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Next: Motion in Central Field Up: Orbital Angular Momentum Previous: Rotation Operators

Eigenfunctions of Orbital Angular Momentum

In Cartesian coordinates, the three components of orbital angular momentum can be written

$\displaystyle L_x$ $\displaystyle = -{\rm i}\,\hbar\left(y\,\frac{\partial}{\partial z} - z\,\frac{\partial} {\partial y}\right),$ (4.74)
$\displaystyle L_y$ $\displaystyle = -{\rm i}\,\hbar\left(z\,\frac{\partial}{\partial x} - x\,\frac{\partial} {\partial z}\right),$ (4.75)
$\displaystyle L_z$ $\displaystyle = -{\rm i}\,\hbar\left(x\,\frac{\partial}{\partial y} - y\,\frac{\partial} {\partial x}\right),$ (4.76)

using the Schrödinger representation. Transforming to standard spherical coordinates,

$\displaystyle x$ $\displaystyle = r \,\sin\theta\, \cos\varphi,$ (4.77)
$\displaystyle y$ $\displaystyle = r\, \sin\theta\, \sin\varphi,$ (4.78)
$\displaystyle z$ $\displaystyle = r\,\cos\theta,$ (4.79)

we obtain

$\displaystyle L_x$ $\displaystyle = {\rm i}\,\hbar\,\left(\sin\varphi\, \frac{\partial}{\partial \theta} + \cot\theta \cos\varphi\,\frac{\partial}{\partial \varphi}\right),$ (4.80)
$\displaystyle L_y$ $\displaystyle = -{\rm i} \,\hbar\,\left(\cos\varphi\, \frac{\partial}{\partial\theta} -\cot\theta \sin\varphi \,\frac{\partial}{\partial \varphi}\right),$ (4.81)
$\displaystyle L_z$ $\displaystyle = -{\rm i}\,\hbar\,\frac{\partial}{\partial\varphi}.$ (4.82)

(See Exercise 2.) Note that Equation (4.82) accords with Equation (4.57). The ladder operators $ L^\pm = L_x \pm {\rm i} \,L_y$ become

$\displaystyle L^\pm = \pm \hbar\,\exp(\pm{\rm i}\,\varphi)\left(\frac{\partial}{\partial\theta} \pm{\rm i} \,\cot\theta\,\frac{\partial}{\partial\varphi}\right).$ (4.83)

(See Exercise 2.) Now,

$\displaystyle L^2 = L_x^{\,2}+L_y^{\,2}+L_z^{\,2} = L_z^{\,2} + (L^+\, L^- + L^- \,L^+) /2,$ (4.84)


$\displaystyle L^2 = - \hbar^{\,2}\left( \frac{1}{\sin\theta}\frac{\partial}{\pa...
...} + \frac{1}{\sin^2\theta}\frac{\partial^{\,2}} {\partial\varphi^{\,2}}\right).$ (4.85)

(See Exercise 2.)

The eigenvalue problem for $ L^2$ takes the form

$\displaystyle L^2 \,\psi = \lambda \,\hbar^{\,2}\, \psi,$ (4.86)

where $ \psi(r, \theta, \varphi)$ is the wavefunction, and $ \lambda$ is a dimensionless number. Let us write

$\displaystyle \psi(r, \theta, \varphi) = R(r) \,Y(\theta, \varphi).$ (4.87)

Equation (4.86) reduces to

$\displaystyle \left( \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\, \sin...
...\theta}\frac{\partial^{\,2}} {\partial\varphi^{\,2}}\right)Y + \lambda \,Y = 0,$ (4.88)

where use has been made of Equation (4.85). As is well known, square-integrable solutions to this equation only exist when $ \lambda$ takes the values $ l\,(l+1)$ , where $ l$ is an integer (which we can take to be non-negative, without loss of generality) [76]. These solutions are known as spherical harmonics [67], and can be written

$\displaystyle Y_{l\,m}(\theta, \varphi) = \sqrt{ \frac{2\,l+1}{4\pi} \frac{(l-m)!}{(l+m)!}}\, P_{l\,m}(\cos\theta)\,{\rm e}^{\,{\rm i} \,m\,\varphi}.$ (4.89)


$\displaystyle P_{l\,m}(\xi)= \frac{(-1)^{l+m}}{2^l\,l!}\,(1-\xi^{\,2})^{m/2}\,\frac{d^{\,l+m}}{d\xi^{\,l+m}} (1-\xi^{\,2})^l.$ (4.90)

is an associated Legendre function [67], satisfying the equation

$\displaystyle \frac{d}{d\xi}\! \left[ (1-\xi^{\,2})\,\frac{dP_{l\,m}}{d\xi}\right] - \frac{m^{\,2}}{1-\xi^{\,2}}\, P_{l\,m} + l\,(l+1)\,P_{l\,m} = 0.$ (4.91)

It follows that

$\displaystyle P_{l\,-m}(\xi)= (-1)^m\,\frac{(l-m)!}{(l+m)!}\,P_{l\,m}(\xi),$ (4.92)

and, hence, that

$\displaystyle Y_{l\,\,-m} = (-1)^m\, Y_{l\,m}^{\,\ast}.$ (4.93)

Of course, $ m$ must be an integer, so as to ensure that the $ Y_{l\,m}$ are single valued in $ \varphi$ . Moreover, it is clear from Equations (4.90) and (4.92) that $ P_{l\,m}=0$ unless $ -l\leq m\leq l$ . The spherical harmonics are orthogonal functions, and are properly normalized with respect to integration over the entire solid angle [67]:

$\displaystyle \int_0^{2\pi} d\varphi \int_0^\pi d\theta \, \sin\theta\,Y_{l\,m}...
...eta,\varphi)\, Y_{l'\,m'}(\theta, \varphi) = \delta_{l \,l'} \,\delta_{m \,m'}.$ (4.94)

The spherical harmonics also form a complete set for representing general functions of $ \theta$ and $ \varphi$ . The first few spherical harmonics are [67]:

$\displaystyle Y_{0\,0}(\theta,\varphi)$ $\displaystyle =\frac{1}{\sqrt{4\pi}},$ (4.95)
$\displaystyle Y_{1\,0}(\theta,\varphi)$ $\displaystyle = \sqrt{\frac{3}{4\pi}}\,\cos\theta,$ (4.96)
$\displaystyle Y_{1\,\pm1}(\theta,\varphi)$ $\displaystyle = \mp \sqrt{\frac{3}{8\pi}}\,\sin\theta\,{\rm e}^{\pm{\rm i}\,\varphi},$ (4.97)
$\displaystyle Y_{2\,0}(\theta,\varphi)$ $\displaystyle = \sqrt{\frac{5}{16\pi}}\,(3\,\cos^2\theta - 1),$ (4.98)
$\displaystyle Y_{2\,\pm 1}(\theta,\varphi)$ $\displaystyle =\mp\sqrt{\frac{15}{8\pi}}\,\sin\theta\,\cos\theta\,{\rm e}^{\pm{\rm i}\,\varphi},$ (4.99)
$\displaystyle Y_{2\,\pm 2}(\theta,\varphi)$ $\displaystyle = \sqrt{\frac{15}{32\pi}}\,\sin^2\theta\,{\rm e}^{\pm 2\,{\rm i}\,\varphi},$ (4.100)
$\displaystyle Y_{3\,0}(\theta,\varphi)$ $\displaystyle = \sqrt{\frac{7}{16\pi}}\,(5\,\cos^3\theta-3\,\cos\theta),$ (4.101)
$\displaystyle Y_{3\,\pm 1}(\theta,\varphi)$ $\displaystyle =\mp\sqrt{\frac{21}{64\pi}}\,\sin\theta\,(5\,\cos^2\theta-1)\,{\rm e}^{\pm{\rm i}\,\varphi},$ (4.102)
$\displaystyle Y_{3\,\pm 2}(\theta,\varphi)$ $\displaystyle =\sqrt{\frac{105}{32\pi}}\,\sin^2\theta\,\cos\theta\,{\rm e}^{\pm{\rm i}\,2\,\varphi},$ (4.103)
$\displaystyle Y_{3\,\pm 3}(\theta,\varphi)$ $\displaystyle =\mp\sqrt{\frac{35}{64\pi}}\,\sin^3\theta\,{\rm e}^{\pm{\rm i}\,3\,\varphi}.$ (4.104)

By definition,

$\displaystyle L^2 \,Y_{l\,m} = l\,(l+1)\,\hbar^{\,2}\,Y_{l\,m},$ (4.105)

where $ l$ is an integer. It follows from Equations (4.82) and (4.89) that

$\displaystyle L_z \,Y_{l\,m} = m\,\hbar\,Y_{l\,m},$ (4.106)

where $ m$ is an integer lying in the range $ -l\leq m\leq l$ . Thus, the wavefunction $ \psi(r, \theta, \varphi) = R(r) \,Y_{l\,m}(\theta, \phi)$ , where $ R$ is a general function, has all of the expected features of the wavefunction of a simultaneous eigenstate of $ L^2$ and $ L_z$ belonging to the quantum numbers $ l$ and $ m$ . The well-known formula [1]

$\displaystyle \frac{d P_{l\,m}}{d\xi}$ $\displaystyle = -\frac{1}{\sqrt{1-\xi^{\,2}}}\,P_{l\,\,m+1} - \frac{m\,\xi}{1-\xi^{\,2}}\, P_{l\,m}$    
  $\displaystyle = \frac{(l+m)\,(l-m+1)}{\sqrt{1-\xi^{\,2}}}\,P_{l\,\,m-1} + \frac{m\,\xi} {1-\xi^{\,2}}\, P_{l\,m}$ (4.107)

can be combined with Equations (4.83) and (4.89) to give

$\displaystyle L^+ \,Y_{l\,m}$ $\displaystyle = [l\,(l+1)- m\,(m+1)]^{1/2}\,\hbar\,Y_{l\,\,m+1},$ (4.108)
$\displaystyle L^- \,Y_{l\,m}$ $\displaystyle = [l\,(l+1) - m \,(m-1)]^{1/2} \,\hbar \,Y_{l\,\,m-1}.$ (4.109)

(See Exercise 4.) These equations are equivalent to Equations (4.55)-(4.56). Note that a spherical harmonic wavefunction is symmetric about the $ z$ -axis (i.e., independent of $ \varphi$ ) whenever $ m=0$ , and is spherically symmetric whenever $ l=0$ (because $ Y_{0\,0} = 1/\sqrt{4\pi}$ ).

In summary, by solving directly for the eigenfunctions of $ L^2$ and $ L_z$ in the Schrödinger representation, we have been able to reproduce all of the results of Section 4.2. Nevertheless, the results of Section 4.2 are more general than those obtained in this section, because they still apply when the quantum number $ l$ takes on half-integer values.

next up previous
Next: Motion in Central Field Up: Orbital Angular Momentum Previous: Rotation Operators
Richard Fitzpatrick 2016-01-22