# 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.

**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.

--- ## 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 $g(E) \propto \sqrt E$.

--- ## 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. --- ← Prev: [The Harmonic Oscillator](../c05-bound-states/03-harmonic-oscillator.md) · Up: [Pre-Term](../README.md) · Next: [The Radial Equation & the Hydrogen Atom](02-radial-equation-hydrogen.md) →