Can the Matrix Element of a+a^+ be Suppressed in Resonance Raman Measurements?

[hesky]

Please help me, I need to derive exciton-photon interaction.
Here, we are using second quantization. Please refer to this paper http://prb.aps.org/abstract/PRB/v75/i3/e035405
Hamiltonian of electron-photon is
H_{el-op}=\sum_k D_k c^+_{kc}c_{kv}(a+a^+)

c^+_{kc}c_{kv} are creation of electron to conduction band and annihilation electron in valence band, respectively. (a+a^+) are photon annihilation and creation operator.
Exciton wave function is
|\Psi^f\rangle=\sum_k Z^n_{k_{c},k_v}c^+_{k_{c}}c_{k_v}|0\rangle
Where Z is weighting coefficient, kc is electron state (conduction band), kv is hole state (valence band), and |0\rangle is ground state (all electrons occupy valence) band.

Matrix element of exciton-photon is defined as
M_{ex-op}=\langle\Psi^f|H_{el-op}|0\rangle

M_{ex-op}=\sum_k Z^{n*}_{k_{c},k_v}D_k\langle 0|a+a^+|0\rangle
My question is, how can we prove that \langle 0|a+a^+|0\rangle=1 to get

M_{ex-op}=\sum_k Z^{n*}_{k_{c},k_v}D_k

 

[DrDu]

My question is, how can we prove that \langle 0|a+a^+|0\rangle=1 to get

This depends on the photon state you use. With a photon vacuum, the expectation value would vanish. I suppose they assume some coherent state, however, I have no access to that article.

 

[Cthugha]

How did you get from

M_{ex-op}=\langle\Psi^f|H_{el-op}|0\rangle

to

M_{ex-op}=\sum_k Z^{n*}_{k_{c},k_v}D_k\langle 0|a+a^+|0\rangle

?

Where have all the operators acting on the electrons gone?

 

[hesky]

This depends on the photon state you use. With a photon vacuum, the expectation value would vanish. I suppose they assume some coherent state, however, I have no access to that article.

http://flex.phys.tohoku.ac.jp/~rsaito/rsj/j1180.pdf

 

[hesky]

How did you get from

to
?

Where have all the operators acting on the electrons gone?

Just put all the operators, and then take anti-commutation relations several times
\left \langle i|c^+_k c_l|i \right \rangle=\delta_{kl} n_i

\left \langle i|c_k c^+_l|i \right \rangle=\delta_{kl} (1-n_i)

\langle\Psi^f|H|0\rangle=\sum_{k,k'} Z^{n*}_{k_{c},k_v}\langle 0|c^+_{kv}c_{kc}c^+_{k'c}c_{k'v}(a+a^+)|0\rangle

=\sum_{k,k'} Z^{n*}_{k_{c},k_v}\langle 0|c^+_{kv}\delta_{k,k'}(1-n_{kc})c_{k'v}(a+a^+)|0\rangle

with n_{kc}=0 since at initial state no electron occupies conduction band

=\sum_{k} Z^{n*}_{k_{c},k_v}\langle 0|c^+_{kv}c_{kv}(a+a^+)|0\rangle

=\sum_{k} Z^{n*}_{k_{c},k_v}\langle 0|n_{kv}(a+a^+)|0\rangle

n_{kv}=1

 

[DrDu]

My question is, how can we prove that \langle 0|a+a^+|0\rangle=1 to get

M_{ex-op}=\sum_k Z^{n*}_{k_{c},k_v}D_k

I think that they are sloppy here and suppressed the matrix element of a+a^+. Resonance Raman involves rather high power fields and the expectation value of a+a^+ in that field can be expressed in terms of the squareroot of the intensity of the laser field (and maybe some constant involving it's frequency). Basically, the photon field can be treated classically.
For the consideration of the relative intensity of different Raman lines, the absolute intensity does not matter.