For a particle in a circular box (with radius $R$) with zero potential inside the circle and infinitely high potential outside of the circle, the SchrÃ¶dinger equation in polar coordinates is:

$$-\frac{\hbar^2}{2m}\Big(\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}\Big)=Eu$$

Where $u(r,\phi)$ is the wave function and $u(R,\phi)=0$ the boundary condition.

With Ansatz $u(r,\phi)=R(r)\Phi(\phi)$ and substitution into the above PDE we get complete separation and the angular equation is:

$$-\frac{\Phi''}{\Phi}=m^2$$

Which has the solution:

$$\Phi(\phi)=c_1e^{im\phi}+c_2e^{-im\phi}$$

But without boundary conditions, **how to determine** $c_1$ and $c_2$?

I then read somewhere that $c_2=0$ but without explanation.

Because of the requirement:

$$\Phi_m(\phi)=\Phi_m(\phi+2\pi )$$

$$e^{im\phi}=e^{im(\phi+2\pi)} \implies m=0, \pm 1, \pm 2, \pm 3,…$$

And the normalised angular function is said to be:

$$\Phi(\phi)=\frac{1}{\sqrt{2\pi}}e^{im\phi}$$

**But I don't get that result**. The normalisation requirement is:

$$\int_0^{2\pi}\Phi^2d\phi=1$$

But that doesn't yield $c_1=\frac{1}{\sqrt{2\pi}}$?

But for $\Phi=c_1\cos (m\phi)$ and $m \neq 0$, I get $c_1=\frac{1}{\sqrt{\pi}}$.

## Best Answer

To normalize this function you have $$ 1~=~\int_0^{2\pi}\Phi^*\Phi d\phi ~=~|C|^2\int_0^{2\pi} e^{-im\phi}e^{im\phi}d\phi~=~|C|^2\int_0^{2\pi}d\phi~=~2\pi|C|^2. $$ The result is then obvious.