Angular Momentum Algebra

Angular momentum is where quantum mechanics turns symmetry into spectra. Rotational invariance does not merely say that space has no preferred direction; it forces commutation relations, multiplets, selection rules, and the labels used for atoms, nuclei, and spin systems.

Sakurai presents angular momentum as the representation theory of rotations, with spin and orbital angular momentum treated under the same algebra. Ballentine gives a rigorous operator and rotation treatment, including finite rotations and tensor operators. The Gottfried-named notes develop the algebra after spin-1/2 and symmetries. Schiff's older coordinate style emphasizes orbital angular momentum, spherical harmonics, and central-potential wave functions.

Definitions

An angular momentum operator $\mathbf J=(J_x,J_y,J_z)$ satisfies

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

The square is

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

It commutes with all components:

$$[J^2,J_i]=0.$$

Therefore one can choose simultaneous eigenstates of $J^2$ and $J_z$:

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

and

$$J_z|j,m\rangle=\hbar m|j,m\rangle.$$

The ladder operators are

$$J_+=J_x+iJ_y,\qquad J_-=J_x-iJ_y.$$

They obey

$$[J_z,J_\pm]=\pm\hbar J_\pm,$$

and

$$[J_+,J_-]=2\hbar J_z.$$

For orbital angular momentum,

$$\mathbf L=\mathbf R\times \mathbf P,$$

with coordinate-space components such as

$$L_z=-i\hbar{\partial\over \partial \phi}.$$

Spin angular momentum obeys the same algebra but is not generated by a particle's spatial orbit. For spin-1/2,

$$\mathbf S={\hbar\over 2}\boldsymbol{\sigma}.$$

Key results

The allowed quantum numbers are

$$j=0,{1\over 2},1,{3\over 2},\ldots, \qquad m=-j,-j+1,\ldots,j-1,j.$$

The ladder action is

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

This formula encodes both the spacing of $m$ and the endpoint condition $J_+\vert j,j\rangle=0$, $J_-\vert j,-j\rangle=0$.

The components cannot all be simultaneously sharp. In $\vert j,m\rangle$,

$$\langle J_x\rangle=\langle J_y\rangle=0,$$

and

$$\langle J_x^2+J_y^2\rangle=\hbar^2[j(j+1)-m^2].$$

This is the quantum version of a vector with known length and known $z$ projection but uncertain transverse direction.

Finite rotations are generated by angular momentum:

$$U(R_{\mathbf n}(\theta))=\exp\left(-{i\theta\over \hbar}\mathbf n\cdot\mathbf J\right).$$

For orbital angular momentum, the angular wave functions are spherical harmonics:

$$L^2Y_\ell^m(\theta,\phi)=\hbar^2\ell(\ell+1)Y_\ell^m(\theta,\phi),$$

and

$$L_zY_\ell^m(\theta,\phi)=\hbar mY_\ell^m(\theta,\phi).$$

Sakurai's algebraic treatment explains why $\ell$ and $m$ have their allowed values before solving any central-potential differential equation. Schiff's wave-mechanics treatment shows the same restrictions emerging from single-valued angular functions.

Visual

Algebraic object	Formula	Physical role
Commutator	$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k$	defines angular momentum
Casimir	$J^2$	labels irreducible multiplet
Projection	$J_z$	labels state inside multiplet
Ladder	$J_\pm$	moves between neighboring $m$ values
Rotation	$e^{-i\theta\mathbf n\cdot\mathbf J/\hbar}$	changes spatial or spin orientation

Worked example 1: Ladder coefficients for j = 1

Problem. Compute $J_-\vert 1,1\rangle$ and $J_-\vert 1,0\rangle$.

Method.

1. Use

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

2. For $j=1$, $m=1$:

$$\begin{aligned} J_-|1,1\rangle &=\hbar\sqrt{1(1+1)-1(1-1)}\,|1,0\rangle\\ &=\hbar\sqrt{2}\,|1,0\rangle. \end{aligned}$$

3. For $j=1$, $m=0$:

$$\begin{aligned} J_-|1,0\rangle &=\hbar\sqrt{1(1+1)-0(0-1)}\,|1,-1\rangle\\ &=\hbar\sqrt{2}\,|1,-1\rangle. \end{aligned}$$

4. At the bottom,

$$J_-|1,-1\rangle=\hbar\sqrt{2-(-1)(-2)}\,|1,-2\rangle=0,$$

so there is no physical $\vert 1,-2\rangle$ state.

Checked answer. The $j=1$ multiplet has exactly three states: $m=1,0,-1$.

Worked example 2: Constructing the spin-1 matrices

Problem. In the ordered basis $\{\vert 1,1\rangle,\vert 1,0\rangle,\vert 1,-1\rangle\}$, find $J_x$.

Method.

1. Build $J_+$ from ladder actions:

$$J_+|1,-1\rangle=\hbar\sqrt2|1,0\rangle, \qquad J_+|1,0\rangle=\hbar\sqrt2|1,1\rangle, \qquad J_+|1,1\rangle=0.$$

2. Therefore

$$J_+=\hbar\begin{pmatrix} 0&\sqrt2&0\\ 0&0&\sqrt2\\ 0&0&0 \end{pmatrix}.$$

3. Since $J_-=J_+^\dagger$,

$$J_-=\hbar\begin{pmatrix} 0&0&0\\ \sqrt2&0&0\\ 0&\sqrt2&0 \end{pmatrix}.$$

4. Use

$$J_x={1\over 2}(J_++J_-).$$

5. Thus

$$J_x={\hbar\over \sqrt2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}.$$

Checked answer. The matrix is Hermitian, as an observable must be, and it couples only neighboring $m$ states.

Code

import numpy as np

def angular_momentum_matrices(j, hbar=1.0):
    m_vals = np.arange(j, -j - 1, -1)
    dim = len(m_vals)
    jp = np.zeros((dim, dim), dtype=complex)
    for col, m in enumerate(m_vals):
        mp = m + 1
        if mp <= j:
            row = np.where(np.isclose(m_vals, mp))[0]
            if len(row):
                jp[row[0], col] = hbar * np.sqrt(j * (j + 1) - m * (m + 1))
    jm = jp.conj().T
    jx = 0.5 * (jp + jm)
    jy = (jp - jm) / (2j)
    jz = np.diag(hbar * m_vals)
    return jx, jy, jz

jx, jy, jz = angular_momentum_matrices(1)
print(jx)
print(np.allclose(jx @ jy - jy @ jx, 1j * jz))

Common pitfalls

• Treating $\mathbf J$ as an ordinary vector with simultaneously known components. Only $J^2$ and one component can be sharp together.
• Forgetting half-integer $j$ values. Spin permits representations that orbital angular momentum alone does not.
• Losing endpoint conditions. The ladder formulas determine allowed $m$ values by requiring nonnegative norms.
• Confusing orbital $\ell$ with total $j$. In atoms with spin, total angular momentum combines orbital and spin parts.
• Using $J_\pm$ without the square-root coefficient. Transition strengths depend on it.
• Assuming $L_z=-i\hbar\partial_\phi$ applies to spin. Spin lives in an internal space, not in ordinary coordinate dependence.
• Ignoring conventions for spherical harmonics and phases. Clebsch-Gordan tables depend on consistent phase choices.

The cleanest way to avoid angular-momentum confusion is to identify which angular momentum is being discussed before writing equations. Orbital angular momentum $\mathbf L$ acts on spatial wave functions. Spin $\mathbf S$ acts on internal spin degrees of freedom. Total angular momentum $\mathbf J$ may mean $\mathbf L+\mathbf S$ or the sum of several subsystem angular momenta. They obey the same algebra but act on different spaces. The shared algebra is why Sakurai can treat them uniformly; the different physical meanings are why notation must stay explicit.

The ladder derivation also teaches a general quantum habit: norms must be nonnegative. The allowed $j$ and $m$ values are not guessed from geometry; they follow because repeated raising or lowering must stop before a state with negative norm appears. This is one of the places where Hilbert-space structure directly produces quantization. The same type of reasoning appears in the harmonic oscillator, where $a^\dagger a$ is positive and the ladder must terminate at the ground state.

For orbital angular momentum, the connection to wave mechanics is made through spherical harmonics. The equation $L_z=-i\hbar\partial_\phi$ implies angular dependence $e^{im\phi}$, and single-valuedness under $\phi\mapsto\phi+2\pi$ forces integer $m$ for orbital wave functions. Spin-1/2 does not arise from such single-valued scalar spatial functions; it comes from projective representations of rotations. This is the conceptual reason half-integer spin is not just "orbital angular momentum with a smaller unit."

In applications, angular momentum is often more valuable as a labeling system than as a set of matrices. Once $j$ and $m$ are good quantum numbers, many Hamiltonian blocks decouple, selection rules appear, and degeneracies become organized. Atomic physics, scattering partial waves, molecular rotations, and nuclear spin all use this bookkeeping. A calculation that ignores angular momentum symmetry usually becomes longer and less transparent.

A final useful check is rotational covariance. If a result claims to describe a rotationally invariant Hamiltonian but depends on a particular $m$ value, either a direction has been introduced or the calculation has broken symmetry by mistake. Conversely, when an external magnetic field is present, dependence on $m$ is expected because the field defines a preferred axis. Asking "what picked the $z$ axis?" is a simple way to catch incorrect degeneracy splitting.

In longer calculations, keep commutators and eigenvalue equations separate. The commutators define the algebra and therefore the representation structure; the eigenvalue equations label a particular basis. You can rotate to another basis without changing the algebra. This distinction is useful when comparing matrix spin calculations with spherical-harmonic wave-function calculations.

Connections

• Spin-1/2 systems
• Addition of angular momentum
• Central potentials and the hydrogen atom
• Symmetries and conservation laws
• Scattering theory