Notes 6

1

Lecture 6 + 7 - Landau theory of phase transitions II I.

FIRST ORDER PHASE TRANSITIONS

To be able to account for first order phase transitions we add a third order symmetry-breaking term: FL = a(T − Tc )m2 − Cm3 + bm4 . . . − m · h

(1)

Note that even though m4 is less important than m3 at small m’s , we must keep it - otherwise the free energy is not bounded from below. To find the minimas of the Landau-free energy we differentiate it with respect to m and obtain: 2a(T − Tc )m − 3Cm2 + 4bm3 = 0

(2)

so apart from m = 0 we have: m=

3C ±

p

9C 2 − 32b2 a(T − Tc ) 8b

(3)

if: b2 a(T − Tc ) <

9 C 32

(4)

we get another minimum (and another maximum). At a slightly lower temperature, the second minimum hits the zero free energy - this is the point of transition, and then at lower temperatures, this minimum is the leading one. The free energy indeed does not jump, but the entropy does, which gives latent heat. There is a way of getting a first order phase transition with symmetry breaking - ie without symmetry breaking terms in the free energy. Although it seems that it requires special tuning this actually happens. Suppose that b the coefficient of m4 is negative. Then it seems that the free energy is not bounded from below. So we include the 6’th order into account. Then we will have: FL = a(T − Tc )m2 − bm4 + α6 m6 . . . − m · h

(5)

Now you can already see what’s going on - there are satellite minimas that come down as T drops to T c , until they hit the axis. then the order parameter jumps - and we have a phase transition of first order. II.

OTHER UNIVERSALITY CLASSES

So far we treated the simplest symmetries of them all - up down, or Z2 . Here we consider more complicated symmetries. A.

X-Y symmetry - SO(2)

Let us consider a different magnetic system than the Ising - magnets that live on the X-Y plane, The Hamiltonian is of the sort X ˆ XY = −J cos(φi − φj ) H

(6)

hi, ji

where the φ’s describe the angle of the magnets about the x-axis. The order parameter is a vector whose magnitude is a measure of order, and its orientation gives the direction of the magnetization: m ~ = (ψx , ψy ) = ψx x ˆ + ψy yˆ

(7)

2 The Hamiltonian is symmetric with respect to global rotations about the z axis. Rθ φ = θ + φ We can see that each term in the Hamiltonian has: X ˆ XY ({Rθ φ}) = −J ˆ XY ({φ}) H cos(φi − φj + θ − θ) = H

(8)

(9)

hi, ji

The order parameter we chose is indeed a representation. It is the vector representation of SO(2): ¶ µ ¶µ cos θ − sin θ ψx Rθ m ~ = sin θ cos θ ψy

(10)

What are the invariant terms of a free energy that we can write out of ψx and ψy ? The only only thing that doesn’t change under rotations is the size of m: ~ m ~ 2 = ψx2 + ψy2

(11)

This means that the most general free energy for the X-Y magnet is: 1 FL = a(T − Tc )(ψx2 + ψy2 ) + β m ~ 4 + h x ψx + h y ψy 2

(12)

We can continue with higher powers. Note that |m| ~ does not appear there - it is a non analytic function. SO(2) is Abelian: all elements in the group commute with each other. This implies that all irreducible representations of the group are one-dimensional. Which means that here we could find a single complex number which will correspond to the order parameter. Indeed: ψ = ψx + iψy

(13)

Rθ ψ = eiθ ψ

(14)

1 FL = a(T − Tc )ψψ ∗ + β(ψψ ∗ )2 + h∗ ψ + hψ ∗ 2

(15)

would do the trick.

The free energy would become:

where h = hx + ihy . And note that the symmetry breaking field transforms like ψ under the symmetry group, hence the last term is invariant. Now we can consider phase transitions. When T > TC nothing happens. ψ = 0 is a stable minimum, and no order ensues. On the other hand, if T < TC , ψψ ∗ has two stable non-zero minimas. But for ψ to be non-zero it has to choose a direction in the complex plane - m ~ has to choose a direction in the x-y plane. This is the symmetry breaking. ψ is constrained to be in the ring which is at the bottom of the Mexican hat potential. it has to choose a direction. Goldstone modes correspond to motion of the order parameter along the minimum of the free energy - changing the orientation. B.

SO(N)

Another example is the Heisenberg model, here the Hamiltonian is: ˆ = −J S ~i · S ~j H

(16)

where S is now an N-dimensional vector. This Hamiltonian is clearly invariant under rotation in the N dimensional space. The group of these rotation is called SO(N ). Everything is essentially the same as before, the invariants are the squares of the N tuple vectors - the only thing that stays constant under rotation is the vector itself.

3 C.

SU(N)

If you allow the numbers to be complex than the model written above is SU(N) - special unitary. III.

LATTICE SYMMETRY BREAKING IN 1-D

Consider a 1-d solid with lattice constant a. (DRAW A CHAIN) the solid can be distorted in many ways. Let’s concentrate on atoms moving up or down. Presumably, there is some Hamiltonian that involves this function y(x). Let us identify the symmetries. The 1-d lattice is invariant under translations by any number of lattice constants: T(na) f (x) = Qn f (x) = f (x + na)

(17)

where I defined Q as the translation by a single lattice constant. Let’s assume that up and down are not the same because the chain lies on some surface. When this symmetry is intact, we know that: Qn y(x) = y(x)

(18)

But this is not necessarily always so. For instance, the chain may buckle, in any number of ways. To describe such a buckling transition we need to go to the next step of the Landau scheme, and identify the representations of the symmetry group. This, again, is an Abelian group, and, in fact, one that we are very familiar with from quantum mechanics. Here are some representations of this. Consider the function: 2π

fm (x) = ei ma x

(19)

we have: 2π

Qfm (x) = ei m fm (x)

(20)

in fact, any complex number with modulus 1 will do, but let’s concentrate on just 2 of them - one would be exactly the same as the Ising model. Let’s consider m = 2 and m = 4. This implies, that we assume that: y(x) = y2 ei a x + y4 ei 2a x + C.C. π

π

(21)

By operating on y(x) with Q, we see: Qy(x) = y2 eiπ ei a x + y4 eiπ/2 ei 2a x = Qy2 ei a x + Qy4 ei 2a x

(22)

Qy2 = −y2 Qy4 = iy4

(23)

π

π

π

π

and we read off:

and the complex conjugate. Which are the invariants? Note that complex rotation is NOT asymmetry here, and we can very well assume that y2 is real, and that y4 is either real or imaginary. The invariants are: y22 , y4 y4∗ and their powers. But also, these two order parameters may couple to each other: y2 y42

(24)

Qy2 (Qy4 )2 = −y2 · (iy4 ) = y2 y42

(25)

is invariant under Q:

4 And higher powers of that can also appear. The free energy is thus: FL = y22 a(T − Tc ) + b2 y24 + h2 y2 +a4 |y4 |2 + b4 |y4 |4 −cy2 y42

(26)

I assumed that the phenomena we want to describe is first dimerization, and then maybe buckling at lower temperature. The dimerization term here is straight forward, just like the Ising. But now look at the four buckling - there is feedback from the dimerization to the buckling. it depends on whether y4 is real or imaginary. Do we understand this? Suppose y2 is positive, this chooses this dimerization. Now, if we choose y4 to be real, then: y4 eiπx/4a + CC = y4 cos(πx/4a).

(27)

y4 eiπx/4a + CC = iy4 sin(πx/4a)

(28)

If y4 is imaginary :

one lifts the atoms that are already up. But the other lifts the one that are down, closer to the surface on which the chain rests. Note that up-down symmetry, which would take y2 → −y2 , will preclude this term.