Does Bose-Einstein condensation depend on boundary conditions?

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.

• Why should it be possible to reconcile two different situations? Apr 10, 2021 at 11:31
• Usually it is said that boundary conditions don't affect the final answer, so I thought that should be the case here too. Apr 10, 2021 at 11:48
• Usually this is only true as you take the system size to infinity. For finite systems, boundaries make a difference. Apr 10, 2021 at 11:51
• 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. Apr 10, 2021 at 12:11
• 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. Apr 10, 2021 at 13:53

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.

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.

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