What Spin Is and Is Not

Spin is intrinsic angular momentum. It obeys the angular momentum algebra, can contribute to magnetic moments, and transforms under rotations, but it is not literal mechanical rotation of a small extended body.

The key idea is:

$$\text{spin} \quad = \quad \text{an internal quantum degree of freedom carrying angular momentum}.$$

Why Spin Was Needed

Spin entered quantum mechanics because experiments and spectra showed angular-momentum-like structure that could not be explained by orbital motion alone. The Stern–Gerlach experiment showed discrete two-valued outcomes for neutral silver atoms; its modern spin-analyzer interpretation is developed in Stern–Gerlach Revisited. Atomic spectra and magnetic splitting required additional angular momentum labels. Relativistic theory organizes particles by mass and spin (or helicity for massless particles), but it also permits spin-zero particles; relativity does not require every particle to have nonzero spin.

This page focuses on the nonrelativistic formalism: spin is an internal Hilbert-space degree of freedom with its own angular momentum operators.

Spin as a Quantum Degree of Freedom

Spin operators satisfy

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

For a spin-$s$ system,

$$S^2\lvert s,m\rangle = \hbar^2s(s+1)\lvert s,m\rangle,$$

and

$$S_z\lvert s,m\rangle = \hbar m\lvert s,m\rangle.$$

The allowed values are

$$m=-s,-s+1,\ldots,s.$$

The spin Hilbert space has dimension $2s+1$: this counts mutually orthogonal basis states, not all possible superpositions. Here we use a fixed-spin nonrelativistic degree of freedom; massless helicity states require a different description.

Spin and Rotations

A spin state transforms under rotations by a unitary representation. For spin-$1/2$,

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

The appearance of the half-angle is not cosmetic. It reflects the relation between $SU(2)$ and $SO(3)$, developed in SU(2) versus SO(3), and leads to the sign change under a $2\pi$ rotation.

Spin Versus Orbital Angular Momentum

Orbital angular momentum is

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

Spin is not of this form. It acts on internal spin states, not on the spatial coordinates of a scalar wavefunction.

For a particle with both orbital and spin angular momentum, the total angular momentum is

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

The same angular momentum algebra appears, but the operators act on different factors of the Hilbert space.

Spin and Magnetic Moment

Spin often couples to magnetic fields. For a spin-$1/2$ particle, a common effective Hamiltonian is

$$H=-\boldsymbol\mu\cdot\mathbf B,$$

with

$$\boldsymbol\mu=\gamma\mathbf S$$

for gyromagnetic ratio $\gamma$ in a given model. The algebra alone does not determine $\gamma$ or guarantee a nonzero magnetic moment. When present, this coupling makes spin visible in magnetic resonance, Zeeman splitting, and Stern–Gerlach-type measurements. The corresponding Hamiltonian and sign conventions are developed in Spin in Magnetic Fields.

Common Misconceptions

• Spin is not a tiny ball spinning about an axis.
• A spinor is not an ordinary three-dimensional vector.
• A $2\pi$ spinor sign change does not contradict the ray nature of states.
• Spin and orbital angular momentum obey the same algebra but have different physical origins.
• The word “up” in spin up means an eigenstate of a chosen spin component, not a little arrow permanently pointing upward in space.

Exercises

1. How many spin states does a spin-$3/2$ system have?

Solution

The dimension is $2s+1=4$. The possible $m$ values are $-3/2,-1/2,1/2,3/2$.

2. Why is $\mathbf S$ not generally equal to $\mathbf R\times\mathbf P$?

Solution

$\mathbf R\times\mathbf P$ is orbital angular momentum and acts on spatial wavefunctions. Spin acts on internal spin degrees of freedom. They obey the same angular momentum algebra, but they act on different parts of the Hilbert space and have different physical origins.

3. A spin-$1/2$ particle is in a spatial $s$ orbital. Is the result of measuring its total angular momentum necessarily zero?

Solution

No. The orbital factor has $\ell=0$, so $\mathbf L$ annihilates it, but the spin factor remains. Thus $J^2=S^2=3\hbar^2/4$, while a component measurement has outcomes $\pm\hbar/2$. Zero orbital motion does not imply zero intrinsic angular momentum.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.