# 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** $g(E) \propto \sqrt E$.
---
## 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](../c01-theoretical-mechanics/04-hamiltonian-mechanics.md)).
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](../c05-bound-states/03-harmonic-oscillator.md).
---
## 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](02-radial-equation-hydrogen.md); 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](../c01-theoretical-mechanics/04-hamiltonian-mechanics.md) 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](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md)).
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](../c05-bound-states/03-harmonic-oscillator.md),
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](03-magnetic-fields-stern-gerlach-spin.md).
### 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](../c05-bound-states/03-harmonic-oscillator.md), 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](../c05-bound-states/02-step-well-barrier.md), 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](02-radial-equation-hydrogen.md). (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](../c02-radiation-old-quantum/01-blackbody-radiation.md); 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.
---
## 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.
---
## Hands-on (Python)
Visualize $|Y_\ell^m|^2$ for $\ell = 0,1,2$ as polar surface plots (radius = magnitude):
```python
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:
```python
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
```
---
## 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.
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.
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.
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 $m\ne m'$. ✓
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.
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.
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 $g(E) \propto \sqrt E$.