# Periodic boundary condition on a Wave Function of a Particle in a Box

Until now, solving the Schrodinger Equation for a particle in a box was relatively easy because the boundaries conditions imposed zero value on the wave function at the boundaries. But now I must find the normalized wave function of the same problem imposing just these periodic boundaries conditions:

$$Y(x,y,z)=Y(x+L,y,z),\\ Y(x,y,z)=Y(x,y+L,z),\\ Y(x,y,z)=Y(x,y,z+L).$$

I got stuck in the normalization process. Before, using the boundary condition (one dimension) $Y(0)=Y(L)=0$, I could get just one constant to solve for, in

$$Y(x)= A\sin{kx} + B\cos{kx}.$$

Applying the conditions above, I get

$$Y(x)= A\sin{kx}\quad \text{where}\quad k=\pi n/L$$

which is easy normalize. But now, with this periodic boundary condition, $Y(0)=Y(L)$. How could I find it?

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Where exactly are you running into trouble? Right now you just have the definition of periodic boundary conditions in your question and we have to guess why you're having difficulties. Please show us the work you've done already and indicate where you got stuck. – Wouter May 18 '13 at 19:01
Hint (which might be obvious): Think of it as a particle on a "ring". That gives you a natural coordinate to talk about and things might be easier. – Siva May 18 '13 at 19:04
Both instances of "$Y(x) = Y(L)$" should be "$Y(0) = Y(L)$"? – Chris White May 23 '13 at 2:32

When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate $x$ must match that at coordinate $x+L$, so we have:

$$\Psi(x)=\Psi(x+L)$$

In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be helpful to remember that this can also be expressed as an exponential with the form:

$$\Psi(x)\propto e^{ikx}$$

Hopefully that should get you off the starting block.

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In general, what we use for periodic boundary conditions is defined by $Y(0)=Y(L)$ and $\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition $Y(0) = Y(L)$ I believe is not sufficient to impose conditions on $k$.

This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the schrodinger equation is of order 2 in derivative, there should be at least 2 boundary conditions.

By imposing the two stated periodc boundary conditions, you obtain the following system : \begin{align} A = A\cos(kL)+B\sin(kL) \\ B = -A\sin(kL)+B\cos(kL) \end{align} Rearraging \begin{align} A(1-\cos(kL))+B(-\sin(kL)) = 0\\ A(\sin(kL))+B(1-\cos(kL)) = 0 \end{align}

Now, the system yields non trivial solutions (trivial solution being A=B=0) only if it's determinant is 0. This gives the conditions $\cos(kL)=1$ which gives the desired $k=\frac{2n\pi}{L}$

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