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Exciton photon matrix element

  1. Jan 12, 2012 #1
    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
    [itex]H_{el-op}=\sum_k D_k c^+_{kc}c_{kv}(a+a^+)[/itex]

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

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

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

    [itex]M_{ex-op}=\sum_k Z^{n*}_{k_{c},k_v}D_k [/itex]
     
  2. jcsd
  3. Jan 12, 2012 #2

    DrDu

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    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.
     
  4. Jan 12, 2012 #3

    Cthugha

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    How did you get from
    to
    ?

    Where have all the operators acting on the electrons gone?
     
  5. Jan 12, 2012 #4
    Please access the paper from here
     
  6. Jan 12, 2012 #5
    Just put all the operators, and then take anti-commutation relations several times
    [itex]\left \langle i|c^+_k c_l|i \right \rangle=\delta_{kl} n_i[/itex]

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

    [itex]\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[/itex]

    [itex]=\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[/itex]

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

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

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

    [itex]n_{kv}=1[/itex]
     
  7. Jan 13, 2012 #6

    DrDu

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    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.
     
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