Quantum Mechanics — Lecture notes for PHYS223 XIII Three-dimensional examples and applications XV Angular momentum

XIV Central potentials

The isotropic harmonic oscillator is an example of a spherical symmetric potential, where the potential energy only depends on the radial distance from the origin, $r=|{\bf r}|$ . These potentials are called central potentials. Another example is the Coulomb potential

V_{c}=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r}

(235)

of two charges $q_{1}$ and $q_{2}$ . For central potentials the Schrödinger equation

E\psi({\bf r})=-\frac{\hbar^{2}}{2\mu}\Delta\psi({\bf r})+V(r)\psi({\bf r})

(236)

can also be separated in three one-dimensional problems, but only when one works in spherical polar coordinates. Note: Here mass is denoted by the symbol $\mu$ since we need $m$ later for a different purpose.

XIV.1 Spherical polar coordinates

The three spherical coordinates are the radial coordinate $r=\sqrt{x^{2}+y^{2}+z^{2}}$ , the azimuthal angle $\phi=\arctan(y/x)$ , and the polar angle $\theta=\arccos(z/r)$ .

The cartesian coordinates can be written as

$\displaystyle x$	$\displaystyle=$	$\displaystyle r\sin\theta\cos\phi,$	(237)
$\displaystyle y$	$\displaystyle=$	$\displaystyle r\sin\theta\sin\phi,$	(238)
$\displaystyle z$	$\displaystyle=$	$\displaystyle r\cos\theta.$	(239)

The Laplace operator is given by

\Delta=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial}{% \partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}% \left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}% \theta}\frac{\partial^{2}}{\partial\phi^{2}}.

(240)

We abbreviate the angular part as

\Delta_{\theta,\phi}=\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(% \sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac% {\partial^{2}}{\partial\phi^{2}}.

(241)

XIV.2 Separation of variables in spherical polar coordinates

We separate the radial dependence from the angular dependence in the wavefunction, $\psi(x,y,z)=R(r)Y(\theta,\phi)$ . From the Schrödinger equation (236) one finds that the angular part solves the equation $\Delta_{\theta,\phi}Y=-l(l+1)Y$ , i.e.,

\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{% \partial Y}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}Y% }{\partial\phi^{2}}=-l(l+1)Y,

(242)

while the radial part solves the equation

-\frac{\hbar^{2}}{2\mu r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)+\left% (V(r)+\frac{\hbar^{2}l(l+1)}{2\mu r^{2}}\right)R=ER,

(243)

where $l$ is a constant.

XIV.3 Angular part

Since Eq. (242) for the angular part does not depend on the potential $V(r)$ we can solve this equation once and for all. The solutions are the spherical harmonics

Y_{lm}(\theta,\phi)=(-1)^{m}\sqrt{\frac{(2l+1)}{4\pi}\frac{(l-m)!}{(l+m)!}}P^{% m}_{l}(\cos\theta)e^{im\phi},

(244)

for $m\geq 0$ , $Y_{l,-m}(\theta,\phi)=(-1)^{m}Y_{l,m}^{*}(\theta,\phi)$ for $m<0$ .

Here $l=0,1,2,3,\ldots$ is called the azimuthal quantum number and $m=0,\pm 1,\pm 2,\ldots,\pm l$ is called the magnetic quantum number. Note that for each $l$ there are $2l+1$ values of $m$ .

For $m=0$ the functions $P^{m}_{l}(s)$ are the Legendre polynomials

P^{0}_{l}(s)=\frac{1}{2^{l}l!}\frac{d^{l}}{ds^{l}}(s^{2}-1)^{l},\quad l=0,1,2,% 3,\ldots.

(245)

Hence

$\displaystyle P^{0}_{0}(s)$	$\displaystyle=$	$\displaystyle 1,$	(246)
$\displaystyle P^{0}_{1}(s)$	$\displaystyle=$	$\displaystyle s,$	(247)
$\displaystyle P^{0}_{2}(s)$	$\displaystyle=$	$\displaystyle(3s^{2}-1)/2,$	(248)
$\displaystyle P^{0}_{3}(s)$	$\displaystyle=$	$\displaystyle(5s^{3}-3s)/2.$	(249)

For $m\neq 0$ the functions $P^{m}_{l}(s)$ are the associated Legendre polynomials

P^{m}_{l}(s)=P^{-m}_{l}(s)=(1-s^{2})^{|m|/2}\frac{d^{|m|}}{ds^{|m|}}P_{l}^{0}(% s).

(250)

The first spherical harmonics are

$\displaystyle Y_{00}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{1}{4\pi}}\quad\mbox{for $l=0$, $m=0$},$	(251)
$\displaystyle Y_{10}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{3}{4\pi}}\cos\theta\quad\mbox{for $l=1$, $m=0$},$	(252)
$\displaystyle Y_{1,\pm 1}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\mp\sqrt{\frac{3}{8\pi}}\sin\theta e^{\pm i\phi}\quad\mbox{for $l% =1$, $m=\pm 1$},$	(253)
			(253)
$\displaystyle Y_{20}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{5}{16\pi}}(3\cos^{2}\theta-1),$	(254)
$\displaystyle Y_{2,\pm 1}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\mp\sqrt{\frac{15}{8\pi}}\sin\theta\cos\theta e^{\pm i\phi},$	(255)
$\displaystyle Y_{2,\pm 2}(\theta,\phi)$	$\displaystyle=$	$\displaystyle\mp\sqrt{\frac{15}{32\pi}}\sin^{2}\theta e^{\pm 2i\phi},$	(256)

XIV.4 Quantisation of angular momentum

The angular momentum operator is given by $\hat{\bf L}=\hat{\bf r}\times\hat{\bf p}$ , hence $\hat{L}_{x}=\hat{y}\hat{p}_{z}-\hat{z}\hat{p}_{y}$ etc. The squared length is given by $\hat{\bf L}^{2}=\hat{L}_{x}^{2}+\hat{L}_{y}^{2}+\hat{L}_{z}^{2}$ . One can show that

\hat{\bf L}^{2}Y_{lm}(\theta,\phi)=\hbar^{2}l(l+1)Y_{lm},

(257)

\hat{L}_{z}Y_{lm}(\theta,\phi)=\hbar mY_{lm}.

(258)

Hence the spherical harmonics are eigenfunctions of $\hat{\bf L}^{2}$ and $\hat{L}_{z}$ , with eigenvalues $\hbar^{2}l(l+1)$ and $\hbar m$ , respectively.

XIV.5 Degeneracy of the radial part

Equation (243) for the radial part $R(r)$ depends on the potential and hence has to be solved separately for every problem in order to find the energies $E$ . However, since this equation only depends on the azimuthal quantum number $l$ but not on the magnetic quantum number $m$ , each energy level is at least $2l+1$ fold degenerate, since for each $l$ there are $2l+1$ solutions $Y_{lm}$ of the angular equation with different values of $m$ .