Spin and the Dirac equation

Non-relativistic quantum mechanics treats spin as an add-on. Orbital angular momentum $\hat{\mathbf{L}}=\hat{\mathbf{r}}\times\hat{\mathbf{p}}$ falls out of the Schrödinger equation automatically, but the two-valuedness seen in the Stern–Gerlach experiment has to be stapled on by hand — Pauli's two-component spinor and his three matrices, introduced because the data demanded them, not because the theory produced them. The Dirac equation changes that completely. In seeking a relativistic wave equation first order in time — the fix for the negative probabilities of Klein–Gordon — Dirac was forced to a four-component wavefunction, and those extra components are not bookkeeping. As Thomson stresses, they are spin: the internal structure needed to make the equation Lorentz-covariant turns out to carry exactly one unit of intrinsic angular momentum with quantum number $s=\tfrac12$ . The proof is a conservation argument, short and beautiful. An observable is conserved when its operator commutes with the Hamiltonian, since $\tfrac{d}{dt}\langle\hat{O}\rangle=i\langle[\hat{H}_D,\hat{O}]\rangle$ . For a free Dirac particle $\hat{\mathbf{L}}$ does not commute with $\hat{H}_D$ — orbital angular momentum alone is not conserved, a jarring departure from the non-relativistic world. Dirac's four-by-four spin operator $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ fails to commute either, but by an equal and opposite amount. Their anomalies cancel, and only the total $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$ is conserved. Spin is what the equation must supply so that angular momentum is conserved at all. As a bonus, coupling the same equation to a magnetic field predicts the electron's gyromagnetic ratio $g=2$ — a number that had to be measured in the old theory now falling out of a single line of algebra.

Key takeaways

The Dirac Hamiltonian $\hat{H}_D=\boldsymbol{\alpha}\cdot\hat{\mathbf{p}}+\beta m$ does not commute with orbital angular momentum: $[\hat{H}_D,\hat{\mathbf{L}}]=-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}\neq0$ , so $\hat{\mathbf{L}}$ alone is not conserved.
A four-by-four operator $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ , built block-diagonally from the Pauli matrices, obeys $[\hat{H}_D,\hat{\mathbf{S}}]=+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ — equal and opposite to the orbital term.
The two anomalies cancel, so only $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$ commutes with $\hat{H}_D$ ; conservation of angular momentum forces the existence of $\hat{\mathbf{S}}$ .
Because $[\hat{S}_i,\hat{S}_j]=i\epsilon_{ijk}\hat{S}_k$ and $\hat{\mathbf{S}}^2=\tfrac34\mathbb{1}=s(s+1)\mathbb{1}$ , the particle carries spin $s=\tfrac12$ .
Minimally coupling the same equation to a field gives the Pauli term $-\tfrac{e}{2m}\boldsymbol{\sigma}\cdot\mathbf{B}$ , predicting the gyromagnetic ratio $g=2$ — measured today as $g=2.00231930436$ , the rest being QED.
Key idea: Neither orbital ( $\hat{\mathbf{L}}$ ) nor spin ( $\hat{\mathbf{S}}$ ) angular momentum is conserved alone for a Dirac particle, but their commutators with $\hat{H}_D$ are equal and opposite, so only the sum $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$ is conserved — and that requirement forces spin-half into existence.

Where this lives today

The chain that begins with "spin is forced by the Dirac equation" ends at the most precise confrontation between theory and experiment ever achieved. The electron gyromagnetic ratio, predicted at $g=2$ by Dirac and corrected by QED, is now measured in a single-electron cyclotron (the Harvard–Northwestern experiments of Gabrielse and Fan) to give $a_e=1.15965218059(13)\times10^{-3}$ — a fractional precision near one part in $10^{10}$ , agreeing with a QED calculation carried to five loops. Its sibling the muon tells a livelier story: the Fermilab Muon g-2 experiment (E989) has measured $a_\mu$ to sub-part-per-million precision, and whether it agrees with the Standard Model hinges on a still-unsettled hadronic contribution — so the muon's moment remains a live hunt for new physics. Beyond metrology, the spin-half structure Dirac derived organizes the whole Standard Model: every quark and lepton is a spinor field, and helicity — the good quantum number this article isolates — is the language of parity violation and of every high-energy scattering amplitude.

The physical picture

Start with a free electron of definite momentum and ask a naive question: is its orbital angular momentum a constant of the motion, as it would be for any Schrödinger central-force problem? The answer is no. Compute $[\hat{H}_D,\hat{\mathbf{L}}]$ and a residue survives, $-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ , because the Dirac Hamiltonian mixes the upper and lower spinor components through the $\boldsymbol{\alpha}$ matrices. Something is leaking angular momentum out of the orbital ledger, and Dirac's insight is that the missing piece lives in the spinor index itself — an internal angular momentum $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ whose own commutator with $\hat{H}_D$ is exactly $+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ . The first figure shows the two ledgers balancing.

Figure

The two anomalous commutators are equal and opposite: −iα×p from the orbital part L and +iα×p from the spin part S cancel exactly, so only the total J = L + S commutes with the Dirac Hamiltonian H_D and is therefore conserved

The second figure is the downstream payoff that made the result famous. Feed the same Dirac equation a magnetic field and the non-relativistic reduction hands you a magnetic moment $\boldsymbol{\mu}=g\,\tfrac{e}{2m}\hat{\mathbf{S}}$ with the gyromagnetic ratio pinned at $g=2$ — not the $g=1$ of a classical spinning charge, and not a number fitted to data. Experiment agrees to the first four digits for free; the small remainder is the fingerprint of quantum-field-theoretic loops.

Figure

Dirac's equation predicts the electron g-factor at tree level to be exactly 2, while QED radiative corrections lift it to the measured g = 2.00231930436; the gap is the anomaly a = (g−2)/2 = 0.00115965, whose leading term is Schwinger's α/2π = 0.00116141

The core equations

The starting point is the Dirac Hamiltonian, obtained by writing the Dirac equation $i\partial_t\psi=\hat{H}_D\psi$ in Schrödinger form. It is linear in momentum — that is the whole point of Dirac's construction — and it acts on a four-component spinor through the matrices $\boldsymbol{\alpha}$ and $\beta$ . We work in natural units $\hbar=c=1$ and the Dirac (standard) representation.

\hat{H}_D=\boldsymbol{\alpha}\cdot\hat{\mathbf{p}}+\beta m,\qquad \alpha^i=\begin{pmatrix}0&\sigma^i\\ \sigma^i&0\end{pmatrix},\quad \beta=\begin{pmatrix}\mathbb{1}&0\\ 0&-\mathbb{1}\end{pmatrix} \qquad (1)

where: $\boldsymbol{\alpha},\beta$ — four-by-four Dirac matrices satisfying $\{\alpha^i,\alpha^j\}=2\delta^{ij}$ , $\{\alpha^i,\beta\}=0$ , $\beta^2=1$ ; $\sigma^i$ — the $2\times2$ Pauli matrices; $\hat{\mathbf{p}}=-i\nabla$ ; $m$ — particle mass. The off-diagonal $\sigma^i$ in $\boldsymbol{\alpha}$ couple the upper and lower spinor blocks — the source of every non-commutation below. The central result is the pair of commutators of $\hat{H}_D$ with orbital and spin angular momentum. Each is nonzero — a shock coming from non-relativistic quantum mechanics, where $\hat{\mathbf{L}}$ commutes with a free Hamiltonian — but the two right-hand sides are negatives of each other.

[\hat{H}_D,\hat{\mathbf{L}}]=-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}},\qquad [\hat{H}_D,\hat{\mathbf{S}}]=+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}} \qquad (2)

where: $\hat{\mathbf{L}}=\hat{\mathbf{r}}\times\hat{\mathbf{p}}$ — orbital angular momentum; $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ — the Dirac spin operator of equation (3); $\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ — the anomalous term, present only because $\hat{H}_D$ is not rotationally inert on the spinor index. Read it as bookkeeping: orbital rotation feeds angular momentum into the internal index at exactly the rate the spin operator receives it. The spin operator is the Pauli triplet promoted to four dimensions by stacking it on the diagonal. It obeys the same $\mathfrak{su}(2)$ algebra as ordinary spin, which is why it deserves the name.

\hat{\mathbf{S}}=\tfrac{1}{2}\hat{\boldsymbol{\Sigma}},\qquad \Sigma^i=\begin{pmatrix}\sigma^i&0\\ 0&\sigma^i\end{pmatrix},\qquad [\hat{S}_i,\hat{S}_j]=i\,\epsilon_{ijk}\hat{S}_k \qquad (3)

where: $\hat{\boldsymbol{\Sigma}}$ — the block-diagonal spin matrices; $\epsilon_{ijk}$ — the Levi-Civita symbol. The commutation relations match those of $\hat{\mathbf{L}}$ , so $\hat{\mathbf{S}}$ is a genuine angular momentum; the next equation fixes $s$ . Adding the two operators gives the total angular momentum, which is conserved, and squaring the spin operator fixes the spin quantum number once and for all.

\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}},\qquad [\hat{H}_D,\hat{\mathbf{J}}]=0,\qquad \hat{\mathbf{S}}^2=\tfrac{3}{4}\,\mathbb{1}=s(s+1)\,\mathbb{1}\ \Rightarrow\ s=\tfrac{1}{2} \qquad (4)

where: $\hat{\mathbf{J}}$ — total angular momentum, the good quantum number for a rotationally invariant system; $\hat{\mathbf{S}}^2=\tfrac34\mathbb{1}$ because each $\Sigma_i^2=\mathbb{1}$ . Solving $s(s+1)=\tfrac34$ gives $s=\tfrac12$ : the Dirac field is intrinsically spin-half. Finally, the equation's most celebrated corollary. Couple it minimally to an electromagnetic field and reduce to the non-relativistic limit; a magnetic-moment term appears with a fixed gyromagnetic ratio.

\boldsymbol{\mu}=g\,\frac{e}{2m}\,\hat{\mathbf{S}},\qquad g=2\quad(\text{Dirac, tree level}) \qquad (5)

where: $\boldsymbol{\mu}$ — intrinsic magnetic moment; $e$ — charge magnitude; $g$ — gyromagnetic (Landé) ratio, equal to $1$ for classical orbital motion but $2$ for Dirac spin. The observed $g=2.00231930436$ exceeds $2$ by the QED anomaly, whose leading term is Schwinger's $a=\alpha/2\pi=0.00116$ .

Derivation AOrbital angular momentum is not conserved for a Dirac particle

The goal is to compute $[\hat{H}_D,\hat{\mathbf{L}}]$ from first principles and show it does not vanish, producing the anomalous term $-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ . This breaks the non-relativistic intuition: for a free Schrödinger particle $\hat{\mathbf{L}}$ commutes with $\hat{H}=\hat{\mathbf{p}}^2/2m$ and is a constant of the motion, but the linear-in-momentum Dirac Hamiltonian spoils this, and tracking exactly how it fails tells us what must be added back to restore conservation.

Step 1. Only the kinetic piece matters. Write $\hat{H}_D=\alpha_l\hat{p}_l+\beta m$ . The mass term $\beta m$ is a constant matrix times the identity in position space, so it commutes with every component of $\hat{\mathbf{L}}=\hat{\mathbf{r}}\times\hat{\mathbf{p}}$ . Hence $[\hat{H}_D,L_i]=[\alpha_l\hat{p}_l,L_i]=\alpha_l[\hat{p}_l,L_i]$ , since the numerical matrix $\alpha_l$ commutes with the spatial operators. Why: isolating the one non-commuting piece keeps the algebra honest — the mass term carries no orbital angular momentum content.

Step 2. Evaluate the momentum commutator. With $L_i=\epsilon_{ijk}\hat{x}_j\hat{p}_k$ and the canonical relation $[\hat{p}_l,\hat{x}_j]=-i\delta_{lj}$ , we get $[\hat{p}_l,L_i]=\epsilon_{ijk}[\hat{p}_l,\hat{x}_j]\hat{p}_k=\epsilon_{ijk}(-i\delta_{lj})\hat{p}_k=-i\,\epsilon_{ilk}\hat{p}_k$ . Why: orbital angular momentum generates rotations of momentum, so its commutator with a momentum component simply rotates that component — the source of the cross product to come.

Step 3. Assemble the vector. Substituting back, $[\hat{H}_D,L_i]=\alpha_l(-i\epsilon_{ilk}\hat{p}_k)=-i\,\epsilon_{ilk}\alpha_l\hat{p}_k=-i(\boldsymbol{\alpha}\times\hat{\mathbf{p}})_i$ , so $[\hat{H}_D,\hat{\mathbf{L}}]=-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ . This is nonzero whenever the particle moves, so $\hat{\mathbf{L}}$ is not conserved. Why: the residue is proportional to $\boldsymbol{\alpha}$ , the very matrix that mixes spinor components — a strong hint that the missing angular momentum has gone into the internal (spin) index.

Step 4. Sanity check against the non-relativistic limit. Replace $\hat{H}_D$ by $\hat{H}_{SE}=\hat{\mathbf{p}}^2/2m$ ; then $[\hat{H}_{SE},L_i]=\tfrac{1}{2m}[\hat{p}^2,L_i]=0$ because $\hat{p}^2$ is a rotational scalar. Orbital angular momentum is conserved there. Why: confirming that the anomaly is a genuinely relativistic effect — it is the linear, not quadratic, dependence on $\hat{\mathbf{p}}$ that produces it.

Go deeper · Derivation B · 1 minThe spin operator restores conservation and fixes s = 1/2

Derivation BThe spin operator restores conservation and fixes s = 1/2

The goal is to construct the operator whose commutator with $\hat{H}_D$ is exactly $+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ , cancelling the orbital anomaly, and then to show that this same operator squares to $\tfrac34$ , forcing spin one-half. We are reverse-engineering the missing conserved quantity: the anomaly in Derivation A tells us its commutator, and the $\mathfrak{su}(2)$ algebra tells us its spectrum, and the two together identify $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ uniquely as intrinsic spin.

Step 1. Commute $\hat{\boldsymbol{\Sigma}}$ with the kinetic term. Since $\beta$ is block-diagonal like $\Sigma_i$ , it commutes with $\Sigma_i$ , so $[\hat{H}_D,\Sigma_i]=\hat{p}_l[\alpha_l,\Sigma_i]$ . Using the block forms, $[\alpha_l,\Sigma_i]$ has the commutator $[\sigma_l,\sigma_i]=2i\epsilon_{lik}\sigma_k$ in both off-diagonal slots, giving $[\alpha_l,\Sigma_i]=2i\,\epsilon_{lik}\alpha_k$ . Why: the Pauli algebra inside each block is what promotes $\hat{\boldsymbol{\Sigma}}$ to a rotation generator on the spinor index, mirroring what $\hat{\mathbf{L}}$ does on the spatial index.

Step 2. Form the vector. Then $[\hat{H}_D,\Sigma_i]=\hat{p}_l\cdot2i\epsilon_{lik}\alpha_k=2i\,\epsilon_{lik}\alpha_k\hat{p}_l=2i(\boldsymbol{\alpha}\times\hat{\mathbf{p}})_i$ , using $\epsilon_{lik}=\epsilon_{ikl}$ . Dividing by two for $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ gives $[\hat{H}_D,\hat{\mathbf{S}}]=+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ . Why: the factor of $\tfrac12$ is not cosmetic — it is exactly what makes the spin anomaly equal and opposite to the orbital one, so their sum cancels.

Step 3. Add and cancel. Therefore $[\hat{H}_D,\hat{\mathbf{J}}]=[\hat{H}_D,\hat{\mathbf{L}}]+[\hat{H}_D,\hat{\mathbf{S}}]=-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}=0$ . The total angular momentum $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$ is conserved. Why: this is the physical statement — rotational invariance of a free relativistic particle guarantees a conserved $\hat{\mathbf{J}}$ , and the only way to build it is to include the internal $\hat{\mathbf{S}}$ .

Step 4. Extract the spin quantum number. Each $\Sigma_i^2=\mathrm{diag}(\sigma_i^2,\sigma_i^2)=\mathbb{1}$ , so $\hat{\mathbf{S}}^2=\tfrac14(\Sigma_x^2+\Sigma_y^2+\Sigma_z^2)=\tfrac34\mathbb{1}$ . Setting this equal to $s(s+1)$ gives $s(s+1)=\tfrac34$ , whose physical root is $s=\tfrac12$ . Combined with $[\hat{S}_i,\hat{S}_j]=i\epsilon_{ijk}\hat{S}_k$ , the operator is a bona fide spin-half. Why: this is the punchline — the Dirac field is not merely "compatible with" spin one-half, it is forced to be spin one-half by the structure of its own operators.

Go deeper · Derivation C · 1 minDirac predicts g = 2 from the non-relativistic reduction

Derivation CDirac predicts g = 2 from the non-relativistic reduction

The goal is to show that coupling the Dirac equation to an electromagnetic field and taking the slow-particle limit reproduces the Pauli equation with a magnetic-moment term whose gyromagnetic ratio is $g=2$ , not the classical $g=1$ . This is the second great consequence of spin-from-Dirac: the same four-component structure that supplied intrinsic angular momentum also supplies a definite intrinsic magnetic moment — the quantity the Stern–Gerlach experiment measures.

Step 1. Minimal coupling. Replace $\hat{\mathbf{p}}\to\boldsymbol{\pi}=\hat{\mathbf{p}}-e\mathbf{A}$ in the Dirac equation and split the four-spinor into upper and lower two-component pieces $\varphi$ and $\chi$ . In the non-relativistic regime the lower components are small, $\chi\approx\dfrac{\boldsymbol{\sigma}\cdot\boldsymbol{\pi}}{2m}\,\varphi$ , driven by the kinetic energy over the rest mass. Why: separating large and small components is the standard route to the Schrödinger-like equation hiding inside the Dirac equation.

Step 2. Eliminate the small components. Substituting $\chi$ back gives an equation for $\varphi$ alone whose kinetic operator is $(\boldsymbol{\sigma}\cdot\boldsymbol{\pi})^2/2m$ . Why: the whole $g=2$ result is buried in the difference between $(\boldsymbol{\sigma}\cdot\boldsymbol{\pi})^2$ and the naive $\pi^2$ — the spin sees the field where a spinless particle would not.

Step 3. Use the Pauli identity. For any vector operator, $(\boldsymbol{\sigma}\cdot\mathbf{a})(\boldsymbol{\sigma}\cdot\mathbf{b})=\mathbf{a}\cdot\mathbf{b}+i\,\boldsymbol{\sigma}\cdot(\mathbf{a}\times\mathbf{b})$ . With $\mathbf{a}=\mathbf{b}=\boldsymbol{\pi}$ , the cross term is $\boldsymbol{\pi}\times\boldsymbol{\pi}=-e(\hat{\mathbf{p}}\times\mathbf{A}+\mathbf{A}\times\hat{\mathbf{p}})=ie\,\mathbf{B}$ , so $(\boldsymbol{\sigma}\cdot\boldsymbol{\pi})^2=\pi^2-e\,\boldsymbol{\sigma}\cdot\mathbf{B}$ . Why: $\boldsymbol{\pi}$ does not commute with itself in a field — its self-cross-product is the magnetic field — and that is precisely what conjures the spin–field coupling out of the kinetic term.

Step 4. Read off the moment. The reduced Hamiltonian is $\dfrac{\pi^2}{2m}-\dfrac{e}{2m}\boldsymbol{\sigma}\cdot\mathbf{B}+e\phi$ ; writing $\hat{\mathbf{S}}=\tfrac12\boldsymbol{\sigma}$ the spin term is $-\dfrac{e}{m}\hat{\mathbf{S}}\cdot\mathbf{B}=-g\dfrac{e}{2m}\hat{\mathbf{S}}\cdot\mathbf{B}$ with $g=2$ . Why: the coefficient is twice the orbital $e/2m$ , so the electron's spin is twice as effective a magnet per unit angular momentum as its orbital motion — a purely relativistic prediction that the old spin models could not explain.

Limiting cases

Case	Result	Meaning
Non-relativistic ( $\hat{H}=\hat{p}^2/2m$ )	$[\hat{H},\hat{\mathbf{L}}]=0$	orbital $\hat{\mathbf{L}}$ conserved alone; spin must be added by hand (Pauli), not derived
Rest frame ( $\hat{\mathbf{p}}=0$ )	$\boldsymbol{\alpha}\times\hat{\mathbf{p}}=0$	both anomalies vanish; $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ are each separately conserved and spinors are pure $S_z$ states
Motion along $\hat{z}$	$\hat{S}_z$ commutes with $\hat{H}_D$	helicity $\hat{h}=\hat{\mathbf{S}}\cdot\hat{\mathbf{p}}/\|\mathbf{p}\|$ is a good quantum number; the plane-wave spinors are $S_z$ eigenstates
General motion	only $\hat{\mathbf{J}}$ and $\hat{h}$ good	$\hat{S}_z$ is not conserved; label states by helicity or total angular momentum, not $S_z$
Massless limit ( $m\to0$ )	helicity = chirality	$\hat{h}$ is Lorentz-invariant; left- and right-handed fields decouple (Weyl spinors)
Weak field, slow limit	$g=2$ , $\boldsymbol{\mu}=\tfrac{e}{m}\hat{\mathbf{S}}$	Pauli equation recovered; spin magnetic moment twice the orbital value per unit angular momentum

Common trapPlane-wave Dirac spinors are eigenstates of S_z

Students expect the free-particle solutions $u^{(s)}(\mathbf{p})$ to be clean eigenstates of $\hat{S}_z$ — pure spin-up and spin-down. In general they are not. Because $[\hat{H}_D,\hat{S}_z]\neq0$ unless the motion is along $\hat{z}$ , the label $s$ on $u^{(s)}$ is only a rest-frame spin that gets scrambled by the boost: a spinor built for momentum in a slanted direction is a superposition of $S_z$ up and down. The genuinely good quantum numbers are helicity (spin projected on the direction of motion, which does commute with $\hat{H}_D$ ) and total angular momentum $\hat{\mathbf{J}}$ . The mirror-image trap is just as common: do not assume $\hat{\mathbf{L}}$ is conserved as in non-relativistic quantum mechanics — for a Dirac particle it is not.

Worked examples

1 · The electron magnetic moment and its anomaly (fully worked)

The Dirac equation predicts $g=2$ . Compute the spin magnetic moment this implies in Bohr magnetons, then compare with the measured $g=2.00231930436$ and identify the leading correction.

Step 1. The moment along the field for a spin-up electron is $\mu_z=g\dfrac{e}{2m}S_z$ with $S_z=+\tfrac{\hbar}{2}$ . In Bohr magnetons $\mu_B=\dfrac{e\hbar}{2m}=9.274\times10^{-24}\ \mathrm{J/T}$ , this is $\mu_z=g\dfrac{\mu_B}{2}$ . With Dirac's $g=2$ the factors of two cancel: $\mu_z=\mu_B$ exactly. A Dirac electron has a spin moment of precisely one Bohr magneton.

Step 2. The measured value is slightly larger. Using $g=2.00231930436$ , $\mu_z=\dfrac{g}{2}\mu_B=1.00115965\,\mu_B$ . The excess over the Dirac prediction is the anomalous magnetic moment, $a=\dfrac{g-2}{2}=\dfrac{0.00231930436}{2}=\mathbf{0.00115965}$ . The tree-level Dirac result is right to four significant figures on its own.

Step 3. Where the excess comes from. Dirac's calculation is the tree-level (no-loop) result of QED; virtual photons and electron–positron pairs shift $g$ . Schwinger's one-loop calculation gives $a=\dfrac{\alpha}{2\pi}$ . With $\alpha=1/137.036$ , $\dfrac{\alpha}{2\pi}=\dfrac{0.00729735}{6.28319}=\mathbf{0.00116141}$ . This alone reproduces the measured anomaly to better than 0.2%; the remaining $0.00000176$ is higher-order QED plus tiny hadronic and weak contributions. This twelve-digit agreement rests on the spin the Dirac equation forced into existence.

2 · Why S_z is not conserved but helicity is

Show explicitly that for a Dirac particle moving in a general direction $\hat{S}_z$ fails to commute with $\hat{H}_D$ , while the helicity $\hat{h}=\hat{\boldsymbol{\Sigma}}\cdot\hat{\mathbf{p}}/|\mathbf{p}|$ does.

Step 1. From equation (2), $[\hat{H}_D,\hat{S}_z]=+i(\boldsymbol{\alpha}\times\hat{\mathbf{p}})_z=i(\alpha_x p_y-\alpha_y p_x)$ . This vanishes only when $p_x=p_y=0$ , i.e. motion along $\hat{z}$ . For a slanted momentum $\hat{S}_z$ is not conserved, so the plane-wave label "spin-up along z" is not preserved in time.

Step 2. Now project spin onto the momentum direction. Helicity commutes because $[\hat{H}_D,\hat{\boldsymbol{\Sigma}}\cdot\hat{\mathbf{p}}]=[\hat{H}_D,\hat{\boldsymbol{\Sigma}}]\cdot\hat{\mathbf{p}}=2i(\boldsymbol{\alpha}\times\hat{\mathbf{p}})\cdot\hat{\mathbf{p}}=0$ , since the cross product $\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ is orthogonal to $\hat{\mathbf{p}}$ and $\hat{\mathbf{p}}$ commutes with itself. Helicity is a good quantum number for any free Dirac particle, whatever its direction of motion.

Step 3. The physical moral. The eigenvalues of $\hat{h}$ are $\pm1$ (spin parallel or antiparallel to $\mathbf{p}$ ), the two physical polarization states of a massive spin-half particle. Label relativistic states by helicity, not $S_z$ — which is why particle physics is written in helicity and chirality, and it sets up the following article on spin and helicity states.

Physics of the portrait

Spin-from-Dirac is one person's result more than almost any other in modern physics. Paul Dirac wrote down his equation in a matter of weeks in late 1927, chasing a purely formal goal — a relativistic wave equation linear in time — and found spin and antimatter waiting inside it, neither one put in by hand.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · relativistic quantum mechanics and the birth of QED

Paul Adrien Maurice Dirac, trained as an electrical engineer in Bristol before turning to physics at Cambridge, was the most austere of the quantum pioneers — he trusted the beauty of an equation over any physical picture. In 1928 he sought a first-order relativistic wave equation to cure the negative probabilities of Klein–Gordon, and the price of first-order-in-time was a four-component wavefunction built on his anticommuting $\alpha$ and $\beta$ matrices. He then found that his equation automatically described a particle of spin one-half with $g=2$ — quantities that Uhlenbeck, Goudsmit and Pauli had introduced only three years earlier to fit spectra. He had not been trying to explain spin; it fell out unbidden, which Dirac took as evidence the mathematics was correct. The same equation's negative-energy solutions led him by 1931 to predict the positron, found by Anderson in 1932, and he shared the 1933 Nobel Prize with Schrödinger. The lesson of this article is his: intrinsic spin is not an assumption bolted onto quantum mechanics but a theorem of relativistic quantum mechanics, forced by the demand that a first-order, Lorentz-covariant equation conserve angular momentum.

History

1922

— Stern and Gerlach split a silver-atom beam in two, revealing space quantization

1925

— Uhlenbeck and Goudsmit propose that the electron carries an intrinsic spin of one-half

1927

— Pauli introduces two-component spinors and the Pauli matrices to describe spin non-relativistically

1928

— Dirac's relativistic equation predicts spin-half and g = 2 with no extra assumptions

1948

— Schwinger computes the anomalous moment a = α/2π, launching precision QED

Going further

Texts. Thomson, Modern Particle Physics, §4.4 (pp. 104–106), Eqs. 4.18–4.29 — the commutator derivation followed here · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7, §7.1–7.2, Eqs. 7.17–7.51 — the solution-side view that plane waves are not $S_z$ eigenstates · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.2–3.4 — spin and the Lorentz representation of the Dirac field · Halzen & Martin, Quarks and Leptons, §5.3–5.4 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 10–11 — the $g=2$ prediction and its QED corrections. The five key results in this article are: (1) $[\hat{H}_D,\hat{\mathbf{L}}]=-i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ , which follows from $[\hat{p}_l,L_i]=-i\epsilon_{ilk}\hat{p}_k$ ; (2) $[\hat{H}_D,\hat{\mathbf{S}}]=+i\,\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ , which follows from $[\alpha_l,\Sigma_i]=2i\epsilon_{lik}\alpha_k$ ; (3) $\hat{\mathbf{S}}^2=\tfrac34\mathbb{1}\Rightarrow s=\tfrac12$ ; (4) $(\boldsymbol{\sigma}\cdot\boldsymbol{\pi})^2=\pi^2-e\,\boldsymbol{\sigma}\cdot\mathbf{B}\Rightarrow g=2$ ; (5) the numerical anomaly $a=(g-2)/2=0.00115965$ , compared with the Schwinger one-loop value $\alpha/2\pi=0.00116141$ .

Prerequisites. The Dirac Equation · Klein–Gordon Equation · Dirac Equation Solutions. Leads to. Spin and Helicity States · Antiparticles · Covariant Form of the Dirac Equation.

Ask the tutor: Ask me to walk through the commutator algebra step by step, or to show explicitly why a boosted Dirac spinor stops being an $S_z$ eigenstate.

Does the Dirac equation really "explain" spin, or just contain it?

It does more than contain it — it forces it. In non-relativistic quantum mechanics you can build a perfectly consistent theory with no spin at all; Pauli had to add spinors and matrices because experiments (Stern–Gerlach, spectral doublets) demanded them. In the Dirac equation you have no such freedom: the requirement of a first-order, Lorentz-covariant equation dictates a four-component wavefunction, and the requirement that angular momentum be conserved then dictates that those components carry exactly $s=\tfrac12$ . You cannot write down a spinless Dirac equation. In that strong sense spin is derived, not assumed — which is why the result felt like magic to Dirac's contemporaries.

Why is the spin exactly one-half and not some other value?

Because $\hat{\mathbf{S}}^2$ comes out to $\tfrac34$ times the identity, and $s(s+1)=\tfrac34$ has physical solution $s=\tfrac12$ . The deeper reason is representation theory: the Dirac field transforms in the $(\tfrac12,0)\oplus(0,\tfrac12)$ representation of the Lorentz group, the smallest one that includes both parities, and that representation is spin-half by construction. Higher-spin relativistic fields exist (the Rarita–Schwinger equation describes spin- $\tfrac32$ , and the graviton is spin-2), but they require more components and more constraints. The four-component Dirac spinor is the minimal, most economical relativistic field, and it is spin-half.

What is the difference between Σ and the Pauli matrices σ?

The Pauli matrices $\sigma_i$ are $2\times2$ and act on the two-component spinors of non-relativistic quantum mechanics. The Dirac spin operator needs to act on four-component objects, so $\Sigma_i$ is the $4\times4$ block-diagonal matrix with $\sigma_i$ in each diagonal block. Physically, $\Sigma_i$ applies the same spin rotation to the upper and lower two-spinors simultaneously, and because each block obeys the Pauli algebra, $\hat{\mathbf{S}}=\tfrac12\hat{\boldsymbol{\Sigma}}$ inherits it: $[\hat{S}_i,\hat{S}_j]=i\epsilon_{ijk}\hat{S}_k$ and $\hat{\mathbf{S}}^2=\tfrac34$ , exactly as ordinary spin does.

Why is g = 2 and not g = 1?

A classical charge whirling in a loop has $\boldsymbol{\mu}=\tfrac{e}{2m}\mathbf{L}$ , i.e. $g=1$ — the moment proportional to angular momentum with the "obvious" coefficient. When Uhlenbeck and Goudsmit proposed electron spin, matching atomic spectra forced them to a spin moment twice as large, $g=2$ , which looked arbitrary. The Dirac equation resolves it: in the non-relativistic reduction the spin–field coupling arises from $(\boldsymbol{\sigma}\cdot\boldsymbol{\pi})^2$ , and the Pauli identity delivers $-\tfrac{e}{2m}\boldsymbol{\sigma}\cdot\mathbf{B}=-\tfrac{e}{m}\mathbf{S}\cdot\mathbf{B}$ with the factor of two already present. So $g=2$ is not a coincidence but a theorem — the first quantitative triumph of the theory.

Is spin really the electron "spinning" on its axis?

No. Model the electron as a charged ball of the classical electron radius and demand angular momentum $\tfrac{\hbar}{2}$ , and its surface must move faster than light — the picture is untenable. Spin is an intrinsic, purely quantum degree of freedom with no classical analogue: the label of how the wavefunction transforms under rotations, encoded in the spinor index the Dirac equation forces to exist. The word "spin" is historical. What is physically real is that the electron carries angular momentum $\tfrac{\hbar}{2}$ and magnetic moment $g\tfrac{e}{2m}\mathbf{S}$ that couple to fields and add into $\hat{\mathbf{J}}$ — all of which the Dirac equation gets right without any rotating charge.

Why isn't S_z a good quantum number for a moving electron?

Because $\hat{S}_z$ does not commute with the Dirac Hamiltonian unless the momentum points along $\hat{z}$ : from equation (2), $[\hat{H}_D,\hat{S}_z]=i(\alpha_x p_y-\alpha_y p_x)$ , nonzero for a slanted momentum. A boost mixes the spin projection with orbital motion, so a state prepared as "spin-up along $z$ " does not stay that way. The projection that does survive is spin along the direction of motion — the helicity — because the anomalous term $\boldsymbol{\alpha}\times\hat{\mathbf{p}}$ is perpendicular to $\hat{\mathbf{p}}$ and drops out when you dot with $\hat{\mathbf{p}}$ . This is why relativistic particle physics classifies states by helicity (and, for massless particles, chirality) rather than by $S_z$ .