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?
 A: 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.
A: 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 $k=\frac{2n\pi}{L}$
