[Physics] Normalisation of angular wave function: particle in a circular box

hilbert-spacenormalizationquantum mechanicsschroedinger equationwavefunction

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 Schrödinger 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:


Which has the solution:


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


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.

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