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# $$\S16.1$$ General Properties of Spherically Symmetric Potential Problems

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### Introduction

We will briefly recall some results from classical mechanics on motion in spherically symmetric potential problems in three dimensions.

The Lagrangian for a particle in three dimensions is
$$\label{EQ01} L = \frac{1}{2}m \dot{\vec{r}}^{\,2} - V(\vec{r})$$
The corresponding Hamiltonian is
$$\label{EQ02} H = \frac{\vec{p}^{\,2}}{2m} + V(\vec{r})$$
A  potential is spherically symmetric if it depends in $$r$$ alone; it does not depend on $$\theta, \phi$$. All the three components of angular momentum, $$\vec{L}= \vec{r}\times\vec{p}$$, are constants of motion.
For spherically symmetric potentials, it turns out to be useful to work in spherical polar coordinates. In polar coordinates, $$(r,\theta, \phi)$$ the Lagrangian takes the form
$$\label{EQ03} L = \frac{1}{2}m\big(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2\sin^2\theta \dot{\phi}^2 \big) - V(r)$$
The Hamiltonian in polar coordinates is
$$\label{EQ04} H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + \frac{p_\phi^2}{2mr^2\sin^2\theta} + V(r)$$

Properties of classical motion

As a consequence of angular momentum conservation the motion of the particle is confined to a plane. Taking this plane to be $$X-Y$$ plane,  one can use plane polar coordinates $$r, \theta$$ to describe the motion. The corresponding Hamiltonian is
$$\label{EQ05} H = \frac{p_r^2}{2m} +\frac{p_\theta^2}{2mr^2} + V(r)$$
Here $$\theta$$ is seen to be a  cyclic coordinate and corresponding momentum, $$p_\theta$$, is a constant of motion. $$p_\theta$$ can be seen to be the total angular momentum. The radial motion is then completely described by the effective potential

$$\label{EQ06} V_\text{eff} = V(r) + \frac{L^2}{2mr^2}$$

Here $$L$$ denotes total angular momentum $$p_\theta$$.

In quantum mechanics all the features of motion in spherically symmetric potential, except one, described above remain true. In quantum mechanics, the orbits are not well defined and correspondence  with the motion being confined to a plane can be established only in the limit of large angular momentum.

### General Properties of Motion

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

$$H = \frac{p^2}{2m} +V(r) \label{T2E1}$$

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

$$H, "\vec{L}^2, L_z \mbox{\rm and } P$$

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.

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

Schrodinger Equation for Spherically Symmetric Potentials

The Schrodinger equation for a spherically symmetric potential is
$$\left[-\frac{\hbar^2}{2m} \nabla^2 + V(r) \right]\psi = E \psi \label{T2E2}$$
The Laplacian $\nabla^2$ in spherical polar coordinates is given by $$\nabla^2 = \frac{1}{r^2}\pp{r}\left( r^2\pp{r}\right) +\frac{1}{r^2\sin\theta}\pp{\theta} \left(\sin\theta\pp{\theta}\right) + \frac{1}{r^2\sin^2\theta}\PP{\phi} \label{T2E3}$$
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}

Separation of Variables

Substitute
$$\psi(r,\theta,\phi) =R(r)Y(\theta,\phi) \label{T2E5}$$
in Eq,\eqref{T2E4} and divide by $R(r)Y(\theta,\phi)$ to get
$$\frac{1}{R(r)}\frac{1}{r^2}\left(\pp{r} r^2 \pp[R]{r}\right) + \frac{1}{Y} \frac{1}{r^2\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{Y} \frac{1}{r^2\sin^2\theta}\PP[Y]{\phi} + \frac{2m}{\hbar^2}(E-V(r))=0 \label{T2E6}$$
Multiply by $r^2$ and rearrange to get
$$\frac{1}{R(r)}\pp{r}\left( r^2 \pp[R]{r}\right)+ \frac{2m}{\hbar^2}(E-V(r))r^2 = - \frac{1}{Y} \left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}=0 \label{T2E7}$$
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
$$\frac{1}{R(r)}\pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}(E-V(r)) r^2 =\lambda \label{T2E8}$$
and
$$\frac{1}{Y} \left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta}\right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}= -\lambda \label{T2E9}$$
On rearranging Eq,\eqref{T2E8} we get the radial Schrodinger equation
$$\pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}\left(E-V(r)-\frac{\lambda}{r^2} \right) R(r)=0 \label{T2E10}$$and Eq,\eqref{T2E9} can be rewritten as
$$-\left\{ \frac{1}{\sin\theta}\pp{\theta}\left(\sin\theta \pp[Y]{\theta} \right) + \frac{1}{\sin^2\theta}\PP[Y]{\phi}\right\}=\lambda Y(\theta,\phi) \label{T2E11}$$
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 $Y_{\ell m}(\theta, \phi)$.

Summary of Results on Spherically Symmetric Potentials

The solutions of the Schrodinger equation
$$\left[-\frac{\hbar^2}{2m} \nabla^2 + V(r) \right]\psi = E \psi \label{T2E13}$$
for a spherically symmetric potential $V(r)$ are of the form
$$\psi(r,\theta,\phi) =R_\ell(r)Y_{\ell m}(\theta,\phi) \label{T2E14}$$
where $R_\ell(r)$ is called the radial wave function and satisfies the radial Schrodinger equation
$$\pp{r}\left( r^2 \pp[R]{r}\right) + \frac{2m}{\hbar^2}\left(E-V(r)-\frac{\lambda}{r^2} \right) R(r)=0 \label{T2EQ15}$$
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
$$-\frac{\hbar^2}{2m}\DD[\chi]{r} +\left( V(r) + \frac{\ell(\ell+1)\hbar^2}{2mr^2} \right)\chi = E\chi \label{T2EQ20}$$
This equation looks like one dimensional Schrodinger equation with potential $V(r)$ replaced with
$$V(r) + \frac{\ell(\ell+1)\hbar^2}{2mr^2} \equiv V_{\mbox{\rm eff}}(r). \label{T2E21}$$
The second term in $V_{\mbox{\rm eff}}(r)$ 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
$$\chi(r) \to 0 \quad \mbox{\rm as} \quad r \to 0 , \label{T2E22}$$
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 $(-\infty,\infty).$

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