Why Rotation Symmetry Quantizes Angular Momentum

Status: an interactive companion to an independent pedagogical writeup. The paper organizes standard results; it does not claim a new spectrum or a new representation-theoretic theorem.

For $j=\tfrac12$, the formula

$$J^2|j,m\rangle=\hbar^2j(j+1)|j,m\rangle$$

returns $\tfrac34\hbar^2$. The largest projection you can measure along any axis is $\hbar/2$, so the naive guess for $J^2$ would be $\tfrac14\hbar^2$. The formula is three times larger. Most first courses hand the result over to be memorized, and if that is where you met it, the $j(j+1)$ probably looked like something read off a spectroscopic table.

It follows instead from three structural facts: spatial rotations do not commute, observable generators are Hermitian, and Hilbert-space norms cannot be negative. The derivation is short and uses only the commutators.

Hydrogen adds a second layer on top of that one. Rotation symmetry determines the angular functions and their allowed quantum numbers, while the Coulomb potential and radial boundary conditions quantize the energy. The two layers are worth keeping apart, because the $n^2$ degeneracy of hydrogen turns out to be two separate facts and only one of them is a statement about symmetry.

Rotations carry an algebra

An infinitesimal rotation by a small vector $\boldsymbol\theta$ acts on a quantum state through

$$U(\boldsymbol\theta) =I-\frac{i}{\hbar}\boldsymbol\theta\cdot\mathbf J +O(|\boldsymbol\theta|^2).$$

The operators $J_x,J_y,J_z$ are the generators of rotations. Turn a book 90° about $x$ and then 90° about $y$, then start over and reverse the order: it lands in a different orientation. The same failure survives all the way down to infinitesimal rotations, where the mismatch is second order in the angle and is itself a small rotation about $z$. Algebraically,

$$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k.$$

This is the Lie algebra $\mathfrak{so}(3)$, equivalently $\mathfrak{su}(2)$ at the infinitesimal level. First courses usually gloss it as “the components cannot be measured together,” which is true and stops short of the interesting part: each component generates a motion on state space, and the commutator records the curvature of composing those motions.

The natural rotational scalar is

$$J^2=J_x^2+J_y^2+J_z^2.$$

Using $[AB,C]=A[B,C]+[A,C]B$, one finds $[J^2,J_i]=0$. Thus $J^2$ is the Casimir operator: within an irreducible representation it has one constant value, while a chosen component such as $J_z$ distinguishes states inside the multiplet. We may therefore seek simultaneous eigenstates

$$J^2|\lambda,m\rangle=\lambda|\lambda,m\rangle, \qquad J_z|\lambda,m\rangle=\hbar m|\lambda,m\rangle.$$

Ladder operators are eigen-operators

We want an operator $A$ that changes the $J_z$ eigenvalue by a fixed amount:

$$[J_z,A]=\lambda A.$$

Trying $A=\alpha J_x+\beta J_y$ and inserting the rotation commutators gives $\lambda=\pm\hbar$ and $\beta=\pm i\alpha$. Up to normalization, the only solutions are

$$J_\pm=J_x\pm iJ_y, \qquad [J_z,J_\pm]=\pm\hbar J_\pm.$$

The only input was $[J_z,A]=\lambda A$. Solving it hands back $J_\pm$, the eigenvectors of the adjoint action $A\mapsto[J_z,A]$. Applied to a state, they shift $m$ by one and leave $J^2$ alone.

Two identities make the spectrum visible:

$$J_-J_+=J^2-J_z^2-\hbar J_z, \qquad J_+J_-=J^2-J_z^2+\hbar J_z.$$

Take the norm of a raised state:

$$\|J_+|\lambda,m\rangle\|^2 =\lambda-\hbar^2m(m+1)\ge0.$$

Worth pausing on that inequality before reading on. As we climb, $\lambda$ is fixed and $m$ grows by one at every step. What stops it?

If the ladder continued upward forever while $\lambda$ stayed fixed, the right-hand side would eventually become negative. Therefore the ladder must end at some $m=j$, where $J_+|j,j\rangle=0$. At that endpoint,

$$0=\hbar^2j(j+1)-\hbar^2j(j+1),$$

which fixes the Casimir eigenvalue to $\hbar^2j(j+1)$. Repeating the argument at the bottom gives $m=-j$. Because the spacing is one, $2j$ must be a nonnegative integer:

$$j=0,\frac12,1,\frac32,\ldots, \qquad m=-j,-j+1,\ldots,j.$$

The normalized action is

$$J_\pm|j,m\rangle =\hbar\sqrt{j(j+1)-m(m\pm1)}\,|j,m\pm1\rangle.$$

Take $j=\tfrac32$ as a worked case. The ladder has four rungs, and at the top rung $m=\tfrac32$ both $j(j+1)$ and $m(m+1)$ equal $\tfrac{15}{4}$, so the coefficient under the square root is exactly zero.

Representation explorer

One algebra, two physical realizations

representation j =

1.5

selected m =

-0.5

m = -1.5m = -0.5m = 0.5m = 1.5

J squared / hbar squared3.75

raising coefficient / hbar2.000

lowering coefficient / hbar1.732

multiplet dimension4

hydrogen n =

2

orbital ell =

1

energy-3.400 eV

radial nodes0

allowed m values3

orbital states at this n4

The upper panel uses only the rotation algebra. The lower panel adds the Coulomb Hamiltonian, boundary conditions, and normalizability. Energy depends only on n in the nonrelativistic Coulomb problem; fine structure, Lamb shifts, and spin are not included.

Use the upper panel to move through a multiplet. The raising coefficient is largest away from the top and vanishes exactly at $m=j$, where the state that $J_+$ would produce has zero norm.

Why orbital angular momentum excludes half-integers

The algebra permits both integer and half-integer $j$, and so far we have seen nothing that prefers either. Which ones occur is decided by the concrete realization.

For a scalar wavefunction on ordinary space, rotations act by moving the argument:

$$(U(R)\psi)(\mathbf r)=\psi(R^{-1}\mathbf r).$$

Expanding an infinitesimal rotation gives the orbital generator

$$\mathbf L=-i\hbar(\mathbf r\times\boldsymbol\nabla).$$

In spherical coordinates,

$$L_z=-i\hbar\frac{\partial}{\partial\phi},$$

so an $L_z$ eigenfunction has azimuthal dependence $e^{im\phi}$. A scalar wavefunction must return to the same value under $\phi\mapsto\phi+2\pi$; therefore $e^{2\pi im}=1$ and $m\in\mathbb Z$. Half-integers fail at the first turn: $m=\tfrac12$ gives $e^{i\pi}=-1$. The orbital multiplet is left with $\ell=0,1,2,\ldots$.

The remaining angular equation is

$$L^2Y_{\ell m}=\hbar^2\ell(\ell+1)Y_{\ell m},$$

whose regular, single-valued solutions on the sphere are the spherical harmonics $Y_{\ell m}(\theta,\phi)$. These are the functions behind the $s$, $p$ and $d$ orbital pictures in a chemistry textbook, which makes them easy to file away as shapes. They are matrix elements of rotation representations realized as functions on $S^2$, and they form the natural Fourier basis for angular data.

Spin lives in an internal representation

A spinor takes values in an internal vector space, which is what lets it carry the half-integer $j$ that scalar functions on space just ruled out. Rotations can therefore act both on position and on components:

$$\Psi(\mathbf r)\longmapsto D(R)\Psi(R^{-1}\mathbf r).$$

The orbital part is generated by $\mathbf L$; the internal matrix representation is generated by $\mathbf S$. On the tensor product,

$$\mathbf J=\mathbf L\otimes I+I\otimes\mathbf S.$$

For spin one-half, the smallest nontrivial representation is two-dimensional. The traceless Hermitian $2\times2$ matrices are spanned by the Pauli matrices, and

$$S_i=\frac{\hbar}{2}\sigma_i$$

satisfies the angular-momentum algebra. A $2\pi$ rotation sends a spinor to minus itself; you need $4\pi$ to get back. If that sounds like bookkeeping, look at where the sign survives: probabilities are blind to it, interference is not. $SO(3)$ and $SU(2)$ share a Lie algebra and differ globally, and this is the place that difference becomes measurable.

Hydrogen: symmetry separates the problem

The relative electron-proton motion has Hamiltonian

$$H=-\frac{\hbar^2}{2\mu}\nabla^2-\frac{e^2}{4\pi\epsilon_0r},$$

with reduced mass $\mu=m_em_p/(m_e+m_p)$. Because the potential depends only on $r$, $H$ commutes with $L^2$ and $L_z$. Write

$$\psi(r,\theta,\phi)=R_{n\ell}(r)Y_{\ell m}(\theta,\phi).$$

The angular equation has already been solved by representation theory. With $u(r)=rR(r)$, the remaining radial equation becomes

$$-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} +\left[-\frac{e^2}{4\pi\epsilon_0r} +\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\right]u=Eu.$$

The second term in brackets is the centrifugal barrier, and at this point it reasonably looks like a force inserted to make the radial problem come out. It is the angular kinetic energy showing up as an effective radial potential, and the $\ell(\ell+1)$ sitting in it is the same $\ell(\ell+1)$ the ladder argument produced. Increasing $\ell$ suppresses probability near the origin, which the lower interactive plot makes visible.

For a bound state, $u(r)$ must be regular at the origin and decay at infinity. After extracting those endpoint behaviours, the remaining power series becomes an associated Laguerre polynomial only if it terminates. That termination condition gives

$$E_n=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\frac1{n^2} \simeq-\frac{13.6\ \mathrm{eV}}{n^2},$$

with $n=1,2,\ldots$, $\ell=0,1,\ldots,n-1$, and $m=-\ell,\ldots,\ell$.

The number of orbital states at fixed $n$ is

$$\sum_{\ell=0}^{n-1}(2\ell+1)=n^2.$$

Rotation symmetry explains the $(2\ell+1)$ degeneracy within each $\ell$ multiplet. The larger fact that the nonrelativistic Coulomb energy depends on $n$ but not $\ell$ is an additional, “accidental” degeneracy associated with the special Coulomb problem; at $n=2$ it is what puts $2s$ and $2p$ at the same energy. Spin, relativistic corrections, and the Lamb shift further split levels that this model leaves degenerate.

What this route buys us

One could solve the angular differential equation first and arrive at the same integer labels. Going through the algebra first buys three things that route leaves buried.

Quantization begins before coordinates are chosen — the ladder argument never mentions a sphere. The split between integer and half-integer $j$ turns out to be global rather than local, decided by which group is being represented, since the commutators are identical either way. And spherical harmonics and spinors are one generator algebra realized on two different spaces, which is why the $\ell(\ell+1)$ of the angular equation and the $j(j+1)$ of the ladder are the same expression.

Hydrogen is where the method reaches its limit, and that is worth saying as plainly as the successes. Its energies need the radial dynamics; symmetry alone gets you the angular labels and stops there. Symmetry organizes the states, and the Hamiltonian and the boundary conditions decide which of them nature permits at a given energy.

• Read the complete pedagogical writeup (PDF)
• See the project and source repository