Infinite Spherical Potential Well
Considering a potential V(x) such as:
V(r) = \begin{cases} 0 & \text{ if } 0 \leq r \leq r_0 \\ \infty & \text{ if } r > r_0 \end{cases}
A particle in this potential is completely free within the sphere, but since the potential is infinite on the border, the wavefunction of any energy solution must go to zero.
Since the problem has a radial symmetry, the natural approach is to use spherical coordinates.
Let’s start then by considering the time-independent Schrödinger equation in spherical coordinates:
\left[-\frac{\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\mathrm d}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right) + V(r)\right]\Psi(r,\theta,\phi) = E\Psi(r,\theta,\phi)
For the given potential V(r), inside the sphere (for 0 \leq r \leq r_0), the potential is zero, and the Schrödinger equation reduces to:
\left[-\frac{\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right)\right]\Psi(r,\theta,\phi) = E\Psi(r,\theta,\phi)
Outside the sphere (for r > r_0), the potential is infinite, which means that the wavefunction must be zero. At the boundary r = r_0, the wavefunction and its derivative must be continuous, which leads to the boundary condition:
\left.\Psi(r,\theta,\phi)\right|_{r=r_0} = 0
This boundary condition ensures that the wavefunction smoothly goes to zero at the boundary of the sphere, as required by the infinite potential barrier.
For a central potential the partial differential equation can be separated into these ordinary differential equation:
\begin{aligned} & -\frac{\hbar^2}{2m}\frac{\mathrm d^2 \chi}{\mathrm d r^2} + \left(V(r) + \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)\chi = E\chi \\ & \frac{1}{Y(\theta,\phi)}\left(\frac{1}{\sin(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right) =\ell(\ell+1) \\ \end{aligned}
The derivation is available here and here.
The radial equation
Since V(r) =0 inside the sphere, the radial equations simplifies to:
-\frac{\hbar^2}{2m}\frac{\mathrm d^2 \chi}{\mathrm d r^2} + \left(\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)\chi = E\chi
Defining:
k \equiv \frac{\sqrt{2mE}}{\hbar}
This is the standard form of the radial equation for a spherically symmetric potential and the solution inside the sphere (0 \leq r \leq r_0) can be written as:
\chi(\rho) = Ar j_\ell(\rho) + Br n_\ell(\rho)
where j_\ell(\rho) and n_\ell(\rho) are the spherical Bessel functions and Neumann functions. To satisfy the boundary condition u(r_0) = 0, we need to set B = 0, as the spherical Neumann function n_\ell(\rho) diverges at the origin.
Therefore, the radial wavefunction inside the sphere is given by:
\chi = Ar j_\ell\left(kr\right)
Returning to the R variable:
R = Aj_\ell\left(kr\right)
The constant A can be determined by normalization, and the allowed energy levels are found by applying the boundary condition u(r_0) = 0, which leads to a set of quantized values for the energy E:
j_\ell(kr_0) = 0
so that kr_0 is the zero of the \ell^{th} order of the spherical Bessel function, which can be numerically be computed as there is no analytical solution. The boundary condition requires that:
k = \frac{\zeta_{n,\ell}}{r_0}
where \zeta_{n,\ell} is the n^{th} zero of the \ell^{th} order spherical Bessel function, with allowed energies equal to:
E_{n,\ell} = \frac{\hbar^2}{2mr_0^2}\zeta_{m,\ell}
So the solution within the sphere is:
R(r) = A_{n,l}j_\ell\left(\zeta_{n,l}\frac{r}{r_0}\right)
When l = 0 the equation is simpler and can be solved analytically:
\frac{\mathrm d^2\chi}{\mathrm d r^2} = -k^2 \chi
or:
-\frac{\hbar}{2m}\frac{\mathrm d^2\chi}{\mathrm d r^2} = E \chi
This is equivalent to the particle in a box problem for a box of length r_0, and the solution is:
\chi = A \sin\left(\frac{n\pi r}{r_0}\right)
Applying the normalization condition (the same integral as the particle in a box case) the solution is:
\chi = \sqrt{\frac{2}{r_0}} \sin\left(\frac{n\pi r}{r_0}\right)
There is a minor difference point to consider: the general solution is of the form:
\chi = A \sin(kr) + B \cos(kr)
In the particle in a box case, there is a potential on left; in the spherical potential, while \frac{\cos(r)}{r} \to \infty (since R = \frac{\chi}{r}) the wavefunction could be normalizable (as there is a r^2 in the normalization condition), there are literature argument to assume B=0, they will not be described here.
Apply the boundary condition u(r_0) = 0 analytically and to the solution is requiring a quantization for the energy levels by setting the argument of the sine function to integer multiples of \pi:
E_n = \frac{\hbar^2}{2m}\left(\frac{n\pi}{r_0}\right)^2, \quad n = 1, 2, 3, \ldots
The angular equation
For the angular equation:
\frac{1}{Y(\theta,\phi)}\left(\frac{1}{\sin(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right) =\ell(\ell+1)
The assumption for solving this equation is to apply the separation of variables technique, looking for a solution of type:
Y(\theta,\phi) = \Theta(\theta)\Phi(\phi)
Substituting into the equation, it gives:
\frac{\Phi(\phi)}{\sin(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial}{\partial \theta}\right)\Theta(\theta) + \frac{\Theta(\theta)}{\sin^2(\theta)} \frac{\partial^2\Phi(\phi)}{\partial \phi^2} = -\ell(\ell+1)\Theta(\theta)\Phi(\phi)
Multiplying by:
\frac{\sin^2(\theta)}{\Theta(\theta)\Phi(\phi)}
and rearranging:
\frac{1}{\Phi(\phi)} \frac{\partial^2\Phi(\phi)}{\partial \phi^2} = -l(l+1)\sin^2(\theta) - \frac{\sin(\theta)}{\Theta(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial}{\partial \theta}\right)\Theta(\theta) = m^2
The equation for \Phi(\phi) is straightforward to solve:
\Phi(\phi) = e^{im\phi}
The periodicity condition \Phi(\phi) = \Phi(\phi + 2\pi) requires m to be an integer.
The equation for \Theta(\theta) is the familiar Legendre equation:
\sin(\theta)\frac{\mathrm d}{\mathrm d\theta}\left(\sin\theta\frac{\mathrm d\Theta(\theta)}{\mathrm d\theta}\right) + \left[\ell(\ell+1)\sin^2(\theta) - m^2\right]\Theta(\theta) = 0
The solutions to this equation are the associated Legendre functions of the first kind, P_\ell^m(\cos\theta), where \ell is the degree of the Legendre polynomial, and m is the order (with |m| \leq \ell). A more in-depth derivation of the spherical harmonics is available here.
Therefore, the complete solution for the angular part of the wavefunction is:
Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) = e^{im\phi}P_\ell^m(\cos\theta)
The values of \ell and m are determined by the boundary conditions and the quantization of angular momentum.
Complete solution
By combining the solutions for the radial part (\chi) and the angular part (Y(\theta,\phi)), we can construct the complete wavefunction for the particle confined within the spherical potential well:
\Psi(r,\theta,\phi) = R(r)Y(\theta,\phi) = A_{n,l}j_\ell\left(\zeta_{n,l}\frac{r}{r_0}\right)e^{im\phi}P_\ell^m(\cos\theta)
This wavefunction satisfies the Schrödinger equation and the boundary conditions for the given potential, and the allowed energy levels and angular momentum values are quantized due to the boundary conditions and the requirements of well-behaved solutions.
Sperical Bessel and Neumann functions
The spherical Bessel function j_n(r) and is given by:
j_n(r) = \sqrt{\frac{\pi}{2r}} J_{n+\frac{1}{2}}(r)
with J_{n+\frac{1}{2}}(r) is the Bessel function of the first kind of order n+\frac{1}{2}:
J_{n+\frac{1}{2}}(r) = \sqrt{\frac{2}{\pi r}} \sin\left(r - \frac{n\pi}{2}\right)
An explicit expressions for the first five is:
\begin{aligned} j_0(r) &= \frac{\sin r}{r} \\ j_1(r) &= \frac{\sin r}{r^2} - \frac{\cos r}{r} \\ j_2(r) &= \left(\frac{3}{r^3} - \frac{1}{r}\right) \sin r - \frac{3 \cos r}{r^2} \\ j_3(r) &= \left(\frac{15}{r^4} - \frac{6}{r^2}\right) \sin r - \left(\frac{15}{r^3} - \frac{1}{r}\right) \cos r \\ j_4(r) &= \left(\frac{105}{r^5} - \frac{45}{r^3} + \frac{1}{r}\right) \sin r - \left(\frac{105}{r^4} - \frac{10}{r^2}\right) \cos r \end{aligned}
They are plotted below:
The Neumann functions, n_n(r), are given by:
n_n(r) = \frac{J_n(r) \cos(n \pi) - J_{-n}(r)}{\sin(n \pi)}
For integer orders n, this simplifies to:
n_n(r) = \lim_{\nu \to n} \frac{J_\nu(r) \cos(\nu \pi) - J_{-\nu}(r)}{\sin(\nu \pi)}
where, J_n(r) and J_{-n}(r) are the Bessel functions of the first kind of order n and -n, respectively.
An explicit expressions for the first five is:
\begin{aligned} y_0(r) &= -\frac{\cos r}{r} \\ y_1(r) &= -\frac{\cos r}{r^2} - \frac{\sin r}{r} \\ y_2(r) &= \left(-\frac{3}{r^3} + \frac{1}{r}\right) \cos r - \frac{3 \sin r}{r^2} \\ y_3(r) &= \left(-\frac{15}{r^4} + \frac{6}{r^2}\right) \cos r - \left(\frac{15}{r^3} - \frac{1}{r}\right) \sin r \\ y_4(r) &= \left(-\frac{105}{r^5} + \frac{45}{r^3} - \frac{1}{r}\right) \cos r - \left(\frac{105}{r^4} - \frac{10}{r^2}\right) \sin r \end{aligned}
They are plotted below:
It is possible to note that they diverge for r\to 0.