplitude is the functional integral (2.3)
Z < xf , tf | xi , ti > =
DxeiS[x(t)] .
The extremum of S, xc (t), satis es (2.4)
δS[x] = 0. δx x=xc (t) We now proceed to expand the action around xc (t). This indicates that xc (t) is the classical trajectory connecting both space-time points of the amplitude and therefore it satis es the Euler-Lagrange equation (2.5)
x ¨c + ω 2 xc = 0. The solution to the equation above with conditions xc (ti ) = xi and xc (tf ) = xf is (2.6)
xc (t) = (sin ωT )−1 [xf sin ω(t − ti ) + xi sin ω(tf − t)], where T = tf − ti . We next plug this solution into the action
Sc := S[xc ] =
S (2.7)
mω [(x2 + x2i ) cos ωT − 2xf xi ]. 2 sin ωT f
As we intended originally, we now expand S[x] around x = xc to obtain (2.8)
Z S[xc + z] = S[xc ] +
Z δS[x] δ 2 S[x] 1 dtz(t) dt1 dt2 z(t1 )z(t2 ) + δx(t) x=xc 2! δx(t1 )δx(t2 ) x=xc
26 where z(t) satis es the boundary condition (2.9)
z(ti ) = z(tf ) = 0. The expansion ends at second order because the action is in second order in x. Noting that
δS[x]/δx = 0 at x = xc we are left with the rst and last terms only. Because the expansion is nite (2.10)
1 S[xc + z] = S[xc ] + 2!
Z
δ 2 S[x] dt1 dt2 z(t1 )z(t2 ) δx(t1 )δx(t2 ) x=xc
the problem can be solved analytically. Let us now compute the second order functional derivative in the integrand; this can be accomplished as follows (2.11)
δ δx(t1 )
Ztf
1 1 ˙ 2 − mω 2 x(t)2 dt mx(t) 2 2
= −m
ti
d2 2 + ω x(t1 ). dt21
Next, using the rule (2.12)
δx(t1 ) = δ(t1 − t2 ) δx(t2 ) we obtain the following expression for the second order functional derivative (2.13)
δ 2 S[x] = −m δx(t1 )δx(t2 )
d2 2 + ω δ(t1 − t2 ). dt21
Plugging this back into the equation for the expansion, we can use the delta function to get rid of the
t variables (2.14)
S[xc +z] = S[xc ]−
m 2!
Z dt1 dt2 z(t1 )z(t2 )δ(t1 − t2 )
d2 + ω2 dt21
= S[xc ]+
m 2!
Z
dt(z˙ 2 − ω 2 z 2 ),
where we have simpli ed the expression by using (2.9). A crucial point to be made is that because Dx is invariant, we may replace it by Dz and this will give us (2.15)
< xf , tf | xi , ti > = eiS[xc ]
Z
m Dz exp i 2
Ztf
dt(z˙ 2 − ω 2 z 2 ).
ti
z(ti )=z(tf )=0
The uctuation part (integral at z(0) = z(T ) = 0) (2.16)
Z If := z(0)=z(T )=0
m Dz exp i 2
ZT
dt(z˙ 2 − ω 2 z 2 )
0
is computed as follows. First, let us shift the variables so that time start at 0 and ends at
27
T. We Fourier expand z(t) as (2.17) z(t) =
∞ X
nπt . T
an sin
n=1
This choice satis es the (2.9). The integral in the exponential gives (2.18)
ZT 0
∞ T X 2 nπ 2 2 a −ω . dt(z˙ − ω z ) = 2 n=1 n T 2
2 2
In order to have a well-de ned transformation we must check that the number of variables is the same before and after the transformation. Indeed, the Fourier transformation from
y(t) to an may be thought of as a change of variables in the integration. To check this, take the number of the time slice to be N + 1, including both t = 0 and t = T , for which there exist N − 1 independent zk variables. Therefore, we must set an = 0 for all n > N − 1. Next, we compute the corresponding Jacobian. Denote by tk the k th time slice when the interval [0, T ] is split into
N in nitesimal parts, then (2.19) JN
∂zk nπtk = det . = det sin ∂an T
We evaluate the Jacobian for the easiest possible case, which is that of the free particle. Therefore, let us make a digression to evaluate the above mentioned probability amplitude. For a free particle, the Lagrangian is L =
1 ˙ 2. 2 mx
Carefully derived solutions to the free
particle can be found in Chapter 2 of Kleinert as well as in Chapter 3 of Grosche and Steiner. The amplitude is computed by rst noting that the Hamiltonian is given (2.20)
H = px˙ − L =
p2 , 2m
so that (2.21) ˆ
< xf , tf | xi , ti > = < xf |e−iHT | xi > = Z =
dp ip(xf −xi ) −iT p2 /(2m) e e = 2π
r
Z
ˆ
dp < xf |e−iHT | p >< p| xi >
m exp 2πiT
im(xf − xi )2 2T
,
where T = tf − ti as noted before and ε here denotes the discretization of time ε = T /N . Similarly to the theory of functional integration we have amplitude (2.22)
< xf , tf | xi , ti > =
lim n→∞
" n # m n/2 Z X m xk − xk−1 2 dx1 · · · dxn−1 exp iε . 2πiε 2 ε k=1
We now change the coordinates to zk = (m/(2ε))1/2 xk so that the amplitude becomes (2.23)
< xf , tf | xi , ti > =
lim n→∞
" n # m n/2 2ε (n−1)/2 Z X 2 dz1 · · · dzn−1 exp i (zk − zk−1 ) . 2πiε m k=1
28 In the appendix, we prove by induction that (2.24)
"
Z
dz1 · · · dzn−1 exp i
n X
# (zk − zk−1 )
2
=
k=1
(iπ)n−1 n
1/2
2
ei(zn −z0 )
/n
.
The expression reduces to (2.25)
hxf , tf | xi , ti i =
lim n→∞
r m n/2 2πiε (n−1)/2 1 m im im(xf −x0 )2 /(2nε) √ e = exp (xf − xi )2 . 2πiε m 2πiT 2T n
Taking (2.25) into account we arrive at the (2.26)
< xf , T | xi , 0 > =
1/2
1 2πiT
1/2 im 1 2 exp eiS[xc ] . (xf − xi ) = 2T 2πiT
When we write this in terms of a path integral we obtain (2.27)
Z
eiS[xc ]
m Dz exp i 2
Ztf
dtz˙ 2 .
ti
z(0)=z(T )=0
Now, from (2.18) we have (2.28)
m 2
ZT
dt z˙ 2 → m
0
N X a2n n2 π 2 4T n=1
and when we compare both path integral expressions one has the equality (2.29)
1 2πiT
1/2
Z
m Dz exp i 2
=
Ztf
dtz˙
2
=
lim N →∞ JN
1 2πiε
1/2 Z da1 . . . daN −1 exp im
n=1
ti
z(0)=z(T )=0
N −1 X
a2n n2 π 2 4T
Now comes the process of evaluating the Gaussian integrals (2.30)
1 2πiT
1/2 =
lim N →∞ JN
1 2πiε
N/2 NY −1 n=1
1 n
4πiT π2
1/2 =
lim N →∞ JN
1 2πiε
N/2
1 (N − 1)!
4πiT π2
(N −1)/2
and from this we obtain a formula for the Jacobian (2.31)
JN = 2−(N −1)/2 N −N/2 π N −1 (N − 1)! → ∞. N →∞
The Jacobian is divergent, but this divergence is not relevant because JN is combined with other divergent factors. Let us now return to the original problem of the probability amplitude of a harmonic oscillator. The amplitude was (2.32)
< xf , T | xi , 0 > =
lim N →∞ JN
1 2πiT
N/2 e
iS[xc ]
Z
# N −1 mT X 2 nπ 2 2 −ω . da1 . . . daN −1 exp i a 4 n=1 n T "
! .
29 As we did with the free particle, we carry out the computation of the Gaussian integrals using the formula (2.33)
Z
1/2 " 2 #−1/2 ωT imT 2 nπ 2 4iT 1− dan exp − ω2 . a = 4 n T πn2 nπ
This breaks the amplitude (2.32) into smaller parts (2.34)
hxf , tf | xi , ti i =
lim N →∞ JN
=
lim N →∞
N 2πiT
N/2 e
iS[xc ]
N −1 Y k=1
1 2πiT
1/2 e
iS[xc ]
" " 1/2 # NY 2 #−1/2 −1 1 4iT ωT 1− k π nπ n=1
N −1 Y
"
1−
n=1
ωT nπ
2 #−1/2 .
It can be shown that his product is equal to (2.35) lim N →∞
N Y
"
1−
n=1
ωT nπ
2 # =
sin ωT . ωT
The divergence of Jn cancels the divergence of the other terms and therefore we are left with a nite value con rming what we stated above concerning the irrelevance of J∞ . When we insert the value of the product we arrive at the nal result (2.36)
1/2 1/2 2 ω ω iω iS[xc ] 2 (xf + xi ) cos ωT − 2xi xf . < xf , tf | xi , ti > = e = exp 2πi sin ωT 2πi sin ωT 2 sin ωT
ˆ whose spectrum is bounded from below then, by adding a If we have a Hamiltonian H ˆ positive de nite, i.e. (2.37) positive constant to the Hamiltonian, we can make H ˆ = {0 < E0 ≤ E1 ≤ · · · ≤ En ≤ · · ·} . specH Also we assume that the ground state is not degenerate. The spectral decomposition of ˆ
e−iHt is (2.38) ˆ
e−iHt =
X
e−iEn t |n >< n|
n
ˆ > = and this decomposition is analytic in the lower half-plane of t, where we have H|n En |n >. As we have done when evaluating the Gaussian integrals we introduce the Wick rotation ˆ
ˆ
t = −iτ where τ is real and positive, this gives us x˙ = idx/dτ and e−iHt = e−Hτ so that (2.39)
Ztf i ti
# # 2 2 Zτf " Zτf " 1 1 dx 1 dx dt mx˙ 2 − V (x) = i(−i) dt − m − V (x) = − dt m + V (x) 2 2 dτ 2 dτ
τi
τi
30 Consequently, the path integral becomes (2.40)
ˆ
< xf , tf | xi , ti > = < xf |e−H(τf −τi ) | xi > =
Z
¯ exp − Dx
Zτf
"
1 m dτ 2
dx dτ
#
2
+ V (x) ,
τi
¯ is the integration measure in the imaginary time τ . Equation (2.40) shows the where Dx connection between the functional approach and statistical mechanics.
ˆ as Z(β) = Let us now de ne partition function [see Kleinert, p.77] of a Hamiltonian H ˆ
Tre−β H , where β is a positive constant and the trace is over the Hilbert space associated ˆ . This partition can be written in terms of eigenstates of energy {|En >} with (2.41) with H ˆ = En |En >, H
< Em | En >= δmn .
In this case (2.42)
Z(β) =
X
ˆ
< En |e−β H | En > =
X
< En |e−βEn | En > =
n
n
X
e−βEn ,
n
or in terms of the eigenvector |x > of the position operator x ˆ, (2.43)
Z
ˆ
dx < x|e−β H | x > .
Z(β) =
Initially we had an arbitrary β but if we set it to be β = iT we nd that (2.44) ˆ
ˆ
< xf |e−iHT | xi > = < xf |e−β H | xi >, and from this we have the path integral expression of the partition function, (2.45)
Z Z(β) =
Z dz
¯ exp − Dx
Zβ
dτ
1 mx˙ 2 + V (x) = 2
0
x(0)=x(β)=z
Z periodic
¯ exp − Dx
Zβ
dτ
1 mx˙ 2 + V (x) , 2
0
where the periodic integral indicates that the integral is over all paths which are periodic in the interval [0, β]. When we apply this to the harmonic oscillator, the partition function is simply (2.46) ˆ −β H
Tre
=
∞ X
e−β(n+1/2)ω .
n=0
Although there are many ways of evaluating the partition function, here we choose one where the use of the zeta regularization is illustrated. Proceed as follows. Set imaginary time τ = iT to obtain the path integral (2.47)
Z z(0)=z(T )=0
Z i d2 2 Dz exp dt z − 2 − ω z → 2 dt
Z
Z 1 d2 2 ¯ Dz exp − dτ z − 2 + ω z , 2 dt
z(0)=z(β)=0
¯ indicates the path integration measure with imaginary time. Suppose we in this case Dz
31 have an n × n Hermitian matrix
M with positive-de nite eigenvalues λk where 1 ≤ k ≤ n
then we show in the appendix that (see Rajantie) (2.48) n Y
Z∞
k=1
"
1 dxk exp − 2
# X
−∞
xp Apq xq = π n/2
p,q
n Y
π n/2 1 √ = . det M λk k=1
This is a matrix generalization of the scalar Gaussian integral
Z∞
1 dx exp − λx2 2
r
=
2π , λ
λ > 0.
−∞
The next task is to de ne the determinant of an operator O by the in nite product of its Q eigenvalues λk . This is accomplished by setting DetO = k λk . Note that Det with capital d denotes the determinant of an operator, whereas with a small d it denotes the determinant of a matrix, same applies for Tr and tr. Using this we can we can write the integral over imaginary time as (2.49)
Z
−1/2 Z 2 2 ¯ exp − 1 dτ z − d + ω 2 z = DetD − d + ω 2 Dz 2 dt2 dτ 2
z(0)=z(β)=0
here the
D denotes the Dirichlet boundary condition z(0) = z(β) = 0.
Similarly to what we did with (2.18) we see that the general solution (2.50) ∞ 1 X nπτ z(τ ) = √ zn sin . β β n=1
We are restricted to having the coe cients zn real as
z is a real function. We are now in
a position to write formal expressions. Knowing that the eigenvalues of the eigenfuction
sin(nπτ /β) are λn = (nπ/β)2 + ω 2 we may write the determinant of the operator as (2.51) DetD
d2 − 2 + ω2 dτ
" " # 2 Y 2 # ∞ ∞ ∞ 2 Y Y nπ nπ βω 2 1+ λn = +ω = . = β β mπ n=1 n=1 m=1 n=1 ∞ Y
It is now time to identify the rst in nite product with the functional determinant, i.e. (2.52)
2 ∞ Y d2 nπ DetD − 2 ↔ . dτ β n=1
Here is where the ζ function comes into play. Suppose O is an operator with positive-de ne eigenvalues λn . Following Grosche and Steiner, in this case, we take the log (2.53)
log DetO = log
Y
λn = Tr log O =
n
n
Now we de ne the spectral ζ function as (2.54)
ζO (s) :=
X
X n
λ−s n .
log λn .
32 The sum converges for su ciently large Re(s ) and ζO (s) is analytic in Additionally, it can be analytically continued to the whole
s in this region.
s plane except at a possible
nite number of points. The derivative of the spectral zeta-function is linked to functional determinant by (2.55)
X dζO (s) = − log λn . ds s=0 n
And therefore the expression for DetO is (2.56)
dζO (s) DetO = exp − . ds s=0 The operator we are interested in is O = −d2 /dτ 2 so this yields (2.57)
ζ−d2 /dτ 2 (s) =
−2s ∞ X nπ β
n=1
2s β ζ(2s). π
=
As we proved in Chapter 1, the zeta function is analytic over the whole complex
s plane
except at the simple pole at s = 1. The values
ζ(0) = −
1 2
1 ζ 0 (0) = − log(2π) 2
were also calculated, and we can use them now to obtain (2.58) 0 ζ−d 2 /dτ 2 (0)
β = 2 log ζ(0) + 2ζ 0 (0) = − log(2β). π
Putting this into the expression for the determinant with Dirichlet conditions we have (2.59)
d2 DetD − 2 = elog(2β) = 2β dτ and nally (2.60)
DetD
d2 − 2 + ω2 dτ
= 2β
∞ Y
"
1+
p=1
βω pπ
2 # .
Note how the in nite product now becomes convergent due to ζ(0) and ζ 0 (0)! Let us go back to the partition function (2.61)
Tre
ˆ −β H
=
∞ X
" e
−β(n+1/2)ω
n=0
=
2β
= 2β
∞ Y p=1
π sinh βω βω
−1/2
( 1+
βω pπ
π ω tanh(βω/2)
2 )#−1/2
1/2 =
π ω tanh(βω/2)
1/2
1 . 2 sinh(βω/2)
It is important to note that there is a more direct way of computing this partition function and which is more satisfying for solvable cases such as the harmonic oscillator but which fails under more complicated Lagrangians.
33 This can be done by computing (2.43)
Z
ˆ
Z(β) = Tre−β H =
ˆ
dx < x|e−β H | x >.
Recall that
hxf , tf | xi , ti i =
1/2 2 ω iω exp (xf + x2i ) cos ωT − 2xi xf , 2πi sin ωT 2 sin ωT
so that
Z(β) =
ω 2πi(−i sinh βω) =
ω 2π sinh βω
1/2 Z
1/2
ω 2 2 dx exp i (2x cosh βω − 2x ) −2i sinh βω
π ω tanh βω/2
1/2 =
1 . 2 sinh(βω/2)
The quantisation of bosonic particles is done by using commutation relations, however, the quantisation of fermionic particles require a more dierent approach, namely that of anticommutation relations. This in turn requires anti-commuting numbers which are called Grassmann numbers. In analogy with the bosonic harmonic oscillator which was described by the Hamiltonian, (2.62)
H = 21 (a† a + aa† ), where
a and a† satisfy the commutation relations (2.63) [a, a† ] = 1
[a, a] = [a† , a† ] = 0.
The Hamiltonian has eigenvalues (n+ 21 )ω where n is an integer with eigenvector |n > (2.64)
H|n >= (n + 12 )ω|n > . From now we drop the hat notation in the operators whenever there is no risk for confusion with the eigenvalue. The prescription for the fermionic Hamiltonian is to set (2.65)
H = 12 (c† c − cc† )ω. This may be thought of as a Fourier component of the Dirac Hamiltonian, which describes relativistic fermions. However, it is evident that if
c and c† were to satisfy commutation
relations then the Hamiltonian would be a constant, and therefore it is more appropriate to consider anti-commutation relations (2.65)
c, c† = 1
{c, c} = c† , c† = 0.
In this case the Hamiltonian becomes (2.66)
H=
1 2
† c c − (1 − c† c) ω = N − 21 ω,
34
N are either 0 or 1 since
where N = c† c. The eigenvalues of
N 2 = c† cc† c = N ⇔ N (N − 1) = 0. Next we need a description of the Hilbert space of the Hamiltonian. To this end, let |n > be an eigenvector of
H with eigenvalue n (necessarily n = 0, 1). Then the follow equations
hold (2.67)
H |0i = −
ω |0i 2
H |1i =
ω |1i 2
For the sake of convenience we introduce the spin-notation (2.68)
0
|0i = c† |0i = |1i
1
! ,
|1i =
1
! .
0
c† |1i = 0 c |0i = 0 c |1i = |0i .
When the basis vectors of the space have this form then the operators take the matrix representations (2.69)
c=
0
0
1
0
! †
c =
0
1
0
0
! N=
1
0
0
0
!
ω H= 2
1
0
0
−1
! .
Instead of having the bosonic commutation relation [x, p] = i we now have [x, p] = 0. The anti-commutation relation {c, c† } = 1 is replaced by {θ, θ∗ } = 0 where θ and θ∗ are anticommuting numbers, i.e. Grassmann numbers which we proceed to develop in further detail in the appendix. The Hamiltonian of the fermionic harmonic oscillator is H = (c† c − 1/2)ω, with eigenvalues
±ω/2. From our previous discussion of the partition function and Grassmann numbers we know that (2.105)
Z(β) = Tre−βH =
1 X
< n|e−βH | n > = eβω/2 + e−βω/2 = 2 cosh(βω/2).
n=0
As with the bosonic case, we can evaluate Z(β) in two dierent ways using the path integral formalism. Let us start with some preliminary propositions. Let
H be the Hamiltonian of
a fermionic harmonic oscillator, its partition function is written as (2.106)
Tre
−βH
Z =
∗
dθ∗ dθ < −θ|e−βH | θ > e−θ θ .
We can show this by the inserting (2.104) into the partition function (2.105) this yields (2.107)
Z(β) =
X n=0,1
=
XZ n
< n|e−βH | n > =
XZ n
dθ∗ dθ|θ >< θ|e−θ
∗
θ
< n|e−βH | n > =
XZ
dθ∗ dθe−θ
n
dθ∗ dθ(1 − θ∗ θ) (< n | 0 > + < n | 1 > θ) < 0|e−βH | n > +θ∗ < 1|e−βH | n >
∗
θ
< n | θ >< θ|e−βH | n
35
=
XZ
dθ∗ dθ(1 − θ∗ θ)
n
× < 0|e−βH | n >< n | 0 > −θ∗ θ < 1|e−βH | n >< n | 1 > +θ < 0|e−βH | n >< n | 1 > +θ∗ < 1|e−βH | n >< n | 0 > . Note now that the last term of the integrand does not contribute to the integral and therefore we may substitute θ∗ to −θ∗ which implies that (2.108)
Z(β) =
XZ
dθ∗ dθ(1 − θ∗ θ)
n
× < 0|e−βH | n >< n | 0 > −θ∗ θ < 1|e−βH | n >< n | 1 > +θ < 0|e−βH | n >< n | 1 > −θ∗ < 1|e−βH | n >< n | 0 > Z ∗ = dθ∗ dθe−θ θ < −θ|e−βH | θ > . Unlike the bosonic case, we have to impose an anti-periodic boundary over [0, β] in the trace since the Grassmann variable is θ when τ = 0 and it is −θ when τ = β . By invoking the expression (2.109)
e−βH = lim (1 − βH/N )N N →∞
and inserting the completeness relation (2.104) at each step one has the following expression for the partition function (2.110)
Z Z(β) = lim
N →∞
Z = lim
N →∞
= lim
N →∞
dθ∗ dθ
N −1 Y
" dθk∗ dθk exp − "
dθk∗ dθk
N −1 X
∗
θ
−θ (1 − βH/N )N θ
# θn∗ θn ×h−θ | (1 − εH)| θN −1 i hθN −1 | · · · | θ1 i hθ1 | (1 − εH)| θi
n=1
k=1
Z Y N
dθ∗ dθe−θ
exp −
N X
# θn∗ θn
×hθN | (1 − εH)| θN −1 i hθN −1 | · · · | θ1 i hθ1 | (1 − εH)| − θN i ,
n=1
k=1
where we have the usual conventions and we have been using all along (2.111)
ε = β/N
and θ = −θN = θ0 ,
∗
∗
θ∗ = −θN = −θ0 .
Matrix elements are evaluated (up to rst order) as (2.112)
< θk |H| θk−1 > < θk |(1 − εH)| θk−1 > = < θk | θk−1 > 1 − ε < θk | θk−1 > ∗
∗
∗
= hθk | θk−1 i exp(−ε < θk |H| θk−1 > / < θk |θk−1 >) = eθk θk−1 e−εω(θk θk−1 −1/2) = eεω/2 e(1−εω)θk θk−1 . In terms of the path integral the partition function becomes (2.113)
Z(β) = lim e N →∞
βω/2
N Z Y k=1
" dθk∗ dθk
exp −
N X n=1
# θn∗ θn
" exp (1 − εω)
N X n=1
# θn∗ θn
36
=e
βω/2
N Z Y
lim
N →∞
" dθk∗ dθk
exp −
N X
# {θn∗ (θn
− θn−1 ) +
εωθn∗ θn−1 }
n=1
k=1
= eβω/2 lim
N →∞
N Z Y
†
dθk∗ dθk exp(−θ · B · θ),
k=1
where we have the following vector and matrix elements (2.114)
θ1 . . θ= . , θN
†
θ =
θ1∗
···
∗ θN
,
BN
1
0
···
0
=
y
1
0
···
0 .. .
y .. .
1 .. .
··· .. .
0
0
···
y
−y
0 0 , .. . 1
where y = −1 + εω . The computation is ended by recalling that from the de nition of the Grassmann Gaussian integral one had (2.115)
βω 1 + (1 − βω/N )N = eβω/2 (1+eβω ) = 2 cosh . N →∞ 2
Z(β) = eβω/2 lim det BN = eβω/2 lim N →∞
As with the bosonic partition function, we can arrive to the same result using the ζ function (a generalization of it), and this will prove useful later. Recall that we showed that (2.115)
Z(β) = eβω/2 lim
N →∞
N Z Y
†
dθk∗ dθk exp(−θ · B · θ) = eβω/2
Z
Dθk∗ Dθk exp −
k=1
Zβ
dτ θ∗
d + ω θ (1 − εω) dτ
0
=e
βω/2
d Detθ(β)=−θ(0) (1 − εω) +ω . dτ
The subscript θ(β) = −θ(0) indicates that the eigenvalue should be evaluated for the solutions of the anti-periodic boundary condition θ(β) = −θ(0). First, we expand the orbit θ(τ ) in the Fourier modes. The eigenmodes and the corresponding eigenvalues are (2.116)
exp where
πi(2n + 1)τ β
,
(1 − εω)
πi(2n + 1) + ω, β
n runs as n = 0, ±1, ±2, · · · . The number of degrees of freedom is N (= β/ε) so the
coherent states are (over)complete. The presence ε in operator will account for the fact that the in nite contribution of the eigenvalues is nite. Since one complex variable has two real degrees of freedom, we need to truncate the product at −N/4 ≤ k ≤ N/4. Following this prescription, one has (2.117) N/4
Z(β) = e
βω/2
lim
N →∞
Y k=−N/4
" # 2 ∞ Y π(2n − 1) 2π(n − 1/2) βω/2 −βω/2 2 i(1 − εω) +ω =e e +ω β β k=1
37
=
2 ∞ ∞ Y π(2k − 1) Y β
k=1
"
1+
n=1
βω π(2n − 1)
2 # .
The trouble comes from the rst in nite product, Ξ, which is divergent and as such it is in need of ζ values to become nite. This can be accomplished as follows (2.118)
log Ξ = 2
∞ X k=1
2π(k − 1/2) , log β
and we de ne the corresponding ζ function by (which is the Hurwitz ζ function) (2.119)
−s ∞ X 2π(k − 1/2)
ζfermion (s) =
β
k=1
=
β 2π
s ζ(s, 1/2),
where (see Chapter 1, Eq 1.13) (2.120)
ζ(s, a) =
∞ X
(k + a)−s ,
k=0
where 0 < a < 1. This gives (2.121) 0 Ξ = exp(−2ζfermion (0)).
So now we are left with the issue of dierentiating the ζfermion function at s = 0 which is done as follows (2.122) 0 ζfermion (0) = log
β 2π
1 ζ(0, 1/2) + ζ 0 (0, 1/2) = − log 2, 2
since we showed in Chapter 1 (Theorem 10) that (2.123)
ζ(0, 12 ) = 0,
ζ 0 (0, 12 ) = − 12 log 2.
Putting this together, we obtain the surprising result (2.124) 0 Ξ = exp(−2ζfermion (0)) = elog 2 = 2.
This result indicates that Ξ is independent of β once the regularization is performed. Finally, the partition function is evaluated using all these facts (2.125) ∞ Y
Z(β) = 2
"
1+
n=1
βω π(2n − 1)
2 # = 2 cosh
βω , 2
by the virtue of the formula (2.126) ∞ Y n=1
" 1+
x π(2n − 1)
2 #
x = cosh . 2
This is in agreement with partition function we computed earlier (2.35) and (2.61).
38 Richard Feynman was an advocate of using solutions of known problems in unknown problems, quoting him 'The
same equations have the same solutions '. The rationale behind this
statement is once we solve a mathematical problem, we can re-use the solution in another physical situation. Feynman was skilled in transforming a problem into one that he could solve. This is precisely what we have done in this case. We can prove (2.126) using path integrals.
REFERENCES The discussion of the role of the zeta function in quantum mechanics and zeta function regularization as explained in the above chapter can be traced back to the Handbook of Feynman Path Integrals C. Grosche and F. Steiner [pages 37 to 44, 55 to 59 for the fermionic case], Peskin and Schroeder [pages 299 to 301] and the Advanced Quantum Field Theory course by Dr Rajantie. It is also complemented by the treatise on path integrals in quantum mechanics from Hagen Kleinert [pages 81 to 83 and 161 to 163].
39
Path Integrals in Field Theory From the quantum mechanical case we can build a generalization with several degrees of freedom, a eld theory. We will exclusively deal with φ4 , where φ(x) is a real scalar eld. Let us summarize standard results from eld theory from Peskin and Schroder as well as Rajantie. The action is built from the Lagrangian (3.1)
Z S= where it is understood that
dx L(φ(x), ∂µ φ(x)),
L is the Lagrangian density. The equations of motion (EOM)
are given by the Euler-Lagrange equation (3.2)
∂ ∂L ∂L =− . ∂xµ ∂(∂µ φ(x)) ∂φ(x) From the free scalar eld Lagrangian (3.3)
1 L0 (φ(x), ∂µ φ(x)) = − (∂µ ∂ µ φ + m2 φ2 ) 2 we can derive the Klein-Gordon equation (3.4)
(∂µ ∂ µ − m2 )φ = 0. When there is a source J present the vacuum amplitude has functional representation (3.5)
Z < 0, ∞|0, −∞ >J = Z[J] = N
Z i Dφ exp i dx L0 + Jφ + εφ2 , 2
with the arti cial iε is added to make sure the integral converges. Integrating by parts we obtain (3.6)
Z Z[J] =
Z Z Z i 1 Dφ exp i dx L0 + Jφ + εφ2 = Dφ exp i dx φ(∂µ ∂ µ φ − m2 )φ + iεφ2 + Jφ . 2 2
In this case, the Klein-Gordon becomes the slightly more generalized equation (3.7)
(∂µ ∂ µ − m2 + iε)φc = −J. Working in
d dimensions and de ning the Feynman propagator as (3.8) ∆(x − y) =
−1 (2π)d
Z
dd k
k2
eik(x−y) + m2 − iε
the solution to the generalized Klein-Gordon equation becomes (3.9)
Z φc (x) = −
dy∆(x − y)J(y).
40 The Feynman propagator obeys (3.10)
(∂µ ∂ µ − m2 + iε)∆(x − y) = δ d (x − y). Hence the vacuum amplitude can be written in terms of the source
J as (3.11)
Z i dxdyJ(x)∆(x − y)J(y) < 0, ∞|0, −∞ >J = N exp − 2 or (3.12)
Z i Z0 [J] = Z0 [0] exp − dxdyJ(x)∆(x − y)J(y) . 2 by setting < 0, ∞|0, −∞ >J := Z0 [J]. The (Feynman) propagator is also computed by the functional derivative of Z0 [J] (3.13)
∆(x − y) =
i δ 2 Z0 [J] . Z0 [0] δJ(x)δJ(y) J=0
In order to evaluate Z0 [0] (which is the vacuum to vacuum amplitude when there is no source) we need to introduce imaginary time x4 = t = ix0 and operator ∂¯µ ∂¯µ = ∂ 2 + ∇2 so τ
that (3.14)
Z Z0 [0] =
Z 1 ¯ exp 1 dxφ(∂¯µ ∂¯µ − m2 )φ = q Dφ , 2 Det(∂¯µ ∂¯µ − m2 )
with capital d, the determinant is the product of eigenvalues with corresponding boundary condition. With term sources, the Lagrangian of the free complex scalar eld takes the form (3.15) 2
L0 = −∂µ φ∗ ∂ µ φ − m2 |φ| + Jφ∗ + J ∗ φ, and consequently the generating functional becomes (3.16) ∗
Z0 [J, J ] =
Z
Z Z Z 2 ∗ ∗ µ 2 ∗ ∗ DφDφ exp i dx(L0 − iε |φ| ) = DφDφ exp i dxφ (∂µ ∂ − m − iε)φ + Jφ + J φ , ∗
dierentiating we obtain the propagator (3.17)
∆(x − y) =
δ 2 Z0 [J, J ∗ ] i . Z0 [0, 0] δJ ∗ (x)δJ(y) J=J ∗ =0
We may split the function by virtue of the Klein-Gordon equations (∂µ ∂ µ − m2 )φ = −J and (∂µ ∂ µ − m2 )φ∗ = −J ∗ (3.18)
Z ∗ Z0 [J, J ] = Z0 [0, 0] exp −i dxdyJ (x)∆(x − y)J(y) ∗
and by using another Wick rotation we have (3.19)
Z Z0 [0, 0] =
Z ∗ µ 2 DφDφ exp −i dxφ (∂µ ∂ − m − iε)φ = ∗
1 . ¯ ¯ Det(∂µ ∂ µ − m2 )
41 The presence of a potential in the Lagrangian (3.20)
L(φ, ∂µ φ) = L0 (φ, ∂µ φ) − V (φ) comes at a double price: the form of the potential is limited by symmetry and renormalization of the theory but this theory needs to be handled perturbatively. The potential is usually of the form V (φ) =
α n n! φ
where α is a real number that sets the strength of the
interaction and n > 2 is an integer. As with the free theory, the generating functional is (3.21)
Z 1 Dφ exp i dx φ(∂µ ∂ µ − m2 )φ − V (φ) + Jφ 2 Z Z Z = Dφ exp −i dxV (φ) exp i dx (L0 (φ, ∂µ φ) + Jφ) Z
Z[J] =
Z = exp −i dxV −i
=
∞ Z X k=0
Z dx1 · · ·
δ δJ(x)
Z
(−i)k dxk V k!
Z Dφ exp i dx (L0 (φ, ∂µ φ) + Jφ)
δ −i δJ(x1 )
· · · V −i
δ Z0 [J]. δJ(xk )
The Green function (which is the vacuum expectation of the order time product of eld operators) (3.22)
Gn (x1 , · · · , xn ) := h0 | T [φ(x1 ) · · · φ(xn )]| 0i =
(−i)n δ n Z[J] δJ(x1 ) · · · δJ(xn ) J=0
is generated by the generating functional Z[J]. However, we can see that this is the
n th functional derivative of Z[J] around J = 0 and
therefore we may plug it into the Taylor expansion of the exponential above and we obtain (3.23)
" n Z # Z ∞ X 1 Y dxi J(xi ) h0 | T [φ(x1 ) · · · φ(xn )]| 0i = 0 T exp dxJ(x)φ(x) 0 . Z[J] = k! i=1 k=1
Connected n -point functions are generated by (3.24)
Z[J] = exp(−W [J]), and the eective action is de ned by a Legendre transformation as follows (3.25)
Z Γ[φcl ] := W [J] −
dτ dxi Jφcl
where (3.26)
φcl := hφiJ =
δW [J] . δJ
We will also see that Γ[φcl ] generates 1-particle irreducible diagrams (see Ramond and Bailin and Love). It is convenient now to derive the above discussion in a formal manner and with a closer
42 analogy to statistical mechanics. The generating functional of correlation functions for a eld theory with Lagrangian
L is given by (3.5) (3.27)
Z Dφ exp i d4 x(L + Jφ) ,
Z Z[J] =
where the time variable is contained between -T and
T, with T → ∞(1 − iε). Furthermore
we have the following (3.28)
R h0 | T φ(x1 )φ(x2 )| 0i =
lim
T →∞(1−iε)
= Z[J]−1 −i
δ δJ(x1 )
h R i T Dφφ(x1 )φ(x2 ) exp i −T d4 xL h R i R T Dφ exp i −T d4 xL
−i
δ δJ(x2 )
Z[J]
. J=0
Let us do some manipulations on the time variable; when we derived the path integral formulation of quantum mechanics (see AQFT) it is shown that the time integration was tilted into the complex plane in the direction that would allow the contour of integration to be rotated clockwise onto the imaginary axis. We assume that the original in nitesimal rotation gives the correct imaginary in nitesimal to produce the Feynman propagator. Now, the wick rotation of the time coordinate t → −ix0 yields a Euclidean 4-vector product 2
2
2
x2 = t2 − |x| → −(x0 )2 − |x| = − |xE | , and similarly we assume that the analytic continuation of the time variables in any Green's function of a quantum eld theory produces a correlation function
invariant under the
rotational symmetry of four-dimensional Euclidean space. Let us now apply this to the φ4 theory. As we know the action in this case is (3.29)
Z S=
Z
4
d x(L + Jφ) =
4
d x
1 1 2 2 λ 4 2 (∂µ φ) − m φ − φ + Jφ , 2 2 4!
and performing the Wick rotation
Z i
Z
4
d xE (LE − Jφ) = i
4
d xE
1 1 2 2 λ 4 2 (∂Eµ φ) + m φ − φ − Jφ , 2 2 4!
which in turn gives the Wick-rotated generating functional (3.30)
Z Z[J] =
Z 4 Dφ exp − d xE (LE − Jφ) .
The functional LE [φ] is bounded from below and when the eld φ has large amplitude or large gradient the functional becomes large. These two facts would imply that LE [φ] has the form of an energy and consequently it a is a possible candidate for a statistical weight for the uctuations of φ. Within this light, the Wick rotated functional Z[J] is the partition function describing the statistical mechanics of a macroscopic system when approximating the uctuating variable as a eld.
43 Finally, let us push this analogy between eld theory and statistical mechanics further by presenting the Green's function of φ(xE ) (3.31)
Z < φ(xE1 )φ(xE2 ) > =
d4 kE eikE ·(xE1 −xE2 ) , 2 + m2 (2π)4 kE
which is in fact the Feynman propagator evaluated in the spacelike region and this falls o
exp(−m|xE1 − xE2 |). This correspondence between quantum eld theory and statistical mechanics will play an important part in understanding ultraviolet divergences. Recalling the generating functional of correlation functions, we de ne an energy functional
E[J] by (3.32) Z Z[J] = exp(−iE[J]) =
Z 4 Dφ exp i d x(L + Jφ) = < Ω|e−iHT |Ω >,
with the constraints on time explained above. Note the −i factor in the exponential in contrast to (3.24). The functional E[J] is, as we have said before, the vacuum energy as
a function of the external source J. Let us perform the functional derivative of E[J] with respect to J(x)
δ δ E[J] = i log Z = − δJ(x) δJ(x)
Z
Z −1 Z Z Dφ exp i d4 x(L + Jφ) Dφφ(x) exp i d4 x(L + Jφ)
and set (3.33)
δ E[J] = − < Ω|φ(x)|Ω >J δJ(x) the vacuum expectation value in the presence of a source J. Next, we de ne - the classical eld as (3.34)
φcl (x) = < Ω|φ(x)|Ω >J , a weighted average over all possible uctuations, and dependent on the source J. - the eective action as the Legendre transformation of E[J] i.e. as in (3.25) (3.35)
Z Γ[φcl ] := −E[J] −
d4 x0 J(x0 )φcl (x0 ).
By virtue of (3.33) we have the following
δ δ Γ[φcl ] = − E[J] − δφcl (x) δφcl (x) Z =−
δJ(x0 ) δE[J] d x − δφcl (x) δJ(x0 ) 4 0
Z
d4 x0
Z
d4 x0
δJ(x0 ) φcl (x0 ) − J(x) δφcl (x)
δJ(x0 ) φcl (x0 ) − J(x) = −J(x) δφcl (x)
which means that when the source is set to zero, the equation (3.36)
δ Γ[φcl ] = 0 δφcl (x) is satis ed by the eective action. This equation has solutions which are the values of
44
hφ(x)i in the stable states of the theory. It will be assumed that the possible vacuum states are invariant under translation and Lorentz transformations. This implies a substantial simpli cation of (3.36) as for each possible vacuum state the corresponding solution φcl will
be independent of x, and hence it is just solving an ODE of one variable.
Thermodynamically, Γ is proportional to the volume of the spacetime region over which the functional integral is taken, and therefore, it can be a large quantity. Consequently, in terms of volume
V and associated T of the region we may write (3.37) Γ[φcl ] = −(V T )Veff (φcl ),
where Veff is the eective
potential. In order that Γ[φcl ] have an extremum we need the
following to hold (3.38)
δ Veff (φcl ) = 0. δφcl Each solution of (3.38) is a translation invariant state without source J = 0. Therefore, the eective action (3.35) is −E in this case (Γ = −E) and consequently Veff (φcl ) evaluated at the solution of (3.38) is the energy density of the corresponding state. The eective potential de ned by (3.37) and (3.38) yields a function whose minimization de nes the exact vacuum sate of the eld theory including all eects of quantum corrections. The evaluation of Veff (φcl ) will follow from the path integral formulation. In order to accomplish this we will follow Peskin and Schroeder's method which in turn follows from R. Jackiw and dates back to 1974. The idea is to compute the eective action Γ directly from its path integral de nition and then obtain Veff by focusing on constant values of φcl . Because we are using renormalized perturbation theory, the Lagrangian
L=
1 λ0 1 (∂µ φ)2 − m20 φ2 − φ4 2 2 4!
ought to be split as
L = L1 + δL, which is analogous to the split of the Lagrangian in renormalized φ4 theory done in AQFT (see Rajantie and Bailin and Love 7.4), i.e. rescaling the eld φ = Z 1/2 φr where residue in the LSZ reduction formula Z d4 x hΩ | T φ(x)φ(0)| Ωi eip·x = with
Z is the
iZ + (terms regular at p2 = m2 ), p2 − m2
m being the physical mass. This rescale changes the Lagrangian into L=
1 1 λ0 1 1 δλ (∂µ φr )2 − m20 φ2r − φ4r + δZ (∂µ φr )2 − δm φ2r − φ4r , 2 2 4! 2 2 4!
by the use of δZ = Z − 1, δm = m20 Z − m2 and δλ = λ2 Z − λ where
m and λ are
physically measured.. The last three terms are known as the counter terms and they take into account the in nite and unobservable shifts between the bare parameters and the physical parameters. At the lowest order in perturbation theory the relationship between the source and the
45 classical eld is (3.39)
δL + J(x) = 0. δφ φ=φcl
Because the functional Z[J] depends on φcl through its dependence on
J and our goal is to
compute Γ as a function of φcl . This will be the starting point. Next, we de ne J1 to be the function that exactly satis es the classical eld equation above for higher orders, i.e. when L = L1 (3.40)
δL1 + J1 (x) = 0, δφ φ=φcl and the dierence between both sources
J and J1 will be written as (see Peskin and
Schroeder 11.4) (3.41)
J(x) = J1 (x) + δJ(x), where δJ has to be determined, order by order in perturbation theory by use of (3.34), that is by using the equation hφ(x)iJ = φcl (x). We may now write (3.32) as (3.42)
e
−iE[J]
Z Z 4 4 Dφ exp i d x(L1 [φ] + J1 φ) exp i d x(δL[φ] + δJφ) ,
Z =
where all the counter terms are in the second exponential. Let us concentrate on the rst exponential rst. Expanding the exponential about φ(x) = φcl (x) + η(x) yields (3.43)
Z
Z
4
d x(L1 [φ] + J1 φ) = 1 + 3!
Z
Z Z δL1 δ 2 L1 1 4 4 4 d x(L1 [φcl ] + J1 φcl )+ d xη(x) + J1 + d xd yη(x)η(y) δφ 2! δφ(x)δφ(y) 4
4
4
4
d xd yd zη(x)η(y)η(z)
δ 3 L1 δφ(x)δφ(y)δφ(z)
+ ···
and it is understood that the functional derivatives of L1 are evaluated at φcl . The second integral on the RHS vanishes by (3.40) and therefore the integral over η is a Gaussian integral, where the perturbative corrections are given by the cubic and higher terms. Let us assume that the coe cients of (3.43) (i.e. the successive functional derivatives of L1 ) give well-de ned operators. If we keep only terms up to quadratic order in η and we only focus of the rst integral of (3.42) we nd that there is a pure Gaussian integral which can be evaluated in terms of a functional determinant as we have computed in the Appendix (3.44)
Z 1 δ 2 L1 4 4 4 d xd yη η Dη exp i d x(L1 [φcl ] + J1 φcl + 2 δφ(x)δφ(y) Z −1/2 δ 2 L1 4 = exp i d x(L1 [φcl ] + J1 φcl ) det − , δφ(x)δφ(y)
Z
the lowest-order quantum correction to the eective action is given by the determinant. If we now consider the second integral of (3.42) which consists of the counter terms of the Lagrangian and expanding as we have done before we have (3.45)
(δL[φcl ] + δJφcl ) + (δL[φcl + η] − δL[φcl ] + δJη).
46 The cubic and higher order terms in η in (3.43) produce Feynman diagram expansion of the functional integral in (3.42) in which the propagator is the operator invert (3.46)
−i
δ 2 L1 δφ(x)δφ(y)
−1 ,
and hence when the second term in (3.45) is expanded as a Taylor series in η the successive terms give counter term vertices which can be included in the above mentioned Feynman diagrams. The rst term is a constant with respect to the integral over thus it gives additional terms in the exponent of (3.44). Taking (3.44) with the contributions from higher order vertices and counter terms together we obtain an expression for the functional integral (3.42). Feynman diagrams representing the higher order terms can be arranged in such a way that they yield the exponential of the sum of the connected diagrams, obtaining the expression for E[J] (3.47)
Z −iE[J] = i
d4 x(L1 [φcl ] + J1 φcl ) − Z +i
1 δ 2 L1 log det − + (connected diagrams) 2 δφ(x)δφ(y)
d4 x(δL[φcl ] + δJφcl ),
and nally by virtue of (3.41) and (3.35) we nally have (3.48)
Z Γ[φcl ] =
d4 xL1 [φcl ] +
Z δ 2 L1 i log det − − i(connected diagrams) + d4 x(δL[φcl ]). 2 δφ(x)δφ(y)
This indeed the expression we were seeking since Γ is a function of φcl , taking away the
J dependence. The Feynman diagrams in the expression for Γ have no external lines and they all contain at least two loops. The last term of (3.48) gives the counter terms that are needed for the renormalization conditions on Γ and cancel the divergences that appear in the evaluation of the determinant and the diagrams. We shall ignore any one-particle irreducible one-point diagram (these diagrams are cancelled by the adjustment of δJ ). We now go back to the Euclidean space de nition of the generating functional (3.49)
Z ZE [J] = NE
Z 1 1 Dφ exp − d4 x ¯ ∂¯µ φ∂¯µ φ + m2 φ2 + V (φ) − Jφ . 2 2
This will be evaluated by expanding the action. In order to do this, set φ0 to be a eld con guration, then (3.50)
Z SEuc [φ, J] :=
d4 x ¯
1 ¯ ¯µ 1 ∂µ φ∂ φ + m2 φ2 + V (φ) − Jφ 2 2 Z
SEuc [φ, J] = SEuc [φ0 , J] + 1 + 2
Z
4
4
d x ¯1 d x ¯2
δSEuc d x ¯ (φ − φ0 ) δφ 4
δ 2 SEuc (φ(¯ x1 ) − φ0 (¯ x1 ))(φ(¯ x2 ) − φ0 (¯ x2 )) + · · · δφ(¯ x1 )δφ(¯ x2 )
It is understood that the functional derivatives are evaluated at φ0 . We know from AQFT
47 that the classical limit can be recovered, by taking SEuc to be stationary at φ0 and this implies that φ0 satis es the classical EOM with the source term (3.51)
δSEuc = −∂¯µ ∂¯µ φ(x0 ) + m2 φ(x0 ) + V 0 [φ(x0 )] − J = 0. δφ x0 Integrating by parts gives (3.52)
1 SEuc [φ(x0 ), J] = 2
Z
d x ¯ 2 − φ(x0 ) 4
d [−Jφ(x0 ) + V 0 [φ(x0 )]], dφ(x0 )
whereas the second derivative is an operator (3.53)
δ2 SEuc = δ(¯ x1 − x ¯2 ) −∂¯µ ∂¯µ + m2 + V 00 [φ(x1 )] . δφ(x1 )δφ(x2 ) We need to make regression now on how to evaluate integrals of the sort (3.54)
Z I :=
dx exp(−α(x))
This can be evaluated by expanding the exponential around a point x0 where α is stationary, i.e. (3.55)
The
1 α(x) = α(x0 ) + (x − x0 )200 α00 (x0 ) + O(x3 ). 2
I integral is then approximated (3.56) Z I = exp(−α(x0 ))
1 dx exp − (x − x0 )2 α00 (x0 ) , 2
and we can recognize this as a Gaussian integral (when the higher derivatives are ignored)! The degree of this approximation depends obviously on α. When α is smallest then the integrand is largest and the points away from the minimum do not add a substantial contribution. Equipped with this method we can apply it to the functional (see Ramond 3.4) we have just arrived at a crucially important result (3.57)
Z ZE [J] = NE exp {−SEuc [φ(x0 ), J]}
= NE0 q
Z δ 2 SEuc 1 d4 x ¯ 1 d4 x ¯2 φ(¯ x1 ) φ(¯ x2 ) Dφ exp − 2 δφ(¯ x1 )δφ(¯ x2 )
exp {−SEuc [φ(x0 ), J]} , det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
where we have ignored the higher order terms. The precise derivation is in the Appendix. Let us do some re-write to make this expression easier to handle, rst the determinant, which we will call
M can be taken care of by using the identity (3.58) det M = exp(tr log M),
48 (3.59)
ZE [J] =
NE0
1 µ 2 00 ¯ ¯ exp −SEuc [φ(x0 ), J] − tr log(−∂µ ∂ + m + V [φ(x0 )])δ(x1 − x2 ) , 2
the delta term accounts for the quantum perturbations (or corrections) to Z[J], whereas the rst term accounts for the classical contribution. Also note that we set by convention the determinant of an operator to be the product of its eigenvalues and that because φ0 satis es δSEuc /δφ|x0 = 0 then it is a functional of J.
The concluding remark is to nd the equivalence of these results in terms of the classical eld φcl and explore the corresponding eective action. When we work in Euclidean space, the classical eld was de ned as (3.60)
φcl (¯ x) = −
δZE δSE ≈− + O(~), δJ(¯ x) δJ(¯ x)
and by using (3.57) and (3.58) we can have φcl as a function (functional, rather) of J, however at the cost of doing it order by order in λ. The term O(~) stands for quantum corrections. This relationship can be inverted and we can nd J(¯ x) as a function of φcl . This inversion can be carried out to give (see Ramond) (3.61)
λ x). J(¯ x) = (∂¯2 − m2 )φcl (¯ x) − φ3cl (¯ 3! An attractive (and indispensable) feature is that there are no higher terms in λ, comparison with
0=
δSEuc = −∂¯µ ∂¯µ φ(x0 ) + m2 φ(x0 ) + V 0 [φ(x0 )] − J δφ x0
gives
φcl (¯ x) = φ0 (¯ x) + O(~). By integrating J(x) = −δΓ[φcl ]/δφcl we see that to this order the eective action is (3.62)
Z Γeff [φcl ] = −
1 λ 4 2 2 ¯ d x ¯ φcl (∂ − m )φcl − φcl (¯ x) , 2 4! 4
which is a eective action.
REFERENCES The above summary on quantum eld theory has its roots several sources, overall it follows the presentation given by Peskin and Schroeder [275 to 294, 306 to 308 and 365 to 372] as well as Ramond and section 4.4 of Bailin and Love. At times it is complemented by the courses of QED and AQFT from the MSc QFFF. However, it is broad enough to be explained in most QFT books.
49
Zeta Regularization in Field Theory At this point we can put together some of the concepts acquired in the second and third chapters. Firstly we will further develop the theory of quantum corrections by evaluating the determinant of the M matrix and then see how this is related to the use of the zeta function
in quantum theory as we explained in Chapter 2. In what follows, we will extrapolate the results found in the quantum mechanical section to eld theory, this section borrows some its contents from 3.5 and 3.6 from Ramond as well as 22.1 to 22.3 of Har eld. As we mentioned earlier, the determinant in the expression (3.57) (4.1)
ZE [J] = q
NE0 exp {−SEuc [φ(x0 ), J]} det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
must be interpreted as the product of the eigenvalues of the operator. In order to discretize these eigenvalues we truncate the space (by use of a box). We then multiply the resulting eigenvalues and then let the size of the box increase to in nity. First we state some preliminary results that will become useful later on. As it is shown in a course on partial dierential equations, the heat function
G(¯ x, y¯, t) :=
X
exp(−λn t)ψn (¯ x)ψn∗ (¯ y)
n
satis es the heat equation
Ax¯ G(¯ x, y¯, t) = −
∂ G(¯ x, y¯, t). ∂t
Also, similar to our expression (1.15) we have the following result (4.2)
Z∞ ζA (s)Γ(s) =
dt ts−1
Z
d4 x ¯G(¯ x, x ¯, t),
0
this in fact follows from the orthogonality of eigenfunctions ∗ hψn , ψm i=
Z
dxψn∗ (x)ψm (x) = δnm ,
X
ψn∗ (x)ψn (x0 ) = δ(x − x0 )
n
By this same orthogonality we have (4.3)
G(¯ x, y¯, t = 0) = δ(¯ x − y¯). This is the analytic representation of the zeta function we are seeking. We can explain this when we take the trace of
G
Z Tr(G) =
dxG(x, x, t) =
X
e−λn t
n
(remember Tr(G) is a function of t ) multiply it by e−λt and integrate it with respect to
t
50 and then with respect to λ to obtain
Z∞
Z dλ
dtTr(G)e
−λt
dλ
n
0
The determinant of
=
XZ
X 1 = log(λn + λ). λn + λ n
A then is written as (*) Z Z∞ det A = exp dλ dt Tr(G)e−λt λ=0 0
ignoring the factor that would show from the λ integration. Swapping the integrals yields
Z∞
det A = exp −
Tr(G) dt . t
0
The relationship between the trace of the heat kernel TrG and the ζ function is given by a Mellin transformation
1 ζA (s) = Γ(s)
Z∞
dtts−1 Tr(G),
0
which justi es (4.2). A solution of the heat equation (4.4)
−∂¯x2 G0 (¯ x, y¯, t) = −
∂ G0 , ∂t
with the boundary condition G0 (¯ x, y¯, t = 0) = δ(¯ x − y¯) is (4.5)
1 1 2 G0 (¯ x, y¯, t) = exp − (¯ x − y¯) , 16π 2 r2 4r also a classic result from partial dierential equations.
A i runs from 1 to n and its eigenfunctions
Let us generalize some of the techniques we used in Chapter 2. Consider an operator with positive real discrete eigenvalues ai where
are fn (x) i.e. Afi (x) = ai fi (x). From here we set (4.6)
ζA (s) =
X
a−s n ,
n
which we call the zeta function associated to the operator eigenvalues and
A. The sum is over all the
A is a real variable. If the operator A were the one-dimensional harmonic
oscillator Hamiltonian, then ζA would be the Riemann zeta function (excluding the singular zero-point energy). The rst observation is that (4.7)
X Y d ζA (s) =− log asn exp(−s log asn )|s=0 = − log an , ds s=0 n n
51 which gives the determinant (4.8)
det A =
Y
0 an = exp(−ζA (0)).
n
As we know from our discussion of chapter 2. The zeta-function ζA is not always singular at s = 0 for physically interesting operators and hence the convenience of writing this representation for det A. Algorithmically we have a procedure to compute the determinant of
A and it can be done
as follows. First we need to nd the solution to the heat equation subject to the deltafunction initial condition (4.3). Second, once this is done we can insert the solution into the
ζ representation above and we have ζA . Finally evaluate at s = 0 and compute exp(−ζA0 (0)). In our case, our operator is (4.9)
λ A = −∂¯2 + m2 + φ20 (¯ x), 2 where φ0 (¯ x) is a solution of the classical equations with source J as we discussed in Chapter 3. Also note that in (4.1) we have a V 00 factor in the determinant which takes the form
V 00 = (λ/2)φ20 in the
A operator.
As we have pointed out a solution of the heat equation with the boundary condition
G0 (¯ x, y¯, t = 0) = δ(¯ x − y¯) is (4.10) 1 1 2 exp − (¯ x − y¯) . G0 (¯ x, y¯, t) = 16π 2 r2 4r However, this is only a part of the operator as we want to nd G0 (¯ x, y¯, t) subject to the initial condition (4.3) which obeys the whole
A operator (4.11)
∂ λ 2 2 2 ¯ x) G(¯ x, y¯, t) = − G(¯ x, y¯, t). −∂ + m + φ0 (¯ 2 ∂t For arbitrary elds φ0 we need to expand the eective action as (0)
(1)
ΓE [φcl ] = ΓE [φcl ] + ~ΓE [φcl ] + · · · , the ~ indicates quantum terms. Given the fact that the eective potential at 1 loop is of the form (Ramond) (1)
ΓE [φcl ]
Z
d4 x =
1 log det(−∂ 2 + m20 + 2
λ0 2 2 φcl )
−
1 log det(−∂ 2 + m20 ), 2
as we also argued in Chapter 3 Eq 3.48, allows us to write (4.12)
1 (1) ΓE [φcl ] = − ζ 0 ¯2 2 λ 2 (0), 2 [−∂ +m + 2 φ0 (¯x)] by use of (4.1) and (4.8) combined with (4.9). Note that replacing φ0 by φcl is not a problem as there are no new quantum errors, that is errors up to O(~). By the expansion of the
52 eective potential we can swap m and m0 . Also the eective action can be written as (4.13)
Z ΓE [φcl ] =
d4 x ¯ V (φcl (¯ x)) + F (φcl )∂¯µ φcl (¯ x)∂¯µ φcl (¯ x) + · · ·
constant v is a constant independent of x¯.
To compute the quantum O(~) contribution to V (φcl (¯ x)) we need to consider a eld con guration: i.e. suppose we take φcl (¯ x) = v where Then (4.14)
Z
d4 x ¯V (v),
ΓE [φcl ] =
and it is proportional to the in nite volume element
d4 x ¯, since the Euclidean space R4 is unbounded. This can be temporarily solved by taking the space to be the sphere S4 then R
the volume is just that of the 5-dimensional sphere and hence nite. While the radius is nite we need not worry about the infrared divergence. We then let the radius of the sphere tend to in nity. Taking
V out of the integral we have [see Hart eld] (4.15) Z V (v)
1 d4 x ¯ = − ζ 0 ¯2 2 λ 2 (0). 2 [−∂ +m + 2 v ]
We can proceed to integrate (4.11) when
G(¯ x, y¯, t) =
v is constant this yields (4.16)
µ4 µ2 (¯ x − y¯)2 λ 2 t 2 v exp − exp −m + , 16π 2 t2 4t 2 µ2
where the µ factor needs to be explained: it has dimensions of mass so that t is dimensionless. Using the Mellin transform (4.2) we obtain (4.17)
1 ζ(s) = Γ(s)
Z∞ dt t
s−1
Z
µ4 d x ¯ exp 16π 2 t2 4
λ −m + v 2 2 2
m2 + λ2 v 2 µ2
t µ4 = 2 µ 16π 2 t2
!2−s
Γ(s − 2) Γ(s)
Z
d4 x ¯,
0
the volume element
R
d4 x ¯ is present because it is in (4.15) and here
t has been rescaled
since the integration over t is valid when s > 2 however the ζ function is de ned everywhere by analytic continuation as we know from Chapter 1. Comparing these two equations we have (4.18)
µ4 d 1 V (v) = − 2 32π ds (s − 2)(s − 1)
m2 + λ2 v 2 µ2
Equipped with this functional form of
!2−s
1 = 64π 2
λ m + v2 2 2
2
m2 + λ2 v 2 3 log − 2 µ 2
s=0
V the eective potential can be written as (4.19)
1 λ ~ V (φcl ) = m2 φ2cl (¯ x) + φ4cl (¯ x) + 2 4! 64π 2
2 λ m + φ2cl (¯ x) 2 2
m2 + λ2 v 2 3 log − µ2 2
! ,
ignoring terms of order ~2 . We make a pause now to examine this result. Super cially the rst striking observation is that there is a strong dependence on the unknown and arbitrary scale µ2 . This would seem to imply that the potential is therefore arbitrary. However
V
depends on the parameters m and λ which are unde ned, except for the fact that they are 2
! .
53 included in the classical Lagrangian. Let us take the special massless case. This yields the following (4.20)
d2 V = 0. dφ2 φ=0 Now we
de ne the mass squared as the coe cient of the terms φ2 in the Lagrangian eval-
uated at φ = 0. To rst quantum corrections the coe cient is zero, if it is classically zero. The λ term is de ned to be the coe cient of the fourth derivative of
V evaluated at some
constant point φ = M, i.e. (4.21)
d4 V λ := . dφ4 φ=M We cannot take φ = 0 as with the mass squared factor because of the infrared divergence coming from the logarithm. When we dierentiate (4.19), set m2 = 0 and use (4.21) we see that the above condition requires (4.22)
log
λM 2 8 =− . 2µ2 3
In this case, we may use M 2 instead of 2µ2 /λ and write the result as (4.23)
λ λ2 φ4cl V (φcl ) = φ4cl (¯ x) + 4! 256π 2
φ2cl 25 log 2 − . M 6
This was proved by Coleman and Weinberg in 1973. The main result that can be extracted is that we need to be careful with how we de ne the input parameters in the Lagrangian if we are to take into account quantum corrections. Again, super cially it seems that (4.23) depends on another arbitrary scale M 2 , but in fact it does not. Given the normalization condition, if we change the scale from M 2 to M 02 we simultaneously have to change at the same time λ to λ0 by use of (4.21) (4.24)
λ0 = λ +
3λ2 M0 log . 16π 2 M
Therefore the potential (see Ramond) (4.25)
V (φcl ) =
λ0 4 λ02 φ4cl φcl (¯ x) + 4! 256π 2
log
φ2cl 25 − M 02 6
is indeed invariant under the representation V (λ0 , M 0 ) = V (λ, M ). This proves that the physics behind remains unchanged but our way of interpreting the coe cients changes. Let us now look more closely to the scaling of determinants and the coupling constants. The ζ function technique just used allows us derive scaling properties for determinants.
(0). λ [−∂¯2 + 2 φ2cl ]
First, we will need the computation of ζ function ζ
This can be accomplished
54 by taking the asymptotic expansion of G(¯ x, y¯, t) at µ2 = 1, (4.26)
G(¯ x, y¯, t) = e−εt
2 ∞ e−(¯x−¯y) /(4t) X an (¯ x, y¯)tn , 16π 2 t2 n=0
with ε > 0 as a convergence factor. The boundary condition (4.3) sets the condition (4.27)
a0 (¯ x, x ¯) = 1. Additionally, when we insert (4.26) into the PDE (4.11) we nd recursion relations for the
an coe cients (4.28) (¯ x − y¯)µ
∂ a0 (¯ x, y¯) = 0 ∂x ¯µ
and for n = 0, 1, 2, · · · (4.29)
∂ λ (n + 1) + (¯ x − y¯)µ an+1 (¯ x, y¯) = ∂¯x2 − φ2cl + ε an (¯ x, y¯). ∂x ¯µ 2 When we compute the rst terms we have (4.30)
λ a1 (¯ x, x ¯) = − φ2cl + ε 2 a2 (¯ x, x ¯) =
λ2 4 λ ε ε2 φcl (¯ x) − ∂¯x2 φcl (¯ x) − λφ2cl (¯ x) + . 8 4 2 2
Let us now use these results. We work under a scale change A → A0 = ead A, where the natural dimension of
A. By the de nition of the zeta function we have (4.31)
d is
ζA0 (s) = e−ads ζA (s) which implies that (4.32)
det(ead A) = eadζA (0) det A. Let us illustrate this with an example. Under the transformation (4.33)
xµ → x0µ = ea xµ
φcl → φ0cl = e−a φcl
the massless classical action (4.34)
Z SE [φcl ] = −
d4 x ¯
λ 1 ¯2 φcl ∂ φcl − φ4cl 2 4!
is unchanged. However, the path integral corresponding to this action is not scale invariant since in the steepest descent approximation the change in the eective action to quantum order is as follows (4.35) eff eff 0 eff SE [φcl ] → SE [φcl ] = SE [φcl ] − ~aζ
(0). λ [−∂¯2 + 2 φ2cl ]
Plugging (4.26) into (4.2) and with the assistance of the rst two terms a1 and a2 and
55 integrating out the ∂¯2 with the divergence theorem yields (4.36)
1 ζ(0) = 16π 2
Z
d4 x ¯
λ2 4 φ (¯ x). 8 cl
A small digression is now required to further explain this. In 4 dimensions and in the presence of mass the heat kernel is
2 e−(¯x−¯y) /(4t) λ0 φ2cl 2 G(x, x, t) = exp − m0 + t 16π 2 t2 2 and like in (4.17) and (4.18)
ζ(s) =
1 16π 2
m20 +
λ0 φ2cl 2
2−s
Γ(s − 2) Γ(s)
Z
d4 x ¯.
Furthermore, note that in the more restricted case B = −∂ 2 + ω 2 we have kernel
(x − x0 )2 1 exp − exp(−ω 2 t) G(x, x t) = √ 4t 4πt 0
and the determinant of
B is
1 log det B = − √ 4π
Z∞
dtt−3/2 exp(−ω 2 t)
Z
ω d4 x ¯ = −√ 4π
Z
d4 x ¯
0
Z∞
dte−t t−3/2
0
ω −√ 4π
Z
∞ h i d4 x ¯ t−1/2 e−t − 2π 1/2 0
integration by parts was used in the last line. Ignoring the divergent factor we have the simpli cation
Z det B = exp ω d4 x ¯ .
This can nd justi cation the integral in log det B is expressed in terms of Γ functions in this divergence zone by analytic continuation. Finally, the divergence of Γ(s − 2) at s = 0 cancels that the divergence of Γ(0), i.e.
1 Γ(s − 2) = Γ(s) (s − 2)(s − 1) and this process makes ζ(0) regular. In terms of the eective action (4.35) (4.37) eff 0 SE [φcl ]
=
eff SE [φcl ]
λ2 − ~a 128π 2
Z
d4 x ¯φ4cl (¯ x).
Consequently the eect of the transformation to quantum order has been to change the coupling constant λ by the following (4.38)
λ λ0 λ λ2 3λ2 → = − ~a ⇔ λ0 = λ − ~a. 2 4! 4! 4! 128π 16π 2
56 What this is means is that
the dimensionless coupling constant λ evolves as a result of
quantum eects. This evolution is in term of scale dependence. At large scales the coupling constant decreases indicating that the non-interaction theory is a good approximation for the asymptotic states. On the other hand, if the scale decreases, the coupling increases. Independently of how small λ was at the beginning this increment might throw away the results obtained in the perturbation of λ. Moreover, this scaling law is like the one we found earlier, and they are both correct to quantum orders.
t a quantum mechanical system with one degree of freedom, q and canonically conjugate momentum p, is described in terms of the spectrum We recall that for any given time
of its Hamiltonian H(p, q). From the path integral formulation we know that if the system at an initial time ti is measured to be in the state q i , then the probability that the system will be found in the state q f , at a nal time tf is exactly (4.39)
E E D D qtff qtii = q f e−i(tf −ti )H q i , and it can be written in terms of path integrals as (4.40)
tf Z Z E Z D q f e−i(tf −ti )H q i = Dq Dp exp i dt[pq˙ − H(p, q)], ti
the factor Dq denotes integration between the initial and nal con gurations q i and q f ; the dot over
q denotes the derivative of q with respect to time.
Quantum eld theory and statistical mechanics share certain common elements and precisely this analogy will allow us to apply path integrals to the description of dynamics systems at nite temperature (see Peskin and Schroeder, and Kleinert). The rst step, as we did in Chapter 2, is to compute the partition function (4.41)
Z = Tr[e−βH ], where the constant is (4.42)
β = (kT )−1 , taking into account that the trace is taken to be the sum over all the possible con gurations the system is allowed to take. Note that the time is singled out. Now, the probability for the system to be in state of energy
E is identi ed with (4.43) P = Z −1 e−βH .
The value of any function for the dynamical variable f (q, p) is given by (4.44)
hf i = Tr(f P ) = Z −1 Tr(f e−βH ). This does show a special similarity with zero temperature quantum mechanics and QFT but the degree of this similarity is not fully understood. However, we can push the analogy to calculate partition functions, specially this one. Let us start with a system which can be
57 regarded as a eld theory in zero space dimensions. We can compare this expression to the partition function for the same system at temperature
β −1 (4.45) Z = Tr[e−βH ] =
X q e−βH q . q
Let us draw comparisons between (4.40) and (4.45). If we set i(tf − ti ) = β or alternatively set ti = 0 and then itf = β since the origin of time is arbitrary. Next, set q f = q i which means that the initial and nal con gurations are the same, and since the dierence is a β factor, the only requirement is that the relevant con guration is periodic in the functional integrals (4.46)
q(β) = q(0). Thus the functional integration Dq is over the space of sum over
periodic functions. In this case, the
q in (4.40) is implicit. When we do the comparison we can write (4.47) Z = Tr(e−βH ) =
Z
Z Dq
β Z dq Dp exp dτ ip − H , dτ 0
bearing in mind again that Dq is over periodic functions. If we take a well-behaved potential V (q) we could scale the temperature dependence purely into the
q integral. To do this, we make the following transformations (4.48) τ¯ = τ β −1 ,
q¯ = qβ −1/2 ,
p¯ = pβ 1/2 ,
then the exponent of the integrand becomes (4.49)
Z1
p¯2 d¯ q − − βV (β 1/2 q¯) . d¯ τ i¯ p d¯ τ 2
0
Furthermore we can drop all the bars because the path integral measure is invariant under the changes (4.48) thus we can write the partition function as (4.50)
Z Z=
Z Dq
1 Z 1 Dp exp dτ ipq˙ − p2 − βV (qβ 1/2 ) . 2 0
Next, set p0 = p − iq˙, so that the measures are equal (4.51)
Dp0 = Dp, which in turn, by competing the square in the exponent, allows us to write (4.52)
Z Z=
1 Dp0 exp − 2
Z1 0
dτ p02
Z
Z1
Dq exp −
dτ
1 2 q˙ + βV (qβ 1/2 ) . 2
0
As we have done repeatedly in AQFT we can ignore the p0 integral (even though it is
58 in nite) because it is independent of β . We call this integral be ignored is because
N, and the reason why it can
N is usually always present in the numerator and denominators of
correlation functions. The only example we could tackle is that of an integral that can be evaluated, i.e. a Gaussian integral. The integral becomes Gaussian when we take potential V (q) = 12 ω 2 q 2 , the partition function is (4.53)
Z1
Z Dq exp −
Z=N
dτ
1 2 1 2 2 2 q˙ + β ω q . 2 2
0
It is one of the very few types that can actually be integrated. By virtue of (4.46) we have (4.54)
Z1
1 2
dτ
dq dτ
2
1 =− 2
0
Z1 dτ q
d2 q, dτ 2
0
since the extra surface term is eliminated due to (4.46) and therefore (4.55)
Z
1 Dq exp − 2
Z=N
Z1
d2 dτ q − 2 + ω 2 β 2 q . dτ
0
If we proceed by analogy with the discrete case we have (4.56)
Z Dq exp −
1 2
Z1
dτ q − 0
where N 0 is a constant, and
N0 d + ω2 β 2 q = √ , 2 dτ det A 2
A is the operator (4.57) A=−
d2 + ω2 β 2 dτ 2
with positive de nite eigenvalues (it must not contain zero eigenvalues as these would create in nities which have to be removed). In order to prove (4.56) we need to express q(τ ) in terms of its Fourier components (QFT course) then transform into the normal modes of
A
and integrate each one using (4.58)
Z∞
r 1 2π dqn exp − an qn2 = . 2 an
0
A operates on periodic functions with unit period, which can all be expanded in terms of the complete Fourier set {e2πinτ }. The eigenvalues of A are (4.59) The operator
(4π 2 n2 + ω 2 β 2 );
n ∈ Z.
Note the analogy with the eigenvalues of the quantum
mechanical operator (2.116).
59 Hence multiplying we have the determinant of the operator
Y
det A =
A, (4.60)
(4π 2 n2 + ω 2 β 2 ).
n∈Z
Setting x2 = ω 2 β 2 yields the following (4.61)
X X 1 1 1 d log det A = = 2 +2 . 2 2 2 2 2 2 dx 4π n + x x 4π n + x2 n∈Z
n≥1
Substituting the formula that was shown in the Appendix (4.62)
1 2x X 1 + 2 πx π x + n2
coth πx =
n≥1
1 d x log det A = coth dx2 2x 2 and from here we can integrate to nd (4.63)
log
det A = C
Z dx coth
x x ωβ = 2 log sinh = 2 log sinh . 2 2 2
Removing the logs we have the formula for the determinant (4.64)
det A = C sinh2
ωβ . 2
When we clean the expression we arrive to (4.65)
F =− where
D 1 1 1 log Z = − + ω + log(1 − e−ωβ ), β β 2 β
D is another constant, the zero-point energy is identi ed at 1 = e−ωβ . This formula
is called the thermodynamic potential. Let us go back to our discussion of the ζ function. The heat equation associated with the
A operator (4.57) is (see Ramond, Elizalde, Odintsov, Romeo, Bytsenk and Zerbini) (4.66) X
G(t, t0 , σ) =
exp 2πin(t − t0 ) − (ω 2 β 2 + 4π 2 n2 )σ ,
n∈Z
and recalling our expression (4.2) for the ζ function (4.67)
1 ζA [s] = Γ(s)
Z∞ dσσ
s−1
0
Z1 dt 0
X
exp −(ω 2 β 2 + 4π 2 n2 )σ .
n∈Z
Scaling σ by ω 2 β 2 + 4π 2 n2 leads nowhere as we simply come back to the expression (4.68)
ζA [s] =
X n∈
1 . (ω 2 β 2 + 4π 2 n2 )s
The technique we need to use is somewhat messier, it involves expanding in powers of ωβ
60 then integrating and re-arranging the sums. With this in mind we have (4.69)
X
e−4π
2
n2 σ
=1+2
n∈Z
X
e−4π
2
n2 σ
,
n≥1
and integrating we have (4.70) ∞
ζA [s] = (ωβ)−2s +
∞ ∞ Z X 2 2 2 X (ωβ)2k (−1)−k dσσ s+k−1 e−4π n σ . Γ(s) k! n=1 k=0
0
Now it is the time when we can rescale σ by 4π 2 n2 and we can also identify the sum over P −2s n with the Riemann ζ function ζ(2s) = ∞ , which gives (4.71) n=1 n ∞
ζA [s] = (ωβ)−2s +
2 2 X (ωβ)2k (−1)−k Γ(s + k)ζ(2s + 2k). ζ(2s) + (4π 2 ) Γ(s) k! (4π 2 )s+k k=1
In order to dierentiate ζA at s = 0 we note that the sum is well behaved at s = 0 and as
s → 0 a non zero term arises from the derivative of Γ−1 (s). Recalling our formulas from Chapter 1, (4.72)
1 ζ(0) = − , 2
1 ζ 0 (0) = − log 2π, 2
(4.73) 0 ζA [0] = −2 log(ωβ) − 2 log 2π + 2 log 2π + 2
∞ X (ωβ)2k (−1)k ζ(2k) k=1
k(4π 2 )k
.
Recalling the formula for the even values of the Riemann ζ function in terms of Bernoulli numbers (4.74)
ζ(2k) =
(−1)k+1 (2π)2k B2k , 2(2k)!
we can continue simplifying (4.73) (4.75) 0 ζA [0] = −2 log(ωβ) −
∞ X (ωβ)2k 1 B2k . (2k)! k
k=1
Using another formula from the appendix (4.76) ∞
coth x =
X (2x)2k−1 1 +2 B2k , x (2k)! k=1
and integrating yields (4.77) ∞
Z dx coth x = log x +
1 X (2x)2k B2k , 2 k(2k)! k=1
and nally by comparing with (4.76) and setting x = 12 ωβ , we obtain (4.78) 0 ζA [0] = −2 log(ωβ) + 2 log
ωβ ωβ − 2 log sinh = −ωβ − 2 log(1 − e−ωβ ). 2 2
61 Solving for
Z yields (4.79) log Z =
1 1 0 ζ [0] = − ωβ − log(1 − e−ωβ ), 2 A 2
which is similar to what we found earlier on the quantum mechanical case. Although this technique is lengthy it has enabled us to show the connection between the ζ function and quantum eld theory. It is desirable to be able to obtain (4.19) using another regularization and see how these two compare. In the Uni cation course, it was shown that spontaneous symmetry breaking requires the development of a vacuum expectation value (VEV) from the scalar eld. Furthermore, as we have shown in this Chapters 3 and 4, the VEV is determined by the minimisation of the eective potential, (4.80)
dV = 0. dϕcl The eective potential is given exclusively by the potential
V of the Lagrangian if no
quantum eects are taken into account. Perturbation theory allows us to place the quantum terms, however this would clash with the non-perturbative nature of spontaneous symmetry breaking. The alternative parameter is the loop expansion which we now describe. Let us re-write the theory around (3.24) as follows. With a generating functional
X of the
connected Green functions in a scalar eld theory, (4.81)
Z[J] = exp(i~−1 X[J]) = N 0
Z
Z Dϕ exp i~−1 d4 x(L + Jϕ)
with normalization constant N 0 chosen so that (as we pointed out in Chapter 3) (4.82)
Z[0] = 1 X[0] = 0. As we have also pointed out, the one-particle-irreducible Green functions Γ(n) are generated by the eective action Γ[ϕcl ] by the use of (4.83)
Γ[ϕcl ] =
Z ∞ n Z X i d4 x1 · · · d4 xn Γ(n) (x1 , · · · , xn )ϕcl (x1 ) · · · ϕcl (xn ). n! n=1
In (4.81) the factor ~−1 multiplies the whole Lagrangian (not just the interaction part) each
of the V vertices in any diagram will carry a ~−1 factor and each of the I internal lines will ˜ (E) has a propagator. carry a ~ factor. Each of the E external lines in Green functions G With this information, there overall factor is (4.84)
~−V +I+E = ~L+1−E , by virtue of L = I − V + 1. Furthermore, there is a factor of ~L−E in any diagram in ˜ (E) have no propagators, expansion of ~−1 X . The one-particle-irreducible Green functions Γ hence the multiplying factor is only ~L . This means that the power of the ~ indicates the
62 number of
loops. When there are no loops (L = 0) the only non-vanishing Γ˜ (E) are (4.85) ˜ (2) (p, −p) = p2 − µ2 Γ
and (4.86)
˜ (4) (p1 , p2 , p3 , p4 ) = −λ, Γ which gives the approximation to the potential (4.87)
V0 (ϕcl ) =
1 1 2 2 µ ϕcl + λϕ4cl , 2 4!
as classically expected without quantum corrections. Note, however, that the role of ~ is not central to the discussion as no assumptions about its size have been made, it is only an expansion parameter. Bailin and Love [201] give same eective action Γ[ϕcl ] with (4.88)
1 Γ[ϕcl ] = −i log N − λ 8
Z
1 − λi∆F (0) 4
Z
dx[i∆F (0)]2 − dxϕ2cl (x)
1 2
1 − λ 24
Z
Z
dxϕcl (x)(∂ µ ∂µ + µ2 )ϕcl (x) dxϕ4cl (x) + O(λ2 )
The terms that contain a Feynman propagator ∆F (0) come from divergent loop integrals and hence they do not contribute in zeroth order, consequently we may write (4.89)
Γ0 [ϕcl ] = −
1 2
Z
dxϕcl (x)(∂ µ ∂µ + µ2 )ϕcl (x) −
1 λ 4!
Z
dxϕ4cl (x)
Again taking into account that N = 1 implying Γ[0] = 0. The rst order accuracy of ~, the loop expansion that is, of the eective potential
V and
eective action Γ can be computed by writing (4.90)
ϕ(x) = ϕ0 (x) + ϕ(x), ˜ With ϕ0 being the zeroth order approximation to ϕcl . Hence this shift in the functional integration variable must satisfy the following EOM (4.91)
(∂ µ ∂µ + µ2 )ϕ0 (x) +
λ 3 ϕ (x) = J(x). 6 0
When we plug this change into the Lagrangian density (4.92)
L(ϕ) =
1 µ 1 1 (∂ ϕ)(∂µ ϕ) − µ2 ϕ2 − λϕ4 2 2 4!
we have the following integral (4.93)
Z
Z +
d4 x(L + Jϕ) =
Z
d4 x(L(ϕ0 (x)) + Jϕ0 )
Z 1 1 1 d4 x (∂ µ ϕ)(∂ ˜ µ ϕ0 ) − µ2 ϕϕ ˜ 0 − λϕϕ ˜ 30 + J ϕ˜ + d4 x L2 (ϕ, ˜ ϕ0 ) − λϕ˜3 ϕ0 − λϕ˜4 6 6 4
63 with L2 accounting for all the quadratic leftovers in ϕ˜ when the shift is performed, i.e. (4.94)
L2 =
1 µ 1 1 (∂ ϕ)(∂ ˜ µ ϕ) ˜ − µ2 ϕ˜2 − λϕ2 ϕ˜2 . 2 2 4
Now, since ϕ˜ minimises the classical action then the linear term in ϕ˜ of (4.93) disappears. Our next step is to re-scale the integration variable as follows (4.95)
ϕ˜ = ~1/2 ϕ And plug this back into (4.81) yielding (4.96)
Z[J] = N 0 exp i~−1
Z Z 1 λ d4 x[L(ϕ0 ) + Jϕ0 ]× Dϕ exp i d4 x L2 (ϕ, ϕ0 ) − ~1/2 ϕ3 ϕ0 − λ~ϕ4 6 24
Z
We can recover a Gaussian integral out of this by noting that we are only interested in the rst-order corrections, hence terms proportional to ~1/2 and ~ may be discarded. This procedure gives (4.97)
Z
1 d xL2 (ϕ, ϕ0 ) = − 2 4
Z
d4 xd4 x0 ϕ(x0 )A(x0 , x, ϕ0 )ϕ(x)
with, as expected from our previous discussions (4.98)
1 2 µ 2 A(x , x, ϕ0 ) = −∂x0 µ ∂x + µ + λϕ0 δ(x0 − x), 2 0
nally obtaining for the value of 0
−1
Z[J] ≈ N exp i~
Z
Z (4.99) 1 0 d x[L(ϕ0 ) + Jϕ0 ] exp − tr log A(x , x, ϕ0 ) . 2 4
By our conditions (4.82) and ϕ0 [0] = 0 we then obtain (4.100)
i Z[0] = 1 ≈ N 0 exp − tr log A(x0 , x, 0) . 2 Comparison with (4.81) shows that by keeping the same de nition of A we then have (4.101)
Z X0 [J] =
d4 x[L(ϕ0 ) + Jϕ0 ]
X1 [J] =
A(x0 , x, ϕ0 ) i tr log . 2 A(x0 , x, 0)
(4.102)
The eective action is computed by expanding (4.102)
Γ[ϕcl ] = Γ0 [ϕcl ] + ~Γ1 [ϕcl ] + O(~2 ).
64 Now, since ϕcl is a functional of J, then (4.103)
ϕcl (x) =
δX[J] , δJ(x)
we have the following expansion (4.104)
Z Γ0 [ϕcl ] = X0 [J]−
Z
4
d xJϕ0 =
Z
4
4
d xL(ϕ0 ) =
d x
1 µ 1 2 2 1 4 (∂ ϕcl )(∂µ ϕcl ) − µ ϕcl − ϕcl , 2 2 4!
which is the same as (4.89). The additional term in the expanded eective action is (4.105)
Z ~Γ1 [ϕcl ] = X0 [J] − Γ0 [ϕcl ] − Z
Z
4
d x[L(ϕ0 ) + Jϕ0 ] −
=
d4 xJϕcl + ~X1 [J]
d4 x[L(ϕcl ) + Jϕcl ] +
A(x0 , x, ϕ0 ) i~ tr log . 2 A(x0 , x, 0)
Because ϕ0 is a solution of (4.91) the dierence of the two integrals in (4.105) is of order
(ϕ0 − ϕcl )2 = O(~2 ), and we can interchange ϕ0 and ϕcl at this level of accuracy. Therefore (4.106)
Γ1 [ϕcl ] =
A(x0 , x, ϕcl ) i tr log . 2 A(x0 , x, 0)
Next, the eective potential V (ϕcl ) can be derived from Γ[ϕcl ] by setting ϕcl constant, in which case (4.107)
Z Γ[ϕcl ] = −
d4 xV (ϕcl ).
Delta functions allow us to diagonalise A(x0 , x, ϕcl ) which is a prerequisite to properly de ne the logarithmic part of (4.106). This is done as follows (4.108)
0
A(x , x, ϕcl ) = Z =
d4 k (2π)4
−∂x0 µ ∂xµ
1 2 + µ + λϕcl δ(x0 − x) 2 2
Z 0 1 d4 k 1 2 2 2 ik(x0 −x) λϕ −∂x0 µ ∂xµ + µ2 + λϕ2cl eik(x −x) = −k + µ + cl e 4 2 (2π) 2
Hence performing log and trace operation (4.109)
log A(x0 , x, ϕcl ) =
Z
d4 kd4 k 0
0 0 1 2 e−ixk eix k 0 2 2 log −k + µ + λϕ cl δ(k − k) 2 (2π) 2 (2π)2
(4.110)
Z tr log A =
4
4 0
0
0
Z
d xd x δ(x − x) log A(x , x, ϕcl ) =
4
d x
Z
d4 k 1 2 2 2 log −k + µ + λϕcl . (2π)4 2
Summarizing we can now state the following: the one-loop order contribution to the eective potential is (4.111)
−i V1 (ϕcl ) = 2
Z
d4 k log (2π)4
−k 2 + µ2 + 21 λϕ2cl −k 2 + µ2
65 and while at this order, the eective potential is approximated by (4.112)
1 1 i V (ϕcl ) ≈ V0 (ϕcl ) + V1 (ϕcl ) = µ2 ϕ2cl + λϕ4cl − ~ 2 4! 2
1 λϕ2cl d4 k log 1 − , (2π)4 2 k 2 − µ2
Z
where, the Greek parameters are bare parameters, i.e. the ones present in the Lagrangian. The integral term in the potential is ultraviolet divergent and it requires dimensional regularization to deal with i.e. (see for instance Bailin and Love, p.80) (4.113)
d2ω k 2 1 (k − µ2B + iε)−1 = (2π)2ω i
Z I(ω, µB ) =
iµ2B = (M 2 )ω−2 16π 2
d2ω 6 k 2 (6 k + µ2B )−1 (2π)2ω
Z
1 µ2B 0 + O(ω − 2) . + Γ (1) + 1 − log 2−ω 4πM 2
The derivative of the gamma function at -1 is Γ0 (1) = −γ, as it can be seen by taking the logarithmic derivative of Weierstrass product (1.12)
− log Γ(s) = log s + γs +
∞ h X k=1
∞ X Γ0 (s) 1 1 s si 1 − ⇒− = +γ+ − log 1 + k k Γ(s) s k+s k
k=1
and hence 0
Γ (1) = −1 − γ −
∞ X k=1
1 1 − = −γ. k+1 k
By the use of the following transformations for the bare parameters (4.114)
ϕB (x) = Z 1/2 ϕ(x) Zµ2B = µ2 + δµ2
Z 2 λB = λ + δλ,
we can transform the above potential (4.112) to (4.115)
V (ϕcl ) =
1 2 2 1 1 1 i µ ϕcl + λϕ4cl + δµ2 ϕ2cl + δλϕ4cl − ~ 2 4! 2 4! 2
Z
1 λϕ2cl d4 k log 1 − , (2π)4 2 k 2 − µ2
with the new Greek parameters now being the renormalized parameters rather than the bare. It is important to note, however, that these parameters have constraints imposed by the MS scheme. In terms of Green functions, these develop singularities in ω − 2 and these are neutralized by the counter terms which are (4.116)
δλ = M
4−2ω
ˆ M/µ, ω) + a0 (λ,
∞ X ˆ M/µ) ak (λ, k=1
2
2
δµ = µ
ˆ M/µ, ω) + b0 (λ,
ˆ M/µ, ω) + δZ = c0 (λ,
(2 − ω)k
∞ X ˆ M/µ) bk (λ, k=1
!
!
(2 − ω)k
∞ X ˆ M/µ) ck (λ, k=1
(2 − ω)k
ˆ = λM 2ω−4 . Because we have only simple poles, with a0 , b0 and c0 regular as ω → 2 and λ the k -integrals only have simple poles at 2 − ω . This means that ∀k > 1 ⇒ ak = bk = 0.
66 The one-point-irreducible Green functions of the renormalized theory behave as (4.117)
ˆ 2 1 µ2 0 ˜ 2 (p, −p) = p2 (1 + δZ1 ) − µ2 − δµ21 + λµ Γ + Γ + O(ω − 2) . (1) + 1 − log 32π 2 2 − ω 4πM This can be shown by using (4.118)
Z I4 (µB ) =
d4 k 2 (k − µ2B + iε)−1 (2π)4
and (4.113) on the Feynman propagator (4.119)
Z ∆F (0) =
2 d4 k 2 2 −1 2ω−4 iµ (k − µ + iε) = M B (2π)4 16π 2
1 µ2 0 + Γ (1) + 1 − log + O(ω − 2) . 2−ω 4πM
This propagator, in turn, is present in the Green function as (4.120)
1 2 2 2 2 ˜ Γ (p, −p) = p (1 + δZ1 ) + µ + λi∆F (0) + δµ1 + O(λ2 ), 2 as it can be seen by writing the Feynman diagrams to leading order. ˜ 4 is lengthier but follows the same lines, eventually, since Γ ˜ 4 is nite The computation of Γ in any renormalisation scheme we have that as ω → 2 the following holds (4.121)
ˆ 1 3λλ − δλ2 → const, 32π 2 2 − ω with (4.122)
δλ =
∞ X
δλk .
k=2
The nite part of δλ2 is arbitrary. Going back to our discussion of (4.116) now we see that the poles in 2 − ω are neutralized by setting (4.123)
a1 =
3 ˆ2 λ , 32π 2
b1 =
1 ˆ λ. 32π 2
Therefore putting this back together in the potential one has (4.124)
λ 3 1 2 2 0 + Γ (1) + log 4π V (ϕcl ) = µ ϕcl 1 + b0 − 2 32π 2 2 1 4 3λ2 3 0 λϕcl 1 + a0 − + Γ (1) + log 4π 4! 32π 2 2 " # 2 µ2 + 12 λϕ2cl 1 1 µ2 2 4 + µ + λϕcl log − µ log 2 . 64π 2 2 M2 M +
In turns out that in the MS scheme we may choose (see Rajantie or Bailin and Love) (4.125) MS
δλ
ˆ 3λλ = 32π 2
1 0 + Γ (1) + log 4π + O(λ3 ) 2−ω
67
aMS = 0
ˆ2 3λ [Γ0 (1) + log 4π] + O(λ3 ), 32π 2
aMS = 1
ˆ2 3λ + O(λ3 ), 32π 2
This will get rid of the Γ0 (1) + log 4π factors. However, and here is the main objective of this comparison with the zeta function regularization, we could also renormalize V (ϕcl ) by writing it as a function of physical mass and coupling constant. This can be accomplished by setting aphys and bphys such that the following hold (4.126) 0 0
d2 V = µ2 , dϕ2cl ϕcl =0
d4 V = λ, dϕ4cl ϕcl =M
as we did in (4.20) and (4.21). In the case that µ2 is small there could be radiative corrections that generate spontaneous symmetry breaking. Performing the second and fourth derivatives (4.126) along with the discussion in (4.20) to (4.22) leads to the same potential (4.25), namely (4.127)
V (ϕcl ) =
1 2 2 λ λ2 ϕ4cl µ ϕcl + ϕ4cl + 2 4! 256π 2
ϕ2 25 log cl2 − . M 6
Eectively we have compared the zeta function regularization and the one-loop order expansion and have seen that both yield the same results in dierent fashions. The zeta function technique makes use of more powerful tools (specially, more sophisticated functions) whereas the one-loop expansion mostly requires Γ function and its derivative.
REFERENCES The above discussion of the role of the zeta function in quantum eld theory follows from Hart eld [611 to 616], Ramond [81 to 93] and the rst three chapters from Elizalde, Odintsov, and Romeo. Additionally, the exposition is the progression from the ideas presented in Chapter 2 (zeta functions in quantum mechanics) combined with Chapter 3 (path integrals in quantum elds). Finally, the discussion of the eective potential in one-loop order follows from Bailin and Love [200 - 207] and other remarks
68
Appendix Claim 1: The following holds
Z
s
Z
dxn exp iλ (x1 − a)2 + (x2 − x1 )2 + · · · + (b − xn )2 =
dx1 · · ·
Proof.
in π n iλ 2 exp (b − a) (n + 1)λn n+1
(by induction)
Assume it is true for n and show it is true for n + 1 Z Z dx1 · · · dxn+1 exp iλ (x1 − a)2 + (x2 − x1 )2 + · · · + (b − xn )2
s = =
in π n (n + 1)λn
in π n (n + 1)λn
Z dxn+1 exp
2 Z
iλ (xn+1 − a)2 exp[iλ(b − xn+1 )2 ] n+1
dxn+1 exp iλ
1 (xn+1 − a)2 + (b − xn+1 )2 n+1
,
the exponential in the integrand can be worked out as follows
1 n+2 2 = (xn+1 − a)2 + (b − xn+1 )2 y − 2y(b − a) + (b − a)2 xn+1 −a=y n + 1 n+1 2 n+1 1 n+2 y− (b − a) + (b − a)2 . = n+1 n+2 n+2 Finally, let λ − ((n + 1)/(n + 2))(b − a) = z so that the integral becomes
s
in π n (n + 1)λn
Z
s n+2 2 iλ iλ in+1 π n+1 2 2 dz exp iλ z + (b − a) = exp (b − a) , n+1 n+2 (n + 1 + 1)λn+1 n+2
and this concludes the proof by induction. Claim 3: If
J are N
M
is a symmetric N × N matrix with real-valued elements Mij and
and
component vectors with components qi and Ji respectively, then
Z
1 d q exp − qT Mq + JT q 2 N
Z(J) =
Proof.
q
The process is to diagonalize the matrix
relations hold
(2π)N/2 =√ exp det M
M
ΛT Λ = 1
det Λ = 1
Z
˜ T where the following as M = ΛMΛ
m ˜1 . ˜ = . M . 0
··· .. .
0 .. .
···
m ˜N
,
the integral becomes
Z(J) =
1 T −1 J M J 2
1 ˜ T q + JT q , dN q exp − qT ΛMΛ 2
69
˜ = ΛT J from which it follows next we de ne the following q ˜ = ΛT q and J dN q = dN q˜ det Λ = dN q˜. " # Z X 1 1 T ˜ T N 2 ˜ ˜ ˜ M˜ ˜ i q˜i + Ji q˜i Z(J) = d q exp − q q+J q ˜ = d q˜i exp − m 2 2 i ∞ " " #! ! # !−1/2 r Z N N X 1 Y X J2 Y Y ˜2 2π J i i N 2 N/2 = d q˜i exp − m ˜ i q˜i + J˜i q˜i = exp = (2π) m ˜i exp . 2 m ˜i 2m ˜i 2m ˜i i i i i=1 i=1 Z
N
−∞
Finally we note that the inverse of the diagonal matrix is
˜ −1 M
m ˜ −1 1 . = .. 0
··· .. .
0 .. .
···
m ˜ −1 N
and therefore the last product is the determinant of the matrix
Y
˜ = det M, m ˜ i = det M
i
and consequently the result follows
(2π)N/2 Z(J) = √ exp det M
1 ˜T −1 ˜ J M J 2
(2π)N/2 =√ exp det M
1 T −1 J M J . 2
Claim: One has the following
Z WE [J] = NE exp {SEuc [φ(x0 ), J]}
= NE0 q
Z 1 δ 2 SEuc d4 x1 d4 x2 φ(x1 ) φ(x2 ) Dφ exp − 2 δφ(x1 )δφ(x2 )
exp {SEuc [φ(x0 ), J]} . det (−∂¯µ ∂¯µ + m2 + V 00 [φ(x0 )])δ(x1 − x2 )
Proof. This follows from Claim 3 with the action Z 1 1 SE [φ, J] = d4 x ¯ ∂¯µ φ∂¯µ φ + m2 φ2 + V (φ) − Jφ 2 2 as expanded on (3.54) and using
δ2 SEuc = δ(¯ x1 − x ¯2 ) −∂¯µ ∂¯µ + m2 + V 00 [φ(x1 )] δφ(x1 )δφ(x2 ) Z WE [J] = NE exp {SEuc [φ(x0 ), J]}
Z 1 δ 2 SEuc Dφ exp − d4 x1 d4 x2 φ(x1 ) φ(x2 ) 2 δφ(x1 )δφ(x2 )
70 The formulas ∞
coth x =
X (2x)2k 1 B2k , +2 x (2k)k
coth πx =
1 2x X 1 + πx π x2 + n2 n≥1
k=1
are almost always quoted as Gradshteyn and Ryzhik, p.35. The rst one represents the Laurent series. There is very little added value in reproducing the proofs here. Grassman Numbers Let us introduce some notation rst. For convenience, we will the same equation labelling as in Chapter 2 to ease the read. The following presentation about Grassmann numbers follows the notes of A. Rajantie and Peskin and Schroeder. Ordinary commuting numbers will be denoted c-numbers (these can be real or complex). Now let n generators {θ1 , · · · , θn } satisfy the anti-commutation relations (2.70)
{θi , θj } = 0 ∀i, j. Then the set of the linear combinations of {θi } with the c-number coe cient is called the Grassmann number and the algebra generated by {θi } is called the Grassmann algebra, denoted by Λn . Let us taken an arbitrary element
g(θ) = g0 +
n X i=1
gi θi +
X
gij θi θj + · · · =
i<j
g of this algebra expand it as (2.71) X
0≤k≤n
1 X gi1 ,··· ,ik θi1 · · · θik , k! {i}
where g0 , gi , gij , · · · and gi1 ,··· ,ik are c-numbers that are anti-symmetric under the exchange of two indices. Additionally, we can write
g(θ) =
X
g as (2.72)
g˜k1 ,··· ,kn θ1k1 · · · θnkn .
ki =0,1
It is impossible for the set of Grassmann numbers to be an ordered set because the generator
θk does not have a magnitude. The only number that is both c-number and Grassmann number is zero, moreover, a Grassmann number commutes with a c-number. From the discussion above it follows that (2.73)
θk2 = 0 θk1 θk2 · · · θkn = εk1 k2 ···kn θ1 θ2 · · · θn θk1 θk2 · · · θkm = 0
(m > n).
The tensor εk1 k2 ···kn is the Levi-Civita symbol, de ned as
εk1 k2 ···kn
+1 if{k1 · · · kn }isanevenpermutationof{1 · · · n} = −1 if{k1 · · · kn }isanoddpermutationof{1 · · · n} . 0 otherwise
Functions of Grassmann numbers are de ned in terms of Taylor expansions of the function.
71 If n = 1 we have the simple expression (2.74)
eθ = 1 + θ since terms O(θ2 ) are zero. Our next step is to develop the theory of dierentiation and integration of Grassmann variables, this theory has a few surprising facts, for instance dierentiation is the same process as integration. We assume that the dierential operator acts on a function from the left, let θi and θj be two Grassmann variables then (2.74)
∂θj ∂ = θj = δij . ∂θi ∂θi Similarly, we assume that the dierential operator anti-commutes with θk . The product rule has the slightly dierent form (2.75)
∂ ∂θj ∂θk (θj θk ) = θk − θj = δij θk − δik θj . ∂θi ∂θi ∂θi Moreover, the following properties hold (2.76)
∂ ∂ ∂ ∂ ∂2 + = 0 ⇒ 2 = 0, ∂θi ∂θj ∂θj ∂θi ∂θi the last equation is termed nil-potency and nally (2.77)
∂ ∂ θj + θj = δij . ∂θi ∂θi Let us now move to integration. To this end, we adopt the notation with respect to a Grassmann variable and
D for dierentiation
I for integration. Let us suppose that these
operations satisfy the relations (2.78)
ID = DI = 0, where
D(A) = 0 ⇒ I(BA) = I(B)A,
A and B are arbitrary functions of Grassmann variables. The rst part of the rst
equation implies that the integral of a derivative gives a surface term and it is set to zero, whereas the second part implies that the derivative of an integral vanishes. The last equation implies that if the derivative of the function is zero then it can be taken out the integral. The relations are satis ed when
D is proportional to I and for normalization purposes we
set I = D and write (2.79)
Z dθg(θ) =
∂f (θ) . ∂θ
From the previous de nition it follows that (2.80)
Z dθ =
∂1 = 0, ∂θ
Z dθ θ =
∂θ = 1. ∂θ
72 We can generalize this for
n generators {θi } as (2.81)
Z dθ1 dθ2 · · · dθn g(θ1 θ2 · · · θn ) =
∂ ∂ ∂ ··· g(θ1 θ2 · · · θn ). ∂θ1 ∂θ2 ∂θn
In a theory where dierentiation is equivalent to integration we would expect some strange behaviour. In order to see this behaviour we consider the simpler case when we have only one generator and we change variables θ0 = aθ where (2.82)
Z dθg(θ) =
a is a complex number, then one has
∂g(θ) ∂g(θ0 /a) = =a ∂θ ∂θ0 /a
Z
dθ0 g(θ0 /a)
which implies that dθ0 = (1/a)dθ. The extension to the general case with
n generators
yields θi → θi0 = aij θj and hence (2.83)
Z dθ1 dθ2 · · · dθn g(θ) =
n X ∂θk0 1 ∂θ0 ∂ ∂ ∂ ∂ ∂ ··· g(θ) = · · · kn 0 · · · 0 g(a−1 θ0 ) ∂θ1 ∂θ2 ∂θn ∂θ1 ∂θn ∂θk1 ∂θkn ki =1
=
n X
εk1 ···kn ak1 1 · · · akn n
ki =1
∂ ∂ · · · 0 g(a−1 θ0 ) = det a ∂θk0 1 ∂θkn
Z
dθ10 · · · dθn0 g(a−1 θ0 ).
Consequently the measure has the Jacobian (2.84)
dθ1 dθ2 · · · dθn = (det a)dθ10 · · · dθn0 . In the case of a single variable, the delta function of a Grassmann variable is de ned in a similar fashion as with c-numbers de ned as (2.85)
Z dθ δ(θ − z)g(θ) = g(z). However, in the case of Grassmann variables we can obtain a closed expression for the delta function. If we set g(z) = a + bz in the de nition of the delta function we have (2.86)
Z dθ δ(θ − z)(a + bθ) = a + bz and this means that (2.87)
δ(θ − z) = θ − z. Again, we can extend this to
n generators if we are careful about the order of the variables
(2.88)
δ n (θ − z) = (θn − zn ) · · · (θ2 − z2 )(θ1 − z1 ). We can nd the integral of the delta function by considering complex Grassmann variables which we proceed to develop later. Consider (2.89)
Z
dξ eiξθ =
Z dξ (1 + iξθ) = iθ
73 so that we have (2.90)
Z
dξ eiξθ .
δ(θ) = θ = −i
One of the most crucial developments of the Grassmann variables is the Grassmann Gaussian integral which will be fundamental when developing the path integral formalism of fermions. Let us evaluate the following integral (2.91)
Z I=
dθ1∗ dθ1 · · · dθn∗ dθn exp
X
θi∗ Mij θj
ij
where it is important to stress that {θi } and {θi∗ } are two independent sets of Grassmann variables. Since Grassmann variables θi and θi∗ anti-commute we can take the n×n c-number matrix
M to be anti-symmetric.
The formula for the transformation of the measure solves P the problem of the computation. Set θi0 = j Mij θj this yields (2.92)
Z I = det M
! dθ1∗ dθ10
· · · dθn∗ dθn0
exp −
X
θi∗ θi0
Z = det M
n dθ dθ(1 + θ θ ) = det M. ∗
0 ∗
i
Complex conjugation is de ned as (2.93)
(θi )∗ = θi∗ ,
(θi∗ )∗ = θi .
In the case of Grassmann variables we have (2.94)
(θi θj )∗ = θj∗ θi∗ . The reasoning behind (2.94) is that the real c-number θi θi∗ does not satisfy (θi θi∗ )∗ = θi θi∗ .
Let us recall that the annihilation and creation operators c and c† satisfy the anti-commutation relations {c, c† } = 1 and {c, c} = {c† , c† } = 0 and that the number operator N = c† c has eigenvectors |0 > and |1 >. We are now in a position to study the Hilbert space Ω spanned by these vectors, i.e. (2.95)
Ω = span {|0 >, |1 >} . An arbitrary vector |ω > ∈ Ω can be written in the form (2.96)
|ω > = |0 > ω0 + |1 > ω1 , with ωi ∈ C where i = 1, 2. Next we de ne the coherent states (2.97)
|θi = |0 > +|1 > θ
< θ| = < 0| + θ∗ < 1|
where θ and θ∗ are Grassmann numbers. The coherent states are eigenstates of
c and c† respectively, that is (2.98)
c|θ > = |0 > θ = |θ > θ,
< θ| c† = θ∗ < 0| = θ∗ < θ|.
74 It can be shown fairly easily that the following identities hold (2.99)
hθ0 | θi = 1 + θ0∗ θ = eθ
0∗
θ
,
hθ | gi = g0 + θ∗ g1 ,
θ |c† | g = hθ | 1i g0 = θ∗ g0 = θ∗ hθ | gi , hθ |c| gi = hθ | 0i g1 =
∂ hθ | gi . ∂θ∗
Finally, we show how matrix elements are represented and the completeness relation. Let (2.100)
p(c, c† ) = p00 + p10 c† + p01 c + p11 c† c,
pij ∈ C
c and c† . The complex matrix elements of p can be written in terms of scalar products as (2.101) be an arbitrary function of
h0 |p| 0i = p00 ,
h0 |p| 1i = p01 ,
h1 |p| 0i = p10 ,
h1 |p| 1i = p00 + p11 .
From these scalar products we can form the more general product (2.102)
hθ |p| θ0 i = (p00 + θ∗ p10 + p01 θ0 + θ∗ θ0 p11 )eθ
∗ 0
θ
.
Moreover, one has (2.103)
Z
∗
−θ ∗ θ
dθ dθ |θi hθ| e Z =
Z =
dθ∗ dθ (|0i + |1i θ) (h0| + θ∗ h1|) (1 − θ∗ θ)
dθ∗ dθ (|0i h0| + |1i θ h0| + |0i θ∗ h1| + |1i θθ∗ h1|) (1 − θ∗ θ) = |0i h0| + |1i h1| = I,
and therefore the completeness relation is (2.104)
Z
dθ∗ dθ |θi hθ| e−θ
∗
θ
= I.
75
References Apostol, Tom "Introduction to Analytic Number Theory" Bailin and Love "Introduction to Gauge Field Theory" Elizalde, Odintsov, and Romeo "Zeta Regularization Techniques with Applications" Freitag, Eberhard and Busam, Rolf "Complex Analysis" Grosche C. and Steiner F, "Handbook of Feynman Path Integrals" Ingham, A.E. "The Distribution of Prime Numbers" Kleinert, Hagen "Path Integrals in Quantum Mechanics, Statistic, Polymer Physics and Financial Markets" Hat eld, Brian "Quantum Field Theory of Point Particles and Strings" Peskin, Michael E. and Schroeder, Daniel V. "An Introduction to Quantum Field Theory" Rajantie, Arttu "Advanced quantum eld theory Course Notes" Ramond, Pierre "Field Theory: A Modern Primer" Rivers, Ray J. "Path Integral Methods in Quantum Field Theory" Sondow, Jonathan and Weisstein, Eric W. Riemann Zeta Function." From MathWorldA Wolfram Web Resource Titchmarsh, E. C. and Heath-Brown, Roger "The Theory of the Riemann Zeta-Function" Weinberg, Steven "The Quantum Theory of Fields: Volume I Foundations" Wikipedia on the Gammafunction Whitaker and Watson "A Course on Modern Analysis"
