# Periodic boundary condition for a particle in a box

I've been trying to solve the Schrödinger equation for a particle in a box of volume $V$ (length $L$) with periodic boundary conditions. The general solution for the equation yields, for one dimension, $$\Psi (x) = A \mathrm{e}^{i k_x x}+B \mathrm{e}^{-i k_x x}.$$ When using $\Psi(0)=\Psi(L)$ and $\Psi'(0)=\Psi'(L)$, I get $$\begin{cases} A+B = A \mathrm{e}^{i k_x L}+B \mathrm{e}^{-i k_x L} \\ A-B = A \mathrm{e}^{i k_x L}-B \mathrm{e}^{-i k_x L} \end{cases}.$$ Adding both equations yields $\mathrm{e}^{i k_x L}=1$, so $k_x = \frac{2n \pi}{L}$, which according to this Phys.SE post, is correct.

Now, if I rewrite the boundary conditions with the value found for $k_x$, I end up with no equations from which to obtain $A$ and $B$ (apart, of course, from the normalization of the wave function). All this sounds correct, so I don't really know what I'm missing here.

In Statistical Mechanics, section 5.3, Pathria states that the solution for this problem has the form $$\Psi(\mathbf{r}) = \left( \frac{1}{L} \right)^{\frac{3}{2}} \mathrm{e}^{i \mathbf{k} \cdot \mathbf{r}}.$$

• What exactly is your question? – By Symmetry May 20 '17 at 22:06

You have correctly derived your result on $k_x$. The key here is that, once you satisfy the condition on $k_x$ being an integer multiple of $\frac{2\pi}L$, any choice of $A$ and $B$ will not disrupt your boundary conditions, that are automatically satisfied. $A$ and $B$ can be arbitrary, and you only have a condtion relating them (the normalization of the wavefunction), so you have some freedom here. You can build wavefunctions that are not proportional to each other.
This is because you're solving for a free particle in a $1D$ box, so any eigenvalue of the energy $E=\hbar^2 k^2/2m$ is doubly degenerate: particles with impulses $\hbar k$ and $-\hbar k$ have the same enrgy, so the states $|\hbar k\rangle$ and $|-\hbar k\rangle$ are both eigenvectors for the Hamiltonian with the same eigenvalue, and they are linearly independent. The eigenspace relative to a certain eigenvalue $E$ has dimension $2$, and you have the freedom to mix different eigenstates of the impulse.
It is common use (and a convenient choice) to describe this eigenspace through a basis in which another useful operator is diagonal. As an example, one could use a basis of eigenvectors of the impulse (like you are doing in your question), $$\langle x|\hbar k\rangle=\frac{1}{\sqrt L}e^{i kx},\quad\quad \langle x|-\hbar k\rangle=\frac{1}{\sqrt L}e^{-i kx},$$ or a basis of eigenvectors of the parity operator, transforming $x$ into $L-x$: $$\langle x|+\rangle=\frac{1}{\sqrt {2L}}\left(e^{i kx}+e^{-ikx}\right),\quad \langle x|-\rangle=\frac{1}{\sqrt {2L}}\left(e^{i kx}-e^{-ikx}\right).$$ Any way you choose the basis eigenvectors $|1\rangle$ and $|2\rangle$ (assuming that they are orthonormal), you have that any eigenstate of that energy $E$ is given by $$|\Psi\rangle=\alpha|1\rangle+\beta|2\rangle,$$ with the normalization condition $|\alpha|^2+|\beta|^2=1$.
That's correct: neither $A$ nor $B$ can be determined from the boundary conditions. For a particle on a "ring", both the states $\Psi(x) = e^{ikx}$ and $\Psi(x) = e^{-ikx}$ (with $k = 2\pi/L$) are degenerate energy eigenstates, and so any linear superposition of these two states will also be an energy eigenstate.