Question about Quantum + Thermodynamic Perturbation theory

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
mSSM
Messages
31
Reaction score
1
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.
 
Physics news on Phys.org
Split up the original double sum containing only w_n into two and rename the summation indices n<->m in the second one.
 
Oh dear, thank you so much! I was going crazy over this... :D This is such a silly trick.