Angular Momentum, Spin, and Rotations

Quantum angular momentum organizes spatial wave patterns and intrinsic spin because both carry unitary actions of the rotation group. The shared commutator algebra explains their common multiplet labels; their different carrier spaces explain why orbital motion and spin are not the same physical degree of freedom.

Required background. Calculus and Linear Algebra supplies inner products, eigenvectors, tensor products, and infinitesimal transformations; Ordinary Differential Equations supplies separated mode equations; Quantum States, Observables, and Measurement supplies rays, self-adjoint generators, and measurement probabilities. Helpful background. Waves, Green Functions, and Geometric Optics supplies mode decompositions and the distinction between symmetry labels and radial dynamics.

Rotations act unitarily on quantum states

We keep $\hbar$ explicit. A spatial rotation acts on quantum rays, so its unitary representative is defined only up to phase. To impose an exact group law, lift the spatial rotation $R$ to an element $\widetilde R$ of the double cover $SU(2)$ described below. Then

$$U(\widetilde R_1)U(\widetilde R_2) =U(\widetilde R_1\widetilde R_2).$$

For scalar orbital states the two lifts of the same $R$ act identically, so the representation descends to $SO(3)$.

For a rotation through angle $\theta$ about the unit vector $\hat{\mathbf n}$,

$$U(\hat{\mathbf n},\theta) =\exp\!\left(-\frac{i\theta}{\hbar} \hat{\mathbf n}\mathbin{\cdot}\mathbf J\right),$$

where the self-adjoint components of $\mathbf J$ are the generators. On a scalar wavefunction,

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

Expanding this expression for a small angle gives, initially on a common dense domain of smooth functions,

$$\mathbf L=-i\hbar\,\mathbf r\times\nabla =\mathbf r\times\mathbf p.$$

Thus orbital angular momentum is not guessed from classical mechanics: it is the generator of rotations of the spatial argument.

If a Hamiltonian is rotationally invariant in the quantum representation,

$$[H,U(\widetilde R)]=0 \quad\text{for every }\widetilde R\in SU(2),$$

then differentiating at the identity gives $[H,J_i]=0$. The Hilbert space can be decomposed into irreducible rotation multiplets $|\alpha;j,m\rangle$, where $\alpha$ denotes any additional radial, particle, or multiplicity labels. Within each irreducible copy, rotational invariance may still mix equivalent copies with the same $j$. After diagonalizing $H$ in that multiplicity space, each resulting energy is independent of $m$. Rotational invariance conserves total $\mathbf J$; it need not conserve orbital $\mathbf L$ and intrinsic $\mathbf S$ separately.

The angular-momentum algebra builds multiplets

Define $J^2=J_x^2+J_y^2+J_z^2$, and let $\epsilon_{ijk}$ denote the Levi–Civita symbol. The generators obey

$$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k, \qquad [J^2,J_i]=0.$$

Because $J^2$ is nonnegative, parametrize its eigenvalue by $j\geq0$ and choose simultaneous eigenstates of $J^2$ and $J_z$:

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

Define $J_\pm=J_x\pm iJ_y$. The commutators $[J_z,J_\pm]=\pm\hbar J_\pm$ show that $J_\pm$ changes $m$ by one. Positivity supplies the endpoint condition:

$$\left\lVert J_\pm|j,m\rangle\right\rVert^2 =\hbar^2\left[j(j+1)-m(m\pm1)\right].$$

A finite multiplet must terminate above and below. Consequently

$$m=-j,-j+1,\ldots,j, \qquad 2j\in\mathbb Z_{\geq0}, \qquad \dim\mathcal H_j=2j+1.$$

This short ladder argument is why integer and half-integer labels arise. Tong derives the orbital generator, its algebra, and the angular eigenfunctions in Tong, Quantum Mechanics, undated, §4.1, printed pp. 96–103.

Orbital and intrinsic realizations

Fix an orbital label $\ell$ and an intrinsic-spin label $s$. The same algebra can act on different spaces; $L^2(S^2)$ means square-integrable angular functions on the unit sphere.

Feature	Orbital angular momentum	Intrinsic spin
Carrier space	Spatial wavefunctions, with angular part in $L^2(S^2)$	Internal space $\mathcal H_s\simeq\mathbb C^{2s+1}$
Rotation action	Rotates the argument $\mathbf r$	Multiplies internal components by a matrix
Generator	$\mathbf L=\mathbf r\times\mathbf p$	$\mathbf S$, with no $\mathbf r\times\mathbf p$ interpretation
Labels	Integer $\ell$ for single-valued scalar angular functions	Integer or half-integer $s$

The spherical harmonics are angular—not full spatial—eigenfunctions:

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

A bound state in a central potential has the separated form $\psi(\mathbf r)=R_{n\ell}(r)Y_{\ell m}(\theta,\phi)$, where $n$ is a radial bound-state label. A continuum state uses an energy or momentum label instead. The radial factor contains the remaining dynamics. For spin one-half,

$$\mathbf S=\frac{\hbar}{2}\boldsymbol\sigma, \qquad [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k,$$

where $\boldsymbol\sigma$ is the Pauli triple. This matrix acts on a two-component internal state, not on the position argument, and the displayed Pauli identity immediately reproduces the angular-momentum commutator for $\mathbf S$.

Why SU(2) contains more representations than SO(3)

$SO(3)$ is the group of real three-dimensional proper rotation matrices. $SU(2)$ is the group of complex $2\times2$ unitary matrices with determinant one. The map $SU(2)\to SO(3)$ is two-to-one: $U$ and $-U$ describe the same rotation of ordinary vectors. The groups therefore share a Lie algebra but not the same global topology.

For spin one-half,

$$U_{1/2}(\hat{\mathbf n},\theta) =e^{-i\theta\hat{\mathbf n}\cdot\boldsymbol\sigma/2}, \qquad U_{1/2}(\hat{\mathbf n},2\pi)=-I, \qquad U_{1/2}(\hat{\mathbf n},4\pi)=I.$$

The minus sign after $2\pi$ is an unobservable global phase for an isolated ray. It becomes observable only as a relative phase, for example when one interferometer branch is rotated and another is not. Half-integer-spin states thus form linear representations of $SU(2)$ and projective representations of $SO(3)$. Turner states the two-to-one map and representation distinction in Turner 2017, Lecture 21, §21.1, Eqs. (21.1)–(21.2), printed p. 90. The sentence following Eq. (21.1) contains an isolated “$SU(3)$” typo; the section heading, equation, and two-to-one construction all identify the intended group as $SO(3)$.

Adding orbital and spin angular momentum

On $\mathcal H_\ell\otimes\mathcal H_s$, the total generator is

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

The product representation decomposes as

$$\mathcal H_\ell\otimes\mathcal H_s =\bigoplus_{j=|\ell-s|}^{\ell+s}\mathcal H_j,$$

where successive values of $j$ differ by one. The dimensions check the decomposition:

$$(2\ell+1)(2s+1) =\sum_{j=|\ell-s|}^{\ell+s}(2j+1).$$

These are tensor-product statements, not rules for adding two classical vectors. Turner derives the selection rule and dimension identity in Turner 2017, Lecture 21, §21.2, Eqs. (21.3)–(21.18), printed pp. 90–92.

Worked example: coupling ℓ = 1 to spin one-half

Let $S_z|\uparrow\rangle=(\hbar/2)|\uparrow\rangle$ and $S_z|\downarrow\rangle=-(\hbar/2)|\downarrow\rangle$. The product space has dimension $3\times2=6$ and decomposes into $j=3/2$ and $j=1/2$ multiplets of dimensions four and two.

Choose the Condon–Shortley phase convention. The highest-weight state is

$$\left|\frac32,\frac32\right\rangle =|1,1\rangle|\uparrow\rangle.$$

Applying $J_-=L_-+S_-$ and normalizing gives

$$\begin{aligned} L_-|1,1\rangle&=\hbar\sqrt2\,|1,0\rangle,\\ S_-|\uparrow\rangle&=\hbar|\downarrow\rangle,\\ J_-\bigl(|1,1\rangle|\uparrow\rangle\bigr) &=\hbar\sqrt2\,|1,0\rangle|\uparrow\rangle +\hbar|1,1\rangle|\downarrow\rangle. \end{aligned}$$

The two product states are orthogonal, so the last vector has norm $\hbar\sqrt3$. Therefore

$$\left|\frac32,\frac12\right\rangle =\sqrt{\frac23}|1,0\rangle|\uparrow\rangle +\sqrt{\frac13}|1,1\rangle|\downarrow\rangle.$$

Ket juxtaposition here abbreviates a tensor product. The Condon–Shortley convention fixes the phases of the highest state and lowering operations. With that convention, the normalized orthogonal combination is

$$\left|\frac12,\frac12\right\rangle =-\sqrt{\frac13}|1,0\rangle|\uparrow\rangle +\sqrt{\frac23}|1,1\rangle|\downarrow\rangle.$$

It is the top of the lower multiplet: applying $J_+$ makes the two contributions cancel. A second check uses

$$\mathbf L\mathbin{\cdot}\mathbf S =\frac12\left(J^2-L^2-S^2\right).$$

Its eigenvalue is $\hbar^2/2$ in the $j=3/2$ multiplet and $-\hbar^2$ in the $j=1/2$ multiplet. The coupled basis therefore diagonalizes interactions proportional to $\mathbf L\cdot\mathbf S$, whereas the uncoupled product basis makes $m=m_\ell+m_s$ transparent.

Common pitfalls

Calling spin a literal orbit. Intrinsic spin carries the rotation algebra but is not motion around a spatial path. Its carrier is an internal vector space.

Treating a global sign as a measurement outcome. A ray is unchanged by an overall minus sign. Only a branch-dependent $2\pi$ rotation can create a measurable relative phase.

Conserving each contribution separately. Rotational invariance conserves total $\mathbf J$. Spin–orbit coupling can exchange angular momentum between $\mathbf L$ and $\mathbf S$.

Rotation labels in gravitational problems

Spherical symmetry makes $\ell,m$ good labels only when the relevant wave or field operator commutes with rotations. Axial symmetry alone guarantees a Fourier label $m$; a Kerr-like spheroidal $\ell$ label additionally relies on separability. Scalar harmonics are insufficient for vector, tensor, or spinor fields.

Spin weight in gravitational perturbation theory describes how a field component changes under a rotation of a chosen tetrad frame. It is not the same object as intrinsic quantum spin. Local spin and Lorentz structure can remain meaningful on curved spacetime even when no global rotation group, and hence no canonical bulk Noether angular momentum, exists. Asymptotically flat spacetimes may still admit ADM- or Bondi-type angular-momentum charges from additional boundary structure. Spin connections, spin–statistics, quantum fields, and quantum-gravity proposals lie outside this bridge.

The next page develops controlled methods for equations that cannot be solved exactly: Perturbation Theory, Adiabatic Evolution, and WKB.

Exercises

Two spin-one-half systems

Use lowering operators and orthogonality to decompose $\mathcal H_{1/2}\otimes\mathcal H_{1/2}$. Give all coupled states, their $J^2$ eigenvalues, and their behavior under interchange of the two systems.

Solution

The highest state $|\uparrow\uparrow\rangle$ has $j=1,m=1$. Here $S_{1-}$ and $S_{2-}$ act on the first and second tensor factors. Repeatedly applying $J_-=S_{1-}+S_{2-}$ gives the normalized triplet

$$\begin{aligned} |1,1\rangle&=|\uparrow\uparrow\rangle,\\ |1,0\rangle&=\frac{|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle}{\sqrt2},\\ |1,-1\rangle&=|\downarrow\downarrow\rangle. \end{aligned}$$

The remaining normalized state is orthogonal to $|1,0\rangle$:

$$|0,0\rangle =\frac{|\uparrow\downarrow\rangle -|\downarrow\uparrow\rangle}{\sqrt2}.$$

For the triplet, $J^2=1(1+1)\hbar^2=2\hbar^2$; for the singlet, $J^2=0$. The three triplet states are symmetric under interchange, while the singlet is antisymmetric. Their dimensions $3+1=4$ exhaust the product space.

References

• Tong, David. Quantum Mechanics, §4.1, printed pp. 96–103. Cambridge lecture notes, undated; official PDF inspected August 30, 2026. Official PDF.
• Turner, Andrew. Quantum Theory I, Lecture 21, prepared for instructor Senthil Todadri, MIT 8.321, Fall 2017, printed pp. 90–92. MIT OpenCourseWare PDF.