# The relation between boundary conditions and the density of states of a system

The solution for a particle in an one-dimensional box is given by

$$\psi(x) = A\,\sin(k_{x}x)+B\,\cos(k_{x}x).$$

Now we can apply the boundaries conditions. We may state that $\psi(0)=\psi(L)=0$, which will lead us to

$$k_{x}=\frac{\pi}{L} n$$

On the other hand we may apply periodic boundary conditions: $\psi(0)=0$ and $\psi(x+L)=\psi(x)$. In this case

$$\sin(k_{x}x)=\sin(k_{x}(x+L))$$

is satisfied for

$$k_{x}=\frac{2\pi}{L}n$$

In both cases, the wavefunction is zero at the boundaries. I find this weird because the physics seems to be the same in both cases but in the first, the distance between two states is $\Delta k=\pi/L$ and in the second it is $\Delta k=2\pi/L$ which changes, in the three-dimensional case, the number of states in $k$-space with energy $E<(k)$.

What is going on here? Is there a correct choice of expressing the boundary conditions? Thank you very much.

• With the periodic condition, you remove all solution where $n$ was odd because the corresponding wave-function is a partial sinus, which needs two lengths to finish and be periodic. So this condition is more restrictive and corresponds to another physical situation where you constrain you solutions even more. – gingras.ol Jul 3 '17 at 18:36

The density of states per unit energy, per unit length, is the same for both period and rigid-wall boundary conditions. There are two reasons for this. 1) Mathematics: for periodic boundary conditions the eigenstates are $\psi_n(x)=\exp(ik_nx)$ with $k_n= 2\pi n/L$. You need both positive and negative values of the the momentum $k_n$ to have a complete set of eigenstates. For rigid-wall boundary conditions $\psi_n(x)= \sin(k_n x)$ with $k_n = \pi n /L$, but you only have positive values of $k_n$. Thus, although the energy levels are twice as far apart in the periodic case, each periodic state is doubly degenerate ($\pm k_n$ have same energy). When plotted in terms of $E$, therefore, and when $L$ is large, the energy spectra look essentially the same unless you can resolve energy differences of order $1/L$. 2) Physics: In order for a particle in a large region to know whether it is in a periodic box or one with rigid walls, the particle actually has to get the walls and back multiple times to set up a standing-wave wavefunction. The larger the box the longer the time, and so the more precisely you have to measure the energy to tell the difference. Consequently, when you measure something physical like the specific heat per unit volume the exact boundary conditions make little difference.
Note also that for rigid walls the momentum $\hat p$ is not an observable: there are no momentum eigenstates compatible with rigid wall ($\psi(0)=\psi(L)=0$) boundary conditions. Only the energy $E=\hat p^2/2m$ is an observable in that case. As a result, the density of eigenstates per unit energy makes for the most meaningful comparison between the two cases.
The physics is not the same in the two situations. For a particle in a one-dimensional box, the probability for the particle to be outside of the box is zero. Since the wave function should be continuous, we need to impose the boundary condition $\psi(0)=\psi(L)=0$. Therefore the result must be time-independent, and of the general form $$\psi(x) = A\,\sin(kx),$$ with $k=\frac{\pi}{L} n$.
We impose periodic boundary conditions when we want to normalize the plane wave function or set a upper limit for the wave length. Sometimes we even just use it as a technique to solve problems, and set $k\rightarrow\infty$ after solving them. In this case the wave function is defined everywhere (even outside the 'box'). Since there is no infinite potential anywhere, both $\psi(x)$, $\psi'(x)$ need to be continuous. Therefore the boundary conditions read $\psi(x)=\psi(x+L)$ and $\psi'(x)=\psi'(x+L)$. The result can be dependent on time, and of the general form $$\psi(x) = Ae^{i(\omega_kt-kx)}+Be^{-i(\omega_kt-kx)},$$ with $k=\frac{2\pi}{L} n$.
If we keep your periodic boundary condition $\psi(0)=0$ and $\psi(x+L)=\psi(x)$, which I think is inappropriate, we can still see where the problem is. This boundary condition can imply the first boundary condition $\psi(0)=\psi(L)=0$, but the converse is not true. Therefore it is reasonable that the solutions of the second case in contained in the first case.