[NOTES/QM-16002] Spherically Symmetric Potentials — Using Spherical Polar Coordinates

                                                      Contents

1. Schrodinger Equation for Spherically Symmetric Potentials .
2. Separation of Variables .
3. Summary of Results on Spherically Symmetric Potentials . 

We shall discuss energy eigenvalue problem in three dimensions for a spherically symmetric potential given. A spherically symmetric potential depends only on $r$ and does not depend on $\theta$ and $\phi$. The Hamiltonian for such a system is \begin{equation} H = \frac{p^2}{2m} +V(r) \label{T2E1} \end{equation} For a spherically symmetric potential the Hamiltonian commutes with the angular momentum operators $\vec{L} =\vec{r}\times\vec{p}$ and the angular momentum components $L_x,L_y,L_z$ are constants of motion and therefore $H,\vec{L}^2,L_z$ form a commuting set of operators. It is seen that the parity operators $P$ commutes with all these operators and that the set of operators $$ H, "\vec{L}^2, L_z and  P$$ is a complete set of commuting operators. This means that $\vec{L}^2,L_z , P$ are constants of motion and that the energy eigenfunctions can be selected to have definite values of $\vec{L}^2,L_z , P$ also. We shall see these features in the folowing specfic examples to be discussed later.

• Free Particle, $V(r)=$ constant.
• Hydrogen atom, $v(r)=-\dfrac{e^2}{r}$
• Square well and other similar potentials.

1. Schrodinger equation for Spherically Symmetric Potentials.

The Schrodinger equation for a spherically symmetric potential is \begin{equation} \left[-\frac{\hbar^2}{2m} \nabla^2 + V(r) \right]\psi = E \psi \label{T2E2} \end{equation} The Laplacian $\nabla^2$ in spherical polar coordinates is given by \begin{equation} \nabla^2 = \frac{1}{r^2}\pp{r}\left( r^2\pp{r}\right) +\frac{1}{r^2\sin\theta}\pp{\theta} \left(\sin\theta\pp{\theta}\right) + \frac{1}{r^2\sin^2\theta}\PP{\phi} \label{T2E3} \end{equation} Therefore, eq,\eqref{T2E2} takes the form \begin{eqnarray}\label{T2E4} \left\{\frac{1}{r^2}\left(\pp{r} r^2 \pp{r}\right) +\frac{1}{r^2\sin\theta} \pp{\theta}\left(\sin\theta\pp{\theta}\right)\right. &+& \left. \frac{1}{r^2\sin^2\theta} \PP{\phi} \right\} \psi(r,\theta,\phi) \\ &+& \frac{2m}{\hbar^2}(E-V(r)) \psi(r,\theta,\phi) =0 . \end{eqnarray}

2. Separation of Variables.

Substitute \begin{equation} \psi(r,\theta,\phi) =R(r)Y(\theta,\phi) \label{T2E5} \end{equation} in eq,\eqref{T2E4} and divide by $R(r)Y(\theta,\phi)$ to get \begin{equation} \frac{1}{R(r)}\frac{1}{r^2}\left(\pp{r} r^2 \pp[R]{r}\right) + \frac{1}{Y} \frac{1}{r^2\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{Y} \frac{1}{r^2\sin^2\theta}\PP[Y]{\phi} + \frac{2m}{\hbar^2}(E-V(r))=0 \label{T2E6} \end{equation} Multiply by $r^2$ and rearrange to get \begin{equation} \frac{1}{R(r)}\pp{r}\left( r^2 \pp[R]{r}\right)+ \frac{2m}{\hbar^2}(E-V(r))r^2 = - \frac{1}{Y} \left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}=0 \label{T2E7} \end{equation} The left hand side of the above equation is a function of $r$ alone and the right hand side is a function of $\theta$ and $\phi$ only. This is possible only when each side is a constant, say $\lambda$. Thus we get two ordinary differential equations \begin{equation} \frac{1}{R(r)}\pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}(E-V(r)) r^2 =\lambda \label{T2E8} \end{equation} and \begin{equation} \frac{1}{Y} \left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}= -\lambda \label{T2E9} \end{equation} On rearranging eq,\eqref{T2E8} we get the radial Schrodinger equation \begin{equation} \pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}\left(E-V(r)-\frac{\lambda}{r^2} \right) R(r)=0 \label{T2E10} \end{equation} and eq,\eqref{T2E9} can be rewritten as \begin{equation} -\left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta} \right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}=\lambda Y(\theta,\phi) \label{T2E11} \end{equation} is seen to be just the eigenvalue problem for angular momentum operator $\vec{L}^2$. The variables $\theta$ and $\phi$ can be separated in eq,\eqref{T2E11} by writing $$ Y(\theta,\phi) = Q(\theta)E(\phi), $$ resulting partial differential equation \begin{eqnarray} \left\{ \frac{1}{P}\frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[P]{\theta}\right) +\frac{1}{E} \frac{1}{\sin^2\theta}\PP[P] {\phi}\right\}= \lambda \label{T2E12} \end{eqnarray} separates into two ordinary differential equations one of which is just the eigenvalue equation for $L_z$. For these equations physically acceptable solutions are known to exist only when $\lambda=\ell(\ell+1), m=\ell,\ell-1,\cdots,-\ell-1,-\ell$. The solutions for $Y$ are the spherical harmonics $Y_{\ell m}(\theta, \phi)$.

3. Summary of Results on Spherically Symmetric Potentials.

The solutions of the Schrodinger equation \begin{equation} \left[-\frac{\hbar^2}{2m} \nabla^2 + V(r) \right]\psi = E \psi \label{T2E13} \end{equation} for a spherically symmetric potential $V(r)$ are of the form \begin{equation} \psi(r,\theta,\phi) =R_\ell(r)Y_{\ell m}(\theta,\phi) \label{T2E14} \end{equation} where $R_\ell(r) $ is called the radial wave function and satisfies the radial Schrodinger equation \begin{equation} \pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}\left(E-V(r)-\frac{\lambda}{r^2} \right) R(r)=0 \label{T2eq15} \end{equation} The angular part of the wave function $Y_{\ell m}(\theta,\phi)$ is simultaneous eigenfunction of $\vec{L}^2$ and $L_z$ with eigenvalues $\ell(\ell+1)\hbar^2$ and $m\hbar$, respectively. Note that only $\ell$ appears in the radial equation and that it does not contain $m$. Hence

1. The energy eigenvalues are independent of $m$; there are $2(\ell+1)$ linearly independent solutions for each fixed $\ell$ all having the same energy. Thus they are $(2\ell+1)$ fold degenerate.
2. The energy eigenvalues depend on $\ell$ and increase with increasing $\ell$.

For a spherically symmetric potential we need to concentrate only on the radial equation. If we substitute $R(r)=\dfrac{1}{r}\chi(r)$, the radial equation takes the form of one dimensional Schrodinger equation. Using \begin{eqnarray} \frac{d R(r)}{dr}&=&-\frac{1}{r^2} \chi(r) + \frac{1}{r} \chi(r) \label{T2E16}\\ r^2 \frac{d R(r)}{dr}&=&- \chi(r) + r \chi(r) \label{T2E17}\\ \frac{1}{r^2}\pp{r}\left( r^2 \pp[R]{r}\right)&=& \frac{1}{r^2}\left( -\pp[\chi]{r} + r \PP[\chi]{r} + \pp[\chi]{r} \right) \label{T2E18}\\ &=& \frac{1}{r} \PP[\chi]{r} \label{T219} \end{eqnarray} eq,\eqref{T2eq15} takes the form \begin{equation} -\frac{\hbar^2}{2m}\DD[\chi]{r} +\left( V(r) + \frac{\ell(\ell+1)\hbar^2}{2mr^2} \right)\chi = E\chi \label{T2eq20} \end{equation} This equation looks like one dimensional Schrodinger equation with potential $V(r)$ replaced with \begin{equation} V(r) + \frac{\ell(\ell+1)\hbar^2}{2mr^2} \equiv V_{\rm eff}(r). \label{T2E21} \end{equation} The second term in $V_(r)$ is just the centrifugal potential term which also appears in the classical equation for the radial motion. The radial Schrodinger equation eq,\eqref{T2eq20} can be analyzed in the same manner as one dimensional problems. There is one difference however that we must demand \begin{equation} \chi(r) \to 0 \quad \quad r \to 0 , \label{T2E22} \end{equation} so that the radial wave function $R(r)=\dfrac{\chi(r)}{r}$ does not become singular at $r=0$. In addition to above boundary condition on the solutions, another difference between eq,\eqref{T2eq20} and a one dimensional problem is that the variable $r$ takes values in the interval $(0,\infty)$ instead of $(-\infty,\infty).$