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: $$\Phi(\phi)=\frac{1}{\sqrt{2\pi}}e^{im\phi}$$ 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 $c_1=\frac{1}{\sqrt{\pi}}$.

  • $\begingroup$ Can you itemize your questions? I see three subquestions here. $\endgroup$ – knzhou Jun 21 '16 at 23:28
  • $\begingroup$ 1) Depends on what you want to do. There are multiple valid choices, like you could choose $c_1 = \pm c_2$ to have purely real/imaginary solutions, or choose $c_2 = 0$ and $c_1 = 0$ to have solutions with a simple form. 2) I do get $c_1 = 1/\sqrt{2\pi}$? 3) It makes sense you get a different $c_1$ if you're using a different wavefunction... I don't get the question here. $\endgroup$ – knzhou Jun 21 '16 at 23:29
  • $\begingroup$ "[...] or choose $c_2=0$ and $c1=0$". Surely it;s either or? And choose $0$ on what basis? The $\Phi$ DE has no boundary or initial conditions. $\endgroup$ – Gert Jun 22 '16 at 0:12
  • $\begingroup$ I mean, one solution has $c_2 = 0$, and the other solution has $c_1 = 0$, where $\Phi(\phi) = c_1 e^{im\phi} + c_2 e^{-im\phi}$ as in your question. Note that your solution $\Phi = \cos(m\phi)$ is actually a special case of this. $\endgroup$ – knzhou Jun 22 '16 at 0:14

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.

  • $\begingroup$ Oooopsie. Totally overlooked $\Phi$ is a complex function. (Blushes). Thank you. +10. $\endgroup$ – Gert Jun 21 '16 at 23:59

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