I'm solving a problem involving a Fermi gas. There is a specific sum I cannot figure my way around.

A set of equidistant levels, indexed by $m=0,1,2 \ldots$, is populated by spinless fermions with population numbers $\nu_m =0 $ or $1$. I need to compute the following sum over the set of all possible configurations $\{ \nu_l \}$:

$Q(\beta,\beta_c) = \sum_{\{ \nu_l \}} \sum_{l} \prod_m \exp({\beta_c \, l \, \nu_l}-{ [ \beta \, m + i \phi] \, \nu_m} )$.

Any hints on how to deal with this are appreciated. This is not homework, it is a research problem.

It is known that $\beta >0$, $\beta_c>0$, and $\phi \in [0; 2 \pi ]$.

EDIT: corrected with the complex phase (the sum is coming from a generating function)


This is not an answer just some thoughts from playing with the expression. I've read the question before you included the phase, so for now let $\phi = 0$ (sorry it this makes my response useless for you).

I'll simply write $Z$ instead of your $Q(\beta, \beta_c)$ and also drop the arguments where obvious. Denote by $Z_{abc\dots}$ the partition function where we do not include the sites at $a, b, c, \dots$ in the problem. Also denote $f_k = 1 + \exp(-\beta k)$ and $g_k = 1 + \exp((\beta_c - \beta) k)$.

Now (unless I screwed up), by summing over the site at $k$ we can get the relation $$Z = f_kZ_k + g_k \sum_{\nu \setminus k} \prod_{m \neq k} \exp(-\beta m \nu_m) $$ and iterating it $$Z = \left( \prod_{m \in abc\dots z} f_m \right) Z_{abc\dots z} + $$ $$ \left(g_a f_b \dots f_z + f_a g_b \dots f_z + \cdots + f_a f_b \dots g_z) \right) \sum_{\nu \setminus abc\dots z} \prod_{m \neq abc \dots z} \exp(-\beta m \nu_m).$$

It is a simple observation that for the reduces system consisting of a single level $a$ we get $Z_{bc \ldots z} = g_a$ so the first term above gives a similar contribution like the other terms (all but one factors are $f$ and one of them is $g$). Therefore, we can write $$Z = \left( \prod_{m} f_m \right) \left ( \sum_k \frac{g_k} { f_k} \right).$$

These expressions are exact in case we have finite number of states. Otherwise they are just formal and are to be understood as limits only if everything converges.

  • $\begingroup$ This is absolutely fantastic! +1 (I'd give +10 if I could). I've checked every step, arrived to the same result. $\sum_{\nu} \prod_m \exp(-\beta m \nu_m)=\prod_k f_k$ can be identified a bit earlier. The inclusion of $\phi$ just changes $f_k=1+\exp(-\beta k - i \phi)$ and $g_k=1+\exp(-\beta k+ \beta_c k - i \phi)$. Really saved me from frustration. $\endgroup$ – Slaviks Aug 19 '11 at 7:35
  • $\begingroup$ I'm very glad it helped :) $\endgroup$ – Marek Aug 19 '11 at 8:59
  • $\begingroup$ A little mistake needs to be corrected: $Z_{bc\ldotsz}=g_a$, not 1. Therefore in the final answer there is no termm with `1' -- the total number of summands is equal to the number of levels. Shall edit your post to fox this, Marek? $\endgroup$ – Slaviks Aug 19 '11 at 11:35

The answer by Marek was so useful for me, that I wish to share the full problem and the resulting answer (to improve the lasting value of this Q-A):

Problem: In an equilibrium Fermi gas at inverse thermodynamic temperature $\beta$ defined by a set of single particle levels $\epsilon_m$ ($m=0,1,\ldots$) which are populated by $n$ fermions, the canonical averages of an arbitrary single-particle field $\langle h_m(\nu_m) \rangle_n$ (where $\nu_m =0, 1$ is the occupation number) can be computed via the generating function

$ Z[h_m; z] \equiv \sum_{n=0}^{\infty} z^{n} \langle h_m(\nu_m) \rangle_n Z_n =\sum_{ \{ \nu_k\} } \sum_l h_l(\nu_l) \prod_m e^{-\beta \epsilon_m \nu_m} z^{\nu_m} $.

Solution:As was shown by Marek,

$Z[h_m; z]= Z(z) \sum_k \frac{g_k}{f_k}$ where

$f_k = 1 + e^{-\beta \epsilon_k} z$,

$g_k = h_k(0) + h_k(1) e^{-\beta \epsilon_k} z$ and

$ Z(z) \equiv \prod_m f_m$.

The canonical partition functions $Z_n$ are generated by $Z(z)= \sum_{n=0}^{\infty} Z_n z^n$.


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