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 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 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 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 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 at 13:53
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 at 16:27
  • $\begingroup$ @AlexGhorbal Edited the answer to address that. $\endgroup$
    – SuperCiocia
    Apr 11 at 19:19
-1
$\begingroup$

Usually the periodic boundary conditions and the hard wall boundary conditions refer to different situations:

  • Periodic boundary conditions are a mathematical trick to deal with the continuum of states - one usually takes in the end $L\longrightarrow\infty$, recovering the continuous spectrum.
  • Hard wall boundary conditions imply actual walls, such as presence of an atomic trap for the bc condensate.

Note that if we take the limit $L\longrightarrow\infty$ with hard wall boundary conditions, we do not recover the same spectrum as in the case of the periodic bc. The simplest case to convince oneself that thsi is the case is by working through the classical problem of one-dimensional hard wall potential. E.g., if the potential is zero in the interval $[0,L]$ and infinity outside of these interval, we will have as solutions $\sin(\pi x/L)$, but not $\cos(\pi x/L)$, although both should appear in the continuous spectrum, and indeed appear, if we use the periodic boundary conditions.

$\endgroup$
4
  • $\begingroup$ I agree with you, but I don't think it exactly answers my question. If the two boundary conditions are so different, why do they always give the same results, except for the BEC? And does the fact that periodic boundary conditions are a mathematical trick imply it is meaningless to consider a BEC using them? $\endgroup$ Apr 10 at 11:12
  • 1
    $\begingroup$ Why should cos appear in the continuous spectrum if there is a hard wall at 0? $\endgroup$ Apr 10 at 11:32
  • $\begingroup$ @AlexGhorbal if one does the math correctly, then they actually give different results - even in the continuum limit the DOS differs by a factor of 2. This is why, one only continuum limit is of interest, one uses the hard wall and compensates for the factor - it is, e.g., useful in numerical calculations. But when the size of the box is finite, the two are not interchangeable. $\endgroup$ Apr 10 at 12:16
  • $\begingroup$ @AlexGhorbal one often uses continuum limit for finite systems, if the boundary effects are negligeable. E.g., there are no infinite crystals, but one does use the Bloch theorem, since the corrections due to the finite size are smaller than the effects that are of interest. But, there are cases where one cannot do it. $\endgroup$ Apr 10 at 12:19

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.