Part 4 · Spin and angular momentum

4.1 Spin

Source: Centini’s whiteboard of 6 Oct, p. 6 (the cone and the magnets in sequence); the magnetic moment, Stern–Gerlach, the spin-½ operators, Pauli matrices, ladder operators and spin $s$ from the 2024/25 boards #9–#11

Magnetic moment

A small current loop, or a bar magnet, behaves like an arrow $\bar{m}$, its magnetic moment. In a field $\bar{B}$ it feels a torque $\bar{m}\wedge \bar{B}$ that turns it towards $\bar{B}$, like a compass needle, and has energy $$U=-\bar{m}\cdot \bar{B}=-mB\cos \theta ,$$ lowest with $\bar{m}$ along $\bar{B}$. A loop of radius $R$ carrying current $i$ has $\bar{m}=i\pi R^2\hat{n}$, with $\hat{n}$ normal to the loop by the right-hand rule. In a field that changes from place to place, $U$ depends on position and the moment is pushed towards lower $U$: $\bar{F}=\bar{\nabla }(\bar{m}\cdot \bar{B})$. ==A uniform field only turns a magnetic moment; a non-uniform field also pushes it.==

An electron on a circular orbit, charge $-e$, is a current loop. Its moment is set by its orbital angular momentum $\bar{L}=\bar{r}\wedge \bar{p}$: $${\bar{m}}_e=-\frac{e}{2m_e}\bar{L}.$$ The minus sign: the electron is negative, so its current runs against its motion. One unit of angular momentum, $L=\hslash$, gives a moment of one Bohr magneton, $${\mu }_B=\frac{e\hslash }{2m_e}=9.27\times 10^{-24}\text{J/T}=5.79\times 10^{-5}\text{eV/T}.$$

Spin

Besides any orbital motion, the electron carries an angular momentum of its own, spin $\bar{S}$, an intrinsic angular momentum. The board’s picture, an electron spinning on itself like the Earth, is only a mnemonic: a spinning ball would need its surface to move faster than light (beyond the boards). Its magnetic moment is $${\bar{\mu }}_e=g_e\bar{S},g_e=2\left(\frac{-e}{2m_e}\right),{\mu }_z=g_e\left(\pm \frac{\hslash }{2}\right)=\mp {\mu }_B:$$ one full Bohr magneton from half a unit of angular momentum. That is the board’s circled 2. Electron, proton and neutron all have spin ½ and are fermions (§5.1 Identical particles). (The board’s proton and neutron factors, 2.79 and 1.9, are their moments in nuclear magnetons $e\hslash /2m_p$, a different convention from its $g$; nuclear moments are some 1836 times smaller than the electron’s.)

In chemistry this is the fourth quantum number, $m_s=\pm \frac{1}{2}$, the one that lets two electrons share an orbital.

The Stern–Gerlach experiment

Why silver. An Ag atom has 47 electrons: 46 in closed shells with paired spins, whose moments cancel, and one in 5s, with no orbital angular momentum. So the whole atom carries exactly one electron spin, and it is neutral: no Lorentz force, only the force on its moment. (The board writes “5D”; board 10’s recap has 5s.)

The set-up. An oven emits Ag atoms, a slit makes a beam, the beam passes between magnet poles shaped so that $B_z$ grows strongly towards one of them, then hits a screen. Only $B_z$ varies appreciably, so $$F_z={\mu }_z\frac{\partial B_z}{\partial z}.$$ ==Each atom is deflected in proportion to its $S_z$==: the screen is a histogram of $S_z$.

• Classical prediction. The moments point every which way, so ${\mu }_z$ takes every value between $\pm |\mu |$: one continuous band.
• Observed. Two separate spots, $S_z=\pm \frac{\hslash }{2}$, each with probability 0.5.

The $z$ component of spin is quantised. With $\partial B_z/\partial z=10^3$ T/m and atoms at 500 m/s through a 3.5 cm magnet, the spots are 0.13 mm from the axis (beyond the boards): what Stern and Gerlach saw in 1922.

The operators from the measurements

The two beams out of a magnet along $z$ are the eigenstates of ${\hat{S}}_z$, written $|\pm ⟩=|S_z,\pm ⟩$: ${\hat{S}}_z|\pm ⟩=\pm \frac{\hslash }{2}|\pm ⟩$. They are an orthonormal basis of all spin-½ states.

${\hat{S}}_x$. A beam in $|S_x,+⟩$ through a $z$ magnet splits 50/50 (Example 2.2.9), so $|⟨\pm |S_x,+⟩{|}^2=\frac{1}{2}$, and each coefficient is $\frac{1}{\sqrt{2}}$ times a phase. An overall phase changes no probability, so only the relative one, ${\delta }_1$, matters: $$|S_x,\pm ⟩=\frac{1}{\sqrt{2}}\left(|+⟩\pm e^{i{\delta }_1}|-⟩\right),$$ the minus sign making $|S_x,-⟩$ orthogonal to $|S_x,+⟩$. The spectral form, multiplied out, keeps only the cross terms (the board strikes out the diagonal ones): $${\hat{S}}_x=\frac{\hslash }{2}\left(e^{-i{\delta }_1}|+⟩⟨-|+e^{i{\delta }_1}|-⟩⟨+|\right).$$ ${\hat{S}}_x$ flips the spin: ${\hat{S}}_x|+⟩=\frac{\hslash }{2}{e}^{i{\delta }_1}|-⟩$. ==$|+⟩$ is an eigenstate of ${\hat{S}}_z$ but not of ${\hat{S}}_x$.==

${\hat{S}}_y$ is built the same way with a phase ${\delta }_2$. What fixes the phases is one more experiment: $|S_y,\pm ⟩$ through an $x$ magnet also splits 50/50, so $$\frac{1}{4}{\left|1\pm e^{i({\delta }_2-{\delta }_1)}\right|}^2=\frac{1}{2}.$$ The board guesses $1\pm i$, and $|1\pm i{|}^2=2$ works: ${\delta }_2-{\delta }_1=\pm \frac{\pi }{2}$. Without guessing, $|1+e^{i\Delta }{|}^2=2+2\cos \Delta =2$ needs $\cos \Delta =0$ (beyond the boards). The board takes ${\delta }_1=0$, ${\delta }_2=\frac{\pi }{2}$: the first fixes where the $x$ axis points, the second makes $x,y,z$ right-handed.

A spin state is a column, a spinor $\left(\begin{array}{c}{\alpha }_{+}\\ {\alpha }_{-}\end{array}\right)$ with ${\alpha }_{\pm }=⟨\pm |\alpha ⟩$, its bra the conjugate row; $|{\alpha }_{\pm }{|}^2$ are the probabilities of $S_z=\pm \frac{\hslash }{2}$. A ket-bra is a column times a row, a $2\times 2$ matrix: $|+⟩⟨-|=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. The entry in row $a$, column $b$ is $⟨a|\hat{S}|b⟩$. ${\hat{S}}_z$ is diagonal because $|\pm ⟩$ are its eigenstates; ${\hat{S}}_x$ and ${\hat{S}}_y$ are off-diagonal because they flip.

Commutators

==No two components are compatible==: measuring one scrambles the others (§2.3 Commutators and compatible observables). The general bound gives $\Delta S_x\Delta S_y\ge \frac{\hslash }{2}|⟨S_z⟩|$; in $|+⟩$ that is $\frac{{\hslash }^2}{4}$, exactly the $S_x=0\pm \frac{\hslash }{2}$, $S_y=0\pm \frac{\hslash }{2}$ of the cone (Key observation 2.2.8) (beyond the boards). These are the commutators of every angular momentum (§4.2 Angular momentum).

${\hat{S}}^2$ and the cone

${\hat{S}}_z^2=\frac{{\hslash }^2}{4}(|+⟩⟨+|+|-⟩⟨-|)=\frac{{\hslash }^2}{4}\hat{I}$, by completeness, and the same for ${\hat{S}}_x^2$ and ${\hat{S}}_y^2$: any component measured gives $\pm \frac{\hslash }{2}$, whose square is always $\frac{{\hslash }^2}{4}$. So $${\hat{S}}^2={\hat{S}}_x^2+{\hat{S}}_y^2+{\hat{S}}_z^2=\frac{3{\hslash }^2}{4}\hat{I}.$$ ${\hat{S}}^2$ commutes with every component, since $\hat{I}$ commutes with everything. So $S^2$ and one component can be sharp together: every spin-½ state has $S^2=\frac{3}{4}{\hslash }^2$, and $|\pm ⟩$ have $S_z=\pm \frac{\hslash }{2}$ too.

The length $\frac{\sqrt{3}}{2}\hslash =0.87\hslash$ is longer than the largest $S_z$, so ==the spin never points along $z$==: in $|+⟩$ it lies on a cone around $+z$, half-opening $\arccos \frac{1}{\sqrt{3}}=54.7°$ (beyond the boards). Lying exactly on $z$, with $S_x=S_y=0$, would need ${\hat{S}}_x$ and ${\hat{S}}_y$ to commute; the board crosses it out.

Raising and lowering

${\hat{S}}_{+}$ raises $S_z$ by $\hslash$ and cannot raise $|+⟩$, already at the top. Backwards, ${\hat{S}}_x=\frac{1}{2}({\hat{S}}_{+}+{\hat{S}}_{-})$ and ${\hat{S}}_y=-\frac{i}{2}({\hat{S}}_{+}-{\hat{S}}_{-})$. ${\hat{S}}_{\pm }$ are adjoints of each other, not Hermitian, not observables: the spin version of the oscillator’s ladder operators.

Spin $s$. A state of spin $s$ is $|s,m_s⟩$, with $m_s$ from $s$ down to $-s$ in steps of 1, $2s+1$ states: $${\hat{S}}_z|s,m_s⟩=\hslash m_s|s,m_s⟩,{\hat{S}}^2|s,m_s⟩={\hslash }^{2}s(s+1)|s,m_s⟩,$$ $${\hat{S}}_{\pm }|s,m_s⟩=\hslash \sqrt{s(s+1)-m_s(m_s\pm 1)}|s,m_s\pm 1⟩.$$ Spin ½ gives ${\hslash }^2\cdot \frac{1}{2}\cdot \frac{3}{2}=\frac{3}{4}{\hslash }^2$, as found. At the ends the square root vanishes and the ladder stops. Spin $\frac{3}{2}$ has four states; it is a nucleus, say, or spins added together (§4.4 Addition of angular momenta). Where the square root comes from is in §4.2 Angular momentum.

Exercises