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\begin{document}

\section{Angular Momentum by Operator Methods}

The section on the hydrogen atom showed how orbital angular momentum
can be analyzed by separating variables in configuration space.
We found that the eigenvalues of angular momentum squared are $\hbar^2
l(l + 1)$ where $l$ takes on integer values, $l = 0,~1,~2, \ldots$.
The $z$-component of angular momentum takes on values $-\hbar l,~
-\hbar (l - 1), \ldots ,~ \hbar l$.

The Stern-Gerlach experiment was found to yield results in conflict
with this picture of angular momentum in atoms.  In particular,
half-integal values of $l$ were inferred from the number of beams
emerging from the apparatus for certain atoms.  In this section we
analyze angular momentum using operator methods similar to those used
in calculating the eigenvalues and eigenvectors of the harmonic
oscillator.

\subsection{The Angular Momentum Operator}

The angular momentum operator is defined as in classical physics:
\begin{equation}
     \bold{L} = \bold{r} \times \bold{p} ,
\end{equation}
where $\bold{r}$ is the position vector and $\bold{p}$ is the
momentum.  In Cartesian coordinates,
\begin{equation}
    L_x = y p_z - z p_y ,
\label{comps}
\end{equation}
etc.  Using $[ x ,~p_x ] = i$ (henceforth in this section we set $\hbar =
1$ for simplicity) and the corresponding commutation relations for the
$y$ and $z$ components of position and momentum, it is easy to show
that
\begin{equation}
     [ L_x ,~ L_y ] = i L_z ,~~~[ L_y ,~ L_z ] = i L_x ,~~~
     [ L_z ,~ L_x ] = i L_y .
\label{comm1}
\end{equation}
The square of the angular momentum vector is
\begin{equation}
     L^2 \equiv | \bold{L} |^2 = L_x^2 + L_y^2 + L_z^2 .
\label{squared}
\end{equation}
Using the identity $[A,~B^2 ] = [A,~B] B + B [A,~B ]$, it is also easy
to show that
\begin{equation}
     [L_x ,~L^2 ] = [L_y ,~L^2 ] = [L_z ,~L^2 ] = 0 .
\label{comm2}
\end{equation}

From equations (\ref{comm1}) and (\ref{comm2}) we see that the angular
momentum squared and one of its components form a compatible set of
operators.  However, different components of the angular momentum are
incompatible, so only one component may be known at a time.  It is
conventional to select the $z$ component, but since the $z$-axis can
be aligned arbitrarily, this constitutes no loss of generality.

\subsection{Angular Momentum and the Hamiltonian}

We now show that the square of the angular momentum and its $z$
component commute with Hamiltonians that have spherically symmetric
potentials.  Defining $r^2 = x^2 + y^2 + z^2$, such Hamiltonians take
the form
\begin{equation}
     H = p^2 /(2M) + U(r) .
\end{equation}
It is sufficient to prove that $L^2$ and $L_z$ each commute with $p^2$
and $r^2$.  Several preliminary commutation relations are need to
carry out this proof.

It is easy to show that
\begin{equation}
     [x,~L_x ] = 0,~~~ [y,~L_x ] = -iz ,~~~ [z,~L_x ] = iy ,
\end{equation}
with similar commutation relations for $L_y$ and $L_z$ obtained by
cyclic permutation.  In addition,
\begin{equation}
     [p_x ,~L_x ] = 0 ,~~~ [p_y ,~L_x ] = -ip_z ,~~~ [p_z ,~L_x ] =
     ip_y ,
\end{equation}
etc.  From these one can show that
\begin{equation}
     [r^2 ,~L_x ] = [r^2 ,~L_y ] = [r^2 ,~L_z ] = 0
\end{equation}
and
\begin{equation}
     [r^2 ,~L^2 ] = 0 ,
\end{equation}
which means that the spherically symmetric potential energy commutes
with the components and squared magnitude of angular momentum.

It is similarly possible to show that
\begin{equation}
     [p^2 ,~L_x ] = [p^2 ,~L_y ] = [p^2 ,~L_z ] = [p^2 ,~L^2 ] = 0 ,
\end{equation}
which concludes the proof:
\begin{equation}
     [H ,~L_z ] = [H ,~L^2 ] = 0~~~~\mbox{(spherical symmetry)} .
\end{equation}

By substituting equation (\ref{comps}) etc. into equation
(\ref{squared}) and performing a lengthy but routine series of
manipulations, we can show that
\begin{equation}
     L^2 = r^2 p^2 - ( \bold{r} \cdot \bold{p} )^2
     + i \bold{r} \cdot \bold{p} .
\end{equation}
Note that the order of terms containing $\bold{r}$ and $\bold{p}$ and
their magnitudes is important, since these operators don't commute.
Solving for $p^2$, the Hamiltonian can be written
\begin{equation}
     H = \frac{1}{2M r^2} \left[ (\bold{r} \cdot \bold{p} )^2
     -i \bold{r} \cdot \bold{p} + L^2 \right] .
\end{equation}

In configuration space
\begin{equation}
     \bold{r} \cdot \bold{p} = -i \left(
          x \frac{\partial}{\partial x} +
          y \frac{\partial}{\partial y} +
          z \frac{\partial}{\partial z} \right) =
          -i r \frac{\partial}{\partial r} .
\end{equation}
Simplifying the derivatives shows that
\begin{equation}
     p^2 = \frac{L^2}{r^2} - \frac{1}{r^2}
     \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} .
\end{equation}
Comparison with the analysis of the hydrogen atom in configuration
space agrees with the form of the $L^2$ operator deduced there:
\begin{equation}
     L^2 = -\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}
     \sin \theta \frac{\partial}{\partial \theta} -
     \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} .
\end{equation}

\subsection{Determination of Eigenvalues}

The determination of the eigenvalues of $L^2$ and $L_z$ follows from a
rather extended argument.  The operators
\begin{equation}
     L^+ = L_x + i L_y ,~~~~~ L^- = L_x - i L_y .
\label{comm3}
\end{equation}
will play an important role in this argument, and will ultimately be
identified as raising and lowering operators for the $z$ component of
angular momentum.  The left and right products of $L^+$ and $L^-$ can
be written in terms of $L^2$ and $L_z$:
\begin{equation}
     L^+ L^- = L^2 - L_z^2 + L_z ,~~~~ L^- L^+ =
     L^2 - L_z^2 - L_z .
\label{comm4}
\end{equation}
Notice further that the raising and lowering operators commute with
$L^2$,
\begin{equation}
     [ L^{\pm} ,~ L^2 ] = 0 ,
\label{comm5}
\end{equation}
while
\begin{equation}
     [ L^{\pm} ,~ L_z ] = \mp L^{\pm} .
\label{comm6}
\end{equation}

Let us assume that $\psi_{\lambda , m}$ is an eigenstate of $L^2$
(eigenvalue $\lambda$) and $L_z$ (eigenvalue $m$).  At this time we
don't presume to know the values of $\lambda$ and $m$.  From equation
(\ref{comm5}) we see that
\begin{equation}
     L^2 L^{\pm} \psi_{\lambda , m} =
     L^{\pm} L^2 \psi_{\lambda , m} =
     \lambda L^{\pm} \psi_{\lambda , m} ,
\end{equation}
which shows that $L^{\pm} \psi_{\lambda , m}$ is an eigenstate of
$L^2$ with eigenvalue $\lambda$.  Thus, $L^{\pm}$ doesn't alter the
squared magnitude of the angular momentum.

On the other hand, equation (\ref{comm6}) shows that
\begin{equation}
     L_z L^{\pm} \psi_{\lambda , m} =
     L^{\pm} L_z \psi_{\lambda , m} \pm
     L^{\pm} \psi_{\lambda , m} =
     (m \pm 1) L^{\pm} \psi_{\lambda , m} ,
\end{equation}
which shows that $L^{\pm} \psi_{\lambda , m}$ is an eigenstate of
$L_z$ with eigenvalue $m \pm 1$.  Thus, $L^+$ raises the $z$ component
of the angular momentum by $1$, while $L^-$ lowers it by $1$.  This
justifies the identification of $L^+$ and $L^-$ as raising and
lowering operators.  We may thus write
\begin{equation}
     L^{\pm} \psi_{\lambda , m} = C_{\lambda , m}^{\pm}
     \psi_{\lambda , m \pm 1} ,
\end{equation}
where the $C_{\lambda , m}^{\pm}$ are constants.

The combination $L^- L^+$ first raises the value of $m$ by $1$ and
then lowers it by $1$, thus returning the original eigenstate
multiplied by a constant.  Using equation (\ref{comm4}), we find
\begin{eqnarray}
     L^- L^+ \psi_{\lambda , m} & = & C_{\lambda , m + 1}^-
     C_{\lambda , m}^+ \psi_{\lambda , m} =
     \nonumber \\
     & & ( L^2 - L_z^2 - L_z ) \psi_{\lambda , m} =
     ( \lambda - m^2 - m) \psi_{\lambda , m} ,
\end{eqnarray}
from which we infer that $C_{\lambda , m + 1}^- C_{\lambda , m}^+
 = \lambda - m^2 - m$.  A similar exercise with $L^+ L^-$ shows that
$C_{\lambda , m - 1}^+ C_{\lambda , m}^- = \lambda - m^2 + m$.  These
equations have the solutions
\begin{equation}
     C_{ \lambda , m}^+ = ( \lambda - m^2 - m )^{1/2} ,~~~~
     C_{ \lambda , m}^- = ( \lambda - m^2 + m )^{1/2} ,
\label{norm}
\end{equation}
which are unique up to a phase factor.

Further insight into possible values of $m$ may be obtained by noting
that the operator $L_x^2 + L_y^2$ must have non-negative eigenvalues
since it is composed of the sum of the squares of Hermitian operators.
However, $L_x^2 + L_y^2 = L^2 - L_z^2$, has the eigenvalues $\lambda -
m^2$, which implies that $m^2 \le \lambda$.  Thus, $m$ is bounded
above and below.  If $t$ and $b$ are the top and bottom values of $m$,
then $L^+ \psi_{\lambda , t} = 0$ and $L^- \psi_{\lambda , b} = 0$ ---
otherwise states violating the boundedness condition on $m^2$ would
exist.  Equation (\ref{norm}) thus implies
$\lambda = t^2 + t = b^2 - b$, which means that either $b = t + 1$ or
$b = -t$.  The first solution violates the identification of $t$ as
the top value of $m$, so the second solution is the valid one.

The final piece of the puzzle comes from the realization that $t - b =
n$, where $n$ is some non-negative integer.  Combining this with $b =
-t$ results in $t = n/2$.  Recalling that $\lambda = t^2 + t$, we
rename $t$ to $l$ to be consistent with previous terminology, so that
$\lambda = l^2 + l = l(l + 1)$.

To summarize, the eigenvalues of $L^2$ are
\begin{equation}
     \lambda = l(l + 1) ,~~~~l = 0,~1/2 ,~1 ,~3/2 ,~2 ,~\ldots ,
\end{equation}
and the eigenvalues of $L_z$ are
\begin{equation}
     m = -l,~-l + 1,~ \ldots ,~l .
\end{equation}
There are $2l + 1$ values of $m$.

Notice that this analysis has found solutions with half-integral
values of $l$.  These solutions have no configuration space
representation, and must therefore represent a type of angular
momentum that is distinct from that associated with the orbital motion
of a particle.  Note in particular that the solution with $l = 1/2$
has two possible values of $m$, and thus constitutes a possible
explanation for the (hypothesized) behavior of ground state hydrogen
passing through the Stern-Gerlach apparatus.  The evidence indicates
that the electron possesses an {\em intrinsic} angular momentum.  We
now change our terminology to distinguish this case from the orbital
angular momentum case by replacing $l$ with $s$.  This is called the
{\em spin} quantum number, and the electron is said to have a spin
$s = 1/2$.  Possible values of $L_z$ are $\pm 1/2$.

\end{document}
