Comprehensive Study Notes

Before studying the hydrogen atom, we shall consider the more general problem of a single particle moving under a central force. The results of this section will apply to any central-force problem. Examples are the hydrogen atom (Section 6.5) and the isotropic three-dimensional harmonic oscillator (Prob. 6.3).

A central force is one derived from a potential-energy function that is spherically symmetric, which means that it is a function only of the distance of the particle from the origin: $V=V(r)$. The relation between force and potential energy is given by (5.31) as

\(
\begin{equation}
\mathbf{F}=-\nabla V(x, y, z)=-\mathbf{i}(\partial V / \partial x)-\mathbf{j}(\partial V / \partial y)-\mathbf{k}(\partial V / \partial z) \tag{6.1}
\end{equation}
\)

The partial derivatives in (6.1) can be found by the chain rule [Eqs. (5.53)-(5.55)]. Since $V$ in this case is a function of $r$ only, we have $(\partial V / \partial \theta){r, \phi}=0$ and $(\partial V / \partial \phi){r, \theta}=0$. Therefore,

\(
\begin{gather}
\left(\frac{\partial V}{\partial x}\right){y, z}=\frac{d V}{d r}\left(\frac{\partial r}{\partial x}\right){y, z}=\frac{x}{r} \frac{d V}{d r} \tag{6.2}\
\left(\frac{\partial V}{\partial y}\right){x, z}=\frac{y}{r} \frac{d V}{d r}, \quad\left(\frac{\partial V}{\partial z}\right){x, y}=\frac{z}{r} \frac{d V}{d r} \tag{6.3}
\end{gather}
\)

where Eqs. (5.57) and (5.58) have been used. Equation (6.1) becomes

\(
\begin{equation}
\mathbf{F}=-\frac{1}{r} \frac{d V}{d r}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=-\frac{d V(r)}{d r} \frac{\mathbf{r}}{r} \tag{6.4}
\end{equation}
\)

where (5.33) for $\mathbf{r}$ was used. The quantity $\mathbf{r} / r$ in (6.4) is a unit vector in the radial direction. A central force is radially directed.

Now we consider the quantum mechanics of a single particle subject to a central force. The Hamiltonian operator is

\(
\begin{equation}
\hat{H}=\hat{T}+\hat{V}=-\left(\hbar^{2} / 2 m\right) \nabla^{2}+V(r) \tag{6.5}
\end{equation}
\)

where $\nabla^{2} \equiv \partial^{2} / \partial x^{2}+\partial^{2} / \partial y^{2}+\partial^{2} / \partial z^{2}$ [Eq. (3.46)]. Since $V$ is spherically symmetric, we shall work in spherical coordinates. Hence we want to transform the Laplacian operator to these coordinates. We already have the forms of the operators $\partial / \partial x, \partial / \partial y$, and $\partial / \partial z$ in these coordinates [Eqs. (5.62)-(5.64)], and by squaring each of these operators and
then adding their squares, we get the Laplacian. This calculation is left as an exercise. The result is (Prob. 6.4)

\(
\begin{equation}
\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2}} \cot \theta \frac{\partial}{\partial \theta}+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}} \tag{6.6}
\end{equation}
\)

Looking back to (5.68), which gives the operator $\hat{L}^{2}$ for the square of the magnitude of the orbital angular momentum of a single particle, we see that

\(
\begin{equation}
\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}-\frac{1}{r^{2} \hbar^{2}} \hat{L}^{2} \tag{6.7}
\end{equation}
\)

The Hamiltonian (6.5) becomes

\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2}+V(r) \tag{6.8}
\end{equation}
\)

In classical mechanics a particle subject to a central force has its angular momentum conserved (Section 5.3). In quantum mechanics we might ask whether we can have states with definite values for both the energy and the angular momentum. To have the set of eigenfunctions of $\hat{H}$ also be eigenfunctions of $\hat{L}^{2}$, the commutator $\left[\hat{H}, \hat{L}^{2}\right]$ must vanish. We have

\(
\begin{gather}
{\left[\hat{H}, \hat{L}^{2}\right]=\left[\hat{T}, \hat{L}^{2}\right]+\left[\hat{V}, \hat{L}^{2}\right]} \
{\left[\hat{T}, \hat{L}^{2}\right]=\left[-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2}, \hat{L}^{2}\right]} \
{\left[\hat{T}, \hat{L}^{2}\right]=-\frac{\hbar^{2}}{2 m}\left[\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}, \hat{L}^{2}\right]+\frac{1}{2 m}\left[\frac{1}{r^{2}} \hat{L}^{2}, \hat{L}^{2}\right]} \tag{6.9}
\end{gather}
\)

Recall that $\hat{L}^{2}$ involves only $\theta$ and $\phi$ and not $r$ [Eq. (5.68)]. Hence it commutes with every operator that involves only $r$. [To reach this conclusion, we must use relations like (5.47) with $x$ and $z$ replaced by $r$ and $\theta$.] Thus the first commutator in (6.9) is zero. Moreover, since any operator commutes with itself, the second commutator in (6.9) is zero. Therefore, $\left[\hat{T}, \hat{L}^{2}\right]=0$. Also, since $\hat{L}^{2}$ does not involve $r$, and $V$ is a function of $r$ only, we have $\left[\hat{V}, \hat{L}^{2}\right]=0$. Therefore,

\(
\begin{equation}
\left[\hat{H}, \hat{L}^{2}\right]=0 \quad \text { if } V=V(r) \tag{6.10}
\end{equation}
\)

$\hat{H}$ commutes with $\hat{L}^{2}$ when the potential-energy function is independent of $\theta$ and $\phi$.
Now consider the operator $\hat{L}{z}=-i \hbar \partial / \partial \phi[E q . ~(5.67)]$. Since $\hat{L}{z}$ does not involve $r$ and since it commutes with $\hat{L}^{2}$ [Eq. (5.50)], it follows that $\hat{L}_{z}$ commutes with the Hamiltonian (6.8):

\(
\begin{equation}
\left[\hat{H}, \hat{L}_{z}\right]=0 \quad \text { if } V=V(r) \tag{6.11}
\end{equation}
\)

We can therefore have a set of simultaneous eigenfunctions of $\hat{H}, \hat{L}^{2}$, and $\hat{L}_{z}$ for the central-force problem. Let $\psi$ denote these common eigenfunctions:

\(
\begin{gather}
\hat{H} \psi=E \psi \tag{6.12}\
\hat{L}^{2} \psi=l(l+1) \hbar^{2} \psi, \quad l=0,1,2, \ldots \tag{6.13}\
\hat{L}_{z} \psi=m \hbar \psi, \quad m=-l, \quad-l+1, \ldots, l \tag{6.14}
\end{gather}
\)

where Eqs. (5.104) and (5.105) were used.

Using (6.8) and (6.13), we have for the Schrödinger equation (6.12)

\(
\begin{gather}
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}\right)+\frac{1}{2 m r^{2}} \hat{L}^{2} \psi+V(r) \psi=E \psi \
-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}\right)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} \psi+V(r) \psi=E \psi \tag{6.15}
\end{gather}
\)

The eigenfunctions of $\hat{L}^{2}$ are the spherical harmonics $Y{l}^{m}(\theta, \phi)$, and since $\hat{L}^{2}$ does not involve $r$, we can multiply $Y{l}^{m}$ by an arbitrary function of $r$ and still have eigenfunctions of $\hat{L}^{2}$ and $\hat{L}_{z}$. Therefore,

\(
\begin{equation}
\psi=R(r) Y_{l}^{m}(\theta, \phi) \tag{6.16}
\end{equation}
\)

Using (6.16) in (6.15), we then divide both sides by $Y_{l}^{m}$ to get an ordinary differential equation for the unknown function $R(r)$ :

\(
\begin{equation}
-\frac{\hbar^{2}}{2 m}\left(R^{\prime \prime}+\frac{2}{r} R^{\prime}\right)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}} R+V(r) R=E R(r) \tag{6.17}
\end{equation}
\)

We have shown that, for any one-particle problem with a spherically symmetric potential-energy function $V(r)$, the stationary-state wave functions are $\psi=R(r) Y_{l}^{m}(\theta, \phi)$, where the radial factor $R(r)$ satisfies (6.17). By using a specific form for $V(r)$ in (6.17), we can solve it for a particular problem.