Let us assume a single bosonic mode, in equilibrium with a reservoir.

For a non-interacting Bose gas, the partition function becomes

$\mathcal{Z_\text{nonint}}=\sum_{N=0}^\infty e^{-\beta(\epsilon-\mu)N}=\frac{1}{1-e^{-\beta(\epsilon-\mu)}}$.

from which it is easy to obtain a free energy $F=-T\log(\mathcal{Z})$ and particle number statistics $N=-\frac{\partial F}{\partial \mu}=\frac{1}{e^{\beta(\epsilon-\mu)}-1}$ yielding the Bose-Einstein distribution.

How is this modified for finite two-particle interaction, i.e, how to evaluate

$\mathcal{Z_\text{int}}=\sum_{N=0}^\infty e^{-\beta\left[(\epsilon-\mu)N+\frac{U}{2}N^2\right]}$ ?


2 Answers 2


To evaluate the grand canonical partition function, as is done for no interaction, you need the microstates $E_i$. In the noninteracting case the energy of each microstate is simply the sum of the single particle energies, i.e. you could write $E_i=\sum_j \epsilon_j n_{i_j}$, with $n_{i_j}$ the occupation of the single particle state $\epsilon_j$ in the microstate $i$.

This is decomposition of many-body eigenstates into products of single particle eigenstates is no longer accurate when the interaction is nonzero. There is no way around finding all the exact eigenenergies $E_i$ of the many-body problem. Needless to say, this is extremely hard.

Perturbation theory is therefore often used to approximate

  1. the energy of each microstate

  2. the spectrum itself, i.e. the types of possible excitations

The procedure is: compute a many-body ground state, do linear response on it, aka Bogolyubov, evaluate the partition function and hope for the best.

Caveat: There are two main problems with this approach:

  • related to 1: the higher you go in energy, the worse your linear response approximation for the $E_i$ becomes.

  • related to 2: from exactly solvable models it is known that Bogolyubov excitations constitute only a fraction of the full excitation spectrum, see e.g. Exact Analysis of an Interacting Bose Gas. II. The Excitation Spectrum. So you do not evaluate (not even approximately) the full partition function when you rely on Bogolyubov theory. See also the plots of the excitation spectrum on Scholarpedia.

Nevertheless, if you are still interested, this question Partition function of weakly interacting Bose gas discusses a Bogolyubov treatment.

  • $\begingroup$ Not exactly what I was looking for, but very interesting nevertheless, thank you! $\endgroup$
    – Wouter
    Oct 12, 2018 at 8:09

Replacing the sum by an integral, the partition function becomes

$\mathcal{Z}_{int}=\int_0^\infty e^{-\beta\left[(\epsilon-\mu)N+\frac{U}{2}N^2\right]}dN=\frac{e^{\frac{(\epsilon-\mu)^2\beta}{2U}}}{\sqrt{2U\beta/\pi}}Erfc\left[\frac{(\epsilon-\mu)\sqrt{\beta}}{\sqrt{2U}}\right]$,

at least for $U>0$. From this, a general equation of state can be derived as in the noninteracting case, with two particular limits.

In the weakly interacting limit $N^2U/T\rightarrow0$, the argument of the complementary error function goes to $+\infty$ so that an asymptotic expansion can be done, resulting to lowest order in $N\approx\frac{T}{\epsilon-\mu}\left(1-\frac{2TU}{(\epsilon-\mu)^2}\right)$ and $\frac{\sqrt{<N^2>-<N>^2}}{<N>}\approx1-\frac{U}{T}N^2$

In the strongly interacting limit $N^2U/T\rightarrow+\infty$, the argument of the complementary error fuction goes to $-\infty$ and $N\approx\frac{\mu-\epsilon}{U}$ and $\frac{\sqrt{<N^2>-<N>^2}}{<N>}\approx\sqrt{\frac{T}{UN^2}}$.

  • $\begingroup$ What you wrote above approximates the sum by an integral, but the issue is that ${\mathcal Z}_{int}$, as defined in your question is not a grand canonical partition function of an interacting Bose gas. $\endgroup$
    – jkds
    Oct 12, 2018 at 10:21
  • $\begingroup$ @jkds please elaborate? $\endgroup$
    – Wouter
    Oct 12, 2018 at 11:07
  • $\begingroup$ This is the grand canonical partition function $\mathcal{Z}_{int} = Tr e^{-\beta (\hat H -\mu \hat N)}$. $\hat H$ is the many-body Hamiltonian and $\hat N$ the operator for the total number of particles. In the energy basis you get $\sum_i e^{-\beta E_i -\mu N_i}$. You need the $E_i, N_i$ for every microstate. You have added $U N^2$ to the exponent for every microstate. But the interaction affects different microstates differently. What you have written works only for one microstate, $\Psi=const.$ $\endgroup$
    – jkds
    Oct 12, 2018 at 11:59
  • $\begingroup$ Yes, as I indicated I was looking at a single mode (2nd quantization). As $\hat{H}$ is diagonal in the particle number eigenbasis, we obtain the mentioned expression straight away. $\endgroup$
    – Wouter
    Oct 12, 2018 at 12:37
  • $\begingroup$ OK, a many-particle system with only one energy level is unusual, but I get what you mean. Could you add a description of a system that you have in mind? $\endgroup$
    – jkds
    Oct 12, 2018 at 13:22

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