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Summary of Angular Momentum

C. David Sherrill
School of Chemistry and Biochemistry
Georgia Institute of Technology
November 1999

We start from the classical expression for angular momentum, ${\bf L} = {\bf r} \times {\bf p}$, to obtain the quantum mechanical version ${\bf\hat
L} = {\bf\hat R} \times {\bf\hat P}$, where ${\bf\hat L}$, ${\bf\hat
R}$, and ${\bf\hat P}$ are all three-dimensional vectors. This definition leads immediately to expressions for the three components of ${\mathbf L}$:

$\displaystyle {\hat L}_x$ $\textstyle =$ $\displaystyle {\hat Y} {\hat P}_z - {\hat Z} {\hat P}_y$ (1)
$\displaystyle {\hat L}_y$ $\textstyle =$ $\displaystyle {\hat Z} {\hat P}_x - {\hat X} {\hat P}_z$ (2)
$\displaystyle {\hat L}_z$ $\textstyle =$ $\displaystyle {\hat X} {\hat P}_y - {\hat Y} {\hat P}_x.$ (3)

From these definitions, we may easily derive the following commutators
$\displaystyle \left[ {\hat L}_i, {\hat R}_j \right]$ $\textstyle =$ $\displaystyle \epsilon_{ijk} i \hbar {\hat R}_k$ (4)
$\displaystyle \left[ {\hat L}_i, {\hat P}_j \right]$ $\textstyle =$ $\displaystyle \epsilon_{ijk} i \hbar {\hat P}_k$ (5)
$\displaystyle \left[ {\hat L}_i, {\hat L}_j \right]$ $\textstyle =$ $\displaystyle \epsilon_{ijk} i \hbar {\hat L}_k,$ (6)

where the indices $i,j,k$ can be $x$, $y$, or $z$, and where the coefficient $\epsilon_{ijk}$ is unity if $i,j,k$ form a cyclic permutation of $x,y,z$ [i.e., ($x,y,z$), ($y,z,x$), or ($z,x,y$)] and -1 for a reverse cyclic permutation [($z,y,x$), ($x,z,y$), or ($y,x,z$)]. The final commutator indicates that we cannot generally know $L_x$, $L_y$, and $L_z$ simultaneously except if we have an eigenstate with eigenvalue 0 for each of these.

Classically, any component of the angular momentum must be less than or equal to the magnitude of the overall angular momentum vector. Quantum mechanically, the average value of any component of the angular momentum must be less than or equal to the square root of the expectation value of ${\bf\hat L}$ dotted with itself:

<{\hat L}_i> \leq \sqrt{ < {\bf\hat L}^2 > }.
\end{displaymath} (7)

${\bf\hat L}^2$ is simply ${\hat L}_x^2 + {\hat L}_y^2 + {\hat L}_z^2$.

Since ${\bf\hat L}^2$ commutes with any component ${\hat L}_i$, we can have simultaneous eigenfunctions of ${\bf\hat L}^2$ and a given component ${\hat L}_i$. We usually pick the $z$ axis, since the expression for ${\hat
L}_z$ is the easiest of the three when we work in spherical polar coordinates:

$\displaystyle {\hat L}_x$ $\textstyle =$ $\displaystyle \frac{\hbar}{i}
- sin \phi \hspace{0.1cm} \frac{\partial}{...
...\hspace{0.1cm} cot \theta
\hspace{0.1cm} \frac{\partial}{\partial \phi}
\right)$ (8)
$\displaystyle {\hat L}_y$ $\textstyle =$ $\displaystyle \frac{\hbar}{i}
cos \phi \hspace{0.1cm} \frac{\partial}{\p...
...\hspace{0.1cm} cot \theta \hspace{0.1cm}
\frac{\partial}{\partial \phi}
\right)$ (9)
$\displaystyle {\hat L}_z$ $\textstyle =$ $\displaystyle \frac{\hbar}{i} \frac{\partial}{\partial \phi}.$ (10)

Of course it is also possible to express ${\bf\hat L}$ in terms of the unit vectors for spherical polar coordinates,

$\displaystyle {\hat r}$ $\textstyle =$ $\displaystyle sin \theta \hspace{0.1cm} cos \phi \hspace{0.1cm} {\hat i}
+ sin ...
...ce{0.1cm} sin \phi \hspace{0.1cm} {\hat j}
+ cos \theta \hspace{0.1cm} {\hat k}$ (11)
$\displaystyle {\hat \phi}$ $\textstyle =$ $\displaystyle - sin \phi \hspace{0.1cm} {\hat i}
+ cos \phi \hspace{0.1cm} {\hat j}$ (12)
$\displaystyle {\hat \theta}$ $\textstyle =$ $\displaystyle cos \theta \hspace{0.1cm} cos \phi \hspace{0.1cm} {\hat i}
+ cos ...
...e{0.1cm} sin \phi \hspace{0.1cm} {\hat j}
- sin \theta \hspace{0.1cm} {\hat k}.$ (13)

\nabla = {\hat r} \frac{\partial}{\partial r}
+ {\hat \phi...
+ {\hat \theta} \frac{1}{r} \frac{\partial}{\partial \theta}
\end{displaymath} (14)

$\displaystyle {\bf\hat L}$ $\textstyle =$ $\displaystyle {\bf\hat R} \times {\bf\hat P}$  
  $\textstyle =$ $\displaystyle r {\hat r} \times \frac{\hbar}{i} \nabla$  
  $\textstyle =$ $\displaystyle \frac{\hbar}{i} \left(
{\hat \phi} \frac{\partial}{\partial \theta}
- {\hat \theta} \frac{1}{sin \theta}
\frac{\partial}{\partial \phi}
\right).$ (15)

The simultaneous eigenfunctions of ${\bf\hat L}^2$ and ${\hat
L}_z$ are called the spherical harmonics, $Y_l^m(\theta, \phi)$, where $l$ is the total angular momentum quantum number, and $m$ is the so-called magnetic quantum number. The spherical harmonics are defined as

Y_l^m(\theta, \phi) = \sqrt{ \frac{2l+1}{4 \pi} \frac{(l-m)!}{(l+m)!} }
(-1)^m e^{im\phi} P_l^m(cos \theta),
\end{displaymath} (16)

where $P_l^m$ are the associated Legendre polynomials. We require that $-l \leq m \leq l$, and spherical harmonics with $m<0$ are defined in terms of the spherical harmonics with $m>0$ according to $Y_l^m = (-1)^m [
Y_l^{-m}]^*$. The spherical harmonics are normalized over integration of angular coordinates such that
\int_0^{\pi} sin \theta \hspace{0.1cm} d\theta \int_0^{2 \pi...
... \phi)]^* Y_l^m(\theta, \phi) =
\delta_{l,l'} \delta_{m, m'},
\end{displaymath} (17)

and they have the following special properties:
$\displaystyle {\hat L}_z Y_l^m$ $\textstyle =$ $\displaystyle m \hbar Y_l^m$ (18)
$\displaystyle {\bf\hat L}^2 Y_l^m$ $\textstyle =$ $\displaystyle l(l+1) \hbar^2 Y_l^m.$ (19)

It can be useful to define ladder operators for angular momentum. The following ladder operators work not only for straight angular momentum ${\bf\hat L}$, but also for combined angular momenta such as ${\bf\hat J} = {\bf\hat L} + {\bf\hat S}$. If

{\bf\hat J}_{\pm} = {\hat J}_x \pm i {\hat J}_y
\end{displaymath} (20)

{\bf\hat J}_{\pm} \vert jm \rangle =
\hbar \left[ (j \mp m)(j \pm m + 1) \right]^{1/2} \vert j, m \pm 1 \rangle .
\end{displaymath} (21)

We can see that these ladder operators raise or lower the magnetic quantum number $m$ but leave $l$ alone.

One can also show that in spherical polar coordinates

{\bf\hat L}^2 = - \hbar^2 \left[
\frac{1}{sin^2 \theta} \fr...
...pace{0.1cm} \frac{\partial}{\partial \theta}
\right) \right].
\end{displaymath} (22)

By comparing this expression with that for $\nabla^2$ in spherical polar coordinates,
$\displaystyle \nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r}
\left( r^2 \...
...+ \frac{1}{r^2 sin^2 \theta}
\left( \frac{\partial^2}{\partial \phi^2} \right),$     (23)

we can see that the Hamiltonian can be written as
$\displaystyle {\hat H}$ $\textstyle =$ $\displaystyle - \frac{\hbar^2}{2m} \nabla^2 + {\hat V}$  
  $\textstyle =$ $\displaystyle \frac{1}{r^2} \frac{\partial}{\partial r}
\left( r^2 \frac{\partial}{\partial r} \right)
+ \frac{{\bf\hat L}^2}{2 m r^2} + {\hat V}.$ (24)

Clearly ${\bf\hat L}^2$ commutes with the kinetic energy term, ${\bf\hat L}^2$ has no $r$ dependence. Likewise, if ${\hat V} = V(r)$, then ${\bf\hat L}^2$ commutes with the whole Hamiltonian. Hence, for problems where the potential depends only on $r$ (central force problems), we can find simultaneous eigenfunctions of ${\hat H}$, ${\bf\hat L}^2$, and ${\hat

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David Sherrill 2006-02-23