4
$\begingroup$

In a box with sides of length $L$, the energy eigenvalues depend on the boundary conditions. For periodic boundary conditions, they are $$E_{n_x,n_y,n_z}=\frac{\hbar^2}{2m}\left(\frac{2\pi}{L}\right)^2(n_x^2+n_y^2+n_z^2)\tag{1}\label{1}$$ where $(n_x,n_y,n_z)$ are integers, but for reflecting boundary conditions, they are $$E_{n_x,n_y,n_z}=\frac{\hbar^2}{2m}\left(\frac{\pi}{L}\right)^2(n_x^2+n_y^2+n_z^2)\tag{2}\label{2}$$ where $(n_x,n_y,n_z)$ are positive integers. In most cases, this makes no difference, as they both yield the same density of states. However, in a Bose-Einstein condensate, where a macroscopic number of particles are in the ground state, it seems to have an effect.

For example, in $(\ref{1})$, the ground state is $E_{0,0,0}=0$, whereas in $(\ref{2})$ it is $E_{1,1,1}=\frac{3\hbar^2\pi^2}{2mL^2}$. This is just an arbitrary energy shift. Where a difference arises is the energy gap to the first excited state. For $(\ref{1})$, this is $E_{1,0,0}-E_{0,0,0}=\frac{2\hbar^2\pi^2}{mL^2}$, but for $(\ref{2})$ it is $E_{2,1,1}-E_{1,1,1}=\frac{3\hbar^2\pi^2}{2mL^2}$. According to Bose-Einstein statistics, the two otherwise identical gases should therefore have a different proportion of particles in the first excited state. How is this reconciled?

Edit: to derive $(\ref{1})$, you assume wave function (for a box centred on the origin) $\psi(\boldsymbol{r})=\frac{1}{\sqrt{L^3}}e^{i\boldsymbol{k}\cdot\boldsymbol{r}}$ and impose periodic boundary conditions, e.g. $\psi(L/2,y,z)=\psi(-L/2,y,z)$. To derive $(\ref{2})$, you use 3D square well wavefunction (now with a corner at the origin) $\psi(\boldsymbol{r})=\left(\frac{2}{L}\right)^{3/2}\sin(k_xx)\sin(k_yy)\sin(k_zz)$ and impose that it must be zero at the walls.

$\endgroup$
12
  • $\begingroup$ Why should it be possible to reconcile two different situations? $\endgroup$ Apr 10, 2021 at 11:31
  • $\begingroup$ Usually it is said that boundary conditions don't affect the final answer, so I thought that should be the case here too. $\endgroup$ Apr 10, 2021 at 11:48
  • $\begingroup$ Usually this is only true as you take the system size to infinity. For finite systems, boundaries make a difference. $\endgroup$ Apr 10, 2021 at 11:51
  • 3
    $\begingroup$ I'd be surprised if boundary conditions don't matter for finite system size. Do you have a reference for that? E.g., in the example you discuss, the energy spectrum is clearly different. This will have consequences of some kind. $\endgroup$ Apr 10, 2021 at 12:11
  • 1
    $\begingroup$ But the density of states is only well-defined in the limit of an infinite system. Otherwise, it is not a smooth function but a sequence of delta peaks, which do depend on the boundary conditions! Books can be sloppy with these concepts, and e.g. mention such assumptions only in passing. $\endgroup$ Apr 10, 2021 at 13:53

1 Answer 1

2
$\begingroup$

Does Bose-Einstein condensation depend on boundary conditions?

No.

This can be shown rigorously in 'On the Bose-Einstein condensation of an ideal gas' by L. J. Landau and I. F. Wilde (1979).

The proof lies in computing the fugacity $z = \mathrm{e}^{\beta \mu}$ (called activity in the paper) and showing that it exhibits a non-analytic behaviour at some $T = T_{\mathrm{c}}$ regardless of the specific boundary condition. Among others, they consider periodic boundary conditions and reflective walls, the latter I am assuming is what you mean by 'reflective' boundary conditions. Non-analytic behaviour of thermodynamic quantities in the thermodynamic limit is a sign of a phase transition.

So all that matters is the geometry of the system, in this case a free Bose gas enclosed within a cubical box of side $L$. In 3D. (It is known that free systems for $d < 2$ do not exhibit BEC, because of the Mermin-Wagner theorem).

Physical comment

I don't think it's surprising that a BEC does not depend on the boundary condition. Periodic or reflective boundary conditions are usually used for dynamical systems like electrons in the bulk of a crystalline material to study transport properties. BEC is an equilibrium phenomenon, so the geometry of the system (e.g. a box or a harmonic potential) is what dictates the equilibrium physics.

Your equations

In order to directly address your alleged discrepancy with density of states and energy gap, I would like to see derivations of your two equations.

Response to edit

Thanks. Yes in hindsight your derivations were kind of obvious, sorry. I just was not sure what you meant by 'reflective' boundary conditions but I think you mean "repulsive walls" and hence just a normal box trap.

Anyway, I was just starting to develop the maths, based on finding the critical temperature $T_c$ by the usual equation $$N_{\mathrm{excited\,states}} = \sum_{i\in\mathrm{all\,but\,ground\,state}} \frac{1}{\mathrm{e}^{E_i/(kT_c)}-1}$$ and using $E_i$ in your two cases, when I found a paper where they have already done it.

They never published it so it may not have gone through peer-review, so beware of what they say. But it's called 'Finite-size effects with boundary conditions on Bose-Einstein condensation' and they show that different boundary conditions cause a shift in $T_c$, which however becomes negligible and non-existent when you increase system size $L\rightarrow \infty$.

Finite-size or not

I should add that, to be a real phase transition and hence a real BEC, you need to take the system size to infinity or better reach the thermodynamic limit $N/V \rightarrow \infty$, where $N$ is particle number and $V$ is volume.

In a finite system the integral becomes a sum and you can define a critical particle number, then giving you a (quasi)condensate even in situations e.g. 2D space where it could not exist. But this "condensate" would not survive the thermodynamic limit and so it's not a real phase in the statistical mechanical sense.

$\endgroup$
2
  • $\begingroup$ I have added a section explaining where my equations come from if you would like to try and explain the discrepancy. $\endgroup$ Apr 11, 2021 at 16:27
  • $\begingroup$ @AlexGhorbal Edited the answer to address that. $\endgroup$
    – SuperCiocia
    Apr 11, 2021 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy