Nonrelativistic Reduction and the Pauli Equation

Solve the Dirac equation for a slow electron and something remarkable falls out for free: the electron's magnetic moment, with the gyromagnetic ratio pinned at $g=2$ . Nobody put it in. For twenty years spin had been an add-on — Pauli bolted a two-component wavefunction and a $g=2$ moment onto the Schrödinger equation by hand, admitting he did not know where the factor of two came from. The Dirac equation delivers it as a theorem. This article shows exactly how, by reducing the four-component Dirac spinor to its slow, nonrelativistic core — the Pauli equation. The mechanism is a slaving relation. In the standard representation the Dirac spinor splits into an upper pair $\varphi$ and a lower pair $\chi$ ; for a particle moving at $v\ll c$ the lower pair is smaller by $v/c$ and entirely determined by the upper, $\chi\approx(\boldsymbol\sigma\cdot\boldsymbol\pi)/2mc\,\varphi$ . Fold that back in and the four-component problem collapses to a two-component Schrödinger equation. The subtlety that does all the work: the kinetic term arrives squared as $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ , and because the covariant momentum $\boldsymbol\pi$ does not commute with itself in a magnetic field, that square is not $\boldsymbol\pi^2$ . The leftover is the Zeeman term, its coefficient exactly twice the orbital one. We own two machine-proved facts: the identity $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-(q\hbar/c)\,\boldsymbol\sigma\cdot\mathbf B$ and the Pauli Hamiltonian it produces. One honest caveat sets the scope: $g=2$ is the Dirac prediction; the measured $2.0023$ is a QED radiative effect well beyond this article. What the Dirac equation gives, exactly and for free, is the $2$ .

Key takeaways

The Pauli equation is the nonrelativistic ( $v\ll c$ ) limit of the Dirac equation: eliminate the small lower components $\chi\approx(\boldsymbol\sigma\cdot\boldsymbol\pi)/2mc\,\varphi$ in favor of the large $\varphi$ .
Owned identity: $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-\tfrac{q\hbar}{c}\,\boldsymbol\sigma\cdot\mathbf B$ , proved from $[\pi_i,\pi_j]=i\hbar\tfrac{q}{c}\varepsilon_{ijk}B_k$ — the Zeeman term is born from the noncommutativity of $\boldsymbol\pi$ , not added by hand.
Owned result: the reduction gives $H_{\text{Pauli}}=\dfrac{\boldsymbol\pi^2}{2m}-\dfrac{q\hbar}{2mc}\,\boldsymbol\sigma\cdot\mathbf B+q\phi$ , the spin moment carrying $g=2$ — twice the orbital gyromagnetic ratio.
Scope (honest): $g=2$ is the Dirac prediction; the measured $2.0023$ is a QED loop effect beyond this article. Numbers: $\mu_B B=5.79\times10^{-5}\,\text{eV}$ at $1\,\text{T}$ , spin gap $g\mu_B B=1.16\times10^{-4}\,\text{eV}$ , electron-spin Larmor $\approx28\,\text{GHz/T}$ .
The next order in $1/mc$ produces spin-orbit, Darwin, and the $p^4$ correction — but a naive iteration is non-Hermitian, which is why the systematic route is the Foldy-Wouthuysen transformation.
Key idea: reduce the Dirac spinor by slaving its small components, $\chi\approx(\boldsymbol\sigma\cdot\boldsymbol\pi)/2mc\,\varphi$ ; the Dirac kinetic term arrives as $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ , whose magnetic piece $-\tfrac{q\hbar}{c}\boldsymbol\sigma\cdot\mathbf B$ is the Zeeman coupling with $g=2$ , delivered by relativity rather than by hand.

The core insight

Reduce the Dirac spinor by slaving its small components, $\chi\approx(\boldsymbol\sigma\cdot\boldsymbol\pi)/2mc\,\varphi$ ; the Dirac kinetic term appears as $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-\tfrac{q\hbar}{c}\boldsymbol\sigma\cdot\mathbf B$ , whose magnetic piece is the Zeeman coupling with $g=2$ — the electron's magnetic moment delivered by relativity, not inserted by hand.

Where this lives today

The Pauli equation is what "the Schrödinger equation with spin" actually means, and it is the quiet workhorse under an enormous amount of modern physics. Every magnetic-resonance experiment (NMR, MRI, electron-spin resonance) is Pauli precession read off hardware, the $g=2$ moment setting the Larmor frequency the scanner tunes to; spin qubits in quantum dots and donor atoms are governed by exactly the $-\tfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ term, single-qubit gates being literal solutions under engineered pulses. The one number the bare Dirac equation gets slightly wrong — the anomalous $0.0023$ — became, in QED's hands, the most stringent test of any physical theory: measured and computed electron moments agree to better than one part in $10^{12}$ , a triumph that begins exactly where this article's honest $g=2$ ends.

The physical picture

A slow relativistic electron carries baggage it barely uses. The four-component Dirac spinor keeps its lower components alive, but when the electron ambles along at $v\ll c$ those components shrink to order $v/c$ and stop being independent — they become a shadow of the upper pair, rigidly slaved to it. Reducing the Dirac equation is nothing more than writing down that shadow relation and substituting it away, leaving a genuine two-component particle whose state is the Pauli spinor $\varphi$ . The magnetic moment is not inserted into this theory — it is discovered inside it. In the Schrödinger world a charged particle has kinetic energy $\boldsymbol\pi^2/2m$ with $\boldsymbol\pi=\mathbf p-\tfrac{q}{c}\mathbf A$ ; in the Dirac world the same slot is filled by $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2/2m$ . If $\boldsymbol\pi$ commuted with itself these would be identical, but two components of $\boldsymbol\pi$ fail to commute when a magnetic field threads between them, $[\pi_x,\pi_y]\propto B_z$ . That same failure quantizes cyclotron orbits into Landau levels — and here it leaves a residue $-\tfrac{q\hbar}{c}\boldsymbol\sigma\cdot\mathbf B$ , the spin's coupling to the field. The figure below is the whole story on one axis: orbital angular momentum and spin both couple to $\mathbf B$ , but spin couples twice as hard — the factor of two Pauli could name but not explain, handed over by the Dirac reduction as a theorem.

Figure

The g=2 punchline: spin couples to B twice as hard as orbital motion

From four components to two

The reduction is a controlled elimination. Throughout, $q$ is the particle's charge (for the electron $q=-e<0$ with $e>0$ ), $m$ its mass, Gaussian units keeping the factors of $c$ visible. The starting point is the Dirac equation with minimal coupling — a first-order equation for a four-component spinor whose off-diagonal $\boldsymbol\alpha$ matrices carry information between the upper and lower blocks:

\left[\,c\,\boldsymbol\alpha\cdot\boldsymbol\pi+\beta mc^2+q\phi\,\right]\psi=i\hbar\,\partial_t\psi,\qquad \boldsymbol\pi=\mathbf p-\frac{q}{c}\mathbf A \qquad (1)

where: $\psi$ is the four-component spinor; $\boldsymbol\alpha=\begin{pmatrix}0&\boldsymbol\sigma\\\boldsymbol\sigma&0\end{pmatrix}$ (off-diagonal, coupling the blocks) and $\beta=\begin{pmatrix}\mathbb 1&0\\0&-\mathbb 1\end{pmatrix}$ (diagonal, giving $+mc^2$ to the upper block, $-mc^2$ to the lower); $\boldsymbol\pi$ the gauge-covariant momentum; $\phi$ the scalar potential. The off-diagonal $c\,\boldsymbol\alpha\cdot\boldsymbol\pi$ is precisely the term the reduction must tame. Everything hinges on one non-classical fact about $\boldsymbol\pi$ . Its Cartesian components commute for a free particle, but in a magnetic field they do not — and the amount by which they fail is the field itself:

[\pi_i,\pi_j]=i\hbar\,\frac{q}{c}\,\varepsilon_{ijk}B_k,\qquad \mathbf B=\nabla\times\mathbf A \qquad (2)

where: $\varepsilon_{ijk}$ is the Levi-Civita symbol and $B_k$ the magnetic field. It follows from the canonical $[p_i,A_j]=-i\hbar\,\partial_iA_j$ and the curl $\mathbf B=\nabla\times\mathbf A$ . This single nonzero commutator is the seed of the entire magnetic moment; if $\mathbf B=0$ it vanishes and spin becomes a spectator. The bridge from the squared kinetic term to physics is the Pauli vector identity, valid even for non-commuting operators because it never assumes $\mathbf a,\mathbf b$ commute:

(\boldsymbol\sigma\cdot\mathbf a)(\boldsymbol\sigma\cdot\mathbf b)=(\mathbf a\cdot\mathbf b)\,\mathbb 1+i\,\boldsymbol\sigma\cdot(\mathbf a\times\mathbf b) \qquad (3)

where: $\mathbf a,\mathbf b$ are any vector operators. Set $\mathbf a=\mathbf b=\boldsymbol\pi$ and the second term — zero for ordinary vectors — survives because $\boldsymbol\pi\times\boldsymbol\pi=\tfrac12$ (its commutator) $\ne 0$ . That surviving term is the whole game.

Derivation AThe magnetic square — why the Zeeman term is born from non-commuting momentum

We prove the operator identity $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-\tfrac{q\hbar}{c}\,\boldsymbol\sigma\cdot\mathbf B$ — the algebraic heart of the reduction. The strategy: apply the Pauli identity with both vectors equal to $\boldsymbol\pi$ , then recognize that the cross-product of an operator with itself is not zero but half its commutator, which the previous section fixed to be proportional to $\mathbf B$ . Watch the two factors of $i$ conspire to turn the imaginary cross term into a real, Hermitian magnetic energy — if you ever see this identity written with a $+$ sign, a factor of $i$ was dropped.

Step 1. Apply the Pauli identity (3) with $\mathbf a=\mathbf b=\boldsymbol\pi$ :

(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi\cdot\boldsymbol\pi+i\,\boldsymbol\sigma\cdot(\boldsymbol\pi\times\boldsymbol\pi).

Why we can't just drop the second term: for ordinary numbers $\boldsymbol\pi\times\boldsymbol\pi=0$ , but $\boldsymbol\pi$ is an operator that does not commute with itself in a field, so its self-cross-product is an operator, not zero.

Step 2. Evaluate the self-cross-product as an antisymmetrized product, i.e. a commutator:

(\boldsymbol\pi\times\boldsymbol\pi)_k=\varepsilon_{kij}\pi_i\pi_j=\tfrac12\varepsilon_{kij}[\pi_i,\pi_j].

Why the factor of one-half: $\varepsilon_{kij}$ is antisymmetric in $i,j$ , so it picks out only the antisymmetric (commutator) part of $\pi_i\pi_j$ , and the symmetric part cancels.

Step 3. Insert the commutator (2) and contract the Levi-Civita symbols using $\varepsilon_{kij}\varepsilon_{lij}=2\delta_{kl}$ :

(\boldsymbol\pi\times\boldsymbol\pi)_k=\tfrac12\varepsilon_{kij}\,i\hbar\tfrac{q}{c}\varepsilon_{ijl}B_l=i\hbar\tfrac{q}{c}B_k.

So the self-cross-product of the kinetic momentum is literally the magnetic field, dressed with $i\hbar q/c$ .

Step 4. Substitute back and let the two $i$ 's meet:

(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2+i\,\boldsymbol\sigma\cdot\Big(i\hbar\tfrac{q}{c}\mathbf B\Big)=\boldsymbol\pi^2-\frac{q\hbar}{c}\,\boldsymbol\sigma\cdot\mathbf B.

Why the sign is negative: $i\cdot i=-1$ . The result is manifestly Hermitian — $\boldsymbol\sigma\cdot\mathbf B$ is Hermitian and the coefficient is real — which is exactly what a physical energy must be. *Verified numerically: the symbolic identity holds for arbitrary $\mathbf B$ (sympy), and the explicit commutator $[\pi_x,\pi_y]=i\hbar\tfrac{q}{c}B$ was confirmed by acting the differential operators on a test function with the symmetric gauge $\mathbf A=\tfrac12\mathbf B\times\mathbf r$ .*

Go deeper · Derivation B · 1 minSlaving the small components — the Dirac equation becomes the Pauli equation

Derivation BSlaving the small components — the Dirac equation becomes the Pauli equation

We reduce the four-component Dirac equation (1) to a two-component Schrödinger equation for $\varphi$ and read off the magnetic moment. The route: peel off the fast rest-energy phase $e^{-imc^2t/\hbar}$ to expose the slow envelope, note that the lower components then feel a huge $-2mc^2$ offset that pins them to the upper ones, solve that slaving relation algebraically, and substitute back. Watch that this leading-order elimination is not a unitary change of basis, so it is trustworthy only for the Pauli equation itself — pushing it naively to next order breaks Hermiticity, the cliffhanger this article hands to Foldy-Wouthuysen.

Step 1. Write (1) in $2\times2$ block form with $\psi=(\varphi,\chi)^{\mathsf T}$ , using the off-diagonal $\boldsymbol\alpha$ and diagonal $\beta$ :

i\hbar\,\partial_t\begin{pmatrix}\varphi\\\chi\end{pmatrix}=\begin{pmatrix}q\phi+mc^2 & c\,\boldsymbol\sigma\cdot\boldsymbol\pi\\ c\,\boldsymbol\sigma\cdot\boldsymbol\pi & q\phi-mc^2\end{pmatrix}\begin{pmatrix}\varphi\\\chi\end{pmatrix}.

Why the blocks differ by $2mc^2$ : $\beta$ gives $+mc^2$ to $\varphi$ and $-mc^2$ to $\chi$ ; this enormous gap is what will make $\chi$ small.

Step 2. Strip the rest-energy phase, $\psi=e^{-imc^2t/\hbar}(\tilde\varphi,\tilde\chi)^{\mathsf T}$ . The phase cancels $mc^2$ in the upper equation and stacks up to $-2mc^2$ in the lower one:

i\hbar\,\partial_t\tilde\varphi=c\,\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\chi+q\phi\,\tilde\varphi,\qquad i\hbar\,\partial_t\tilde\chi=c\,\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\varphi+(q\phi-2mc^2)\tilde\chi.

\tilde\chi\simeq\frac{c\,\boldsymbol\sigma\cdot\boldsymbol\pi}{2mc^2}\,\tilde\varphi=\frac{\boldsymbol\sigma\cdot\boldsymbol\pi}{2mc}\,\tilde\varphi.

Why $\chi$ is small: it is $\tilde\varphi$ multiplied by $\boldsymbol\sigma\cdot\boldsymbol\pi/2mc\sim p/2mc\sim v/2c$ , so the lower pair is suppressed by $v/c$ — for a hydrogen electron ( $v/c\sim\alpha$ ) that is a $0.4\%$ shadow.

Step 4. Substitute the slaving relation into the upper equation:

i\hbar\,\partial_t\tilde\varphi=\frac{(\boldsymbol\sigma\cdot\boldsymbol\pi)^2}{2m}\,\tilde\varphi+q\phi\,\tilde\varphi.

The squared Dirac kinetic term has appeared exactly where the Schrödinger $\boldsymbol\pi^2/2m$ would sit.

Step 5. Deploy the magnetic-square identity from Derivation A and drop tildes:

i\hbar\,\partial_t\varphi=\left[\frac{\boldsymbol\pi^2}{2m}-\frac{q\hbar}{2mc}\,\boldsymbol\sigma\cdot\mathbf B+q\phi\right]\varphi.

Why this is a triumph: $\boldsymbol\pi^2/2m$ is ordinary kinetic energy, but the new $-\tfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ is a spin-magnetic coupling no nonrelativistic ansatz produced — the moment the Dirac equation predicts. *Verified numerically: the block structure ( $\{\boldsymbol\alpha,\beta\}$ signs, $\alpha_i^2=\mathbb 1$ ) and the $g=2$ read-off were checked symbolically; the Pauli Hamiltonian matches Bjorken–Drell §4.3.*

The magnetic term reads off a moment. Comparing $H_{\text{spin}}=-\tfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ with $-\boldsymbol\mu\cdot\mathbf B$ , $\boldsymbol\mu=g\tfrac{q}{2mc}\mathbf S$ and $\mathbf S=\tfrac{\hbar}{2}\boldsymbol\sigma$ forces the gyromagnetic ratio:

\boldsymbol\mu=\frac{q\hbar}{2mc}\,\boldsymbol\sigma=g\,\frac{q}{2mc}\,\mathbf S\quad\Longrightarrow\quad g=2 \qquad (4)

where: $\boldsymbol\mu$ is the intrinsic moment, $g$ the gyromagnetic ratio, $\mathbf S=\tfrac{\hbar}{2}\boldsymbol\sigma$ the spin. Orbital motion, hidden inside $\boldsymbol\pi^2/2m$ , carries $g_\ell=1$ ; spin couples twice as strongly. This is the Dirac prediction — and only at this order: the measured $g=2.0023$ carries a further $0.1\%$ from QED vacuum fluctuations the Dirac equation alone does not contain.

Limiting cases

case	result	meaning
no field, $\mathbf A=0,\ \phi=0$	$i\hbar\partial_t\varphi=\dfrac{\mathbf p^2}{2m}\varphi$	two decoupled Schrödinger equations; spin is a spectator
uniform weak $\mathbf B$	$H=\dfrac{\boldsymbol\pi^2}{2m}-\dfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B+q\phi$	Landau orbital quantization plus a $g=2$ spin Zeeman split
orbital vs spin coupling	$-\dfrac{q}{2mc}(\mathbf L+2\mathbf S)\cdot\mathbf B$	the factor 2 on $\mathbf S$ is exactly $g=2$ ; the anomalous Zeeman effect
next order in $1/mc$	$-\dfrac{\mathbf p^4}{8m^3c^2}+H_{\text{SO}}+H_{\text{Darwin}}$	fine structure — obtained cleanly only via Foldy-Wouthuysen
$v\sim c$	slaving fails	large and small components comparable; full Dirac theory required

Common trapThe g=2 term is added to the Schrödinger equation by hand — no, the Dirac reduction produces it

The most common conceptual error is to treat the spin-magnetic coupling $-\tfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ as an extra ingredient bolted onto the Schrödinger equation, as Pauli did in 1927. It is not an ingredient — it is a consequence. Once you accept the Dirac equation the term is forced: the kinetic operator arrives as $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ , and the Pauli identity splits that into $\boldsymbol\pi^2$ plus a Zeeman term whose coefficient is fixed, with no freedom to adjust $g$ . A spinless particle with the same minimal coupling has $\boldsymbol\pi^2/2m$ and no such term — the difference is entirely the presence of $\boldsymbol\sigma$ . The tell that you have slipped is treating $g$ as free at this order: at the Dirac level it is $2$ , full stop, and the observed $2.0023$ is a separate, QED story. Confusing "the Dirac equation predicts $g=2$ " with "predicts the measured moment" overclaims by exactly the anomalous $0.0023$ .

Figure

The reduction: four Dirac components collapse to two slaved by v/c

Worked examples

Example 1 — the electron-spin Zeeman splitting at one tesla (confidence builder). A free electron sits in a uniform field $\mathbf B=B\hat z$ , $B=1\,\text{T}$ . Find each spin level and the gap.

Step 1. Isolate the spin term. The spin part of $H_{\text{Pauli}}$ is $H_{\text{spin}}=-\tfrac{q\hbar}{2mc}B\,\sigma_z$ ; the natural unit is the Bohr magneton $\mu_B=\tfrac{e\hbar}{2m_ec}=5.788\times10^{-5}\,\text{eV/T}$ . At $B=1\,\text{T}$ , $\mu_B B=\mathbf{5.79\times10^{-5}\,\text{eV}}$ (machine-verified from CODATA $\mu_B=9.274\times10^{-24}\,\text{J/T}$ ).

Step 2. The eigenvalues. Writing the spin energy as $g\mu_B B\,m_s$ with $g=2,\ m_s=\pm\tfrac12$ , the levels are $E_\pm=\pm\mu_B B=\pm5.79\times10^{-5}\,\text{eV}$ — each shifts by exactly one Bohr magneton because $g=2$ and $m_s=\tfrac12$ multiply to one.

Step 3. The gap. The splitting is $\Delta E=g\mu_B B=2\mu_B B=\mathbf{1.16\times10^{-4}\,\text{eV}}$ (machine-verified). With the orbital value $g=1$ it would be half as large — the doubling is the direct fingerprint of the Dirac $g=2$ . As a frequency this is a Larmor precession of $\approx\mathbf{28\,\text{GHz/T}}$ , the working number of every electron-spin-resonance spectrometer.

Example 2 — the anomalous Zeeman effect and the factor of two (exam level). A valence electron in a $p$ state ( $\ell=1$ ) feels a weak field. Show that the combined Zeeman Hamiltonian carries a factor $2$ on spin, and evaluate the extreme-state shifts.

Step 1. Assemble the coupling. Expanding $\boldsymbol\pi^2/2m$ to first order in $\mathbf A=\tfrac12\mathbf B\times\mathbf r$ gives the orbital term $-\tfrac{q}{2mc}\mathbf L\cdot\mathbf B$ ; Derivation A gave the spin term $-\tfrac{q}{mc}\mathbf S\cdot\mathbf B$ . Together $H_Z=-\tfrac{q}{2mc}(\mathbf L+2\mathbf S)\cdot\mathbf B$ . The $2$ on $\mathbf S$ is $g=2$ made visible; a $g=1$ world would give the symmetric $\mathbf L+\mathbf S=\mathbf J$ and the normal Zeeman effect, which nature does not show.

Step 2. Read the projections. With $\mathbf B=B\hat z$ , $H_Z=\mu_B B\,(m_\ell+2m_s)$ for an electron ( $q=-e$ ), so the level pattern is set by the integer $m_\ell+2m_s$ , not by $m_j=m_\ell+m_s$ — the hallmark of anomalous splitting.

Step 3. Evaluate extremes. Stretched state $m_\ell=+1,\ m_s=+\tfrac12$ : $\Delta E=\mu_B B(2)=\mathbf{1.16\times10^{-4}\,\text{eV}}$ at $1\,\text{T}$ . State $m_\ell=+1,\ m_s=-\tfrac12$ : $\Delta E=\mu_B B(0)=\mathbf{0}$ — a level that does not shift at all despite carrying angular momentum, because orbital and spin cancel. That zero-shift line is impossible in the $g=1$ picture and was one of the sharpest early confirmations that spin's $g$ is $2$ .

The moment that had no explanation

Origins · 1 minpauli matrices and how this was discovered

pauli matrices

introduced 1927 · Zürich, Hamburg

Before the Dirac equation, spin was a fact in search of a theory. In 1925 Uhlenbeck and Goudsmit proposed the electron's intrinsic $\hbar/2$ angular momentum, and spectroscopy demanded its magnetic moment be twice what a classical spinning charge would give — a $g=2$ that fit the data but explained nothing. In 1927 Wolfgang Pauli formalized the two-valued spin by introducing the three $2\times2$ Pauli matrices $\sigma_x,\sigma_y,\sigma_z$ and a nonrelativistic two-component equation with a $-\tfrac{e\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ term inserted by hand — a phenomenological input he was uneasy about. A year later Dirac's equation swept the mystery away: the very same Pauli matrices reappear inside the four-component Dirac spinor, and the reduction here shows the $g=2$ term dropping out with no adjustable coefficient. The anomaly $g=2.0023$ waited two decades for Schwinger's 1948 QED correction $\alpha/2\pi$ ; the electron moment is now the most precisely verified prediction in physics.

History

1925

— Uhlenbeck and Goudsmit propose electron spin ½; spectroscopy demands g=2

1927

— Pauli introduces the 2×2 spin matrices and a two-component equation, g=2 by hand

1928

— Dirac's equation; its nonrelativistic reduction yields g=2 as a theorem

1948

— Schwinger computes the QED anomaly, g = 2 + α/π + ..., giving 2.0023

2008

— Harvard measures the electron g to 0.28 parts per trillion, matching QED

Going further

References. Bjorken & Drell, Relativistic Quantum Mechanics, §4.3 · Sakurai, Advanced Quantum Mechanics, §3.3 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., ch. 8 · Griffiths, Elementary Particles, 2nd ed., ch. 7 · Shankar, Principles of Quantum Mechanics, 2nd ed., ch. 20 · Schwinger, Phys. Rev. 73, 416 (1948) · CODATA 2018 ( $g=2.00231930436$ ).

Provenance. The Dirac equation, its gamma matrices, and the plane-wave spinors are owned by The Dirac Equation, Dirac Gamma Matrices and Clifford Algebra, and Dirac Spinors and Plane-Wave Solutions; the emergence of spin and $g=2$ as a structural fact of the Dirac equation is developed in Spin from the Dirac Equation, and the Pauli matrices in Spin-1/2 and the Pauli Matrices — all recalled by name. Owned and machine-proved here: (1) the identity $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2=\boldsymbol\pi^2-\tfrac{q\hbar}{c}\boldsymbol\sigma\cdot\mathbf B$ , checked symbolically and via $[\pi_x,\pi_y]=i\hbar\tfrac{q}{c}B$ in the symmetric gauge; (2) the Pauli Hamiltonian and the $g=2$ read-off. Numbers recomputed: $\mu_B B=5.79\times10^{-5}\,\text{eV}$ , spin gap $1.16\times10^{-4}\,\text{eV}$ at $1\,\text{T}$ , Larmor $\approx28\,\text{GHz/T}$ . Scope: $g=2$ is the Dirac prediction; $2.0023$ is a QED effect beyond this article. The systematic Hermitian route to the next order is owned by Foldy-Wouthuysen Transformation, the resulting fine-structure operators by Relativistic Corrections to the Hydrogen Atom, both by name.

Builds on: The Dirac Equation · Dirac Spinors and Plane-Wave Solutions · Spin from the Dirac Equation · Dirac Gamma Matrices and Clifford Algebra · Spin-1/2 and the Pauli Matrices

Leads to: Foldy-Wouthuysen Transformation · Relativistic Corrections to the Hydrogen Atom · Charge Conjugation and the Positron

Where does the factor of two in g=2 actually come from — I can see it in the algebra but not physically?

It comes from the fact that the electron's kinetic energy in the Dirac theory is built from $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ rather than $\boldsymbol\pi^2$ : spin is woven into the momentum operator, so squaring the momentum automatically drags in a spin-field coupling. The orbital moment comes from the ordinary $\boldsymbol\pi^2/2m$ cross term $-\tfrac{q}{2mc}\mathbf L\cdot\mathbf B$ , coefficient $q/2mc$ . The spin coupling emerges with coefficient $q/mc$ — exactly twice — because it descends from the commutator $[\pi_i,\pi_j]$ , a genuinely quantum object with no classical analog. The "two" is the ratio of a noncommutator effect to a classical one; there is no way to get spin's moment from a picture of a literally spinning charge.

Is the Pauli equation just the Schrödinger equation with a spin label glued on?

Almost, but the gluing is where the physics is. Attach a two-component spinor to the plain Schrödinger equation and spin is inert — nothing couples to it. The Pauli equation is the Schrödinger equation plus the specific term $-\tfrac{q\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ , whose coefficient the Dirac reduction fixes. So it is "Schrödinger with spin coupled correctly," and the correctness is exactly the $g=2$ . Where $\mathbf B=0$ everywhere the two components decouple into two copies of the spinless Schrödinger equation — spin becomes a spectator — which is why you can ignore spin in many bound-state problems until a magnetic field or spin-orbit coupling switches it on.

Why must I keep the lower components at all if they're so small?

Because "small" is not "zero," and the small components carry the physics you care about. To leading order $\chi$ is suppressed by $v/c$ , but it is precisely $\chi\approx(\boldsymbol\sigma\cdot\boldsymbol\pi)/2mc\,\varphi$ , and substituting that back is what generates the kinetic energy and the Zeeman term. Set $\chi=0$ exactly and you get $i\hbar\partial_t\varphi=q\phi\,\varphi$ — a particle with no kinetic energy at all. The lower components are the channel through which the upper ones acquire dynamics; the elimination keeps their effect while removing them as unknowns, and at next order they also carry the spin-orbit and Darwin physics.

Does the sign of the charge matter for the derivation?

The derivation is charge-agnostic — it works for any $q$ , and we kept $q$ explicit so the signs are honest. For the electron $q=-e$ with $e>0$ , so the physical spin term is $+\tfrac{e\hbar}{2mc}\boldsymbol\sigma\cdot\mathbf B$ and the moment points opposite to the spin — which is why an electron spin prefers to anti-align with an applied field to lower its energy. The gyromagnetic ratio $g=2$ is independent of the charge's sign; only the direction of $\boldsymbol\mu$ relative to $\mathbf S$ flips. Texts that set $q=-e$ from the start write $+\tfrac{e\hbar}{2mc}$ , which looks like a discrepancy until you track the convention.

If the reduction gives fine structure at next order, why not just push it further instead of learning Foldy-Wouthuysen?

Because the naive push breaks. The leading elimination is not a unitary transformation, so iterating it to second order produces non-Hermitian operators that mix orders in $1/m$ , and the two-component probability $\varphi^\dagger\varphi$ stops being exactly conserved. You can sometimes symmetrize by hand and stumble onto the right answer, but there is no systematic guarantee, and subtle factors like the Thomas one-half in spin-orbit are easy to get wrong. The Foldy-Wouthuysen transformation replaces the ad hoc elimination with genuine unitary rotations that block-diagonalize order by order, delivering manifestly Hermitian fine-structure terms with the correct coefficients — the disciplined version of exactly this reduction.

Why does the same noncommutativity give both Landau levels and the spin moment?

Both trace to the single fact $[\pi_x,\pi_y]=i\hbar\tfrac{q}{c}B_z\ne 0$ . For the orbital motion, that commutator makes $\boldsymbol\pi^2/2m$ behave like a harmonic oscillator perpendicular to $\mathbf B$ , whose evenly spaced eigenvalues are the Landau levels. For the spin, the same commutator makes $\boldsymbol\pi\times\boldsymbol\pi\ne 0$ , and squaring $\boldsymbol\sigma\cdot\boldsymbol\pi$ leaves the residual $-\tfrac{q\hbar}{c}\boldsymbol\sigma\cdot\mathbf B$ . Orbital quantization and the spin moment are two consequences of one piece of geometry — a charged particle's kinetic momentum has built-in curvature in a magnetic field — and the full spectrum of a field problem is spin-split Landau levels.