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 givenin Henrickson’s paper [1]. Most of the derivation shown here is provided inthe original paper. The purpose is just to point out some details required tounderstand the derivation.

2 Derivation

2.1 Electromagnetic Interaction Hamiltonian

We recall that the electromagnetic wave can be written in terms of the vectorand scalar potentials.

E = − ∂A

∂t−∇φ (1)

B =∇× A (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

H = H0 + H1 =

(

p2

2m0

+ U

)

+

(

q

m0

A · p)

(3)

where the first bracket is the zeroth-order Hamiltonian, and the secondbracket is the electromagnetic interaction part.

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Now let us write the vector potential in terms of the bosonic annihilationand creation operators

A(x, t) = A0(x)(be−iωt + b†eiωt) (4)

where A0(x) is dtermined by

∇2A0(x) +ω2

c2A0(x) = 0 (5)

The solution to this equation has the general form

A0(x) = A0eik·x (6)

where k is the wave vector. We can substitute this back to (5) to verify it.

−k2A0eik·x +

ω2

c2A0e

ik·x = 0 (7)

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

k =ω

c(8)

In a material it becomesk =

ω

c

√µrǫr (9)

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

E = −A0(x)(

−iωbe−iωt + iωb†eiωt)

= iωA0ei(k·x−ωt − iωA∗

0e−i(k·x−ωt) (10)

B = ik × A0ei(k·x−ωt) − ik ×A∗

0e−i(k·x−ωt) (11)

The energy density of the electromagnetic field is

E =1

2(D ·E + B · H) (12)

from this we can get the time averaged energy density as

Eavg = 2ǫω2A20 (13)

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and integrating over volume, we get the time average total energy as

2ǫω2V A20 (14)

Assuming that the field energy is due to a photon with energy ~ω, we have

~ω =2ǫω2V A20

A0 =

√

~

2ωǫV(15)

We want to reexpressed this in terms of photon flux

Iω ≡ Nc

V√

µrǫr(16)

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

A0 =

(

Iω~√

µrǫr

2ωǫNc

)1/2

(17)

Note that the direction of A0 is determined by the polarization of the field,which we will denote by a. So now the vector potential is given by

A(x, t) = a

(

Iω~√

µrǫr

2ωǫNc

)1/2(

be−iωt + b†eiωt)

(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 interactionHamiltonian in the second quantized Hamiltonian as given by

H1 =∑

lm

〈l|H1|m〉a†l am (19)

where l and m are the site-basis eigenstates. The matrix element is

〈l|H1|m〉 =q

m0A · 〈l|p|m〉 (20)

Now we assume that the field is polarized in the z direction. Then thematrix element becomes

〈l|H1|m〉 = q

(

~√

µrǫr

2NωǫcIω

)1/2

(be−iωt + b†eiωt) ×⟨

l

∣

∣

∣

∣

pz

m0

∣

∣

∣

∣

m

⟩

(21)

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Recall that p = mv, so that p/m = v = dx/dt. Therefore,⟨

l

∣

∣

∣

∣

pz

m0

∣

∣

∣

∣

m

⟩

=

⟨

l

∣

∣

∣

∣

dz

dt

∣

∣

∣

∣

m

⟩

=

⟨

l

∣

∣

∣

∣

i

~

[

H0, z]

∣

∣

∣

∣

m

⟩

(22)

where we have used the following relationship in the last term.

d

dt〈Q〉 =

i

~〈[H, Q]〉 +

⟨

∂Q

∂t

⟩

(23)

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

〈l|H0z − zH0|m〉 = 〈l|H0z|m〉 − 〈l|zH0|m〉= zm〈l|H0|m〉 − zl〈l|H0|m〉 (24)

Hence⟨

l

∣

∣

∣

∣

pz

m0

∣

∣

∣

∣

m

⟩

=i

~(zm − zl)

⟨

l∣

∣H0∣

∣m⟩

(25)

when we substitute this back, we will get

〈l|H1|m〉 = (zm − zl)iq

~

(

~√

µrǫr

2NωǫcIω

)1/2

(be−iωt + b†eiωt)〈l|H0|m〉 (26)

and from (19)

H1 =∑

lm

(zm − zl)iq

~

(

~√

µrǫr

2NωǫcIω

)1/2

(be−iωt + b†eiωt)〈l|H0|m〉a†l am (27)

If we use finite difference, the matrix element is discretized as

Hjk =

~2

m∗∆2 + Uj , j = k

− ~2

2m∗∆2 , j = k ± 10 , otherwise

(28)

Note that if l = m, then zm − zl = 0 and 〈l|H1|m〉 = 0. If m = l + 1, thenzm − zl = ∆. On the other hand, if m = l − 1, then zm − zl = −∆. Hence,the interaction Hamiltonian can be written as

H1 =∑

lm

Mlm(b−iωt + b†eiωt)a†l am (29)

where

Mlm =q~

i2m∗∆

(

~√

µrǫr

2NωǫcIω

)1/2

Plm (30)

and

Plm =

+1 , m = l + 1−1 , m = l − 1

0 , else(31)

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2.2 Electron-Photon Self-Energy

The lowest-order contribution to the less-than and greater-than self-energiesfor electron-photon interactions are given by

Σ>lm(t1, t2) =

∑

pq

G>pq(t1, t2)D

>lp;qm(t1, t2) (32)

Σ<lm(t1, t2) =

∑

pq

G<pq(t1, t2)D

<lp;qm(t1, t2) (33)

where D≷ are the photon propagators

D<lp;qm(t1, t2) ≡ =

⟨

H1qm(t2)H

1lp(t1)

⟩

(34)

D>lp;qm(t1, t2) ≡ =

⟨

H1lp(t1)H

1qm(t2)

⟩

(35)

Assuming that the photon population remains at equilibrium, we can showthat

〈b(t1)b(t2)〉 = 0 (36)

〈b(t1)b†(t2)〉 = (N + 1)e−iω(t1−t2) (37)

〈b†(t1)b(t2)〉 = Neiω(t1−t2) (38)

〈b†(t1)b†(t2)〉 = 0 (39)

Recall that the matrix element for the interaction Hamiltonian is givenby (29), or

H1lp(t1) = Mlp

(

be−iωt1 + b†eiωt1)

(40)

Substituting this into (34) and (35) gives

D>lp;qm(t1, t2) = MlpMqm

(

be−iωt1beiωt2 + b†e−iωt1beiωt2

+be−iωt1b†eiωt2 + b†e−iωt1b†eiωt2)

(41)

= MlpMqm

(

Neiω(t1−t2) + (N + 1)e−iω(t1−t2))

(42)

In steady state, we can Fourier transform with respect to t1 − t2.

D>lp;qm(E) = 2πMlpMqm {Nδ(E + ~ω) + (N + 1)δ(E − ~ω)} (43)

Similarly, we can obtain

D<lp;qm(E) = 2πMlpMqm {Nδ(E − ~ω) + (N + 1)δ(E + ~ω)} (44)

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

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Σ>lm(E) =

∑

pq

MlpMqm {Nδ(E + ~ω) + (N + 1)δ(E − ~ω)} (45)

Σ<lm(E) =

∑

pq

MlpMqm {Nδ(E − ~ω) + (N + 1)δ(E + ~ω)} (46)

Then the retarded self-energy can be obtained from the above equationsthrough

ΣRlm(E) = i

∫

dE ′

2π

Σ<lm(E ′) − Σ<

lm(E ′)

E − E ′ + iη(47)

References

[1] L. E. Henrickson, “Nonequilibrium photocurrent modeling in res-onant 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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