Part 4 · Spin and angular momentum

4.3 Orbital angular momentum

Source: the operators, their commutators, ${\hat{L}}_z$ in spherical coordinates and its eigenfunctions, the spherical harmonics, the rigid rotator and the exercises from the 2024/25 boards #9 (end), #14–#16 and #18; this year’s course has not reached it yet

From classical to quantum

A particle at $\bar{r}$ with momentum $\bar{p}$ has $\bar{L}=\bar{r}\wedge \bar{p}$: perpendicular to $\bar{r}$ and $\bar{p}$, and only the part of the motion going around the origin counts (beyond the boards). As a determinant, $$\bar{L}=\left|\begin{array}{ccc}{\hat{u}}_x& {\hat{u}}_y& {\hat{u}}_z\\ x& y& z\\ p_x& p_y& p_z\end{array}\right|=(y{p}_z-z{p}_y){\hat{u}}_x+(z{p}_x-x{p}_z){\hat{u}}_y+(x{p}_y-y{p}_x){\hat{u}}_z,$$ each component cyclic in $x\to y\to z$. Replace $\bar{p}$ by $-i\hslash \bar{\nabla }$ (Definition 2.4.1): $${\hat{L}}_x=-i\hslash \left(y\frac{\partial }{\partial z}-z\frac{\partial }{\partial y}\right),{\hat{L}}_y=-i\hslash \left(z\frac{\partial }{\partial x}-x\frac{\partial }{\partial z}\right),{\hat{L}}_z=-i\hslash \left(x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x}\right).$$ There is no ordering problem: each component multiplies by one coordinate and differentiates by a different one, and those commute.

==$\hat{\bar{L}}$ depends on motion: its eigenfunctions are functions of position==, unlike spin’s $|\pm ⟩$. That is why its eigenvalue problem is a differential equation, not only a matrix.

Spherical coordinates

Angular momentum is about rotation, so use coordinates that follow rotations: $$x=r\sin \theta \cos \phi ,y=r\sin \theta \sin \phi ,z=r\cos \theta ,$$ with $r>0$, the polar angle $0\le \theta \le \pi$ from the north pole, like a co-latitude, and the azimuth $0\le \phi \le 2\pi$ around $z$, like a longitude. A rotation about $z$ changes only $\phi$.

None contains $r$: ==angular momentum acts only on the angles==. (The board writes the last term with $\frac{1}{\sin \theta }$; it is $\frac{1}{{\sin }^2\theta }$.)

Eigenfunctions of ${\hat{L}}_z$

${\hat{L}}_z$ only involves $\phi$, so look for $u(\phi )$ with $-i\hslash u^{\prime }=L_{z}u$. It is the momentum equation, with $\phi$ for $x$:

==${\hat{L}}_z$ is the momentum conjugate to the angle $\phi$.== So far $m$ could be any real number.

A half-integer $m$ would give $e^{i\pi }=-1$ after one turn: the function would change sign on going once around $z$. It is the periodic box again: the box is the circle, $L=2\pi$ in $\phi$, so $L_z=\frac{2\pi \hslash }{2\pi }m$ (beyond the boards). The normalisation is $2\pi |A{|}^2=1$. (Board 15 writes the exponent as $imk\pi$ and its value as $\pm 1$; it is $e^{2\pi imk}=+1$. Board 14 calls the whole eigenvalue $m_l$, $\hslash$ included; the board 15 notation is used here.)

Integer and half-integer. So orbital $\ell$ is an integer: $m$ is, and it runs from $-\ell$ to $\ell$ in whole steps. Spin has no wavefunction in $\phi$ and no such condition, and both kinds occur: integer $s$ for bosons, half-integer for fermions (§5.1 Identical particles).

Spherical harmonics

${\hat{L}}_z$ only sees $\phi$, so in a product $Y(\theta ,\phi )=\Theta (\theta )u(\phi )$ the $\Theta$ comes out as a constant. ==Whatever $\Theta$ is, the product is an eigenfunction of ${\hat{L}}_z$ as long as $u=e^{im\phi }$.== ${\hat{L}}^2$ fixes $\Theta$.

Putting $Y=\Theta e^{im\phi }$ into ${\hat{L}}^2$ gives the associated Legendre equation, whose solutions stay finite at both poles only for integer $\ell \ge |m|$ (beyond the boards). The first few, in the sign convention board 18 uses (the table is beyond the boards):

They are orthonormal over the sphere, $\int Y_{{\ell }^{\prime }{m}^{\prime }}^{\ast }{Y}_{\ell m}\sin \theta d\theta d\phi ={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}$, with $\sin \theta d\theta d\phi$ the area element; and complete: any function of direction is a sum of them, as any function in a box is a sum of plane waves.

$|Y_{\ell m}{|}^2$ is the angular shape of an atomic orbital. The real $p_x$ and $p_y$ of chemistry are sums of $m=\pm 1$: $\sin \theta \cos \phi =x/r\propto Y_{1,-1}-Y_{1,1}$. They are eigenfunctions of ${\hat{L}}^2$ but not of ${\hat{L}}_z$ (Exercise 4.3.4). The radial part and the quantum number $n$ come with the hydrogen atom, which the boards do not reach.

The rigid rotator

A particle of mass $m$ on a sphere of radius $R$ can only turn. With $\bar{v}\perp \bar{r}$, $L^2=R^2{p}^2$, so $$\frac{p^2}{2m}=\frac{L^2}{2I},I=m{R}^2,$$ the moment of inertia, and $\hat{H}={\hat{L}}^2/2I$.

==$\hat{H}$ is ${\hat{L}}^2$ times a constant, so the eigenfunctions of ${\hat{L}}^2$ are the energy eigenfunctions.== $E_{\ell }$ does not depend on $m$: the energy depends on how fast it turns, not on the direction of the axis (beyond the boards). $\ell =2$ has 5 states, all at $3{\hslash }^2/I$. The levels are $0,\frac{{\hslash }^2}{I},\frac{3{\hslash }^2}{I},\frac{6{\hslash }^2}{I}$, with growing gaps ${\hslash }^2\ell /I$. (The board starts the list at $\ell =1$; $\ell =0$, at rest in a uniform $Y_{00}$, is allowed too.)

Exercises