In Section 3.3 we found the eigenfunctions and eigenvalues for the linear-momentum operator $\hat{p}_{x}$. In this section we consider the same problem for the angular momentum of a particle. Angular momentum plays a key role in the quantum mechanics of atomic structure. We begin by reviewing the classical mechanics of angular momentum.
Classical Mechanics of One-Particle Angular Momentum
Consider a moving particle of mass $m$. We set up a Cartesian coordinate system that is fixed in space. Let $\mathbf{r}$ be the vector from the origin to the instantaneous position of the particle. We have
\(
\begin{equation}
\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z \tag{5.33}
\end{equation}
\)
where $x, y$, and $z$ are the particle's coordinates at a given instant. These coordinates are functions of time. Defining the velocity vector $\mathbf{v}$ as the time derivative of the position vector, we have [Eq. (5.32)]
\(
\begin{gather}
\mathbf{v} \equiv \frac{d \mathbf{r}}{d t}=\mathbf{i} \frac{d x}{d t}+\mathbf{j} \frac{d y}{d t}+\mathbf{k} \frac{d z}{d t} \tag{5.34}\
v{x}=d x / d t, \quad v{y}=d y / d t, \quad v_{z}=d z / d t
\end{gather}
\)
We define the particle's linear momentum vector $\mathbf{p}$ by
\(
\begin{gather}
\mathbf{p} \equiv m \mathbf{v} \tag{5.35}\
p{x}=m v{x}, \quad p{y}=m v{y}, \quad p{z}=m v{z} \tag{5.36}
\end{gather}
\)
The particle's angular momentum $\mathbf{L}$ with respect to the coordinate origin is defined in classical mechanics as
\(
\begin{gather}
\mathbf{L} \equiv \mathbf{r} \times \mathbf{p} \tag{5.37}\
\mathbf{L}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
x & y & z \
p{x} & p{y} & p{z}
\end{array}\right| \tag{5.38}\
L{x}=y p{z}-z p{y}, \quad L{y}=z p{x}-x p{z}, \quad L{z}=x p{y}-y p{x} \tag{5.39}
\end{gather}
\)
where (5.28) was used. $L{x}, L{y}$, and $L_{z}$ are the components of $\mathbf{L}$ along the $x, y$, and $z$ axes. The angular-momentum vector $\mathbf{L}$ is perpendicular to the plane defined by the particle's position vector $\mathbf{r}$ and its velocity $\mathbf{v}$ (Fig. 5.4).
FIGURE 5.4 L $\equiv \mathbf{r} \times \mathbf{p}$.
The torque $\tau$ acting on a particle is defined as the cross product of $\mathbf{r}$ and the force $\mathbf{F}$ acting on the particle: $\tau \equiv \mathbf{r} \times \mathbf{F}$. One can show that $\tau=d \mathbf{L} / d t$. When no torque acts on a particle, the rate of change of its angular momentum is zero; that is, its angular momentum is constant (or conserved). For a planet orbiting the sun, the gravitational force is radially directed. Since the cross product of two parallel vectors is zero, there is no torque on the planet and its angular momentum is conserved.
One-Particle Orbital-Angular-Momentum Operators
Now let us turn to the quantum-mechanical treatment. In quantum mechanics, there are two kinds of angular momenta. Orbital angular momentum results from the motion of a particle through space, and is the analog of the classical-mechanical quantity $\mathbf{L}$. Spin angular momentum (Chapter 10) is an intrinsic property of many microscopic particles and has no classical-mechanical analog. We are now considering only orbital angular momentum. We get the quantum-mechanical operators for the components of orbital angular momentum of a particle by replacing the coordinates and momenta in the classical equations (5.39) by their corresponding operators [Eqs. (3.21)-(3.23)]. We find
\(
\begin{align}
& \hat{L}{x}=-i \hbar\left(y \frac{\partial}{\partial z}-z \frac{\partial}{\partial y}\right) \tag{5.40}\
& \hat{L}{y}=-i \hbar\left(z \frac{\partial}{\partial x}-x \frac{\partial}{\partial z}\right) \tag{5.41}\
& \hat{L}_{z}=-i \hbar\left(x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}\right) \tag{5.42}
\end{align}
\)
(Since $\hat{y} \hat{p}{z}=\hat{p}{z} \hat{y}$, and so on, we do not run into any problems of noncommutativity in constructing these operators.) Using
\(
\begin{equation}
\hat{L}^{2}=|\hat{\mathbf{L}}|^{2}=\hat{\mathbf{L}} \cdot \hat{\mathbf{L}}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}_{z}^{2} \tag{5.43}
\end{equation}
\)
we can construct the operator for the square of the angular-momentum magnitude from the operators in (5.40)-(5.42).
Since the commutation relations determine which physical quantities can be simultaneously assigned definite values, we investigate these relations for angular momentum. Operating on some function $f(x, y, z)$ with $\hat{L}_{y}$, we have
\(
\hat{L}_{y} f=-i \hbar\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right)
\)
Operating on this last equation with $\hat{L}_{x}$, we get
\(
\begin{equation}
\hat{L}{x} \hat{L}{y} f=-\hbar^{2}\left(y \frac{\partial f}{\partial x}+y z \frac{\partial^{2} f}{\partial z \partial x}-y x \frac{\partial^{2} f}{\partial z^{2}}-z^{2} \frac{\partial^{2} f}{\partial y \partial x}+z x \frac{\partial^{2} f}{\partial y \partial z}\right) \tag{5.44}
\end{equation}
\)
Similarly,
\(
\begin{gather}
\hat{L}{x} f=-i \hbar\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \
\hat{L}{y} \hat{L}_{x} f=-\hbar^{2}\left(z y \frac{\partial^{2} f}{\partial x \partial z}-z^{2} \frac{\partial^{2} f}{\partial x \partial y}-x y \frac{\partial^{2} f}{\partial z^{2}}+x \frac{\partial f}{\partial y}+x z \frac{\partial^{2} f}{\partial z \partial y}\right) \tag{5.45}
\end{gather}
\)
Subtracting (5.45) from (5.44), we have
\(
\begin{align}
\hat{L}{x} \hat{L}{y} f-\hat{L}{y} \hat{L}{x} f & =-\hbar^{2}\left(y \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial y}\right) \
{\left[\hat{L}{x}, \hat{L}{y}\right] } & =i \hbar \hat{L}_{z} \tag{5.46}
\end{align}
\)
where we used relations such as
\(
\begin{equation}
\frac{\partial^{2} f}{\partial z \partial x}=\frac{\partial^{2} f}{\partial x \partial z} \tag{5.47}
\end{equation}
\)
which are true for well-behaved functions. We could use the same procedure to find [ $\left.\hat{L}{y}, \hat{L}{z}\right]$ and $\left[\hat{L}{z}, \hat{L}{x}\right]$, but we can save time by noting a certain kind of symmetry in (5.40)-(5.42). By a cyclic permutation of $x, y$, and $z$, we mean replacing $x$ by $y$, replacing $y$ by $z$, and replacing $z$ by $x$. If we carry out a cyclic permutation in $\hat{L}{x}$, we get $\hat{L}{y}$; a cyclic permutation in $\hat{L}{y}$ gives $\hat{L}{z}$; and $\hat{L}{z}$ is transformed into $\hat{L}{x}$ by a cyclic permutation. Hence, by carrying out two successive cyclic permutations on (5.46), we get
\(
\begin{equation}
\left[\hat{L}{y}, \hat{L}{z}\right]=i \hbar \hat{L}{x}, \quad\left[\hat{L}{z}, \hat{L}{x}\right]=i \hbar \hat{L}{y} \tag{5.48}
\end{equation}
\)
We next evaluate the commutators of $\hat{L}^{2}$ with each of its components, using commutator identities of Section 5.1.
\(
\begin{align}
{\left[\hat{L}^{2}, \hat{L}{x}\right]=} & {\left[\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{x}^{2}, \hat{L}{x}\right]+\left[\hat{L}{y}^{2}, \hat{L}{x}\right]+\left[\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{y}^{2}, \hat{L}{x}\right]+\left[\hat{L}{z}^{2}, \hat{L}{x}\right] } \
= & {\left[\hat{L}{y}, \hat{L}{x}\right] \hat{L}{y}+\hat{L}{y}\left[\hat{L}{y}, \hat{L}{x}\right]+\left[\hat{L}{z}, \hat{L}{x}\right] \hat{L}{z}+\hat{L}{z}\left[\hat{L}{z}, \hat{L}{x}\right] } \
= & -i \hbar \hat{L}{z} \hat{L}{y}-i \hbar \hat{L}{y} \hat{L}{z}+i \hbar \hat{L}{y} \hat{L}{z}+i \hbar \hat{L}{z} \hat{L}{y} \
& \quad\left[\hat{L}^{2}, \hat{L}{x}\right]=0 \tag{5.49}
\end{align}
\)
Since a cyclic permutation of $x, y$, and $z$ leaves $\hat{L}^{2}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}_{z}^{2}$ unchanged, if we carry out two such permutations on (5.49), we get
\(
\begin{equation}
\left[\hat{L}^{2}, \hat{L}{y}\right]=0, \quad\left[\hat{L}^{2}, \hat{L}{z}\right]=0 \tag{5.50}
\end{equation}
\)
To which of the quantities $L^{2}, L{x}, L{y}, L_{z}$ can we assign definite values simultaneously? Because $\hat{L}^{2}$ commutes with each of its components, we can specify an exact value for $L^{2}$
FIGURE 5.5 Spherical coordinates.
and any one component. However, no two components of $\hat{\mathbf{L}}$ commute with each other, so we cannot specify more than one component simultaneously. (There is one exception to this statement, which will be discussed shortly.) It is traditional to take $L_{z}$ as the component of angular momentum that will be specified along with $L^{2}$. Note that in specifying $L^{2}=|\mathbf{L}|^{2}$ we are not specifying the vector $\mathbf{L}$, only its magnitude. A complete specification of $\mathbf{L}$ requires simultaneous specification of each of its three components, which we usually cannot do. In classical mechanics when angular momentum is conserved, each of its three components has a definite value. In quantum mechanics when angular momentum is conserved, only its magnitude and one of its components are specifiable.
We could now try to find the eigenvalues and common eigenfunctions of $\hat{L}^{2}$ and $\hat{L}_{z}$ by using the forms for these operators in Cartesian coordinates. However, we would find that the partial differential equations obtained would not be separable. Therefore we transform these operators to spherical coordinates (Fig. 5.5). The coordinate $r$ is the distance from the origin to the point $(x, y, z)$. The angle $\theta$ is the angle the vector $\mathbf{r}$ makes with the positive $z$ axis. The angle that the projection of $\mathbf{r}$ in the $x y$ plane makes with the positive $x$ axis is $\phi$. (Mathematics texts often interchange $\theta$ and $\phi$.) A little trigonometry gives
\(
\begin{align}
x=r \sin \theta \cos \phi, \quad y & =r \sin \theta \sin \phi, \quad z=r \cos \theta \tag{5.51}\
r^{2}=x^{2}+y^{2}+z^{2}, \quad \cos \theta & =\frac{z}{\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}, \quad \tan \phi=\frac{y}{x} \tag{5.52}
\end{align}
\)
To transform the angular-momentum operators to spherical coordinates, we must transform $\partial / \partial x, \partial / \partial y$, and $\partial / \partial z$ into these coordinates. [This transformation may be skimmed if desired. Begin reading again after Eq. (5.64).]
To perform this transformation, we use the chain rule. Suppose we have a function of $r, \theta$, and $\phi: f(r, \theta, \phi)$. If we change the independent variables by substituting
\(
r=r(x, y, z), \quad \theta=\theta(x, y, z), \quad \phi=\phi(x, y, z)
\)
into $f$, we transform it into a function of $x, y$, and $z$ :
\(
f[r(x, y, z), \theta(x, y, z), \phi(x, y, z)]=g(x, y, z)
\)
For example, suppose that $f(r, \theta, \phi)=3 r \cos \theta+2 \tan ^{2} \phi$. Using (5.52), we have $g(x, y, z)=3 z+2 y^{2} x^{-2}$.
The chain rule tells us how the partial derivatives of $g(x, y, z)$ are related to those of $f(r, \theta, \phi)$. In fact,
\(
\begin{align}
\left(\frac{\partial g}{\partial x}\right){y, z} & =\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial x}\right){y, z}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial x}\right){y, z}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial x}\right){y, z} \tag{5.53}\
\left(\frac{\partial g}{\partial y}\right){x, z} & =\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial y}\right){x, z}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial y}\right){x, z}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial y}\right){x, z} \tag{5.54}
\end{align}
\)
\(
\begin{equation}
\left(\frac{\partial g}{\partial z}\right){x, y}=\left(\frac{\partial f}{\partial r}\right){\theta, \phi}\left(\frac{\partial r}{\partial z}\right){x, y}+\left(\frac{\partial f}{\partial \theta}\right){r, \phi}\left(\frac{\partial \theta}{\partial z}\right){x, y}+\left(\frac{\partial f}{\partial \phi}\right){r, \theta}\left(\frac{\partial \phi}{\partial z}\right)_{x, y} \tag{5.55}
\end{equation}
\)
To convert these equations to operator equations, we delete $f$ and $g$ to give
\(
\begin{equation}
\frac{\partial}{\partial x}=\left(\frac{\partial r}{\partial x}\right){y, z} \frac{\partial}{\partial r}+\left(\frac{\partial \theta}{\partial x}\right){y, z} \frac{\partial}{\partial \theta}+\left(\frac{\partial \phi}{\partial x}\right)_{y, z} \frac{\partial}{\partial \phi} \tag{5.56}
\end{equation}
\)
with similar equations for $\partial / \partial y$ and $\partial / \partial z$. The task now is to evaluate the partial derivatives such as $(\partial r / \partial x)_{y, z}$. Taking the partial derivative of the first equation in (5.52) with respect to $x$ at constant $y$ and $z$, we have
\(
\begin{align}
2 r\left(\frac{\partial r}{\partial x}\right){y, z} & =2 x=2 r \sin \theta \cos \phi \
\left(\frac{\partial r}{\partial x}\right){y, z} & =\sin \theta \cos \phi \tag{5.57}
\end{align}
\)
Differentiating $r^{2}=x^{2}+y^{2}+z^{2}$ with respect to $y$ and with respect to $z$, we find
\(
\begin{equation}
\left(\frac{\partial r}{\partial y}\right){x, z}=\sin \theta \sin \phi, \quad\left(\frac{\partial r}{\partial z}\right){x, y}=\cos \theta \tag{5.58}
\end{equation}
\)
From the second equation in (5.52), we find
\(
\begin{align}
& -\sin \theta\left(\frac{\partial \theta}{\partial x}\right){y, z}=-\frac{x z}{r^{3}} \
& \left(\frac{\partial \theta}{\partial x}\right){y, z}=\frac{\cos \theta \cos \phi}{r} \tag{5.59}
\end{align}
\)
Also,
\(
\begin{equation}
\left(\frac{\partial \theta}{\partial y}\right){x, z}=\frac{\cos \theta \sin \phi}{r}, \quad\left(\frac{\partial \theta}{\partial z}\right){x, y}=-\frac{\sin \theta}{r} \tag{5.60}
\end{equation}
\)
From $\tan \phi=y / x$, we find
\(
\begin{equation}
\left(\frac{\partial \phi}{\partial x}\right){y, z}=-\frac{\sin \phi}{r \sin \theta}, \quad\left(\frac{\partial \phi}{\partial y}\right){x, z}=\frac{\cos \phi}{r \sin \theta}, \quad\left(\frac{\partial \phi}{\partial z}\right)_{x, y}=0 \tag{5.61}
\end{equation}
\)
Substituting (5.57), (5.59), and (5.61) into (5.56), we find
\(
\begin{equation}
\frac{\partial}{\partial x}=\sin \theta \cos \phi \frac{\partial}{\partial r}+\frac{\cos \theta \cos \phi}{r} \frac{\partial}{\partial \theta}-\frac{\sin \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \tag{5.62}
\end{equation}
\)
Similarly,
\(
\begin{gather}
\frac{\partial}{\partial y}=\sin \theta \sin \phi \frac{\partial}{\partial r}+\frac{\cos \theta \sin \phi}{r} \frac{\partial}{\partial \theta}+\frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \tag{5.63}\
\frac{\partial}{\partial z}=\cos \theta \frac{\partial}{\partial r}-\frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \tag{5.64}
\end{gather}
\)
At last, we are ready to express the angular-momentum components in spherical coordinates. Substituting (5.51), (5.63), and (5.64) into (5.40), we have
\(
\begin{align}
\hat{L}{x}= & -i \hbar\left[r \sin \theta \sin \phi\left(\cos \theta \frac{\partial}{\partial r}-\frac{\sin \theta}{r} \frac{\partial}{\partial \theta}\right)\right. \
& \left.-r \cos \theta\left(\sin \theta \sin \phi \frac{\partial}{\partial r}+\frac{\cos \theta \sin \phi}{r} \frac{\partial}{\partial \theta}+\frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi}\right)\right] \
\hat{L}{x}= & i \hbar\left(\sin \phi \frac{\partial}{\partial \theta}+\cot \theta \cos \phi \frac{\partial}{\partial \phi}\right) \tag{5.65}
\end{align}
\)
Also, we find
\(
\begin{gather}
\hat{L}{y}=-i \hbar\left(\cos \phi \frac{\partial}{\partial \theta}-\cot \theta \sin \phi \frac{\partial}{\partial \phi}\right) \tag{5.66}\
\hat{L}{z}=-i \hbar \frac{\partial}{\partial \phi} \tag{5.67}
\end{gather}
\)
By squaring each of $\hat{L}{x}, \hat{L}{y}$, and $\hat{L}{z}$ and then adding their squares, we can construct $\hat{L}^{2}=\hat{L}{x}^{2}+\hat{L}{y}^{2}+\hat{L}{z}^{2}[$ Eq. (5.43)]. The result is (Prob. 5.17)
\(
\begin{equation}
\hat{L}^{2}=-\hbar^{2}\left(\frac{\partial^{2}}{\partial \theta^{2}}+\cot \theta \frac{\partial}{\partial \theta}+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right) \tag{5.68}
\end{equation}
\)
Although the angular-momentum operators depend on all three Cartesian coordinates, $x, y$, and $z$, they involve only the two spherical coordinates $\theta$ and $\phi$.
One-Particle Orbital-Angular-Momentum Eigenfunctions and Eigenvalues
We now find the common eigenfunctions of $\hat{L}^{2}$ and $\hat{L}{z}$, which we denote by $Y$. Since these operators involve only $\theta$ and $\phi, Y$ is a function of these two coordinates: $Y=Y(\theta, \phi)$. (Of course, since the operators are linear, we can multiply $Y$ by an arbitrary function of $r$ and still have an eigenfunction of $\hat{L}^{2}$ and $\hat{L}{z}$.) We must solve
\(
\begin{align}
& \hat{L}_{z} Y(\theta, \phi)=b Y(\theta, \phi) \tag{5.69}\
& \hat{L}^{2} Y(\theta, \phi)=c Y(\theta, \phi) \tag{5.70}
\end{align}
\)
where $b$ and $c$ are the eigenvalues of $\hat{L}{z}$ and $\hat{L}^{2}$.
Using the $\hat{L}{z}$ operator, we have
\(
\begin{equation}
-i \hbar \frac{\partial}{\partial \phi} Y(\theta, \phi)=b Y(\theta, \phi) \tag{5.71}
\end{equation}
\)
Since the operator in (5.71) does not involve $\theta$, we try a separation of variables, writing
\(
\begin{equation}
Y(\theta, \phi)=S(\theta) T(\phi) \tag{5.72}
\end{equation}
\)
Equation (5.71) becomes
\(
\begin{align}
-i \hbar \frac{\partial}{\partial \phi}[S(\theta) T(\phi)] & =b S(\theta) T(\phi) \
-i \hbar S(\theta) \frac{d T(\phi)}{d \phi} & =b S(\theta) T(\phi) \
\frac{d T(\phi)}{T(\phi)} & =\frac{i b}{\hbar} d \phi \
T(\phi) & =A e^{i b \phi / \hbar} \tag{5.73}
\end{align}
\)
where $A$ is an arbitrary constant.
Is $T$ suitable as an eigenfunction? The answer is no, since it is not, in general, a singlevalued function. If we add $2 \pi$ to $\phi$, we will still be at the same point in space, and hence we want no change in $T$ when this is done. For $T$ to be single-valued, we have the restriction
\(
\begin{align}
T(\phi+2 \pi) & =T(\phi) \
A e^{i b \phi / \hbar} e^{i b 2 \pi / \hbar} & =A e^{i b \phi / \hbar} \
e^{i b 2 \pi / \hbar} & =1 \tag{5.74}
\end{align}
\)
To satisfy $e^{i \alpha}=\cos \alpha+i \sin \alpha=1$, we must have $\alpha=2 \pi m$, where
\(
m=0, \pm 1, \pm 2, \pm \cdots
\)
Therefore, (5.74) gives
\(
\begin{gather}
2 \pi b / \hbar=2 \pi m \
b=m \hbar, \quad m=\ldots-2,-1,0,1,2, \ldots \tag{5.75}
\end{gather}
\)
and (5.73) becomes
\(
\begin{equation}
T(\phi)=A e^{i m \phi}, \quad m=0, \pm 1, \pm 2, \ldots \tag{5.76}
\end{equation}
\)
The eigenvalues for the $z$ component of angular momentum are quantized.
We fix $A$ by normalizing $T$. First let us consider normalizing some function $F$ of $r, \theta$, and $\phi$. The ranges of the independent variables are (see Fig. 5.5)
\(
\begin{equation}
0 \leq r \leq \infty, \quad 0 \leq \theta \leq \pi, \quad 0 \leq \phi \leq 2 \pi \tag{5.77}
\end{equation}
\)
The infinitesimal volume element in spherical coordinates is (Taylor and Mann, Section 13.9)
\(
\begin{equation}
d \tau=r^{2} \sin \theta d r d \theta d \phi \tag{5.78}
\end{equation}
\)
The quantity (5.78) is the volume of an infinitesimal region of space for which the spherical coordinates lie in the ranges $r$ to $r+d r, \theta$ to $\theta+d \theta$, and $\phi$ to $\phi+d \phi$. The normalization condition for $F$ in spherical coordinates is therefore
\(
\begin{equation}
\int{0}^{\infty}\left[\int{0}^{\pi}\left[\int_{0}^{2 \pi}|F(r, \theta, \phi)|^{2} d \phi\right] \sin \theta d \theta\right] r^{2} d r=1 \tag{5.79}
\end{equation}
\)
If $F$ happens to have the form
\(
F(r, \theta, \phi)=R(r) S(\theta) T(\phi)
\)
then use of the integral identity (3.74) gives for (5.79)
\(
\int{0}^{\infty}|R(r)|^{2} r^{2} d r \int{0}^{\pi}|S(\theta)|^{2} \sin \theta d \theta \int_{0}^{2 \pi}|T(\phi)|^{2} d \phi=1
\)
and it is convenient to normalize each factor of $F$ separately:
\(
\begin{equation}
\int{0}^{\infty}|R|^{2} r^{2} d r=1, \quad \int{0}^{\pi}|S|^{2} \sin \theta d \theta=1, \quad \int_{0}^{2 \pi}|T|^{2} d \phi=1 \tag{5.80}
\end{equation}
\)
Therefore,
\(
\begin{gathered}
\int{0}^{2 \pi}\left(A e^{i m \phi}\right) * A e^{i m \phi} d \phi=1=|A|^{2} \int{0}^{2 \pi} d \phi \
|A|=(2 \pi)^{-1 / 2}
\end{gathered}
\)
\(
\begin{equation}
T(\phi)=\frac{1}{\sqrt{2 \pi}} e^{i m \phi}, \quad m=0, \pm 1, \pm 2, \ldots \tag{5.81}
\end{equation}
\)
We now solve $\hat{L}^{2} Y=c Y$ [Eq. (5.70)] for the eigenvalues $c$ of $\hat{L}^{2}$. Using (5.68) for $\hat{L}^{2}$, (5.72) for $Y$, and (5.81), we have
\(
\begin{gather}
-\hbar^{2}\left(\frac{\partial^{2}}{\partial \theta^{2}}+\cot \theta \frac{\partial}{\partial \theta}+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right)\left(S(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi}\right)=c S(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi} \
\frac{d^{2} S}{d \theta^{2}}+\cot \theta \frac{d S}{d \theta}-\frac{m^{2}}{\sin ^{2} \theta} S=-\frac{c}{\hbar^{2}} S \tag{5.82}
\end{gather}
\)
To solve (5.82), we carry out some tedious manipulations, which may be skimmed if desired. Begin reading again at Eq. (5.91). First, for convenience, we change the independent variable by making the substitution
\(
\begin{equation}
w=\cos \theta \tag{5.83}
\end{equation}
\)
This transforms $S$ into some new function of $w$ :
\(
\begin{equation}
S(\theta)=G(w) \tag{5.8}
\end{equation}
\)
The chain rule gives
\(
\begin{equation}
\frac{d S}{d \theta}=\frac{d G}{d w} \frac{d w}{d \theta}=-\sin \theta \frac{d G}{d w}=-\left(1-w^{2}\right)^{1 / 2} \frac{d G}{d w} \tag{5.85}
\end{equation}
\)
Similarly, we find (Prob. 5.25)
\(
\begin{equation}
\frac{d^{2} S}{d \theta^{2}}=\left(1-w^{2}\right) \frac{d^{2} G}{d w^{2}}-w \frac{d G}{d w} \tag{5.86}
\end{equation}
\)
Using (5.86), (5.85), and $\cot \theta=\cos \theta / \sin \theta=w /\left(1-w^{2}\right)^{1 / 2}$, we find that (5.82) becomes
\(
\begin{equation}
\left(1-w^{2}\right) \frac{d^{2} G}{d w^{2}}-2 w \frac{d G}{d w}+\left[\frac{c}{\hbar^{2}}-\frac{m^{2}}{1-w^{2}}\right] G(w)=0 \tag{5.87}
\end{equation}
\)
The range of $w$ is $-1 \leq w \leq 1$.
To get a two-term recursion relation when we try a power-series solution, we make the following change of dependent variable:
\(
\begin{equation}
G(w)=\left(1-w^{2}\right)^{|m| / 2} H(w) \tag{5.88}
\end{equation}
\)
Differentiating (5.88), we evaluate $G^{\prime}$ and $G^{\prime \prime}$, and (5.87) becomes, after we divide by $\left(1-w^{2}\right)^{|m| / 2}$,
\(
\begin{equation}
\left(1-w^{2}\right) H^{\prime \prime}-2(|m|+1) w H^{\prime}+\left[c \hbar^{-2}-|m|(|m|+1)\right] H=0 \tag{5.89}
\end{equation}
\)
We now try a power series for $H$ :
\(
\begin{equation}
H(w)=\sum{j=0}^{\infty} a{j} w^{j} \tag{5.90}
\end{equation}
\)
Differentiating [compare Eqs. (4.36)-(4.38)], we have
\(
\begin{aligned}
& H^{\prime}(w)=\sum{j=0}^{\infty} j a{j} w^{j-1} \
& H^{\prime \prime}(w)=\sum{j=0}^{\infty} j(j-1) a{j} w^{j-2}=\sum{j=0}^{\infty}(j+2)(j+1) a{j+2} w^{j}
\end{aligned}
\)
Substitution of these power series into (5.89) yields, after combining sums,
\(
\sum{j=0}^{\infty}\left[(j+2)(j+1) a{j+2}+\left(-j^{2}-j-2|m| j+\frac{c}{\hbar^{2}}-|m|^{2}-|m|\right) a_{j}\right] w^{j}=0
\)
Setting the coefficient of $w^{j}$ equal to zero, we get the recursion relation
\(
\begin{equation}
a{j+2}=\frac{(j+|m|)(j+|m|+1)-c / \hbar^{2}}{(j+1)(j+2)} a{j} \tag{5.91}
\end{equation}
\)
Just as in the harmonic-oscillator case, the general solution of (5.89) is an arbitrary linear combination of a series of even powers (whose coefficients are determined by $a{0}$ ) and a series of odd powers (whose coefficients are determined by $a{1}$ ). It can be shown that the infinite series defined by the recursion relation (5.91) does not give well-behaved eigenfunctions. [Many texts point out that the infinite series diverges at $w= \pm 1$. However, this is not sufficient cause to reject the infinite series, since the eigenfunctions might be quadratically integrable, even though infinite at two points. For a careful discussion, see M. Whippman, Am. J. Phys., 34, 656 (1966).] Hence, as in the harmonic-oscillator case, we must cause one of the series to break off, its last term being $a{k} w^{k}$. We eliminate the other series by setting $a{0}$ or $a_{1}$ equal to zero, depending on whether $k$ is odd or even.
Setting the coefficient of $a_{k}$ in (5.91) equal to zero, we have
\(
\begin{equation}
c=\hbar^{2}(k+|m|)(k+|m|+1), \quad k=0,1,2, \ldots \tag{5.92}
\end{equation}
\)
Since $|m|$ takes on the values $0,1,2, \ldots$, the quantity $k+|m|$ takes on the values $0,1,2, \ldots$ We therefore define the quantum number $l$ as
\(
\begin{equation}
l \equiv k+|m| \tag{5.93}
\end{equation}
\)
and the eigenvalues for the square of the magnitude of angular momentum are
\(
\begin{equation}
c=l(l+1) \hbar^{2}, \quad l=0,1,2, \ldots \tag{5.94}
\end{equation}
\)
The magnitude of the orbital angular momentum of a particle is
\(
\begin{equation}
|\mathbf{L}|=[l(l+1)]^{1 / 2} \hbar \tag{5.95}
\end{equation}
\)
From (5.93), it follows that $|m| \leq l$. The possible values for $m$ are thus
\(
\begin{equation}
m=-l,-l+1,-l+2, \ldots,-1,0,1, \ldots, l-2, l-1, l \tag{5.96}
\end{equation}
\)
Let us examine the angular-momentum eigenfunctions. From (5.83), (5.84), (5.88), (5.90), and (5.93), the theta factor in the eigenfunctions is
\(
\begin{equation}
S{l, m}(\theta)=\sin ^{|m|} \theta \sum{\substack{j=1,3, \ldots \ \text { or } j=0,2, \ldots}}^{l-|m|} a_{j} \cos ^{j} \theta \tag{5.97}
\end{equation}
\)
where the sum is over even or odd values of $j$, depending on whether $l-|m|$ is even or odd. The coefficients $a_{j}$ satisfy the recursion relation (5.91), which, using (5.94), becomes
\(
\begin{equation}
a{j+2}=\frac{(j+|m|)(j+|m|+1)-l(l+1)}{(j+1)(j+2)} a{j} \tag{5.98}
\end{equation}
\)
The $\hat{L}^{2}$ and $\hat{L}_{z}$ eigenfunctions are given by Eqs. (5.72) and (5.81) as
\(
\begin{equation}
Y{l}^{m}(\theta, \phi)=S{l, m}(\theta) T(\phi)=\frac{1}{\sqrt{2 \pi}} S_{l, m}(\theta) e^{i m \phi} \tag{5.99}
\end{equation}
\)
The $m$ in $Y_{l}^{m}$ is a label, and not an exponent.
EXAMPLE
Find $Y{l}^{m}(\theta, \phi)$ and the $\hat{L}^{2}$ and $\hat{L}{z}$ eigenvalues for (a) $l=0$; (b) $l=1$.
(a) For $l=0$, Eq. (5.96) gives $m=0$, and (5.97) becomes
\(
\begin{equation}
S{0,0}(\theta)=a{0} \tag{5.100}
\end{equation}
\)
The normalization condition (5.80) gives
\(
\begin{aligned}
\int{0}^{\pi}\left|a{0}\right|^{2} \sin \theta d \theta=1 & =2\left|a{0}\right|^{2} \
\left|a{0}\right| & =2^{-1 / 2}
\end{aligned}
\)
Equation (5.99) gives
\(
\begin{equation}
Y_{0}^{0}(\theta, \phi)=\frac{1}{\sqrt{4 \pi}} \tag{5.101}
\end{equation}
\)
For $l=0$ and $m=0$, Eqs. (5.69), (5.70), (5.75), and (5.94) give the $\hat{L}^{2}$ eigenvalue as $c=0$ and the $\hat{L}_{z}$ eigenvalue as $b=0$.
(b) For $l=1$, the possible values for $m$ in (5.96) are $-1,0$, and 1 . For $|m|=1$, (5.97) gives
\(
\begin{equation}
S{1, \pm 1}(\theta)=a{0} \sin \theta \tag{5.102}
\end{equation}
\)
$a{0}$ in (5.102) is not necessarily the same as $a{0}$ in (5.100). Normalization gives
\(
\begin{gathered}
1=\left|a{0}^{2}\right| \int{0}^{\pi} \sin ^{2} \theta \sin \theta d \theta=\left|a{0}^{2}\right| \int{-1}^{1}\left(1-w^{2}\right) d w \
\left|a_{0}\right|=\sqrt{3} / 2
\end{gathered}
\)
where the substitution $w=\cos \theta$ was made. Thus $S_{1, \pm 1}=\left(3^{1 / 2} / 2\right) \sin \theta$ and (5.99) gives
\(
\begin{equation}
Y{1}^{1}=(3 / 8 \pi)^{1 / 2} \sin \theta e^{i \phi}, \quad Y{1}^{-1}=(3 / 8 \pi)^{1 / 2} \sin \theta e^{-i \phi} \tag{5.103}
\end{equation}
\)
For $l=1$ and $m=0$, we find (see the following exercise) $S{1,0}=(3 / 2)^{1 / 2} \cos \theta$ and $Y{1}^{0}=(3 / 4 \pi)^{1 / 2} \cos \theta$.
For $l=1$, (5.94) gives the $\hat{L}^{2}$ eigenvalue as $2 \hbar^{2}$; for $m=-1,0$, and 1 , (5.75) gives the $\hat{L}_{z}$ eigenvalues as $-\hbar, 0$, and $\hbar$, respectively.
EXERCISE Verify the expressions for $S{1,0}$ and $Y{1}^{0}$.
The functions $S{l, m}(\theta)$ are well known in mathematics and are associated Legendre functions multiplied by a normalization constant. The associated Legendre functions are defined in Prob. 5.34. Table 5.1 gives the $S{l, m}(\theta)$ functions for $l \leq 3$.
The angular-momentum eigenfunctions $Y_{l}^{m}$ in (5.99) are called spherical harmonics (or surface harmonics).
In summary, the one-particle orbital angular-momentum eigenfunctions and eigenvalues are [Eqs. (5.69), (5.70), (5.75), and (5.94)]
TABLE $5.1 \quad \boldsymbol{S}_{\boldsymbol{l}, \boldsymbol{m}}(\boldsymbol{\theta})$
\(
\begin{array}{ll}
l=0: & S{0,0}=\frac{1}{2} \sqrt{2} \
l=1: & S{1,0}=\frac{1}{2} \sqrt{6} \cos \theta \
& S{1, \pm 1}=\frac{1}{2} \sqrt{3} \sin \theta \
l=2: & S{2,0}=\frac{1}{4} \sqrt{10}\left(3 \cos ^{2} \theta-1\right) \
& S{2, \pm 1}=\frac{1}{2} \sqrt{15} \sin \theta \cos \theta \
& S{2, \pm 2}=\frac{1}{4} \sqrt{15} \sin ^{2} \theta \
l=3: & S{3,0}=\frac{3}{4} \sqrt{14}\left(\frac{5}{3} \cos ^{3} \theta-\cos \theta\right) \
& S{3, \pm 1}=\frac{1}{8} \sqrt{42} \sin \theta\left(5 \cos ^{2} \theta-1\right) \
& S{3, \pm 2}=\frac{1}{4} \sqrt{105} \sin ^{2} \theta \cos \theta \
& S{3, \pm 3}=\frac{1}{8} \sqrt{70} \sin ^{3} \theta
\end{array}
\)
\(
\begin{equation}
\hat{L}^{2} Y{l}^{m}(\theta, \phi)=l(l+1) \hbar^{2} Y{l}^{m}(\theta, \phi), \quad l=0,1,2, \ldots \tag{5.104}
\end{equation}
\)
\(
\begin{equation}
\hat{L}{z} Y{l}^{m}(\theta, \phi)=m \hbar Y_{l}^{m}(\theta, \phi), \quad m=-l,-l+1, \ldots, l-1, l \tag{5.105}
\end{equation}
\)
where the eigenfunctions are given by (5.99). Often the symbol $m{l}$ is used instead of $m$ for the $L{z}$ quantum number. We shall later see that the spherical harmonics are orthogonal functions [Eq. (7.27)].
Since $l \geq|m|$, the magnitude $[l(l+1)]^{1 / 2} \hbar$ of the orbital angular momentum $\mathbf{L}$ is greater than the magnitude $|m| \hbar$ of its $z$ component $L{z}$, except for $l=0$. If it were possible to have the angular-momentum magnitude equal to its $z$ component, this would mean that the $x$ and $y$ components were zero, and we would have specified all three components of $\mathbf{L}$. However, since the components of angular momentum do not commute with each other, we cannot do this. The one exception is when $l$ is zero. In this case, $|\mathbf{L}|^{2}=L{x}^{2}+L{y}^{2}+L{z}^{2}$ has zero for its eigenvalue, and it must be true that all three components $L{x}, L{y}$, and $L_{z}$ have zero eigenvalues. From Eq. (5.12), the uncertainties in angular-momentum components satisfy
\(
\begin{equation}
\Delta L{x} \Delta L{y} \geq \frac{1}{2}\left|\int \Psi^{}\left[\hat{L}{x}, \hat{L}{y}\right] \Psi d \tau\right|=\frac{\hbar}{2}\left|\int \Psi^{} \hat{L}_{z} \Psi d \tau\right| \tag{5.106}
\end{equation}
\)
and two similar equations obtained by cyclic permutation. When the eigenvalues of $\hat{L}{z}, \hat{L}{x}$, and $\hat{L}{y}$ are zero, $\hat{L}{x} \Psi=0, \hat{L}{y} \Psi=0, \hat{L}{z} \Psi=0$, the right-hand sides of (5.106) and the two similar equations are zero, and having $\Delta L{x}=\Delta L{y}=\Delta L{z}=0$ is permitted. But what about the statement in Section 5.1 that to have simultaneous eigenfunctions of two operators the operators must commute? The answer is that this theorem refers to the possibility of having a complete set of eigenfunctions of one operator be eigenfunctions of the other operator. Thus, even though $\hat{L}{x}$ and $\hat{L}{z}$ do not commute, it is possible to have some of the eigenfunctions of $\hat{L}{z}$ (those with $l=0=m$ ) be eigenfunctions of $\hat{L}{x}$. However, it is impossible to have all the $\hat{L}{z}$ eigenfunctions also be eigenfunctions of $\hat{L}_{x}$.
FIGURE 5.6 Orientation of L.
FIGURE 5.7 Orientations of $\mathbf{L}$ with respect to the $z$ axis for $l=1$.
Since we cannot specify $L{x}$ and $L{y}$, the vector $\mathbf{L}$ can lie anywhere on the surface of a cone whose axis is the $z$ axis, whose altitude is $m \hbar$, and whose slant height is $\sqrt{l(l+1)} \hbar$ (Fig. 5.6). The possible orientations of $\mathbf{L}$ with respect to the $z$ axis for the case $l=1$ are shown in Fig. 5.7. For each eigenvalue of $\hat{L}^{2}$, there are $2 l+1$ different eigenfunctions $Y_{l}^{m}$, corresponding to the $2 l+1$ values of $m$. We say that the $\hat{L}^{2}$ eigenvalues are $(2 l+1)$-fold degenerate. The term degeneracy is applicable to the eigenvalues of any operator, not just the Hamiltonian.
Of course, there is nothing special about the $z$ axis. All directions of space are equivalent. If we had chosen to specify $L^{2}$ and $L{x}$ (rather than $L{z}$ ), we would have gotten the same eigenvalues for $L{x}$ as we found for $L{z}$. However, it is easier to solve the $\hat{L}{z}$ eigenvalue equation because $\hat{L}{z}$ has a simple form in spherical coordinates, which involve the angle of rotation $\phi$ about the $z$ axis.