Henrickson’s Derivation For Electron-photon Self-energy

Progress Report: Henrickson’s Derivation for Electron-Photon Self-Energy Oka Kurniawan May 14, 2009

1

Introduction

This report gives a more detail derivation of the self-energy expression given in Henrickson’s paper [1]. Most of the derivation shown here is provided in the original paper. The purpose is just to point out some details required to understand the derivation.

2 2.1

Derivation Electromagnetic Interaction Hamiltonian

We recall that the electromagnetic wave can be written in terms of the vector and scalar potentials. ∂A − ∇φ ∂t B =∇ × A E=−

(1) (2)

where E is the electric field, A is the vector potential, φ is the scalar potential, and B is the magnetic field. The Hamiltonian is given by  2    p q 0 1 H =H +H = +U + A·p (3) 2m0 m0 where the first bracket is the zeroth-order Hamiltonian, and the second bracket is the electromagnetic interaction part. 1

Now let us write the vector potential in terms of the bosonic annihilation and creation operators A(x, t) = A0 (x)(be−iωt + b† eiωt )

(4)

where A0 (x) is dtermined by ∇2 A0 (x) +

ω2 A0 (x) = 0 c2

(5)

The solution to this equation has the general form A0 (x) = A0 eik·x

(6)

where k is the wave vector. We can substitute this back to (5) to verify it. −k 2 A0 eik·x +

ω2 A0 eik·x = 0 c2

(7)

The summation gives zero if k is related to the frequency ω by k=

ω c

(8)

In a material it becomes

ω√ µr ǫr (9) c In free space the square root gives unity. Now let us solve for A0 . Substituting (4) to (1) and (2) as well as using (6) gives us k=

E = −A0 (x) −iωbe−iωt + iωb† eiωt




(10)

A∗0 e−i(k·x−ωt)

(11)

= iωA0ei(k·x−ωt − iωA∗0 e−i(k·x−ωt) i(k·x−ωt)

B = ik × A0 e

− ik ×

The energy density of the electromagnetic field is 1 E = (D · E + B · H) 2

(12)

from this we can get the time averaged energy density as Eavg = 2ǫω 2 A20 2

(13)

and integrating over volume, we get the time average total energy as 2ǫω 2 V A20

(14)

Assuming that the field energy is due to a photon with energy ~ω, we have ~ω =2ǫω 2 V A20 r ~ A0 = 2ωǫV

(15)

We want to reexpressed this in terms of photon flux Iω ≡

Nc √ V µr ǫr

(16)

which is defined as the number of photon per unit time per unit area. Substituting the V in (15) using this expression, we get A0 =



 √ Iω ~ µr ǫr 1/2 2ωǫNc

(17)

Note that the direction of A0 is determined by the polarization of the field, which we will denote by ˆ a. So now the vector potential is given by  √
Iω ~ µr ǫr 1/2 A(x, t) = ˆ a be−iωt + b† eiωt 2ωǫNc 

(18)

where we have made a dipole approximation ek·x ≈ 1. Now we have obtained the expression for the vector potential, and ready to calculate the interaction Hamiltonian. We first write the interaction Hamiltonian in the second quantized Hamiltonian as given by X hl|H 1 |mia†l am (19) H1 = lm

where l and m are the site-basis eigenstates. The matrix element is hl|H 1|mi =

q A · hl|p|mi m0

(20)

Now we assume that the field is polarized in the ˆ z direction. Then the matrix element becomes  √ 1/2   pˆz ~ µ ǫ r r hl|H 1 |mi = q (21) Iω (be−iωt + b† eiωt ) × l m 2Nωǫc m0 3

Recall that p = mv, so that p/m = v = dx/dt. Therefore,       pˆz dz i  0  l m = l m = l H , z m m0 dt ~

where we have used the following relationship in the last term. + * ˆ i ˆ ˆ ∂Q d hQi = h[H, Q]i + dt ~ ∂t

(22)

(23)

ˆ In most cases ∂ Q/∂t = 0. Now we can write the last term as

Hence

hl|H 0z − zH 0 |mi = hl|H 0 z|mi − hl|zH 0 |mi = zm hl|H 0 |mi − zl hl|H 0|mi

  pˆz

 i l m = (zm − zl ) l H 0 m m0 ~ when we substitute this back, we will get  √ 1/2 iq ~ µr ǫr 1 Iω (be−iωt + b† eiωt )hl|H 0|mi hl|H |mi = (zm − zl ) ~ 2Nωǫc

and from (19)  √ 1/2 X iq ~ µr ǫr 1 H = (zm − zl ) Iω (be−iωt + b† eiωt )hl|H 0|mia†l am ~ 2Nωǫc lm If we use finite difference, the matrix element is discretized as  ~2  m∗ ∆2 + Uj , j = k 2 Hjk = − ~ ,j = k ± 1  2m∗ ∆2 0 , otherwise

(24) (25)

(26)

(27)

(28)

Note that if l = m, then zm − zl = 0 and hl|H 1|mi = 0. If m = l + 1, then zm − zl = ∆. On the other hand, if m = l − 1, then zm − zl = −∆. Hence, the interaction Hamiltonian can be written as X Mlm (b−iωt + b† eiωt )a†l am (29) H1 = lm

where

and

 √ 1/2 ~ µr ǫr q~ Mlm = Iω Plm i2m∗ ∆ 2Nωǫc   +1 , m = l + 1 Plm = −1 , m = l − 1  0 , else 4

(30)

(31)

2.2

Electron-Photon Self-Energy

The lowest-order contribution to the less-than and greater-than self-energies for electron-photon interactions are given by X > Σ> G> (32) lm (t1 , t2 ) = pq (t1 , t2 )Dlp;qm (t1 , t2 ) pq

Σ< lm (t1 , t2 ) =

X

< G< pq (t1 , t2 )Dlp;qm (t1 , t2 )

(33)

pq

where D ≷ are the photon propagators

1  < Dlp;qm (t1 , t2 ) ≡ = Hqm (t2 )Hlp1 (t1 )

 > 1 (t2 ) Dlp;qm (t1 , t2 ) ≡ = Hlp1 (t1 )Hqm

(34) (35)

Assuming that the photon population remains at equilibrium, we can show that hb(t1 )b(t2 )i = 0

hb(t1 )b† (t2 )i = (N + 1)e−iω(t1 −t2 )

hb† (t1 )b(t2 )i = Neiω(t1 −t2 )

hb† (t1 )b† (t2 )i = 0

(36) (37) (38) (39)

Recall that the matrix element for the interaction Hamiltonian is given by (29), or
Hlp1 (t1 ) = Mlp be−iωt1 + b† eiωt1 (40)

Substituting this into (34) and (35) gives > Dlp;qm (t1 , t2 ) =

=

Mlp Mqm be−iωt1 beiωt2 + b† e−iωt1 beiωt2
+be−iωt1 b† eiωt2 + b† e−iωt1 b† eiωt2
Mlp Mqm Neiω(t1 −t2 ) + (N + 1)e−iω(t1 −t2 )

(41) (42)

In steady state, we can Fourier transform with respect to t1 − t2 . > Dlp;qm (E) = 2πMlp Mqm {Nδ(E + ~ω) + (N + 1)δ(E − ~ω)}

(43)

Similarly, we can obtain < Dlp;qm (E) = 2πMlp Mqm {Nδ(E − ~ω) + (N + 1)δ(E + ~ω)}

Substituting these into (32) and (33) gives the self-energies 5

(44)

Σ> lm (E) =

X pq

Σ< lm (E)

=

X pq

Mlp Mqm {Nδ(E + ~ω) + (N + 1)δ(E − ~ω)}

(45)

Mlp Mqm {Nδ(E − ~ω) + (N + 1)δ(E + ~ω)}

(46)

Then the retarded self-energy can be obtained from the above equations through Z < ′ ′ dE ′ Σ< R lm (E ) − Σlm (E ) Σlm (E) = i (47) 2π E − E ′ + iη

References [1] L. E. Henrickson, “Nonequilibrium photocurrent modeling in resonant tunneling photodetectors,” Journal of Applied Physics, vol. 91, no. 10, pp. 6273–6281, 2002. [Online]. Available: http://link.aip.org/link/?JAP/91/6273/1

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