From the Bose-Einstein distribution it follows that a non-interacting Bose gas condenses into the Bose-Einstein condensate below a certain critical temperature. What happens when interactions are introduced in a Bose gas is not dealt with in introductory Statistical mechanics courses. Hence my questions are pretty naive and basic.

How is(are) the interaction(s) quantitatively modelled in a Bose gas and how does it change the behaviour (compared to the noninteracting Bose gas) when the temperature is lowered? Is there a way to physical understand the change in the behaviour, if any?

As a minor comment, I have been informed that in case of Fermi gases, the role of interactions shifts the effective mass from $m\to m^*$ and the energy levels (effectively mapping the interacting system to a system of quasiparticles that still obey FD statistics). Does the similar thing happen here too?

  • $\begingroup$ So, should the title of this question then be "Understanding the behaviour of an interacting Bose gas"? $\endgroup$
    – Rococo
    Mar 30, 2018 at 16:16

1 Answer 1


1) First note that a non-interacting Bose gas is an idealization. If the gas was truly non-interacting, then it would be impossible to cool by evaporative cooling (or any other method that removes energy and requires the gas to re-equilibrate).

2) In a Bose gas the short range part of the interaction has to be repulsive (otherwise the gas will collape at low temperature). A typical model is a repulsive delta function $$ V(x_1,x_2)=\frac{4\pi a}{m} \delta(x_1-x_2) $$ controlled by the s-wave scattering length $a$. Indeed, for a dilute gas this is not a model but a systematic description of the low energy properties.

3) In a non-interacting gas Bose condensation takes place at the Einstein temperature $$ T_c = \frac{2\pi}{m}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}. $$ The leading shift due to (repulsive, $a>0$) interactions is $$ \Delta T_c \simeq 1.3 an^{1/3} T_c $$ which, even in a strongly interacting gas like helium, is not a large shift.

4) The systematic study of perturbation theory in $a$ goes back to Bogoliubov. He found, for example, that the dispersion relation of quasi-particle in a Bose condensed fluid is $$ \epsilon_p = \frac{1}{2m}\sqrt{(p^2+8\pi an)^2-(8\pi an)^2} $$ which smoothly interpolates between a Goldstone mode at low $p$, and non-interacting atoms at large $p$.

5) And indeed, as remarked below, you can take the interaction in (2) and treat in a mean field approximation. This leads to a non-linear Schroedinger equation (the Gross-Pitaevski equation) for the condensate wave function. This equation can be used to study cloud profiles, collective modes, etc.

  • $\begingroup$ Shouldn't there must also be an attraction (at least) at long-range because once a degenerate ground state is occupied there is an enhanced affinity for the other bosons to occupy the already filled state? @Thomas $\endgroup$
    – SRS
    Mar 30, 2018 at 16:28
  • $\begingroup$ Bose statistics is of course taken into account in the calculation. What we mean by "interaction" is the interaction Hamiltonian for two particles in free space. It is indeed true that the long range part of the interaction of two neutral particles is almost always attractive (the Casimir-Polder-van-der-Waals force). What matters to the (dilute) Bose condensed gas is the short range s-wave interaction, which has to be repulsive. $\endgroup$
    – Thomas
    Mar 30, 2018 at 16:51
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    $\begingroup$ Let me just say that taking the potential $V(x_1,x_2)$ leads to the Gross-Pitaevskii equation, which is the quantum mechanical starting point in the study of interacting Bose gases: $$i\hbar \frac{\partial \psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(\vec{r}) + gN|\psi|^2 \right]\psi $$ $\endgroup$
    – Matteo
    Mar 30, 2018 at 23:59

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