Lorentz Covariance of the Dirac Equation

A relativistic wave equation is only worth the name if every inertial observer writes it down in exactly the same form. If a boosted observer had to use a different equation, the theory would secretly pick out a preferred frame — precisely what special relativity forbids. For the Klein–Gordon equation this invariance is nearly free, because its field is a Lorentz scalar that simply carries along to the new coordinates. The Dirac equation is far more demanding: it is first order, it is stitched together from the gamma matrices, and its wavefunction is a four-component spinor whose transformation law is not handed to us in advance. We have to find it. The problem reduces to one sharp question. If $\psi(x)$ solves $(i\gamma^\mu\partial_\mu-m)\psi=0$ in one frame, what matrix $S(\Lambda)$ must act on the spinor components so that $\psi'(x')=S(\Lambda)\psi(x)$ solves an equation of identical form in the boosted frame? Demanding form-invariance turns out to force a single algebraic condition, $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ : the gamma matrices, though constant numbers, may be treated as the components of a four-vector provided we simultaneously rotate them in spinor space. Everything physical falls out of that one condition. Its infinitesimal solution builds the spin generators $\sigma^{\mu\nu}=\tfrac{i}{2}[\gamma^\mu,\gamma^\nu]$ ; the spatial part exposes intrinsic spin $\tfrac12$ and the notorious $720^\circ$ periodicity; the boost part is non-unitary, a feature forced by the non-compactness of the Lorentz group rather than a bug. The Dirac adjoint $\bar\psi=\psi^\dagger\gamma^0$ is engineered exactly so that $\bar\psi\gamma^\mu\psi$ is a true four-vector current with a positive-definite time component. Mastering covariance is what upgrades a clever matrix factorization of $\Box+m^2$ into a genuine relativistic theory of spin- $\tfrac12$ matter.

Key takeaways

The free Dirac equation is form-invariant under $x'=\Lambda x$ provided the spinor transforms as $\psi'(x')=S(\Lambda)\psi(x)$ , where $S(\Lambda)$ is a $4\times4$ matrix that is emphatically not the Lorentz matrix $\Lambda$ itself.
Form-invariance forces exactly one algebraic condition, the covariance (intertwining) relation $S^{-1}(\Lambda)\gamma^\mu S(\Lambda)=\Lambda^\mu{}_\nu\gamma^\nu$ .
Its solution is $S=\exp(-\tfrac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu})$ with generators $\sigma^{\mu\nu}=\tfrac{i}{2}[\gamma^\mu,\gamma^\nu]$ — a finite-dimensional, generally non-unitary representation of the Lorentz group.
Rotations give unitary $S$ and $720^\circ$ periodicity ( $S_R(2\pi)=-1$ ); boosts give non-unitary $S$ , so $\psi^\dagger\psi$ alone is not invariant.
Because $\bar\psi=\psi^\dagger\gamma^0$ transforms with $S^{-1}$ , the current $\bar\psi\gamma^\mu\psi$ is a four-vector with $j^0=\psi^\dagger\psi\ge0$ — curing the Klein–Gordon negative-density problem.
Key idea: demanding the Dirac equation keep its form under $\psi\to S\psi$ forces $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ , whose solution is the spinor representation of the Lorentz group — the object that encodes spin $\tfrac12$ , double-valuedness, and the four-vector current in one stroke.

The core insight

Require that $x'=\Lambda x,\ \psi'(x')=S(\Lambda)\psi(x)$ leaves $(i\gamma^\mu\partial_\mu-m)\psi=0$ unchanged in form; this single demand forces $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ , whose solution $S=\exp(-\tfrac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu})$ is the (non-unitary) spinor representation of the Lorentz group.

Where this lives today

Lorentz covariance of the Dirac field is the load-bearing assumption behind every calculation in the Standard Model. When a modern amplitude program contracts a string of gamma matrices between external spinors, it is implicitly using the covariance condition to guarantee that the answer is a Lorentz-invariant number, independent of how any observer sliced up spacetime. The bilinear classification the covariance condition organizes — scalar, pseudoscalar, vector, axial vector, tensor — is exactly the menu of allowed couplings that fixes the electromagnetic and weak currents. The representation theory also has a sharp experimental face. High-precision tests of Lorentz invariance in the fermion sector, carried out in the Standard Model Extension framework, bound hypothetical coefficients that would spoil the covariance condition to parts in $10^{-30}$ or better for the electron. The $720^\circ$ periodicity of the spinor, once a mathematical curiosity, has been demonstrated directly with neutron interferometry, where a rotating magnetic field flips the sign of the neutron amplitude after a single $2\pi$ turn.

The physical picture

Picture the four gamma matrices as a fixed set of arrows welded into spinor space. When you change frame, the components of any four-vector rotate by $\Lambda$ . The covariance condition says something almost magical: instead of physically rotating the gammas, you can equivalently conjugate them by a spinor-space transformation $S$ . The Dirac equation survives the change of frame because these two motions cancel — the $\Lambda$ that acts on the derivative $\partial_\mu$ is exactly undone by the $S$ that conjugates the gammas. That cancellation is Lorentz covariance.

Figure

The two representations meet in the covariance condition

Spin- $\tfrac12$ hides in a factor of $\tfrac12$ . When the tensor generator $\sigma^{ij}$ is repackaged as a rotation generator, a rotation by angle $\theta$ about $\hat n$ exponentiates to $\exp(-\tfrac{i}{2}\theta\,\hat n\cdot\boldsymbol\Sigma)$ . That $\tfrac12$ means a full $2\pi$ turn returns $S=-1$ , not $+1$ : the spinor lives on the double cover $SL(2,\mathbb C)$ of the Lorentz group, and you need $720^\circ$ to come home. Only bilinear observables like $\bar\psi\psi$ , quadratic in $\psi$ , are restored after $2\pi$ — which is why the sign is invisible in any measurement. Boosts feel fundamentally different from rotations. Rotations form a compact subgroup and produce unitary $S$ . Boosts are non-compact: their generator stretches rather than rotates, so $S$ is non-unitary and $\psi^\dagger\psi$ is not a Lorentz scalar. This is not a defect to be patched — it is a theorem. The Lorentz group has no finite-dimensional unitary representations except the trivial one, so any four-component object must transform non-unitarily. The fix is to promote $\psi^\dagger\psi$ to the time component of a four-vector via $\bar\psi\gamma^0\psi$ , and the appearance of $\gamma^0$ in the adjoint is precisely the intertwiner that makes this work.

Figure

Rotations preserve the norm; boosts stretch it

The core relations

The whole subject rests on two statements: how the spinor transforms, and the condition that makes the equation survive. We take the chapter convention throughout — the mostly-minus metric $g_{\mu\nu}=\mathrm{diag}(+1,-1,-1,-1)$ with $\hbar=c=1$ , pinned in "The Klein-Gordon Equation" — and the Dirac representation of the gammas. The transformation law states that a spinor is not a set of four scalars but a single object mixed by a $4\times4$ matrix under a change of frame. It looks deceptively like the vector law $x'^\mu=\Lambda^\mu{}_\nu x^\nu$ , but the matrix is different.

\psi'(x')=S(\Lambda)\,\psi(x),\qquad x'^\mu=\Lambda^\mu{}_\nu\,x^\nu \tag{1}

where: $\psi$ is the four-component Dirac spinor; $S(\Lambda)$ is the $4\times4$ spinor-representation matrix; $\Lambda^\mu{}_\nu$ is the ordinary (defining) representation acting on coordinates. Read physically: the same physical event $x$ acquires new coordinates $x'$ , and its spinor value is rotated in the internal four-dimensional spinor space by $S$ , not by $\Lambda$ . Requiring $(1)$ to leave the equation form-identical produces the master condition. It is one line, and it controls everything downstream.

S^{-1}(\Lambda)\,\gamma^\mu\,S(\Lambda)=\Lambda^\mu{}_\nu\,\gamma^\nu \tag{2}

where: $\gamma^\mu$ are the constant Dirac matrices obeying the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ . The reading: the numerical matrices $\gamma^\mu$ transform "as a four-vector" once you allow the compensating spinor conjugation by $S$ . Conjugating on the left by $S^{-1}$ and the right by $S$ is exactly what un-does the boost of $\partial_\mu$ . Solving $(2)$ infinitesimally hands us the generators. They are the antisymmetrized products of gammas, and they turn out to satisfy the Lorentz Lie algebra.

\sigma^{\mu\nu}=\tfrac{i}{2}\,[\gamma^\mu,\gamma^\nu],\qquad S(\Lambda)=\exp\!\left(-\tfrac{i}{4}\,\sigma_{\mu\nu}\,\omega^{\mu\nu}\right) \tag{3}

where: $\sigma^{\mu\nu}$ is the antisymmetric array of six $4\times4$ generator matrices, and $\omega^{\mu\nu}=-\omega^{\nu\mu}$ packages the six Lorentz parameters (three rotation angles, three rapidities). The exponential builds finite transformations from the infinitesimal ones; closure is guaranteed because $\tfrac12\sigma^{\mu\nu}$ obeys the Lorentz algebra. The adjoint relation is what repairs the non-unitarity of boosts. It is the reason the letter $\gamma^0$ appears in the definition of $\bar\psi$ at all.

\gamma^0\,S^\dagger(\Lambda)\,\gamma^0 = S^{-1}(\Lambda)\quad\Longleftrightarrow\quad \bar\psi'(x')=\bar\psi(x)\,S^{-1}(\Lambda) \tag{4}

where: $\bar\psi=\psi^\dagger\gamma^0$ is the Dirac adjoint. Because $\bar\psi$ transforms contragredient to $\psi$ (with $S^{-1}$ ), the sandwich $\bar\psi\,\Gamma\,\psi$ has a clean tensor character for each matrix $\Gamma$ ; in particular $\bar\psi\psi$ is a scalar and $\bar\psi\gamma^\mu\psi$ is a four-vector. A which-object-when summary keeps the pieces straight:

object	transforms as	invariant / covariant combination
$\psi$	$\psi'=S\psi$	not invariant by itself
$\bar\psi=\psi^\dagger\gamma^0$	$\bar\psi'=\bar\psi S^{-1}$	pairs with $\psi$
$\bar\psi\psi$	scalar	Lorentz invariant
$\bar\psi\gamma^\mu\psi$	four-vector	conserved current $j^\mu$ , $j^0\ge0$
$\bar\psi\sigma^{\mu\nu}\psi$	rank-2 tensor	magnetic-moment coupling

Derivation AExistence of S from the covariance condition

We construct the infinitesimal generator that satisfies $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ and prove it works, then exponentiate. The strategy is to linearize both the Lorentz matrix and $S$ in the six small parameters $\omega^{\mu\nu}=-\omega^{\nu\mu}$ , reduce the covariance condition to a commutator equation, and solve it with the antisymmetric ansatz $\Gamma\propto\sigma_{\mu\nu}\omega^{\mu\nu}$ . The single non-obvious ingredient is the Clifford-algebra identity $[\gamma^\mu,\sigma_{\alpha\beta}]=2i(g^\mu{}_\alpha\gamma_\beta-g^\mu{}_\beta\gamma_\alpha)$ ; everything else is bookkeeping. Watch the placement of indices — the antisymmetry of $\omega$ is what collapses two terms into one.

Step 1. Write $\Lambda^\mu{}_\nu=\delta^\mu{}_\nu+\omega^\mu{}_\nu$ and $S=1+\Gamma$ with $\Gamma$ linear in $\omega$ , so $S^{-1}=1-\Gamma$ to first order.

Step 2. Insert into $(2)$ : $(1-\Gamma)\gamma^\mu(1+\Gamma)=\gamma^\mu+\omega^\mu{}_\nu\gamma^\nu$ , which to first order is the commutator equation $[\gamma^\mu,\Gamma]=\omega^\mu{}_\nu\gamma^\nu$ . Why a commutator: only the linear-in- $\Gamma$ piece survives, and it is precisely $\Gamma\gamma^\mu-\gamma^\mu\Gamma$ moved to one side.

Step 3. Adopt the ansatz $\Gamma=-\tfrac{i}{4}\sigma_{\alpha\beta}\omega^{\alpha\beta}$ with $\sigma_{\alpha\beta}=\tfrac{i}{2}[\gamma_\alpha,\gamma_\beta]$ . The antisymmetry of both $\sigma_{\alpha\beta}$ and $\omega^{\alpha\beta}$ matches the six free Lorentz parameters exactly.

Step 4. Evaluate the needed commutator from the Clifford algebra alone: $[\gamma^\mu,\sigma_{\alpha\beta}]=2i\,(g^\mu{}_\alpha\gamma_\beta-g^\mu{}_\beta\gamma_\alpha)$ . Why this identity is the crux: it is the only place the anticommutator $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ enters, and it is what makes the ansatz solve the condition.

Step 5. Substitute into the commutator equation: $-\tfrac{i}{4}[\gamma^\mu,\sigma_{\alpha\beta}]\omega^{\alpha\beta}=\tfrac12(g^\mu{}_\alpha\gamma_\beta-g^\mu{}_\beta\gamma_\alpha)\omega^{\alpha\beta}=\omega^\mu{}_\nu\gamma^\nu$ , using $\omega^{\alpha\beta}=-\omega^{\beta\alpha}$ to fold the two terms. The condition holds identically.

Step 6. Exponentiate the generator: $S=\exp(-\tfrac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu})$ . Closure of finite transformations follows because $\tfrac12\sigma^{\mu\nu}$ obeys the Lorentz Lie algebra $[\tfrac12\sigma^{\mu\nu},\tfrac12\sigma^{\rho\sigma}]=i(g^{\nu\rho}\tfrac12\sigma^{\mu\sigma}-g^{\mu\rho}\tfrac12\sigma^{\nu\sigma}-g^{\nu\sigma}\tfrac12\sigma^{\mu\rho}+g^{\mu\sigma}\tfrac12\sigma^{\nu\rho})$ . Verified: the key commutator $[\gamma^\mu,\sigma_{\alpha\beta}]=2i(g^\mu{}_\alpha\gamma_\beta-g^\mu{}_\beta\gamma_\alpha)$ was checked directly from $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ for all $\mu,\alpha,\beta$ in the Dirac representation (verify script, block B) — Bjorken & Drell Ch. 2, Peskin & Schroeder eq. (3.23).

Go deeper · Derivation B · 1 minThe covariance condition holds for a finite boost

Derivation BThe covariance condition holds for a finite boost

We stop trusting the infinitesimal argument and check the finite, non-perturbative statement for a concrete transformation: a boost along $\hat z$ with rapidity $\eta$ . The goal is to build the explicit $4\times4$ matrix $S_B$ , the explicit $4\times4$ Lorentz matrix $\Lambda$ , and confirm $S_B^{-1}\gamma^\mu S_B=\Lambda^\mu{}_\nu\gamma^\nu$ as a matrix identity for every $\mu$ . The subtlety is that $S_B$ is built from $\gamma^0\gamma^3$ , whose square is $+1$ , so the exponential closes into hyperbolic functions rather than trigonometric ones — the algebraic fingerprint of a non-compact boost.

Step 1. The boost generator is $-\tfrac{i}{4}\sigma_{03}\omega^{03}$ ; with $\omega^{03}=\eta$ this reduces to $-\tfrac{\eta}{2}\gamma^0\gamma^3$ . Write $K=\gamma^0\gamma^3$ and note $K^2=+\mathbb 1$ .

Step 2. Exponentiate: since $K^2=\mathbb 1$ , $S_B=\exp(-\tfrac{\eta}{2}K)=\cosh\tfrac{\eta}{2}-\sinh\tfrac{\eta}{2}\,K$ , with inverse $S_B^{-1}=\cosh\tfrac{\eta}{2}+\sinh\tfrac{\eta}{2}\,K$ . Why hyperbolic, not trigonometric: $K^2=+1$ (a boost) rather than $-1$ (a rotation) turns the exponential series into $\cosh/\sinh$ — this is what makes $S_B$ non-unitary.

Step 3. Write the corresponding Lorentz boost matrix acting on $(t,x,y,z)$ :

\Lambda=\begin{pmatrix}\cosh\eta & 0 & 0 & -\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\sinh\eta & 0 & 0 & \cosh\eta\end{pmatrix}

Step 4. Compute $S_B^{-1}\gamma^\mu S_B$ for each $\mu$ and match it against $\Lambda^\mu{}_\nu\gamma^\nu$ . The mixed identities give $\gamma^0\mapsto\cosh\eta\,\gamma^0-\sinh\eta\,\gamma^3$ and $\gamma^3\mapsto-\sinh\eta\,\gamma^0+\cosh\eta\,\gamma^3$ , while $\gamma^1,\gamma^2$ are untouched — exactly the rows of $\Lambda$ .

Step 5. All four matrix identities hold to symbolic zero. The finite covariance condition is therefore an exact statement, not merely a first-order one. Verified: $S_B^{-1}\gamma^\mu S_B=\Lambda^\mu{}_\nu\gamma^\nu$ was confirmed as a $4\times4$ symbolic identity for all $\mu$ with $S_B=\cosh\tfrac{\eta}{2}-\sinh\tfrac{\eta}{2}\gamma^0\gamma^3$ (verify script, block C) — Peskin & Schroeder Ch. 3.

Go deeper · Derivation C · 1 minThe current is a four-vector; rotation gives spin one-half

Derivation CThe current is a four-vector; rotation gives spin one-half

Two payoffs share one engine. First we show $j^\mu=\bar\psi\gamma^\mu\psi$ transforms as a four-vector, which is what rescues the probability density; then we reduce the spatial generators to the spin operator and read off spin $\tfrac12$ with its $720^\circ$ periodicity. Both follow from the covariance condition and the adjoint relation $(4)$ ; the only care needed is to keep the boost non-unitarity from leaking into the current — which is exactly what $\gamma^0$ in the adjoint prevents.

Step 1. Assemble the bilinear using $\psi'=S\psi$ and $\bar\psi'=\bar\psi S^{-1}$ : $j'^\mu=\bar\psi\,S^{-1}\gamma^\mu S\,\psi$ .

Step 2. Apply the covariance condition $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ to get $j'^\mu=\Lambda^\mu{}_\nu\,\bar\psi\gamma^\nu\psi=\Lambda^\mu{}_\nu j^\nu$ . The current is a genuine contravariant four-vector, and $j^0=\psi^\dagger\psi\ge0$ in every frame. Why this cures Klein–Gordon: the density is $\psi^\dagger\psi$ , manifestly non-negative, unlike the KG density built from $\partial_t\phi$ that could go negative — see "Klein-Gordon Probability Current and Negative Densities".

Step 3. Now specialize to a pure rotation about $\hat z$ by $\theta$ , i.e. $\omega^{12}=-\theta$ . The spatial part reduces to $S_R=\exp(-\tfrac{i}{2}\theta\,\Sigma^3)$ with $\Sigma^3=\mathrm{diag}(\sigma^3,\sigma^3)$ .

Step 4. Because $(\Sigma^3)^2=\mathbb 1$ , $S_R=\cos\tfrac{\theta}{2}-i\,\Sigma^3\sin\tfrac{\theta}{2}$ . The half-angle is the spin quantum number $\tfrac12$ ; the total angular momentum generator is $\mathbf J=\mathbf L+\tfrac12\boldsymbol\Sigma$ .

Step 5. Evaluate at special angles: $S_R(2\pi)=-\mathbb 1$ and $S_R(4\pi)=+\mathbb 1$ . A $360^\circ$ rotation multiplies the spinor by $-1$ ; only after $720^\circ$ does it return — the double-valuedness of $SL(2,\mathbb C)\to SO^+(1,3)$ . Verified: $S_R(2\pi)=-\mathbb 1$ , $S_R(4\pi)=+\mathbb 1$ from $(\Sigma^3)^2=\mathbb 1$ , and $g^0 S^\dagger g^0=S^{-1}$ for both boost and rotation (verify script, blocks D–F) — Shankar Ch. 20.

Limiting cases

How $S(\Lambda)$ specializes for the elementary transformations, all in the pinned Dirac representation:

case	result	meaning
rotation about $\hat n$ by $\theta$	$S_R=\exp(-\tfrac{i}{2}\theta\,\hat n\cdot\boldsymbol\Sigma)$ , $\boldsymbol\Sigma=\mathrm{diag}(\boldsymbol\sigma,\boldsymbol\sigma)$	unitary; period $4\pi$ ; the $\tfrac12$ encodes spin $\tfrac12$
boost along $\hat n$ , rapidity $\eta$	$S_B=\cosh\tfrac{\eta}{2}-\hat n\cdot\boldsymbol\alpha\,\sinh\tfrac{\eta}{2}$ , $\tanh\eta=\beta$	non-unitary, $S_B^\dagger=S_B=\gamma^0 S_B^{-1}\gamma^0$ ; $\boldsymbol\alpha=\gamma^0\boldsymbol\gamma$
infinitesimal $\Lambda=\delta+\omega$	$S=1-\tfrac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu}$	six parameters: three rotations, three boosts
parity (improper)	$S_P=\gamma^0$ , $\psi'(t,\mathbf x)=\gamma^0\psi(t,-\mathbf x)$	$\gamma^0\gamma^k\gamma^0=-\gamma^k$ flips $\mathbf x$ ; covariance extends to $O(1,3)$
$\beta\to0$	$S_B\approx1-\tfrac12\boldsymbol\beta\cdot\boldsymbol\alpha$	small-boost limit feeds the Pauli reduction; lower components $\sim p/m$

Common trapS(Λ) is unitary — false, and it corrupts the current

It is tempting to assume $S(\Lambda)$ is unitary because rotations feel familiar and unitarity is the norm in ordinary quantum mechanics. It is not. Boost generators produce a non-unitary $S_B$ (its exponent contains $\cosh/\sinh$ , not $\cos/\sin$ ), because the Lorentz group is non-compact and has no non-trivial finite-dimensional unitary representation. The practical consequence is concrete: $\psi^\dagger\psi$ is not Lorentz invariant, and if you try to use it as a scalar density you get frame-dependent nonsense. The invariant is $\bar\psi\psi=\psi^\dagger\gamma^0\psi$ , and the density is the time component $j^0$ of the four-vector $\bar\psi\gamma^\mu\psi$ . Confusing $S$ with the coordinate matrix $\Lambda$ is the same mistake in another guise.

Worked examples

Example 1 — A concrete boost: velocity, rapidity, and the growth of $\psi^\dagger\psi$ .

Step 1. Take a boost of velocity $\beta=0.6$ along $\hat z$ . The rapidity is $\eta=\operatorname{artanh}(0.6)=0.6931$ , so $\cosh\eta=1/\sqrt{1-\beta^2}=\gamma=1.2500$ — the Lorentz factor is recovered as $\cosh\eta$ , as it must be.

Step 2. Build the spinor boost $S_B=\cosh\tfrac{\eta}{2}-\sinh\tfrac{\eta}{2}\,\gamma^0\gamma^3$ . Half-rapidity: $\cosh\tfrac{\eta}{2}=1.0607$ , $\sinh\tfrac{\eta}{2}=0.3536$ .

Step 3. Apply $S_B$ to a rest-frame spinor $\psi_0=(1,0,0,0)^\mathsf T$ and compute $\psi^\dagger\psi$ in the boosted frame. Because $S_B$ is non-unitary, the norm changes: $\psi^\dagger\psi$ grows from $1$ to $\cosh\eta=\mathbf{1.2500}$ .

Step 4. This is the smoking gun of non-unitarity — a unitary $S$ would leave $\psi^\dagger\psi$ fixed. The invariant $\bar\psi\psi$ , by contrast, stays put, which is why physical scalars are built from $\bar\psi$ , not $\psi^\dagger$ . Answer: $\eta=0.693$ , $\gamma=1.25$ , and $\psi^\dagger\psi$ grows by the factor $\gamma=1.25$ .

Example 2 — Boosting a rest spinor: the lower-component ratio.

Step 1. A positive-energy spinor at rest has only upper components; a boost mixes in lower components in the fixed ratio $|\mathbf p|/(E+m)$ . For $\beta=0.6$ , $\gamma=1.25$ , this is $|\mathbf p|/(E+m)=\gamma\beta/(\gamma+1)$ .

Step 2. Evaluate: $\gamma\beta=0.75$ and $\gamma+1=2.25$ , so the ratio is $0.75/2.25=\mathbf{0.3333}$ .

Step 3. Cross-check against the half-rapidity form: the same ratio equals $\tanh(\eta/2)=\tanh(0.3466)=0.3333$ . The two routes agree exactly, as the boost matrix demands.

Step 4. Physically, the lower components measure how relativistic the state is: they vanish as $\beta\to0$ (the Pauli limit) and approach the upper components as $\beta\to1$ (the massless/chiral limit). Answer: the lower/upper ratio is $\tanh(\eta/2)=1/3$ for a $\beta=0.6$ boost.

Origins · 1 minhendrik lorentz and how this was discovered

hendrik lorentz

1853–1928 · Dutch physicist whose transformation equations, written to keep Maxwell's theory consistent across frames, became the coordinate backbone of special relativity — and, once married to the Dirac spinor, the covariance condition of this article.

Lorentz introduced his coordinate transformations (1904) to explain the null result of Michelson–Morley within electromagnetic theory, before Einstein reinterpreted them kinematically in 1905. The spinor representation $S(\Lambda)$ that makes the Dirac equation covariant is a projective representation of Lorentz's group; the systematic classification of all relativistic wave equations by mass and spin — via the unitary irreducible representations of the full Poincaré group — was completed by Eugene Wigner in 1939.

History

1904

— Lorentz publishes the transformation equations that carry his name

1928

— Dirac writes the first-order relativistic equation, forcing a spinor transformation law

1929

— Weyl and Fock formulate spinors on the Lorentz group and clarify their double-valuedness

1939

— Wigner classifies the unitary irreps of the Poincaré group by mass and spin

Going further

References. Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7 · Sakurai, Advanced Quantum Mechanics, Ch. 3 · Bjorken & Drell, Relativistic Quantum Mechanics, Ch. 2 · Peskin & Schroeder, An Introduction to QFT, Ch. 3 (eqs. 3.18–3.30) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 20 · Weinberg, The Quantum Theory of Fields, Vol. I, Ch. 5.

Provenance. All equations standard and cross-checked against Bjorken & Drell (Ch. 2), Peskin & Schroeder (Ch. 3), Griffiths (Elementary Particles, Ch. 7), Sakurai (Advanced QM, Ch. 3), and Shankar (Ch. 20). Machine-verified in this build: (1) the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ in the Dirac representation; (2) the key commutator $[\gamma^\mu,\sigma_{\alpha\beta}]=2i(g^\mu{}_\alpha\gamma_\beta-g^\mu{}_\beta\gamma_\alpha)$ from the algebra alone; (3) the finite covariance condition $S_B^{-1}\gamma^\mu S_B=\Lambda^\mu{}_\nu\gamma^\nu$ for a concrete boost as a $4\times4$ symbolic identity for all $\mu$ ; (4) $S_B$ non-unitary vs $S_R$ unitary; (5) the adjoint intertwiner $\gamma^0 S^\dagger\gamma^0=S^{-1}$ ; (6) $S_R(2\pi)=-1$ , $S_R(4\pi)=+1$ ; (7) parity $\gamma^0\gamma^k\gamma^0=-\gamma^k$ ; (8) the worked-example numbers. Metric convention $(+,-,-,-)$ , $\hbar=c=1$ , Dirac representation throughout; no fabricated numerical constants. One base-equation repair is recorded below (eq. 8 boost-spinor formula).

Base repair. The source skeleton wrote the boost-to-spinor relation as a garbled fraction $S_B=\frac{1+\gamma^0\gamma^\mu\hat p_\mu(E/m)\,\text{terms}}{\sqrt{2m(E+m)}}$ containing an untyped "terms" placeholder. It is corrected here to the clean, machine-checked closed form $u(p)=S_B\,u(0)$ with $S_B=\cosh\tfrac{\eta}{2}-\hat n\cdot\boldsymbol\alpha\,\sinh\tfrac{\eta}{2}$ and the verified lower/upper ratio $\tanh(\eta/2)=|\mathbf p|/(E+m)$ .

Builds on: The Dirac Equation · Dirac Gamma Matrices and Clifford Algebra · The Klein-Gordon Equation · Dirac Spinors and Plane-Wave Solutions. Leads to: Spin from the Dirac Equation · Bilinear Covariants and Dirac Adjoint · Charge Conjugation and the Positron · Nonrelativistic Reduction and the Pauli Equation.

Why is S(Λ) four-dimensional but not the same as the 4×4 Lorentz matrix Λ?

Both are $4\times4$ , but they act on different spaces and are different representations of the same abstract Lorentz transformation. $\Lambda$ acts on spacetime indices (it rotates $t,x,y,z$ ); $S$ acts on the four internal spinor components. They are related only through the covariance condition $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ , which links the two representations via the gamma matrices. Numerically they are almost never equal — for a boost, $\Lambda$ contains $\cosh\eta,\sinh\eta$ while $S_B$ contains $\cosh\tfrac{\eta}{2},\sinh\tfrac{\eta}{2}$ .

Where exactly does spin one-half enter — I never put it in by hand?

It is forced by the covariance condition, not assumed. When you reduce the spatial generators $\tfrac12\sigma^{ij}$ to the rotation generator, you get $\tfrac12\boldsymbol\Sigma$ , and a finite rotation exponentiates to $\exp(-\tfrac{i}{2}\theta\,\hat n\cdot\boldsymbol\Sigma)$ . That half-angle is the spin quantum number: it makes the total angular momentum $\mathbf J=\mathbf L+\tfrac12\boldsymbol\Sigma$ and forces $S(2\pi)=-1$ . So spin $\tfrac12$ is a consequence of demanding relativistic covariance of a four-component first-order equation, one of the deepest results in the subject.

If S is non-unitary, doesn't that break probability conservation?

No — probability conservation survives because it is a statement about the four-current, not about $S$ . The conserved density is $j^0=\psi^\dagger\psi$ , the time component of the four-vector $j^\mu=\bar\psi\gamma^\mu\psi$ . Under a boost, $j^0$ changes exactly as the time component of a four-vector should (it mixes with the spatial current), and $\partial_\mu j^\mu=0$ is a Lorentz-scalar equation valid in every frame. The non-unitarity of $S$ is real but it lives in $\psi^\dagger\psi$ before the $\gamma^0$ is inserted; the physical density $\bar\psi\gamma^0\psi$ transforms correctly.

What is the role of the identity γ⁰S†γ⁰=S⁻¹?

It is the technical heart of why $\bar\psi=\psi^\dagger\gamma^0$ is the right object. Since $S$ is not unitary, $\psi^\dagger$ transforms with $S^\dagger\ne S^{-1}$ , so $\psi^\dagger\psi$ is frame-dependent. Sandwiching a $\gamma^0$ converts $S^\dagger$ into $S^{-1}$ (that is precisely what the identity says), so $\bar\psi=\psi^\dagger\gamma^0$ transforms with $S^{-1}$ . Then $\bar\psi\psi$ has an $S^{-1}S=1$ and is invariant, and $\bar\psi\gamma^\mu\psi$ picks up exactly one factor of $\Lambda$ . Without this identity there would be no covariant bilinears at all.

Does covariance also cover parity, or only continuous transformations?

It covers the full Lorentz group $O(1,3)$ , including the improper elements. Parity is implemented by $S_P=\gamma^0$ : one checks $\gamma^0\gamma^k\gamma^0=-\gamma^k$ and $\gamma^0\gamma^0\gamma^0=+\gamma^0$ , which is exactly the covariance condition for the parity matrix $\Lambda_P=\mathrm{diag}(+,-,-,-)$ . This is why $\bar\psi\psi$ is a true scalar but $\bar\psi\gamma^5\psi$ is a pseudoscalar (it flips sign under parity), and why $\bar\psi\gamma^\mu\gamma^5\psi$ is an axial vector. The discrete symmetries fall out of the same representation theory as the boosts and rotations.

Why does the boost exponential use cosh/sinh instead of cos/sin like the rotation?

It comes down to the square of the generator. A rotation is built from $\Sigma^k$ with $(\Sigma^k)^2=+\mathbb 1$ but appearing as $\exp(-\tfrac{i}{2}\theta\Sigma)$ — the factor of $i$ makes it $\cos-i\sin$ . A boost is built from $\gamma^0\gamma^k$ with $(\gamma^0\gamma^k)^2=+\mathbb 1$ and no factor of $i$ : $\exp(-\tfrac{\eta}{2}\gamma^0\gamma^k)=\cosh-\sinh(\gamma^0\gamma^k)$ . The presence or absence of $i$ traces directly to whether the generator squares with a $+$ (compact, trigonometric, unitary) or acts like a real hyperbolic rotation (non-compact, hyperbolic, non-unitary). This single sign is the whole difference between a rotation and a boost.

How do the boosted spinors u(p) connect to this?

They are literally $S_B$ applied to rest-frame spinors: $u(p,s)=S_B(p)\,u(0,s)$ . Because $S_B$ satisfies the covariance condition, the boosted object automatically solves the momentum-space Dirac equation $(\slashed p-m)u(p)=0$ — you never have to guess the plane-wave spinors, you boost them. The lower components appear in the fixed ratio $|\mathbf p|/(E+m)=\tanh(\eta/2)$ computed in Example 2. This construction is developed in "Dirac Spinors and Plane-Wave Solutions" and feeds directly into the Feynman rules of QED.