# Wave function for an electron in and around a small charged sphere

I am interested in solutions of the Schroedinger equation. For simplicity I started my studies with the $n=1$ ground state of the hydrogen atom. I was particularly interested in the higher moments of the kinetic energy operator $\mathbf{T}$. Much to my surprise I encountered serious problems as soon as I worked on the second moment. It turns out that $(\mathbf{T}\psi)(\mathbf{T}\psi)$ is not the same as $\psi(\mathbf{T}\mathbf{T}\psi)$. In other words, the operator $\mathbf{T}$ is not self-adjoint when acting on $\mathbf{T}\psi$, where $\psi(r) = \exp(-r)$.

On this forum, and previously on another physics forum, I had lengthy discussions about this peculiar result. One can talk about the operator $\mathbf{T}$ (for which stronger versions than the usual one can be constructed), about the domain of the operator, about boundary terms at $r=0$ that appear when performing partial integration, or about the fact that the Coulomb potential arises from a point charge and that therefore the wave function $\psi(r) = exp(-r)$ has a cusp at $r=0$.

It occurred to me, that the last comment was perhaps the most useful; well, I mean for me as a QM layman to use as starting point. It does not require specialistic mathematical skill to go from a point charge model for the Hydrogen atom to a model with a broader charge distribution. The idea is to soften the $1/r$ singularity in the potential $V(r)$ and hence make the wave function more analytic. In turn, this might solve the above mentioned problem with the kinetic energy operator. Or at least highlight the origin of the problem.

The simplest modification seemed to me to model the Hydrogen nucleus as a small conducting sphere of radius $R$ and with charge $+e$. Outside of the sphere ($r > R$) the potential $V(r)$ is exactly the same as in the point charge model [this is a well-known fact from electrostatics]. So we can write down the lowest energy solution as: $\psi(r) = exp(-\frac{r}{r_0})$ for $r > R$. Here $r_0$ is the Bohr radius. Note: the energy spectrum is discrete, because:

1. one must balance three terms in the wave function which is difficult to accomplish, and
2. the solution must be bounded at infinity.

Inside the sphere ($r < R$) the potential $V(r)$ is constant [another well-known fact from electrostatics]. Hence, by continuity, its value must be equal to the potential on the surface of the sphere: $V(r) = V(R)$. Inside the sphere, the electron is "free". Its oscillatory solution is the same as for a particle in a spherical box. The solution is the zeroth order Bessel function $j_0(r)$. So we obtain: $\psi(r) = \frac{A\sin(kr)}{r}$. The wave factor $k$ is fully determined by the energy $E$, potential $V$ and some physical parameters (Planck's constant ans electron mass).

At this stage we have only one adjustable parameter in our model, the amplitude $A$. Its value is determined by demanding continuity of the wave function. This leads to the result $A = \frac{R\exp(-\frac{R}{r_0})}{\sin(kR)}$.

My problem is that there are no adjustable parameters left. Therefore one can not demand smoothness of the wave function (continuity of the first derivative). On the other hand, from the continuity of the potential at $r=R$ one would expect the first derivative of the wave function to be not only continuous and but also smooth at $r=R$ ! So I feel something is wrong. But I don't see what. Hence my question in a previous thread about the admissibility of certain diverging terms (which would give one an extra parameter).

## 2 Answers

Well, actually you have two adjustable parameters. One is just that you called $A$ and the other one is an analogous parameter, say $B$, for the solution of radial equation outside the ball where the potential is constant. That equation admits two linearly independent solutions but one exponentially diverges far from the origin and thus it cannot be accepted for describing bounded states, namely states with $E<0$. You have to impose two linear conditions regarding continuity and continuity of the first derivative at $r=R$ for the solution. It seems that these pair of conditions determine the two constants, and this would be impossible since the normalization of the wavefunctions has to remain a free parameter. Actually, since, for a generic value of $E$, the two conditions give rise to an homogeneous linear system, the only possible solution is that trivial $A=B=0$. Indeed you have also to fix the eigenvalue $E$ in order that the two linear conditions turn out to be linearly dependent. This further requirement on $E$ fixes a discrete set of values of the admitted energies. If you instead are interested in solutions with $E \geq 0$, you cannot drop the divergent solution outside the ball, so that you have 3 free parameters and no constraints on the alowed energies, the spectrum turns out to be continuous in this case.

General solution of a second-order ODE has exactly two adjustable parameters - amplitudes of two linearly independent solutions. Your problem is two ODEs with total of 4 boundary conditions (at origin, at infinity and two at joint of solutions).

Now, first of your ODEs is the one for $r\in[0,R]$. Its general solutions are in terms of spherical Bessel functions $j_n(z)$ and $y_n(z)$. However, boundary condition at origin — regularity of wavefunction — sets amplitude of singular solution $y_n(z)$ to zero.

Second of your ODEs is the usual radial equation for hydrogen. It has general solution in terms of hypergeometric function of the first kind ${}_1 F_1(a,b,z)$ which simplifies to generalized Laguerre polynomials $L_n^l(z)$ for bound states and second kind $U(a,b,z)$, which gives singularity at origin.

When solving your equation for hydrogen, you had only second of your ODEs and its first boundary condition was at origin — so you just threw away the singular solution as incompatible with boundary condition. Now, however, you can't simply do this: you have a boundary condition not at the origin and, in fact, you're not even interested in behavior of the solution at the origin — there you replace it with another function, which corresponds to another potential.

So, you instead use second adjustable parameter of spherical Bessel solution and one (or combination) of parameters for second equation's solution to properly join wavefunction and its derivative at joining point of different potentials.

You use your remaining adjustable parameter (or their combination linearly independent of previous ones) for wavefunction normalization.

As you seem to already know, generally not every energy $E$ will allow a remaining boundary condition at infinity — boundedness of wavefunction — to be satisfied. From this you get the spectrum of energies — which won't generally coincide with spectrum of original hydrogen Hamiltonian. As you've changed the potential, you've changed the Hamiltonian, thus you can't just insert the same energy and hope that it's the eigenvalue of new Hamiltonian.

After all this, your wavefunction fully solves the eigenproblem and it doesn't have any divergent behavior anywhere. Moreover, as your new potential is globally continuous, the wavefunction is differentiable everywhere.