I am taking a course in solid state physics, and I have some trouble with the "hard wall" and the periodic boundary conditions for a particle in a box.

The thing is that we obtain, for a box of length L, $k= n \pi/L$, $n\in \mathbb{N}$ for the hard box, where we have insisted that $\psi(0)=\psi(L)$.

We obtain $k= 2\pi m/L$ , $m\in \mathbb{Z}$, for the case of periodic boundary conditions, i.e. by insisting that $\psi(x)=\psi(x+L)$.

Now, if we solve this for the energies for k, we obtain for a particle of mass M

$E= \frac{\hbar^2 k^2}{ 2 M}$ ,

and because the allowed values of k are different for the two sets of boundary conditions, we end up with different energies as well

$E_n= \frac{\hbar^2 n^2}{ 2 M L^2}$ - for the hard wall

$E_m= 4\frac{ \hbar^2 m^2}{ 2 M L^2}$ - for the periodic bondary conditions

So the choice of boundary conditions seemingly affects the energies. Shouldn't the physical situation be unaffected by the choice of boundary conditions?

For example, in treating the free electron model of metals, some authors solve the allowed energies of the electrons by modelling the metal as a cube of side length L and using periodic boundary conditions. If one used the hard wall boundary conditions instead, we would obtain an energy different by a factor of 4.

What don't I understand?

  • 1
    $\begingroup$ Periodic b.c. allow both left- and right-travelling waves, which are degenerate. But that's only half the story. There really are subtle differences between hard walls and periodic b.c. You might think about doubling the length of the box, with or without a hard wall in the middle. $\endgroup$ Commented Dec 17, 2018 at 12:53

2 Answers 2


The energies for periodic boundary conditions and hard wall boundary conditions are not precisely the same. However, they have basically the same statistical properties. That means that the energy spectra are similar enough that they give the same results for macroscopic phenomena (such as in statistical mechanics).

For a region of length $L$ with hard walls at $x=0$ and $x=L$, the allowed wave functions are of the form $$\psi_{n}(x)=N\sin\left(\frac{n\pi x}{L}\right).$$ The length $L$ of the region is an integer number of half wavelengths, and so there are two states with energies $E\leq\frac{2\pi^{2}\hbar^{2}}{L^{2}}$, the $n=1$ state and the $n=2$ state. More generally, there are $2n$ states with energies less than or equal to $\frac{2n^{2}\pi^{2}\hbar^{2}}{L^{2}}$.

With the periodic boundary conditions, there needs to be an integer number of wavelengths in the region of length $L$, so that $\psi(x+L)=\psi(x)$. However, for each wavelength, there are two orthogonal wave functions $$\psi_{n,1}(x)=N'\exp\left(\frac{2n\pi ix}{L}\right)$$ and $$\psi_{n,2}(x)=N'\exp\left(-\frac{2n\pi ix}{L}\right).$$ (You can also use a sine and cosine basis, formed from linear combinations or $\psi_{n,1}$ and $\psi_{n,2}$.) Because each energy eigenstate is doubly degenerate, it is true once again that the number of states with energies $E\leq\frac{2n^{2}\pi^{2}\hbar^{2}}{L^{2}}$ is $2n$.

For macroscopic purposes, these are similar enough to be equivalent. Yes, there will be slight differences in energies at the microscopic level, but note that those differences will be of order ${\cal O}(1/L^{2})$, which can be neglected when $L$ is of macroscopic size. If you form the density of states, for example, the approximations involved in describing the states by a continuous density will wash out the minuscule differences between the two kinds of boundary conditions.

  • $\begingroup$ This argument doesn't seem to be completely convincing. Could elaborate on how the spectrum will look like and why the number of energy levels will be the same? $\endgroup$
    – Fracton
    Commented Oct 2, 2020 at 16:13

From what I understand of your question, you are treating a quantum problem classically. When you’re dealing with h-bar , you leave behind deterministic-classical physics, and enter the world of probablistic-quantum physics. If you consider your wave-function, with sinusoidal changes within a box, the particle is considered a wave with energies in a probable range. So of course the attributes of a wave will change given the length of your box. The choice of boundary condition affects the physical probabilities. In other words, you change the physical situation, when you change the boundary conditions for particle-wave behaviour.

This was super quick , and dirty, but hopefully helps you to think about it differently



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