Intrinsic Angular Momentum

Classical physics knows only one kind of angular momentum: orbital angular momentum, the $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ that a planet accumulates as it moves around the Sun. Quantum mechanics contains a second kind that has no classical counterpart whatsoever — spin.

What spin is not

It is tempting to picture an electron as a tiny ball spinning about its own axis, the way Earth rotates while orbiting the Sun. That picture is seductive but wrong. An electron is a point particle (or at least is structureless down to length scales far below $10^{-18}\,\text{m}$ ); it has no extended mass distribution to rotate. Moreover, if you try to make the "rotating ball" model self-consistent, the surface would have to move faster than the speed of light to account for the observed magnetic moment — a clear contradiction.

Spin is not rotation. It is a purely quantum-mechanical property, built into the particle at a fundamental level just as mass or electric charge is.

Angular momentum is quantized

Before spin, orbital angular momentum already taught us that quantum mechanics quantizes angular momentum. For orbital motion the magnitude of $\mathbf{L}$ can only take values

$|\mathbf{L}| = \hbar\sqrt{\ell(\ell + 1)}, \qquad \ell = 0, 1, 2, \ldots$

and any chosen component (say along the $z$ -axis) is restricted to

$L_z = m_\ell \hbar, \qquad m_\ell \in \{-\ell, -\ell+1, \ldots, \ell-1, \ell\}.$

Here $\hbar = h/(2\pi) \approx 1.055 \times 10^{-34}\,\text{J\,s}$ is the reduced Planck constant.

Intrinsic angular momentum (spin)

Spin obeys the same algebraic rules as orbital angular momentum, but the quantum number $s$ that characterises it is fixed by the species of particle — you cannot change it by moving the particle differently. For an electron, $s = 1/2$ .

The magnitude of the spin angular momentum vector is

$|\mathbf{S}| = \hbar\sqrt{s(s+1)} = \hbar\sqrt{\tfrac{1}{2}\cdot\tfrac{3}{2}} = \frac{\sqrt{3}}{2}\,\hbar.$

Any chosen component (again, conventionally $z$ ) is restricted to

$S_z = m_s \hbar, \qquad m_s \in \left\{-s, -s+1, \ldots, s\right\}.$

For $s = 1/2$ the only possibilities are $m_s = +1/2$ and $m_s = -1/2$ , giving

$S_z = +\tfrac{\hbar}{2} \quad \text{(spin-up, } |{+z}\rangle \text{)} \qquad \text{or} \qquad S_z = -\tfrac{\hbar}{2} \quad \text{(spin-down, } |{-z}\rangle \text{)}.$

Particles with half-integer spin ( $s = 1/2, 3/2, \ldots$ ) are called fermions; those with integer spin ( $s = 0, 1, 2, \ldots$ ) are called bosons. Electrons, protons, and neutrons are all spin- $1/2$ fermions. Photons carry spin $s = 1$ .

Spin and magnetic moment

Spin angular momentum is accompanied by an intrinsic magnetic dipole moment. For an electron

$\boldsymbol{\mu}_s = -g_s \frac{e}{2m_e} \mathbf{S},$

where $e$ is the elementary charge, $m_e$ is the electron mass, and $g_s \approx 2.002$ is the electron $g$ -factor (the small deviation from 2 is a prediction of quantum electrodynamics). The factor of $\approx 2$ distinguishes the spin magnetic moment from the orbital contribution, for which $g_\ell = 1$ exactly. It is precisely this magnetic moment that lets a non-uniform magnetic field deflect a beam of silver atoms — the key to the Stern–Gerlach experiment you will study next.

Total angular momentum

In atoms, spin and orbital angular momentum combine into a total angular momentum $\mathbf{J} = \mathbf{L} + \mathbf{S}$ . The rules for combining them (Clebsch–Gordan decomposition) govern atomic energy levels and selection rules for optical transitions. For now, the important point is that spin is a permanent, unchangeable contribution to the total angular momentum budget of every particle.