Quantum particle in the ground state of a spherical box

Question

Consider a single non-relativistic particle in a spherical box of radius $R$. I want to find the lowest energy level (or the ground state energy). In this case, the Schrödinger equation is simple, for the wave function $\psi(r)$ (by spherical symmetry, the lowest energy wavefunction depends on $r$ only: $\psi \equiv \psi(r)$, with $0 \le r \le R$): $$\tag{1} E \, \psi = -\, \frac{\hbar^2}{2m} \, \nabla^2 \psi = -\, \frac{\hbar^2}{2 m} \, \frac{1}{r} \, \frac{\partial^2}{\partial r^2} \bigl( r \, \psi(r) \bigr). $$ We may write $\phi(r) = r \, \psi(r)$ and $$\tag{2} k = \sqrt{\frac{2m E}{\hbar^2}} > 0, $$ so that (1) reduces to $$\tag{3} \frac{\partial^2 \phi}{\partial r^2} + k^2 \phi = 0. $$ The solution could be written like this: $$\tag{4} \phi(r) = A \sin (k r) + B \cos (k r), $$ where $A$ and $B$ are arbitrary constants. I could use $B = 0$, so that $$\tag{5} \psi(r) = A \frac{\sin (k r)}{r}. $$ This function should vanish at $r = R$, which imposes $k = n \pi / R$ with $n = 1, 2, 3, \ldots, \infty$. I'll consider $n = 1$ for the lowest energy level: $$\tag{6} E = \frac{\pi^2 \hbar^2}{2 m R^2}. $$ Of course, (5) is square-integrable inside the spherical box: $$\tag{7} \int_0^R \psi^2(r) \, 4 \pi r^2 \, dr = 4 \pi A^2 \int_0^R \sin^2(k r) \, dr = 2 \pi R A^2. $$ All is good for now. But then, I could use (4) with $A = 0$, so that $$\tag{8} \tilde{\psi}(r) = B \frac{\cos (k r)}{r}. $$ This function must vanish at $r = R$, so that $k R = \tfrac{\pi}{2}$ (for the lowest value of $E$). It also diverges at $r = 0$, but it's still square-integrable inside the spherical box: $$\tag{9} \int_0^R \tilde{\psi}^2(r) \, 4 \pi r^2 \, dr = 4 \pi B^2 \int_0^R \cos^2(k r) \, dr = 2 \pi R B^2. $$ In this case, the energy level is $$\tag{10} \tilde{E} = \frac{\pi^2 \hbar^2}{8 m R^2} < E. $$ So my question is which is the "true" solution? (5)-(6) or (8)-(10) ? Why should we reject (8)-(10) because $\tilde{\psi}$ diverges at $r = 0$, while the function is still square-integrable and vanishes at $r = R$? What are the conditions to accept or reject a given solution to the Schrödinger equation, apart the usual normalization condition (square-integrable function) and the vanishing of the wavefunction at the boundaries? I don't understand the "continuity condition" at $r = 0$: $$\tag{11} \lim_{r \rightarrow 0} r \, \psi(r) = 0. $$

Answer 1

The short answer is that $B=0$ and you need a node at $r=0$. Physically, this is like a second hard wall boundary condition. There are a couple ways to see this but it all boils down to the regularity at the origin. I will set $2m=\hbar=R=1$ in the following.

Already, you can start by noticing that there is an issue with your eigenvalue problem. Your equation is regular, second order and you only have one boundary condition at $r=1$. You need another at $r=0$. If not, your groundstate energy is rather zero since any positive energy eigenvalue is obtained using: $$ \psi = \frac{\sin[k(1-r)]}r $$ You therefore need a second boundary condition at $r=0$. The issue with the second boundary condition is especially relevant in resolvent formalism which is the mathematically correct way to define the spectrum for general operators. The inverse is ill defined.

A physical argument is that technically if $B\neq0$ your wavefunction is not an energy eigenvector. Mathematically, you can see this with distributions. Indeed, when computing the Laplacian, you need to invoke a weak derivative if your function is not $\mathcal C^2$. If $$ \psi= \frac{A\sin(kr)+B\cos(kr)}r $$ then you rather have: $$ -\Delta\psi= k^2\frac{A\sin(kr)+B\cos(kr)}r+4\pi B\delta $$ which you might recall from electrostatics. In other words: $$ H\psi=k^2\psi+4\pi B\delta $$ therefore due to extra Dirac delta, $\psi$ is not an eigenvector for $B\neq0$.

Mathematically, there a couple ways to see this. You can give a variational formulation of eigenvalue problem as finding the $\psi$ minimizing the functional: $$ E=\int|\nabla\psi|^2d^3x $$ With the constraint: $$ \int|\psi|^2d^3x=1 $$ As usual the difficult part is to consider the right function space where you want to optimize your problem. In this case, the natural setting is the Sobolev space $H_0^1$ where the functional is well defined. Fundamentally, it all boils down to the fact that you not only want the wave function to be square integrable but also its gradient to be square integrable as well to have a finite, well defined mean kinetic energy.

You can also rephrase the problem to avoid talking about unbounded hermitian operators. By taking the inverse $K = (-\Delta)^{-1}$, you rather have a simpler compact operator which is even a Hilbert-Schmidt operator using the Poisson kernel. Therefore, the correct boundary condition is the one used for the Poisson problem: $$ -\Delta\psi = \phi \\ -\frac1{r^2}\partial_r(r^2\partial_r\psi) = \phi $$ so the correct boundary conditions are: $$ \psi(1) = 0 \quad r^2\partial_r\psi \xrightarrow{r\to0} 0 $$ and not the often quoted "continuity condition at $0$".

Hope this helps.