- #1
mSSM
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The following comes from Landau's Statistical Physics, chapter 32.
Using a Hamiltonian
[tex]\hat{H} = \hat{H}_0 + \hat{V}[/tex]
we get the following expression for the energy levels of a perturbed system, up to second order:
[tex]E_n = E_0^{(0)} + V_{nn} + \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2}{E_n^{(0)} - E_m^{(0)}}[/tex].
The prime signifies that the sum is done over all [itex]m\neq n[/itex].
Substituting this into the equation (from the normalisation of the Gibbs canonical distribution):
[tex]e^{-F/T} = \sum_0 e^{-E_n/T}[/tex].
This expression is then logarithmized and expanded in powers of [itex]V/T[/itex], so that we get:
[tex] F = F_0 + \sum_n V_{nn} w_n + \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 w_n}{E_n^{(0)} - E_m^{(0)}} - \frac{1}{2T} \sum_n V^2_{nn} w_n + \frac{1}{2T} \left( \sum_n V_{nn} w_n \right)^2 [/tex],
where [itex]w_n = \exp\left\{(F_0 - E_n^{(0)}/T\right\}[/itex] is the unperturbed Gibbs distribution.
We can now notice that
[tex] \sum_n V_{nn} w_n \equiv \overline{V}_{nn} [/tex],
i.e., the sum is the mean of [itex]V[/itex] avaraged over the quantum state and the statistical distribution.
And now this is where it gets interesting, and where I fail to see something. Landau writes, that you can rewrite the equation for the free energy above in the following way:
[tex] F = F_0 + \overline{V}_{nn} - \frac{1}{2} \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 (w_m - w_n)}{E_n^{(0)} - E_m^{(0)}} - \frac{1}{2T} \left\langle (V_{nn} - \overline{V}_{nn})^2 \right\rangle[/tex]
That makes perfect sense except for the part with the double-sum, where I don't understand how he obtains it:
[tex]\sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 w_n}{E_n^{(0)} - E_m^{(0)}}= -\frac{1}{2} \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 (w_m - w_n)}{E_n^{(0)} - E_m^{(0)}}[/tex]
Is that even correct, or did I miss something? Can you tell me how he gets there? I have also looked up his derivation of the Quantum Perturbation theory, but it does not help me with this particular problem, unfortunately.
Using a Hamiltonian
[tex]\hat{H} = \hat{H}_0 + \hat{V}[/tex]
we get the following expression for the energy levels of a perturbed system, up to second order:
[tex]E_n = E_0^{(0)} + V_{nn} + \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2}{E_n^{(0)} - E_m^{(0)}}[/tex].
The prime signifies that the sum is done over all [itex]m\neq n[/itex].
Substituting this into the equation (from the normalisation of the Gibbs canonical distribution):
[tex]e^{-F/T} = \sum_0 e^{-E_n/T}[/tex].
This expression is then logarithmized and expanded in powers of [itex]V/T[/itex], so that we get:
[tex] F = F_0 + \sum_n V_{nn} w_n + \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 w_n}{E_n^{(0)} - E_m^{(0)}} - \frac{1}{2T} \sum_n V^2_{nn} w_n + \frac{1}{2T} \left( \sum_n V_{nn} w_n \right)^2 [/tex],
where [itex]w_n = \exp\left\{(F_0 - E_n^{(0)}/T\right\}[/itex] is the unperturbed Gibbs distribution.
We can now notice that
[tex] \sum_n V_{nn} w_n \equiv \overline{V}_{nn} [/tex],
i.e., the sum is the mean of [itex]V[/itex] avaraged over the quantum state and the statistical distribution.
And now this is where it gets interesting, and where I fail to see something. Landau writes, that you can rewrite the equation for the free energy above in the following way:
[tex] F = F_0 + \overline{V}_{nn} - \frac{1}{2} \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 (w_m - w_n)}{E_n^{(0)} - E_m^{(0)}} - \frac{1}{2T} \left\langle (V_{nn} - \overline{V}_{nn})^2 \right\rangle[/tex]
That makes perfect sense except for the part with the double-sum, where I don't understand how he obtains it:
[tex]\sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 w_n}{E_n^{(0)} - E_m^{(0)}}= -\frac{1}{2} \sum_n \sideset{}{'}{\sum}_m \frac{\lvert V_{nm}\rvert^2 (w_m - w_n)}{E_n^{(0)} - E_m^{(0)}}[/tex]
Is that even correct, or did I miss something? Can you tell me how he gets there? I have also looked up his derivation of the Quantum Perturbation theory, but it does not help me with this particular problem, unfortunately.