Part III — The physics

Spin: a two-state quantum system

In 1922 Otto Stern and Walther Gerlach sent a beam of silver atoms between the poles of a specially shaped magnet. If each atom were a tiny compass needle pointing in a random direction, the beam should have smeared into a continuous band. It split cleanly into two, and nothing landed between. Why exactly two?

The answer is a property of the electron that has no classical counterpart: spin, an intrinsic angular momentum whose component along any axis can take only two values. Spin is the simplest quantum system there is, two states and nothing else, and it makes the formalism of the earlier essays completely concrete. This essay builds it from the experiment, computes what happens when atoms pass through magnets set at different angles, and compares the result with the photon polarisation of the representations essay.

The experiment

A silver atom has 47 electrons. All but one are paired off, and their magnetic effects cancel. The last one, in an s state with no orbital angular momentum, leaves the atom a small magnet. In Stern and Gerlach's magnet the field grew sharply stronger towards one pole, so an atom whose magnetic moment pointed up was pushed one way and one pointing down the other, by an amount proportional to the component $\mu_z$. A classical magnet can point anywhere, so $\mu_z$ could take any value between its extremes, and the beam should have spread into a band.

Instead there were two spots, placed symmetrically: $\mu_z$ had one of exactly two values, equal and opposite. Since magnetic moment goes with angular momentum, the component of the atom's angular momentum along the field has exactly two values. By the angular momentum essay, a quantum number $s$ gives $2s + 1$ values, so two values means $s = \tfrac12$: a component of $\pm\tfrac12\hbar$, one of the half-whole values the commutators allow. Orbital angular momentum cannot do this, because a wavefunction that must join up with itself around an axis only allows whole numbers. The half comes from something else: an angular momentum the electron carries intrinsically, wherever it is and however it moves. For the electron,

$$\begin{aligned}S_z &= \pm\tfrac12\hbar,\\[4pt] S^2 &= s(s+1)\hbar^2 = \tfrac34\hbar^2 .\end{aligned}$$

The name "spin" suggests a little ball rotating on its axis, and that picture fails. A ball the size the electron could be would need its surface to move faster than light to carry this much angular momentum, and no rotating body gives half-units. Spin is best taken on its own terms: a two-valued quantum property with the character of angular momentum.

History took a detour here. In 1922 the two spots were read as a confirmation of Bohr's quantised orbits. Only after 1925, when George Uhlenbeck and Samuel Goudsmit proposed that the electron itself carries half a unit of angular momentum, to explain the doubled lines of alkali spectra, did it gradually become clear what Stern and Gerlach had seen; the connection was spelled out explicitly by 1927.

Optional: how far the atoms were deflected

The magnetic moment of the unpaired electron is about the Bohr magneton, $\mu_B = 9.27\times10^{-24}$ J/T. In a field gradient of about 10 tesla per centimetre, $1000$ T/m, the force is $\mu_B\times1000 = 9.3\times10^{-21}$ N. A silver atom has mass $1.79\times10^{-25}$ kg, so its acceleration is $5\times10^{4}$ m/s². Atoms from an oven at about 1300 K travel at roughly 550 m/s and spend $6\times10^{-5}$ s in a magnet 3.5 cm long, which deflects them by $\tfrac12 at^2 \approx 0.1$ mm each way. Stern and Gerlach's two traces, a fraction of a millimetre apart, were at the edge of what could then be seen.

Two states, and magnets at other angles

Call the two results $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ for a magnet along $z$. They form a basis, and any spin state is a combination $a|{\uparrow}\rangle + b|{\downarrow}\rangle$ with $|a|^2 + |b|^2 = 1$. Now pass atoms that came out "up" into a second magnet turned by an angle $\theta$ from the first, and ask how many come out up along the new direction. Quantum mechanics' answer, which the worked example below checks for one angle, is

$$P(\text{up along }\theta \mid \text{up along } z) = \cos^2\frac{\theta}{2}.$$

With the second magnet aligned, $\theta = 0$, every atom passes. Turned by $90°$, half pass and half do not: an atom that is definitely up along $z$ has no preference at all along $x$. Turned by $180°$, none pass, since "up along $-z$" is "down along $z$". The two states $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ are orthogonal, and they are separated by a half-turn.

How often a prepared system passes an analyser turned by $\theta$. Solid: a spin prepared up, $\cos^2(\theta/2)$; the opposite state is reached at $180°$ and the curve repeats only after a full turn. Dashed: a photon prepared horizontal, $\cos^2\theta$, from the representations essay; its orthogonal state, vertical, is only $90°$ away.

The comparison with the photon is instructive. Both are two-state systems described by exactly the same mathematics. What differs is how the states sit in ordinary space. For light, two orthogonal polarisations are at right angles. For spin, two orthogonal states point in opposite directions. The angle in the state space is half the angle in the laboratory, and that is where the $\theta/2$ comes from.

The spin operators

In the $\uparrow/\downarrow$ basis, the operator for the spin component along $z$ is diagonal, with eigenvalues $\pm\tfrac12\hbar$. The components along $x$ and $y$ are fixed by requiring that they behave like angular momentum, with commutators like $[\hat S_x, \hat S_y] = i\hbar\hat S_z$. The result, found by Wolfgang Pauli in 1927, is $\hat S_i = \tfrac12\hbar\,\sigma_i$ with the Pauli matrices

$$\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix},$$ $$\sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

The first and third are the two photon operators of the formalism essay, now doing duty for spin. Each has eigenvalues $+1$ and $-1$; the eigenvectors of $\sigma_x$ are $(|{\uparrow}\rangle \pm |{\downarrow}\rangle)/\sqrt2$, the states "up along $x$" and "down along $x$". No two of the matrices commute, so no state is sharp in two components at once: as with orbital angular momentum, only one component can be known.

Worked example: a magnet at 60°

Atoms prepared up along $z$ enter a magnet whose field lies in the $x$–$z$ plane, turned $60°$ from $z$. The spin component along that direction is $\hat S_\theta = \tfrac12\hbar(\cos\theta\,\sigma_z + \sin\theta\,\sigma_x)$, whose matrix is

$$\begin{aligned} \hat S_{60°} &= \frac{\hbar}{2}\begin{pmatrix}\cos60° & \sin60°\\ \sin60° & -\cos60°\end{pmatrix}\\[6pt] &= \frac{\hbar}{2}\begin{pmatrix}0.500 & 0.866\\ 0.866 & -0.500\end{pmatrix}. \end{aligned}$$

Results: the eigenvalues, $\pm\tfrac12\hbar$, as for any direction.

The "up" eigenvector is $(\cos30°, \sin30°) = (0.866, 0.500)$; multiplying out confirms it: the matrix times this list gives $\tfrac12\hbar$ times the same list. Its overlap with $|{\uparrow}\rangle = (1, 0)$ is $0.866$, so $P(\text{up along }60°) = 0.866^2 = 0.750$, and $P(\text{down}) = 0.250$. That is $\cos^2 30°$, the formula with $\theta/2 = 30°$.

Average: $\langle S_{60°}\rangle = 0.75(\tfrac12\hbar) + 0.25(-\tfrac12\hbar) = \tfrac14\hbar$, which is $\tfrac12\hbar\cos60°$: the component of an arrow of length $\tfrac12\hbar$ along the magnet's direction. The average behaves like a classical arrow; no individual atom does. Each one gives $+\tfrac12\hbar$ or $-\tfrac12\hbar$, and never $\tfrac14\hbar$.

Checks: at $\theta = 0$ the matrix is $\tfrac12\hbar\,\sigma_z$ and $P = 1$; at $\theta = 90°$ it is $\tfrac12\hbar\,\sigma_x$ and the overlap is $0.707$, $P = \tfrac12$.

Optional: turning a spin by 360° changes its sign

Because the state space turns at half the rate of real space, rotating a spin by $360°$ turns its state through only $180°$ in the state space, which multiplies it by $-1$. An overall sign changes no prediction on its own. But if one part of a beam is rotated by $360°$ and the other is not, and the two are recombined, the relative sign turns bright fringes dark. Neutron interference experiments in 1975 saw exactly this, and it takes a $720°$ rotation to bring a spin state fully back to itself. This is one more reason no picture of a little rotating ball can be right.

Two questions

1. Three magnets. Atoms leave a first magnet along $z$; only the "up" beam is kept. They pass a second magnet along $x$, and only its "up" beam is kept. A third magnet is along $z$ again. What fraction of the atoms leaving the first magnet emerge from the third as "down"?

Worked answer

From up along $z$, the chance of up along $x$ is $\cos^2 45° = \tfrac12$. Those atoms are now in the state up along $x$, which has no preference along $z$: half come out down. Overall $\tfrac12\times\tfrac12 = \tfrac14$.

The tempting answer is zero, on the reasoning that the first magnet already removed every "down" atom. That would be right if each atom carried a hidden $z$ label through the middle magnet. It does not: after the $x$ measurement, the atom is in an $x$ eigenstate, and its earlier $z$ value is gone. It is the three-filter photon puzzle of the representations essay, in atoms.

2. Opposite at 90° or at 180°? Horizontal and vertical polarisation are orthogonal photon states, 90° apart. Spin up and spin down are orthogonal spin states, 180° apart. Does this mean spin states are "more different" from each other than polarisation states are?

Worked answer

No. In both cases the two states are orthogonal, their overlap is zero, and each can be distinguished from the other with certainty by the right analyser. As states, they are exactly as different as two states can be. What differs is only the laboratory angle through which you must turn an analyser to go from one to the other: 90° for polarisation, 180° for spin.

The wrong turn is to identify the angle in the laboratory with the angle between the states. The state-space angle is what sets probabilities through $\cos^2$, and for spin it is half the laboratory angle.

Where this leaves us

The silver beam splits into exactly two because the electron carries an intrinsic angular momentum whose component along any axis is $+\tfrac12\hbar$ or $-\tfrac12\hbar$ and nothing else. Spin is a complete two-state quantum system: its states are two-component lists, its measurements are the Pauli matrices, and a magnet turned by $\theta$ passes a prepared atom with probability $\cos^2(\theta/2)$.

Spin doubles the count of electron states in hydrogen, from $n^2$ to $2n^2$ at each level. That doubling is only the start of its importance. When there are many electrons, as in every atom but hydrogen, spin combines with a rule about identical particles that has no classical counterpart, and together they decide the structure of the periodic table. That is the next essay.

Quantum Mechanics, Explained