Angular Momentum

In classical mechanics, the vector angular momentum, L, of a particle of position vector ${\bf r}$ and linear momentum ${\bf p}$ is defined as

$${\bf L} = {\bf r}\times {\bf p}.$$ (C.122)

In other words,

$$L_x = y p_z - z p_y,$$ (C.123)

$$L_y = z p_x - x p_z,$$ (C.124)

$$L_z = x p_y - y p_x.$$ (C.125)

In quantum mechanics, the operators, $p_i$ , that represent the Cartesian components of linear momentum, are represented as the spatial differential operators $-{\rm i} \hbar \partial/\partial x_i$ . [See Equation (C.81)]. It follows that:

$$L_x = -{\rm i} \hbar\left(y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}\right),$$ (C.126)

$$L_y = -{\rm i} \hbar\left(z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z}\right),$$ (C.127)

$$L_z = -{\rm i} \hbar\left(x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}\right).$$ (C.128)

In addition, let

$$L^{ 2} = L_x^{ 2}+L_y^{ 2}+L_z^{ 2}$$ (C.129)

be the magnitude-squared of the angular momentum vector.

It is most convenient to work in terms of the standard spherical coordinates, $r$ , $\theta$ , and $\phi$ . These are defined with respect to our usual Cartesian coordinates as follows:

$$x = r \sin\theta \cos\phi,$$ (C.130)

$$y = r \sin\theta \sin\phi,$$ (C.131)

$$z = r \cos\theta.$$ (C.132)

We deduce, after some tedious analysis, that

$$\frac{\partial}{\partial x} = \sin\theta \cos\phi \frac{\partial}{\partial r} + \frac{\cos\... ...partial\theta} - \frac{\sin\phi}{r \sin\theta} \frac{\partial}{\partial\phi},$$ (C.133)

$$\frac{\partial}{\partial y} = \sin\theta \sin\phi \frac{\partial}{\partial r} + \frac{\cos\... ...partial\theta} + \frac{\cos\phi}{r \sin\theta} \frac{\partial}{\partial\phi},$$ (C.134)

$$\frac{\partial}{\partial z} = \cos\theta \frac{\partial}{\partial r} -\frac{\sin\theta}{r} \frac{\partial}{\partial \theta}.$$ (C.135)

Making use of the definitions (C.126)-(C.129), after more tedious algebra, we obtain

$$L_x = - {\rm i} \hbar\left(-\sin\phi \frac{\partial}{\partial\theta} -\cos\phi \cot\theta \frac{\partial}{\partial\phi}\right),$$ (C.136)

$$L_y = - {\rm i} \hbar\left(\cos\phi \frac{\partial}{\partial\theta} -\sin\phi \cot\theta \frac{\partial}{\partial\phi}\right),$$ (C.137)

$$L_z = -{\rm i} \hbar \frac{\partial}{\partial\phi},$$ (C.138)

as well as

$$L^2 = -\hbar^{ 2}\left[\frac{1}{\sin\theta}\frac{\partial}{\part... ...ight) + \frac{1}{\sin^2\theta}\frac{\partial^{ 2}}{\partial\phi^{ 2}}\right].$$ (C.139)

We, thus, conclude that all of our angular momentum operators can be represented as differential operators involving the angular spherical coordinates, $\theta$ and $\phi$ , but not involving the radial coordinate, $r$ .

Let us search for an angular wavefunction, $Y_{l,m}(\theta,\phi)$ , that is a simultaneous eigenstate of $L^{ 2}$ and $L_z$ . In other words,

$$L^{ 2} Y_{l,m} = l (l+1) \hbar^{ 2} Y_{l,m},$$ (C.140)

$$L_z Y_{l,m} =m \hbar Y_{l,m},$$ (C.141)

where $l$ and $m$ are dimensionless constants. We also want the wavefunction to satisfy the normalization constraint

$$\oint \vert Y_{l,m}\vert^{ 2} d{\mit\Omega} = 1,$$ (C.142)

where $d{\mit\Omega}= \sin\theta d\theta d\phi$ is an element of solid angle, and the integral is over all solid angle. Thus, we are searching for well-behaved angular functions that simultaneously satisfy,

$$\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left( \... ...+ \frac{1}{\sin^2\theta}\frac{\partial^{ 2}}{\partial\phi^{ 2}}\right]Y_{l,m} =-l (l+1) Y_{l,m},$$ (C.143)

$$\frac{\partial Y_{l,m}}{\partial\phi} ={\rm i} m Y_{l,m},$$ (C.144)

$$\int_0^\pi\int_0^{2\pi}\vert Y_{l,m}\vert^{ 2} \sin\theta d\theta d\phi = 1.$$ (C.145)

As is well known, the requisite functions are the so-called spherical harmonics,

$$Y_{l,m}(\theta,\phi) =(-1)^{ m} \left[\frac{2 l+1}{4\pi} \frac{(l-m)!}{(l+m)!}\right]^{1/2} P_{l,m}(\cos\theta) {\rm e}^{ {\rm i} m \phi},$$ (C.146)

for $m\geq 0$ . Here, the $P_{l,m}$ are known as associated Legendre polynomials, and are written

$$P_{l,m}(u) = (-1)^{ l+m} \frac{(l+m)!}{(l-m)!} \frac{\left(1-u... ...m/2}}{2^{ l} l!}\left(\frac{d}{du}\right)^{l-m} \left(1-u^{ 2}\right)^{ l}.$$ (C.147)

for $m\geq 0$ . Note that

$$Y_{l,-m} = (-1)^m Y^{ \ast}_{l,m}.$$ (C.148)

Finally, the constant $l$ is constrained to take non-negative integer values, whereas the constant $m$ is constrained to take integer values in the range $-l\leq m\leq +l$ . Thus, $l$ is a quantum number that determines the value of $L^{ 2}$ . In fact, $L^{ 2}= l (l+1) \hbar^{ 2}$ . Likewise, $m$ is a quantum number that determines the value of $L_z$ . In fact, $L_z=m \hbar$ .

The classical Hamiltonian of an extended object spinning with constant angular momentum ${\bf L}$ about one of its principal axes of rotation is

$$H = \frac{L^{ 2}}{2 I},$$ (C.149)

where $I$ is the associated principal moment of inertia. Let us assume that the quantum-mechanical Hamiltonian of an object spinning about a principal axis of rotation has the same form. We can solve the energy eigenvalue problem,

$$H \psi = E \psi,$$ (C.150)

by writing $\psi(r,\theta,\phi) = R(r) Y_{l,m}(\theta,\phi)$ , where $R(r)$ is arbitrary. It immediately follows from Equation (C.140) that

$$E = \frac{l (l+1) \hbar^{ 2}}{2 I}.$$ (C.151)

In other words, the energy levels are quantized in terms of the quantum number, $l$ , that specifies the value of $L^{ 2}$ .

The classical Hamiltonian of an extended object spinning about a general axis is

$$H = \frac{L_x^{ 2}}{2 I_x} + \frac{L_y^{ 2}}{2 L_y} + \frac{L_z^{ 2}}{2 L_z},$$ (C.152)

where the Cartesian axes are aligned along the body's principal axes of rotation, and $I_x$ , $I_y$ , $I_z$ are the corresponding principal moments of inertia. Suppose that the body is axially symmetric about the $z$ -axis. It follows that $I_x=I_y$ . The previous Hamiltonian can be written

$$H = \frac{L^{ 2}}{2 I_x}+\left(\frac{1}{2 I_z}-\frac{1}{2 I_x}\right)L_z^{ 2}.$$ (C.153)

Let us assume that the quantum-mechanical Hamiltonian of an axisymmetric object spinning about an arbitrary axis has the same form as the previous Hamiltonian. We can solve the energy eigenvalue problem, $H \psi=E \psi$ , by writing $\psi(r,\theta,\phi) = R(r) Y_{l,m}(\theta,\phi)$ , where $R(r)$ is arbitrary. It immediately follows from Equations (C.140) and (C.141) that

$$E = \frac{[l (l+1)-m^{ 2}] \hbar^{ 2}}{2 I_x} + \frac{m^{ 2} \hbar^{ 2}}{2 I_z}.$$ (C.154)

In other words, the energy levels are quantized in terms of the quantum number, $l$ , that specifies the value of $L^{ 2}$ , as well as the quantum number, $m$ , that determines the value of $L_z$ . If $I_z\ll I_x$ (in other words, if the object is highly elongated along its symmetry axis) then the spacing between energy levels corresponding to different values of $m$ becomes much greater than the spacing between energy levels corresponding to different values of $l$ . In this case, it is plausible that the system remains in the $m=0$ state (because it cannot acquire sufficient energy to reach the $m=1$ state), so that

$$E = \frac{l (l+1) \hbar^{ 2}}{2 I_x}.$$ (C.155)