Finding solutions for the free electron model with Born-Von-Karman boundary conditions

Question

I'm trying to solve the Schrodingers equation for the free electron model with the Born-Von-Karman boundary conditions.

I'm aware that at least a possible solution of the problems are plane waves however, I haven't been able to find any proof on how to get there and instead I keep finding proofs verifying that in fact, that solution works.

My take on the problem has been the following: $$ -\frac{\hbar^2}{2m}\nabla^2 \psi_{\vec{k}}(\vec r)=\epsilon _{\vec{k}}\psi_{\vec{k}}(\vec{r}) $$

$$\psi_{\vec{k}}(x,y,z)=\psi_{\vec{k}}(x+L,y,z)=\psi_{\vec{k}}(x,y+L,z)=\psi_{\vec{k}}(x,y,z+L) $$

$$ \int \psi^*(\vec r)\psi(\vec r) dr =1 $$

Now, trying to solve this by assuming the solution is of the form of separated variables i get to the following ODEs ($\psi_{\vec{k}}(x,y,z)=\alpha(x)\beta(y)\gamma(z)$):

$$ \alpha(x)_{xx}+\mu_1\alpha(x)=0 $$ $$ \beta(y)_{yy}+\mu_2\beta(y)=0 $$ $$ \gamma(z)_{zz}+\mu_3\gamma(z)=0 $$

where $\mu_1+\mu_2+\mu_3=\epsilon_{\vec k} \frac{2m}{\hbar^2}$

Now, despite the constants $\mu_i$ differ from one equation to another, the way to solve either of them is identical. So, for example, lets choos the first one.

$$ \alpha(x)_{xx}+\mu_1\alpha(x)=0 \Rightarrow \alpha(x)=Ae^{i\sqrt{\mu_1}x}+Be^{-i\sqrt{\mu_1}x}$$

This is the part I'm stuck at. If I could somehow justify that $B=0$ then, applying the boundary conditions,

$$\alpha(x)=\alpha(x+L) \Rightarrow Ae^{i\sqrt{\mu_1}x}=Ae^{i\sqrt{\mu_1}L}e^{i\sqrt{\mu_1}x} \Rightarrow e^{i\sqrt{\mu_1}x}=1 \Rightarrow \mu_1=\left( \frac{2\pi n_x}{L} \right)^2$$

Hence getting the solution $\psi_{\vec k}=Ce^{i\vec k \cdot \vec r}$ where the only thing left to do is normalize the function with the constant value C.

My problem with this is that, as I explained, I can't find a way to ensure that the constant $B$ that goes with the negative exponential has to be zero and without this I can't see how you could get the final answer.

Thanks in advance

Answer 1

There is no reason directly from the Schrodinger equation for $B$ to be zero. The sum of two eigenfunctions with the same eigenvalue is also an eigenvector with that eigenvalue. It is simply convenient to take $$\psi_{k_x,k_y,k_z}(x,y,z) = L^{-3/2} \exp\{ ik_x x+ik_yy+ik_zz\}$$ as these functions, with the allowed values of $(k_x,k_y,k_z)$, constitute a complete orthonormal set. If you kept a more general sum involving both $A$ and $B$ you would have to fiddle about to get them orthonormal.