Central Potentials & Orbital Angular Momentum

One dimension was training. Real atoms are three-dimensional, and the force an electron feels from a nucleus depends only on distance — a central potential. That symmetry splits the Schrödinger equation into a radial problem and a universal angular one, solved by the same ladder trick that cracked the harmonic oscillator — now climbing angular momentum instead of energy. A cat circling a warm laptop conserves $\mathbf L$; so does an electron circling a proton — but only the electron's comes in steps of $\hbar$.

Learning Objectives

After this lesson you will be able to:

1. Separate variables in the 3D Schrödinger equation for a central potential $V(r)$, identifying the separation constant $\ell(\ell+1)$.
2. Derive the commutator algebra $[\hat L_x,\hat L_y] = i\hbar\hat L_z$ from the canonical commutators and recognize its classical shadow, the Poisson bracket $\{L_x,L_y\} = L_z$.
3. Construct the ladder operators $\hat L_\pm$ and derive the eigenvalues of $\hat L^2$ and $\hat L_z$ purely algebraically.
4. Explain why orbital angular momentum requires integer $\ell$, while the algebra alone also permits half-integer values.
5. Use the spherical harmonics $Y_\ell^m$ — eigenvalue equations, orthonormality, the $2\ell+1$ degeneracy — and critique the vector model via $\langle L_x^2 + L_y^2\rangle > 0$.
6. Reduce the polar equation to the associated Legendre equation, explain how finiteness at the poles quantizes $\ell$, and build $P_\ell^m$ from Rodrigues' formula.
7. Separate the cubic box in Cartesian coordinates, distinguish symmetry from accidental degeneracy, and count states to obtain the density of states $g(E) \propto \sqrt E$.

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Intuition

Classically a central force exerts no torque, so $\mathbf L = \mathbf r\times\mathbf p$ is conserved and orbits flatten into planes (P.1.4 Hamiltonian Mechanics). Quantum mechanics inherits the structure: rotational symmetry peels the angular dependence off the wavefunction once and for all — the angular factors are the same for hydrogen, a 3D oscillator, any $V(r)$. But there is a twist: classically $L_x, L_y, L_z$ are three numbers and you may know them all; quantum mechanically they are noncommuting operators — you may know the length $|\mathbf L|$ and one component, and the other two are then irreducibly indeterminate. The entire spectrum follows from the commutators alone, by the ladder logic of P.5.3 The Harmonic Oscillator.

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Theory

Separation of variables in a central potential

The 3D time-independent Schrödinger equation is $-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi$. For $V = V(r)$ use spherical coordinates, with the Laplacian (a standard chain-rule exercise, quoted):

$$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}.$$

Try $\psi(r,\theta,\varphi) = R(r)\,Y(\theta,\varphi)$; substitute and multiply by $-\frac{2mr^2}{\hbar^2}\frac{1}{RY}$:

$$\underbrace{\left\{\frac{1}{R}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) - \frac{2mr^2}{\hbar^2}\bigl[V(r)-E\bigr]\right\}}_{\text{depends only on } r} + \underbrace{\frac{1}{Y}\left\{\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial Y}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2 Y}{\partial\varphi^2}\right\}}_{\text{depends only on } \theta,\varphi} = 0 .$$

A function of $r$ plus a function of $(\theta,\varphi)$ vanishes identically only if each is constant; write the constant — with foresight — as $\ell(\ell+1)$. The radial piece keeps $V(r)$ and is deferred to P.6.2; the angular piece,

$$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial Y}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2 Y}{\partial\varphi^2} = -\ell(\ell+1)\,Y ,$$

is universal — and, as we now show, it is the eigenvalue problem of angular momentum.

The angular momentum operators and their algebra

Quantizing $\mathbf L = \mathbf r\times\mathbf p$ gives $\hat{\mathbf L} = \hat{\mathbf r}\times\hat{\mathbf p}$, with components $\hat L_x = \hat y\hat p_z - \hat z\hat p_y$, $\hat L_y = \hat z\hat p_x - \hat x\hat p_z$, $\hat L_z = \hat x\hat p_y - \hat y\hat p_x$ (no ordering ambiguity: each product pairs a coordinate with a different momentum component). Using $[\hat x_i,\hat p_j] = i\hbar\delta_{ij}$, expand $[\hat L_x,\hat L_y]$ into four brackets; $[\hat y\hat p_z,\hat x\hat p_z]$ and $[\hat z\hat p_y,\hat z\hat p_x]$ contain only mutually commuting factors and vanish, leaving

$$[\hat L_x,\hat L_y] = [\hat y\hat p_z,\hat z\hat p_x] + [\hat z\hat p_y,\hat x\hat p_z] = \hat y\,[\hat p_z,\hat z]\,\hat p_x + \hat x\,[\hat z,\hat p_z]\,\hat p_y = i\hbar\,(\hat x\hat p_y - \hat y\hat p_x) = i\hbar\,\hat L_z ,$$

and cyclically $[\hat L_y,\hat L_z] = i\hbar\hat L_x$, $[\hat L_z,\hat L_x] = i\hbar\hat L_y$. This mirrors the classical Poisson bracket $\{L_x,L_y\} = L_z$ computed in P.1.4 Hamiltonian Mechanics via Dirac's rule $\{\cdot,\cdot\}\to[\cdot,\cdot]/i\hbar$ — the algebra of rotations was there all along.

No two components commute, so no common eigenbasis exists: $L_x$ and $L_z$ cannot both be sharp (the uncertainty relation behind this is developed in 1.3.2 Expectation & Uncertainty). But $\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2$ commutes with every component: using $[\hat A^2,\hat B] = \hat A[\hat A,\hat B] + [\hat A,\hat B]\hat A$,

$$[\hat L^2,\hat L_z] = -i\hbar(\hat L_x\hat L_y + \hat L_y\hat L_x) + i\hbar(\hat L_y\hat L_x + \hat L_x\hat L_y) + 0 = 0 ,$$

so $\hat L^2$ and $\hat L_z$ do share simultaneous eigenfunctions — the maximal sharp angular data. In spherical coordinates (a rotation about $z$ changes only $\varphi$, so $\partial_\varphi = x\partial_y - y\partial_x$):

$$\hat L_z = -i\hbar\frac{\partial}{\partial\varphi}, \qquad \hat L^2 = -\hbar^2\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\,\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right],$$

so the angular equation above is $\hat L^2 Y = \hbar^2\ell(\ell+1)\,Y$.

Ladder operators: the spectrum from pure algebra

In perfect analogy with the $\hat a_\pm$ of P.5.3, define $\hat L_\pm \equiv \hat L_x \pm i\hat L_y$. Then $[\hat L^2,\hat L_\pm] = 0$, and

$$[\hat L_z,\hat L_\pm] = [\hat L_z,\hat L_x] \pm i[\hat L_z,\hat L_y] = i\hbar\hat L_y \pm i(-i\hbar\hat L_x) = \pm\hbar\,\hat L_\pm .$$

Let $\lvert\lambda,\mu\rangle$ satisfy $\hat L^2\lvert\lambda,\mu\rangle = \lambda\lvert\lambda,\mu\rangle$ and $\hat L_z\lvert\lambda,\mu\rangle = \mu\lvert\lambda,\mu\rangle$. Then $\hat L_z(\hat L_\pm\lvert\lambda,\mu\rangle) = (\hat L_\pm\hat L_z \pm \hbar\hat L_\pm)\lvert\lambda,\mu\rangle = (\mu\pm\hbar)\,\hat L_\pm\lvert\lambda,\mu\rangle$ while $\hat L^2(\hat L_\pm\lvert\lambda,\mu\rangle) = \lambda\,\hat L_\pm\lvert\lambda,\mu\rangle$: a ladder in $L_z$, steps of $\hbar$, fixed total length. It is bounded, since $\lambda - \mu^2 = \langle\hat L_x^2 + \hat L_y^2\rangle = \lVert\hat L_x\lvert\lambda,\mu\rangle\rVert^2 + \lVert\hat L_y\lvert\lambda,\mu\rangle\rVert^2 \ge 0$. So there is a top rung, $\hat L_+\lvert\lambda,\mu_{\max}\rangle = 0$ with $\mu_{\max} \equiv \hbar\ell$. Expanding $\hat L_\mp\hat L_\pm = \hat L_x^2 + \hat L_y^2 \pm i[\hat L_x,\hat L_y] = \hat L^2 - \hat L_z^2 \mp \hbar\hat L_z$ and applying $\hat L_-\hat L_+$ to the top rung: $0 = \lambda - \hbar^2\ell^2 - \hbar^2\ell$, i.e. $\lambda = \hbar^2\ell(\ell+1)$. Likewise a bottom rung $\hbar\bar\ell$ gives $\lambda = \hbar^2\bar\ell(\bar\ell - 1)$, so $\bar\ell = -\ell$ (the root $\ell+1$ would put the bottom above the top). Descending from $\hbar\ell$ to $-\hbar\ell$ in steps of $\hbar$ requires $2\ell$ to be a non-negative integer:

$$\hat L^2 \to \hbar^2\ell(\ell+1), \qquad \hat L_z \to \hbar m, \qquad m = -\ell, \dots, \ell \ \ (\text{integer steps}), \qquad \ell = 0, \tfrac12, 1, \tfrac32, \dots$$

The commutators alone — no wavefunctions, no boundary conditions — fixed the entire spectrum, and they permit both integer and half-integer $\ell$.

The differential route: why orbital $\ell$ is an integer

For angular momentum realized as $-i\hbar\,\mathbf r\times\nabla$ on wavefunctions, more is true. Solving $\hat L_z\Phi = \hbar m\Phi$ gives $\Phi = e^{im\varphi}$; but $\varphi$ and $\varphi + 2\pi$ label the same point of space, and single-valuedness $e^{im(\varphi + 2\pi)} = e^{im\varphi}$ forces $m \in \mathbb Z$ — hence integer $\ell$ for orbital angular momentum. The half-integer solutions are not garbage: they describe angular momentum that is not motion through space — no $\varphi$, no constraint. That loophole stays open until the Stern–Gerlach experiment forces us through it in P.6.3 Magnetic Moments, Stern–Gerlach & Spin.

The polar equation and Legendre functions

The differential route can be finished without any algebra. Separate the angular equation once more, $Y(\theta,\varphi) = \Theta(\theta)\Phi(\varphi)$, and multiply it by $\sin^2\theta/\Theta\Phi$:

$$\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\!\left(\sin\theta\,\frac{d\Theta}{d\theta}\right) + \ell(\ell+1)\sin^2\theta = -\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2} = m^2 .$$

The right side gives $\Phi = e^{im\varphi}$ with integer $m$, as above. On the left, substitute $x = \cos\theta$, so that $\sin\theta\,d/d\theta = -(1 - x^2)\,d/dx$. Dividing by $1 - x^2$ leaves the associated Legendre equation

$$(1 - x^2)\,\Theta'' - 2x\,\Theta' + \left[\ell(\ell+1) - \frac{m^2}{1 - x^2}\right]\Theta = 0, \qquad -1 \le x \le 1 .$$

Its trouble spots are the poles, $x = \pm1$. Just like the Hermite series of P.5.3, a power-series solution generically blows up there, and it stays finite at both poles only if the constant is $\ell(\ell+1)$ with $\ell$ a non-negative integer and $\lvert m\rvert \le \ell$. That is the spectrum the ladder algebra delivered, recovered here from a boundary condition, like every other quantization in this term. For $m = 0$ the finite solutions are the Legendre polynomials, generated by Rodrigues' formula

$$P_\ell(x) = \frac{1}{2^\ell\,\ell!}\,\frac{d^\ell}{dx^\ell}\,(x^2 - 1)^\ell : \qquad P_0 = 1,\quad P_1 = x,\quad P_2 = \tfrac12(3x^2 - 1),\quad P_3 = \tfrac12(5x^3 - 3x).$$

For $m \ne 0$, differentiate further: $P_\ell^m(x) = (1 - x^2)^{\lvert m\rvert/2}\,\dfrac{d^{\lvert m\rvert}}{dx^{\lvert m\rvert}}P_\ell(x)$, giving for example $P_1^1 = \sin\theta$, $P_2^1 = 3\sin\theta\cos\theta$ and $P_2^2 = 3\sin^2\theta$. Since $P_\ell$ has degree $\ell$, $P_\ell^m$ vanishes identically for $\lvert m\rvert > \ell$: the bound $\lvert m\rvert \le \ell$ once more. (Some texts put a factor $(-1)^m$, the Condon–Shortley phase, into $P_\ell^m$; here it sits in $Y_\ell^m$ below. The spherical harmonics come out the same.)

Spherical harmonics

The normalized simultaneous eigenfunctions of $\hat L^2$ and $\hat L_z$ are the spherical harmonics, built from the associated Legendre functions $P_\ell^m$:

$$Y_\ell^m(\theta,\varphi) = (-1)^m\sqrt{\frac{2\ell+1}{4\pi}\frac{(\ell-m)!}{(\ell+m)!}}\, P_\ell^m(\cos\theta)\,e^{im\varphi} \quad (m\ge0;\; Y_\ell^{-m} = (-1)^mY_\ell^{m*}).$$

$\ell$	$m$	$Y_\ell^m(\theta,\varphi)$
0	0	$\sqrt{1/4\pi}$
1	0	$\sqrt{3/4\pi}\,\cos\theta$
1	$\pm1$	$\mp\sqrt{3/8\pi}\,\sin\theta\,e^{\pm i\varphi}$
2	0	$\sqrt{5/16\pi}\,(3\cos^2\theta - 1)$
2	$\pm1$	$\mp\sqrt{15/8\pi}\,\sin\theta\cos\theta\,e^{\pm i\varphi}$
2	$\pm2$	$\sqrt{15/32\pi}\,\sin^2\theta\,e^{\pm2i\varphi}$

They are orthonormal on the sphere — $\int_0^{2\pi}\!\int_0^{\pi} Y_{\ell'}^{m'*}\,Y_\ell^m\,\sin\theta\,d\theta\,d\varphi = \delta_{\ell\ell'}\delta_{mm'}$ — and complete. For each $\ell$ there are $2\ell+1$ values of $m$ with the same $\hat L^2$ eigenvalue: a degeneracy every central potential inherits, since energy cannot depend on the orientation of $\mathbf L$ when the potential has no preferred direction.

The vector model and its limits

A common mnemonic draws $\mathbf L$ as an arrow of length $\hbar\sqrt{\ell(\ell+1)}$ on a cone about $z$, with fixed projection $\hbar m$ and smeared azimuth. The cone's opening is real physics:

$$\langle L_x^2 + L_y^2\rangle = \langle\hat L^2 - \hat L_z^2\rangle = \hbar^2\bigl[\ell(\ell+1) - m^2\bigr] > 0 \quad\text{for } \ell > 0 ,$$

even at $m = \ell$ (where it equals $\hbar^2\ell$): $\mathbf L$ never points exactly along $z$, tilting at best to $\cos\theta_{\min} = \ell/\sqrt{\ell(\ell+1)} < 1$. But don't take the cone literally — nothing secretly precesses at a hidden azimuth.

Caution. $|\mathbf L| = \hbar\sqrt{\ell(\ell+1)} > \hbar\ell \ge \hbar m$. The Bohr-era phrase "the angular momentum is $\ell\hbar$" is shorthand and strictly false — the length exceeds the largest projection, which is exactly why $\mathbf L$ can never align with an axis. And a state with definite $L_z$ has indeterminate $L_x$ and $L_y$: not unknown values — no values.

Another separation: the cubic box

Spherical coordinates suit a spherical potential; a box wants Cartesian ones. Confine a particle to a cube of side $L$ ($V = 0$ inside, $V = \infty$ outside). The product $\psi = X(x)\,Y(y)\,Z(z)$ splits $-\frac{\hbar^2}{2m}\nabla^2\psi = E\psi$ into three copies of the 1D infinite well of P.5.2, with the energies adding:

$$\psi_{n_xn_yn_z} = \Big(\frac2L\Big)^{3/2}\sin\frac{n_x\pi x}{L}\,\sin\frac{n_y\pi y}{L}\,\sin\frac{n_z\pi z}{L}, \qquad E = \frac{\pi^2\hbar^2}{2mL^2}\big(n_x^2 + n_y^2 + n_z^2\big),$$

with each $n_i = 1, 2, 3, \dots$ The ground state $(1,1,1)$ has $E_{111} = 3\pi^2\hbar^2/2mL^2$: one zero-point share per direction. In units of $\varepsilon_0 = \pi^2\hbar^2/2mL^2$ the first levels are:

$n_x^2 + n_y^2 + n_z^2$	States $(n_x, n_y, n_z)$	Degeneracy
3	$(1,1,1)$	1
6	$(2,1,1)$ and permutations	3
9	$(2,2,1)$ and permutations	3
11	$(3,1,1)$ and permutations	3
12	$(2,2,2)$	1
14	$(3,2,1)$ and permutations	6

Most of these degeneracies come from the cube's symmetry: relabeling the axes permutes the quantum numbers without changing the energy. But not all of them. The level $n_x^2 + n_y^2 + n_z^2 = 27$ contains both $(3,3,3)$ and the three permutations of $(5,1,1)$, a four-fold level that no symmetry of the cube relates. Such accidental degeneracy comes from the arithmetic of the spectrum rather than from the geometry, a cousin of hydrogen's $\ell$-degeneracy in P.6.2. (The cube has only its finite set of rotations and reflections, so its symmetry degeneracies stay small, at most 6; full rotational symmetry is what gives every central potential its $2\ell + 1$.)

Counting states. At high energy, individual levels matter less than how many states lie below a given $E$. Each state is a lattice point $(n_x, n_y, n_z)$ of positive integers, and those with energy below $E$ fill one octant of a sphere of radius $R = \sqrt{2mL^2E}/\pi\hbar$. For large $R$ the count is the octant's volume:

$$N(E) \approx \frac18\cdot\frac{4\pi}{3}R^3 = \frac{V}{6\pi^2}\left(\frac{2mE}{\hbar^2}\right)^{3/2}, \qquad g(E) = \frac{dN}{dE} = \frac{V}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt E ,$$

with $V = L^3$. This density of states depends on the volume only, not the shape, once the wavelengths are much smaller than the box. It is the same octant count that gave the Rayleigh–Jeans mode density in P.2.1; only the dispersion relation differs. Photons have $E \propto k$, so $N \propto \nu^3$; massive particles have $E \propto k^2$, so $N \propto E^{3/2}$. Doubled for the two spin states of an electron, $g(E)$ is the starting point of the free-electron theory of metals.

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Worked Examples

Example 1 — Sizing up $\ell = 2$

$|\mathbf L| = \hbar\sqrt{2\cdot3} = \sqrt6\,\hbar \approx 2.449\,\hbar$, with five projections $L_z = m\hbar$, $m\in\{-2,\dots,2\}$. Even at $m = 2$ the vector tilts off-axis by $\cos\theta = 2/\sqrt6 = 0.8165$, i.e. $\theta \approx 35.3^\circ$, with transverse spread $\langle L_x^2 + L_y^2\rangle = \hbar^2(6-4) = 2\hbar^2$. The largest projection $2\hbar$ never reaches the length $2.449\,\hbar$.

Example 2 — $Y_1^0$ through the machinery

$Y_1^0 = \sqrt{3/4\pi}\cos\theta$ has no $\varphi$-dependence, so $\hat L_zY_1^0 = 0$: $m = 0$. ✓ For $\hat L^2$ (the $\varphi$ term drops):

$$\frac{1}{\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\,\frac{d\cos\theta}{d\theta}\right) = \frac{1}{\sin\theta}\frac{d}{d\theta}\bigl(-\sin^2\theta\bigr) = -2\cos\theta ,$$

so $\hat L^2Y_1^0 = 2\hbar^2\,Y_1^0 = \hbar^2\ell(\ell+1)Y_1^0$ with $\ell = 1$. ✓ Lowering with $\hat L_- = -\hbar e^{-i\varphi}(\partial_\theta - i\cot\theta\,\partial_\varphi)$: $\hat L_-Y_1^0 = \hbar\sqrt{3/4\pi}\,\sin\theta\,e^{-i\varphi} = \hbar\sqrt2\,Y_1^{-1}$, with coefficient exactly $\hbar\sqrt{\ell(\ell+1) - m(m-1)} = \hbar\sqrt2$ — the ladder in action.

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Hands-on (Python)

Visualize $|Y_\ell^m|^2$ for $\ell = 0,1,2$ as polar surface plots (radius = magnitude):

import numpy as np
import matplotlib.pyplot as plt
from scipy.special import sph_harm

# NOTE scipy's argument order/naming: sph_harm(m, l, azimuthal, polar).
polar = np.linspace(0, np.pi, 101)          # theta in physics texts
azim  = np.linspace(0, 2*np.pi, 201)        # varphi in physics texts
TH, PH = np.meshgrid(polar, azim)

fig = plt.figure(figsize=(13, 8))
for l in range(3):
    for m in range(-l, l + 1):
        rad = np.abs(sph_harm(m, l, PH, TH))**2          # radius = |Y_l^m|^2
        X, Y, Z = (rad*np.sin(TH)*np.cos(PH), rad*np.sin(TH)*np.sin(PH), rad*np.cos(TH))
        ax = fig.add_subplot(3, 5, 5*l + m + 3, projection='3d')  # center each row
        ax.plot_surface(X, Y, Z, cmap='viridis', rstride=2, cstride=2)
        ax.set_title(f"$\\ell={l},\\ m={m}$"); ax.set_axis_off()
plt.tight_layout(); plt.show()
# Expect: l=0 a sphere; l=1 dumbbells (m=0 along z, |m|=1 donuts); l=2 the
# donut/dumbbell family. |Y|^2 is always symmetric about the z-axis.

Verify orthonormality $\int Y_{\ell'}^{m'*}Y_\ell^m\,d\Omega = \delta_{\ell\ell'}\delta_{mm'}$ by 2D quadrature:

def overlap(l1, m1, l2, m2, n_th=400, n_ph=800):
    th, ph = np.linspace(0, np.pi, n_th), np.linspace(0, 2*np.pi, n_ph)
    TH, PH = np.meshgrid(th, ph, indexing='ij')
    f = np.conj(sph_harm(m1, l1, PH, TH))*sph_harm(m2, l2, PH, TH)*np.sin(TH)
    return np.trapezoid(np.trapezoid(f, ph, axis=1), th)   # np.trapz on older NumPy

labels = [(l, m) for l in range(3) for m in range(-l, l + 1)]   # 9 states
G = np.array([[overlap(*a, *b) for b in labels] for a in labels])
print(np.max(np.abs(G - np.eye(9))))   # ~1e-6: Gram matrix = identity

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Exercises

E1 (easy). For $\ell = 3$: how many $m$ values? Compute $|\mathbf L|$ and the smallest possible angle between $\mathbf L$ and the $z$-axis.

Solution

$2\ell+1 = 7$ values, $m = -3,\dots,3$; $|\mathbf L| = \hbar\sqrt{12} = 2\sqrt3\,\hbar \approx 3.464\,\hbar$. At $m = 3$: $\cos\theta = 3/\sqrt{12} = \sqrt3/2$, so $\theta = 30^\circ$ exactly.

E2 (easy). Derive $[\hat L_z,\hat L_x] = i\hbar\hat L_y$ (a) directly from the canonical commutators, (b) by cyclic relabeling.

Solution

(a) $[\hat x\hat p_y - \hat y\hat p_x,\ \hat y\hat p_z - \hat z\hat p_y]$ has nonvanishing pieces $[\hat x\hat p_y,\hat y\hat p_z] = \hat x[\hat p_y,\hat y]\hat p_z = -i\hbar\hat x\hat p_z$ and $[\hat y\hat p_x,\hat z\hat p_y] = \hat z\hat p_x[\hat y,\hat p_y] = i\hbar\hat z\hat p_x$; sum $= i\hbar(\hat z\hat p_x - \hat x\hat p_z) = i\hbar\hat L_y$. (b) $x\to y\to z\to x$ preserves $[\hat x_i,\hat p_j] = i\hbar\delta_{ij}$ and maps $\hat L_x\to\hat L_y\to\hat L_z\to\hat L_x$, carrying $[\hat L_x,\hat L_y] = i\hbar\hat L_z$ into the other two identities.

E3 (medium). Prove $\hat L_\mp\hat L_\pm = \hat L^2 - \hat L_z^2 \mp \hbar\hat L_z$ and deduce $\hat L_\pm\lvert \ell,m\rangle = \hbar\sqrt{\ell(\ell+1) - m(m\pm1)}\,\lvert \ell,m\pm1\rangle$.

Solution

$(\hat L_x \mp i\hat L_y)(\hat L_x \pm i\hat L_y) = \hat L_x^2 + \hat L_y^2 \pm i[\hat L_x,\hat L_y] = \hat L^2 - \hat L_z^2 \mp \hbar\hat L_z$. Then $\lVert\hat L_\pm\lvert \ell,m\rangle\rVert^2 = \langle \ell,m\rvert\hat L_\mp\hat L_\pm\lvert \ell,m\rangle = \hbar^2[\ell(\ell+1) - m(m\pm1)]$; the standard positive phase gives the stated coefficient, which vanishes at $m = \pm\ell$: the ladder terminates itself.

E4 (medium). By explicit integration, verify that $Y_1^1$ is normalized and orthogonal to $Y_1^0$.

Solution

$\int|Y_1^1|^2d\Omega = \frac{3}{8\pi}\cdot2\pi\int_0^\pi\sin^3\theta\,d\theta = \frac{3}{8\pi}\cdot2\pi\cdot\frac43 = 1$ ✓. Orthogonality: $\int Y_1^{0*}Y_1^1\,d\Omega \propto \int_0^{2\pi}e^{i\varphi}d\varphi = 0$ — the azimuthal integral kills any pair with $m\ne m'$. ✓

E5 (hard). In $\lvert \ell,m\rangle$, show $\langle L_x\rangle = \langle L_y\rangle = 0$ and $\langle L_x^2\rangle = \langle L_y^2\rangle = \tfrac{\hbar^2}{2}[\ell(\ell+1) - m^2]$, then verify $\Delta L_x\,\Delta L_y \ge \tfrac\hbar2|\langle L_z\rangle|$.

Solution

$\hat L_x = \tfrac12(\hat L_+ + \hat L_-)$ maps $\lvert \ell,m\rangle$ to orthogonal states, so $\langle L_x\rangle = 0$ (same for $L_y$). In $\hat L_x^2 = \tfrac14(\hat L_+^2 + \hat L_-^2 + \hat L_+\hat L_- + \hat L_-\hat L_+)$ the $\hat L_\pm^2$ terms shift $m$ by $\pm2$ and drop; E3 gives $\langle\hat L_+\hat L_- + \hat L_-\hat L_+\rangle = 2\hbar^2[\ell(\ell+1) - m^2]$, so $\langle L_x^2\rangle = \langle L_y^2\rangle = \tfrac{\hbar^2}{2}[\ell(\ell+1) - m^2]$. Then $\Delta L_x\Delta L_y \ge \tfrac{\hbar^2}{2}|m| \iff \ell(\ell+1) \ge |m|(|m|+1)$, true since $\ell \ge |m|$ — with equality at $m = \pm\ell$: the end rungs are minimum-uncertainty states.

E6 (medium). (a) An electron is confined to a cube of side 1.0 nm. Find the three lowest energies in eV and their degeneracies. (b) Show that the level $n_x^2 + n_y^2 + n_z^2 = 38$ is nine-fold degenerate, and say how much of that the cube's symmetry explains. (c) Estimate the number of electron states (including spin) with energy below 1 eV in a box of volume $1\ \mathrm{cm^3}$.

Solution

(a) $\varepsilon_0 = \pi^2\hbar^2/2mL^2 = \frac{9.87\times(1.055\times10^{-34})^2}{2\times9.11\times10^{-31}\times10^{-18}} = 6.03\times10^{-20}$ J $= 0.376$ eV, the 1D well's ground energy from P.5.2. Levels: $3\varepsilon_0 = 1.13$ eV (1 state), $6\varepsilon_0 = 2.26$ eV (3), $9\varepsilon_0 = 3.39$ eV (3). (b) $38 = 6^2 + 1^2 + 1^2 = 5^2 + 3^2 + 2^2$. The first gives the 3 permutations of $(6,1,1)$, the second the $3! = 6$ permutations of $(5,3,2)$, for 9 in all. Symmetry explains each family separately (permutations of the axes); that the two families share one energy is accidental. (c) $2mE/\hbar^2 = 2\times9.11\times10^{-31}\times1.60\times10^{-19}/(1.055\times10^{-34})^2 = 2.62\times10^{19}\ \mathrm{m^{-2}}$, so $(2mE/\hbar^2)^{3/2} = 1.34\times10^{29}\ \mathrm{m^{-3}}$ and $N = \frac{10^{-6}}{6\pi^2}\times1.34\times10^{29} = 2.3\times10^{21}$, or $4.5\times10^{21}$ with spin. A metal has about $10^{23}$ conduction electrons per cm³; since $N \propto E^{3/2}$, filling that many states needs energies up to several eV, the scale of real Fermi energies.

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Checkpoint

1. In $\psi = R\,Y$, what is the separation constant, and what operator equation does $Y$ satisfy?
2. State $[\hat L_x,\hat L_y]$ and its classical counterpart. What rule connects them?
3. Why do $\hat L^2$ and $\hat L_z$ share eigenfunctions while $\hat L_x$ and $\hat L_z$ do not?
4. Which step of which argument restricts orbital $\ell$ to integers — and why doesn't it bind spin?
5. Why can $\mathbf L$ never point exactly along the $z$-axis for $\ell > 0$?
6. In the differential route, what condition on the polar equation quantizes $\ell$, and why does it also force $\lvert m\rvert \le \ell$?
7. What is the difference between symmetry degeneracy and accidental degeneracy in the cubic box? How does the number of states below $E$ grow with $E$?

Answers

1. $\ell(\ell+1)$; the angular equation is $\hat L^2Y = \hbar^2\ell(\ell+1)Y$.
2. $[\hat L_x,\hat L_y] = i\hbar\hat L_z$, mirroring $\{L_x,L_y\} = L_z$ via Dirac's rule $\{\cdot,\cdot\}\to[\cdot,\cdot]/i\hbar$.
3. $[\hat L^2,\hat L_z] = 0$ guarantees a common eigenbasis; $[\hat L_x,\hat L_z] = -i\hbar\hat L_y \ne 0$ forbids one.
4. Single-valuedness of $e^{im\varphi}$ under $\varphi\to\varphi+2\pi$ in the differential realization forces $m\in\mathbb Z$. Spin is not a function of spatial angles, so the constraint never applies — the algebra's half-integers survive.
5. $|\mathbf L| = \hbar\sqrt{\ell(\ell+1)}$ strictly exceeds the largest projection $\hbar\ell$; equivalently $\langle L_x^2 + L_y^2\rangle = \hbar^2[\ell(\ell+1) - m^2] > 0$.
6. Finiteness at both poles ($x = \cos\theta = \pm1$): the series solutions of the associated Legendre equation blow up there unless the constant is $\ell(\ell+1)$ with integer $\ell \ge 0$. The finite solutions $P_\ell^m$ are $|m|$ derivatives of the degree-$\ell$ polynomial $P_\ell$, which vanish for $|m| > \ell$.
7. Symmetry degeneracy relates states by a symmetry of the box (permuting axes); accidental degeneracy joins states no symmetry relates, such as $(3,3,3)$ and $(5,1,1)$ at 27. $N(E) \propto VE^{3/2}$, so $g(E) \propto \sqrt E$.

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Further Reading

• [Gri] Griffiths & Schroeter, §4.1 — separation of variables and the angular equation; this lesson's conventions.
• [Gri] Griffiths & Schroeter, §4.3 — the algebraic theory of angular momentum; our ladder derivation follows it closely.
• [Sha] Shankar, Ch. 12 — angular momentum from rotational symmetry; the algebra as the group theory of rotations.
• [ER] Eisberg & Resnick, Ch. 7 — quantum numbers with the experimental context of one-electron atoms.

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