Covariant form of the Dirac equation

The Dirac equation has two faces, and which one you show depends on what you want to see. In its original Hamiltonian guise, $i\partial_t\psi=(-i\boldsymbol\alpha\cdot\nabla+\beta m)\psi$ , the matrices $\boldsymbol\alpha$ and $\beta$ sit in plain view and the electron's spin structure is manifest — but time and space are handled by different objects, so its behaviour under a boost is hidden. The covariant form repackages the same physics so its relativistic character is on the surface. The trick is bookkeeping of the most productive kind: absorb $\beta$ and the three $\alpha_k$ into four constant matrices $\gamma^\mu$ carrying a Lorentz index, and the whole equation collapses to the single line $(i\gamma^\mu\partial_\mu-m)\psi=0$ . This article, resting on the Dirac equation in its $\alpha,\beta$ form, the Klein–Gordon equation, and the Dirac probability current, does three things. It builds the $\gamma$ -matrices from $\gamma^0\equiv\beta$ and $\gamma^k\equiv\beta\alpha_k$ and shows every property they need lives in one relation, the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ . It shows, following Griffiths, that the same algebra is forced the moment you demand the first-order operator square back to the relativistic energy–momentum relation. And it assembles the covariant current $j^\mu=\bar\psi\gamma^\mu\psi$ , whose density $j^0=\psi^\dagger\psi$ is positive definite, curing the disease that had sunk Klein–Gordon. Master this and the plane-wave solutions, antiparticles, and Feynman rules that follow become one-line $\gamma$ -manipulations.

Key takeaways

Pre-multiplying the Hamiltonian-form Dirac equation by $\beta$ and setting $\gamma^0\equiv\beta,\ \gamma^k\equiv\beta\alpha_k$ collapses it to $(i\gamma^\mu\partial_\mu-m)\psi=0$ , written $(i\slashed\partial-m)\psi=0$ in Feynman slash notation.
The entire content reduces to the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ , with $(\gamma^0)^2=I$ , $(\gamma^k)^2=-I$ .
The same algebra is forced by demanding the first-order operator factor the relation $p^2-m^2=(\slashed p+m)(\slashed p-m)$ — Dirac's inspiration that the coefficients must be matrices.
The smallest matrices obeying it are $4\times4$ ; the Dirac–Pauli representation is standard, but no result depends on it.
The adjoint spinor $\bar\psi\equiv\psi^\dagger\gamma^0$ builds the conserved current $j^\mu=\bar\psi\gamma^\mu\psi$ , whose density $j^0=\psi^\dagger\psi\ge0$ is positive.
Key idea: bundling $\beta$ and the $\alpha_k$ into four $\gamma$ -matrices that anticommute to the Minkowski metric, $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ , lets the Dirac equation be written as the single tidy expression $(i\gamma^\mu\partial_\mu-m)\psi=0$ whose entire physics lives in that one algebraic relation.

Where this lives today

The covariant Dirac equation is the working notation of every Standard Model calculation. Each fermion is a Dirac (or Majorana) field obeying $(i\slashed\partial-m)\psi=0$ when free and $(i\slashed D-m)\psi=0$ once gauge interactions promote $\partial_\mu$ to $D_\mu$ . Feynman-diagram amplitudes are strings of $\gamma$ -matrices sandwiched between spinors, $\bar u\gamma^\mu u$ , and evaluating them is pure Clifford-algebra bookkeeping: LHC cross sections are computed by "trace technology", reducing squared amplitudes to traces of $\gamma$ -products by the algebra (3) alone, never touching an explicit matrix. Lattice QCD discretises the same operator $\slashed D$ onto a grid — Wilson and staggered fermions are rival schemes for placing $\gamma^\mu\partial_\mu$ on a lattice without spurious "doubler" modes — and inverting it dominates every hadron-mass calculation. The non-relativistic reduction, giving $g=2$ , is the reference against which the muon's $g-2$ is measured to ten figures at Fermilab. And whether the neutrino is a Dirac or Majorana fermion is the open question neutrinoless-double-beta-decay experiments are running now to settle.

The physical picture

The Hamiltonian form treats time and space differently: a bare $\partial_t$ on the left, spatial derivatives dressed in $\alpha_k$ on the right. That asymmetry is cosmetic, but it hides the equation's Lorentz structure. The covariant reformulation removes it by promoting $\beta$ and the $\alpha_k$ to a single quartet $\gamma^\mu$ indexed like a four-vector, so all four derivatives enter alike. What survives is one relation: the four $\gamma$ -matrices anticommute to twice the metric, and every property the solutions can have is encoded there before any explicit matrix is written. The first figure is that relation as a table.

Figure

The whole equation in one table: the four constant matrices γ⁰, γ¹, γ², γ³ satisfy the Clifford algebra {γ^μ, γ^ν} = 2g^{μν}I, so the anticommutator grid is diagonal — the time entry is +2, the three space entries are −2, and every off-diagonal entry is 0, mirroring the Minkowski metric g^{μν} = diag(1, −1, −1, −1)

The first-order form carries a consistency requirement Dirac imposed from the start: every component of $\psi$ must separately satisfy Klein–Gordon. The Clifford algebra guarantees it — applying the operator twice reassembles the second-order wave operator, $(i\slashed\partial-m)(i\slashed\partial+m)=-(\Box+m^2)$ — so the Dirac equation is the "square root" of Klein–Gordon. The second figure follows that square through and pins it to a real number: an electron's mass shell.

Figure

Squaring the Dirac operator recovers Klein–Gordon. Because γ^μ γ^ν symmetrised gives g^{μν}, the product (i γ·∂ − m)(i γ·∂ + m) equals −(□ + m²), so every component of ψ obeys (□ + m²)ψ = 0. In momentum space this is the mass shell E² − p² = m²; for an electron with p = 1.000 MeV, E = 1.123 MeV gives E² − p² = 1.261 − 1.000 = 0.261 = (0.511 MeV)²

The core equations

The whole reformulation is one move: multiply the Hamiltonian-form equation by $\beta$ and relabel, giving a covariant expression in which $\mu$ is summed against $\partial_\mu$ . The compact form and its slash shorthand build everything downstream.

(i\gamma^\mu\partial_\mu-m)\psi=0\qquad\Longleftrightarrow\qquad (i\slashed\partial-m)\psi=0,\quad \slashed a\equiv\gamma^\mu a_\mu \qquad (1)

where: $\gamma^\mu$ — four constant $4\times4$ matrices; $\partial_\mu=(\partial_t,\nabla)$ — covariant four-gradient; $m$ — mass ( $c=\hbar=1$ ); $\psi$ — four-component Dirac spinor. Reading it: the slash $\slashed a=\gamma^\mu a_\mu=\gamma^0a^0-\boldsymbol\gamma\cdot\mathbf a$ contracts any four-vector with the $\gamma$ 's into one matrix, so the equation reads " $(i\slashed\partial-m)$ annihilates $\psi$ ". The $\gamma$ -matrices are not new physics; they are $\beta$ and the $\alpha_k$ renamed. The definition below bridges from the Hamiltonian form, and explains why $\gamma^0$ and the $\gamma^k$ have different squares.

\gamma^0\equiv\beta,\qquad \gamma^k\equiv\beta\alpha_k\quad(k=1,2,3) \qquad (2)

where: $\beta,\alpha_k$ — the Hermitian Dirac matrices with $\beta^2=\alpha_k^2=I$ , $\{\alpha_i,\alpha_j\}=2\delta_{ij}$ , $\{\alpha_k,\beta\}=0$ . Reading it: because $\beta^2=I$ but $\beta\alpha_k\beta\alpha_k=-I$ , the time matrix squares to $+I$ and each space matrix to $-I$ — the Minkowski signature, inherited automatically. Every property of the solutions follows from one relation: the four $\gamma$ 's anticommute to twice the metric. This is the defining Clifford algebra, and no explicit representation is needed to use it.

\{\gamma^\mu,\gamma^\nu\}\equiv\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}I,\qquad g^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1) \qquad (3)

where: $\{\,,\}$ — anticommutator; $g^{\mu\nu}$ — Minkowski metric; $I$ — the $4\times4$ identity. Reading it: $\mu=\nu$ gives $(\gamma^0)^2=I$ , $(\gamma^k)^2=-I$ ; $\mu\neq\nu$ gives $\gamma^\mu\gamma^\nu=-\gamma^\nu\gamma^\mu$ . This one line, not any matrix table, is the equation's content. The $\gamma$ 's are not all Hermitian — necessarily so, since a Hermitian $\gamma^k$ clashes with $(\gamma^k)^2=-I$ . Their Hermiticity is packaged by the relation below, which makes the adjoint equation and current come out right.

\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0,\qquad\text{i.e.}\qquad \gamma^{0\dagger}=\gamma^0,\quad \gamma^{k\dagger}=-\gamma^k \qquad (4)

where: $\dagger$ — Hermitian conjugate. Reading it: $\gamma^0$ is Hermitian (eigenvalues $\pm1$ ), each $\gamma^k$ anti-Hermitian (eigenvalues $\pm i$ ); the sandwich $\gamma^0\gamma^\mu\gamma^0$ trades a Hermitian conjugate for a $\gamma^0$ -conjugation. For explicit calculation one picks a representation. The Dirac–Pauli (standard) choice writes the $\gamma$ 's in $2\times2$ Pauli blocks, and diagonalises $\gamma^0$ so the rest frame separates energy signs cleanly.

\gamma^0=\begin{pmatrix}I&0\\0&-I\end{pmatrix},\qquad \gamma^k=\begin{pmatrix}0&\sigma^k\\-\sigma^k&0\end{pmatrix},\qquad \gamma^5\equiv i\gamma^0\gamma^1\gamma^2\gamma^3=\begin{pmatrix}0&I\\I&0\end{pmatrix} \qquad (5)

where: $I,0$ — $2\times2$ identity and zero; $\sigma^k$ — Pauli matrices. Reading it: this is one solution of the algebra (3), not the algebra itself — Weyl and Majorana representations are equally valid and no result depends on the choice. The fifth matrix $\gamma^5$ , anticommuting with all four $\gamma^\mu$ , governs chirality. Finally, the adjoint spinor pairs with $\psi$ to make Lorentz scalars and the conserved current — the covariant successor to the probability current, with a positive-definite time component.

\bar\psi\equiv\psi^\dagger\gamma^0,\qquad j^\mu=\bar\psi\gamma^\mu\psi,\qquad \partial_\mu j^\mu=0,\qquad j^0=\psi^\dagger\psi\ge0 \qquad (6)

where: $\bar\psi$ — Dirac adjoint (row spinor); $j^\mu$ — probability four-current; $j^0$ — density. Reading it: pairing $\psi^\dagger$ with $\gamma^0$ makes $\bar\psi\psi$ a Lorentz scalar and $\bar\psi\gamma^\mu\psi$ a four-vector; the density $j^0=\psi^\dagger\psi=\sum_i|\psi_i|^2$ is a sum of squares, never negative — the defect that had killed Klein–Gordon.

Derivation AFrom the Hamiltonian form to the covariant equation and its algebra

The goal is to carry out the repackaging that turns the time-asymmetric Hamiltonian-form Dirac equation into the manifestly Lorentz-structured $(i\gamma^\mu\partial_\mu-m)\psi=0$ , and then to show that the defining Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ is not an extra assumption but a direct consequence of the $\alpha,\beta$ relations we already had. This is Thomson's route, and it makes explicit that the covariant form contains exactly the same physics as the original — nothing is added, only relabelled.

Step 1. Start from the Hamiltonian form $i\partial_t\psi=(-i\alpha^k\partial_k+\beta m)\psi$ and move everything to one side: $i\partial_t\psi+i\alpha^k\partial_k\psi-\beta m\psi=0$ . Multiply on the left by $\beta$ and use $\beta^2=I$ : $i\beta\partial_t\psi+i\beta\alpha^k\partial_k\psi-m\psi=0$ . Why: multiplying by $\beta$ is the decisive move — it converts $\partial_t$ into $\beta\partial_t=\gamma^0\partial_0$ and the $\alpha^k$ terms into $\beta\alpha^k=\gamma^k$ , putting all four derivatives on the same footing.

Step 2. Define $\gamma^0\equiv\beta$ and $\gamma^k\equiv\beta\alpha^k$ . With $\partial_0=\partial_t$ , the three terms become $i\gamma^0\partial_0\psi+i\gamma^k\partial_k\psi-m\psi=0$ , which is $(i\gamma^\mu\partial_\mu-m)\psi=0$ once the repeated index $\mu=0,1,2,3$ is summed. Why: the relabelling is legal because the $\gamma$ 's are just fixed matrices; treating $\mu$ as a Lorentz index is suggestive notation whose covariance still has to be proven separately.

Step 3. Derive the algebra. For time, $(\gamma^0)^2=\beta^2=I=g^{00}I$ . For a space index, $(\gamma^k)^2=\beta\alpha^k\beta\alpha^k=-\beta^2(\alpha^k)^2=-I=g^{kk}I$ , using $\beta\alpha^k=-\alpha^k\beta$ . Why: the minus sign in $(\gamma^k)^2=-I$ is manufactured entirely by pushing one $\beta$ past one $\alpha^k$ — the anticommuting of $\alpha$ and $\beta$ injects the Minkowski signature into the $\gamma$ 's.

Step 4. For the cross terms, $\{\gamma^0,\gamma^k\}=\alpha^k+\beta\alpha^k\beta=\alpha^k-\alpha^k=0$ , and $\{\gamma^i,\gamma^j\}=-(\alpha^i\alpha^j+\alpha^j\alpha^i)=-2\delta^{ij}I=2g^{ij}I$ . Collecting all cases gives $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I$ . Why: every entry of the anticommutator table reproduces a component of $g^{\mu\nu}$ , so the metric emerges from the old $\alpha,\beta$ algebra with no new input — the Clifford algebra is the covariant form's fingerprint.

Go deeper · Derivation B · 1 minDirac's factorisation and why the coefficients must be 4×4 matrices

Derivation BDirac's factorisation and why the coefficients must be 4×4 matrices

The aim is to reach the same Clifford algebra from the opposite direction — Dirac's own — by demanding that a first-order wave operator, when squared, reproduce the relativistic relation $p^2=m^2$ . This forces the anticommutation relations on the coefficients and, crucially, forces them to be matrices of size at least $4\times4$ — the origin of the electron's four-component spinor and its spin and antiparticle degrees of freedom.

Step 1. The relation to respect is $p^\mu p_\mu-m^2=0$ . Try to factor it into first-order pieces, $p^\mu p_\mu-m^2=(\slashed p+m)(\slashed p-m)=\slashed p\,\slashed p-m^2$ , which requires $\slashed p\,\slashed p=p^\mu p_\mu$ . Why: a first-order equation is wanted so probability density comes out positive and time evolution is generated by a genuine Hamiltonian — factoring is the way to descend from Klein–Gordon to a square root.

Step 2. Expand $\slashed p\,\slashed p=\gamma^\mu\gamma^\nu p_\mu p_\nu$ . Since $p_\mu p_\nu$ is symmetric, only the symmetric part of $\gamma^\mu\gamma^\nu$ survives, giving $\tfrac12\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu$ . Requiring this to equal $g^{\mu\nu}p_\mu p_\nu$ for all $p$ forces $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ . Why: the same Clifford algebra as Derivation A, now a demand rather than a consequence — the unique condition under which the first-order operator squares back to the correct relation.

Step 3. No four ordinary numbers can satisfy $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ : numbers commute, so $\gamma^\mu\gamma^\nu=\gamma^\nu\gamma^\mu$ , contradicting anticommutation for $\mu\neq\nu$ . The $\gamma$ 's must be matrices — Dirac's "brilliant inspiration". Why: anticommuting objects squaring to $\pm1$ cannot be scalars; the equation demands a multi-component wavefunction, which is where spin enters unbidden.

Step 4. Fix the minimum size. The $\gamma$ 's are traceless (from $\mathrm{tr}\,\gamma^\mu=-\mathrm{tr}\,\gamma^\mu$ using a chosen $\nu\neq\mu$ ) with eigenvalues $\pm1$ or $\pm i$ , so the dimension is even. A $2\times2$ space holds only three mutually anticommuting matrices (the Paulis), one short of four, so the smallest that works is $4\times4$ . Why: the four-component spinor is not a modelling choice but the minimal carrier of the algebra — its four slots become particle and antiparticle, each with two spin states.

Go deeper · Derivation C · 1 minThe adjoint equation and the conserved positive current

Derivation CThe adjoint equation and the conserved positive current

The purpose is to construct the four-current $j^\mu=\bar\psi\gamma^\mu\psi$ and prove it conserved, $\partial_\mu j^\mu=0$ , using the Hermiticity relation (4) and the adjoint spinor $\bar\psi=\psi^\dagger\gamma^0$ . The prize is that the density $j^0=\psi^\dagger\psi$ is manifestly non-negative — the property Klein–Gordon lacked, and the reason the Dirac equation could at first be read as a consistent single-particle theory.

Step 1. Take the Hermitian conjugate of $(i\gamma^\mu\partial_\mu-m)\psi=0$ : $-i(\partial_\mu\psi^\dagger)\gamma^{\mu\dagger}-m\psi^\dagger=0$ . Multiply on the right by $\gamma^0$ and use (4) as $\gamma^{\mu\dagger}\gamma^0=\gamma^0\gamma^\mu$ : with $\bar\psi=\psi^\dagger\gamma^0$ this becomes $i(\partial_\mu\bar\psi)\gamma^\mu+m\bar\psi=0$ , the adjoint equation. Why: the identity $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$ converts the ugly $\gamma^{\mu\dagger}$ into a clean $\gamma^0\gamma^\mu$ , giving an adjoint equation with the same $\gamma^\mu$ and the derivative acting to the left.

Step 2. Form the divergence: $\partial_\mu j^\mu=(\partial_\mu\bar\psi)\gamma^\mu\psi+\bar\psi\gamma^\mu(\partial_\mu\psi)$ . Why: the current is conserved only if these two pieces cancel, so each is evaluated from its own equation of motion — one Dirac, one adjoint.

Step 3. From the Dirac equation, $\gamma^\mu\partial_\mu\psi=-im\psi$ , so $\bar\psi\gamma^\mu\partial_\mu\psi=-im\bar\psi\psi$ . From the adjoint, $(\partial_\mu\bar\psi)\gamma^\mu=+im\bar\psi$ , so $(\partial_\mu\bar\psi)\gamma^\mu\psi=+im\bar\psi\psi$ . Adding: $\partial_\mu j^\mu=0$ . Why: the two mass terms are equal and opposite by construction, so conservation is exact — a continuity equation guaranteeing probability (later, charge) is neither created nor destroyed.

Step 4. Read off the density: $j^0=\bar\psi\gamma^0\psi=\psi^\dagger\psi=\sum_{i=1}^4|\psi_i|^2\ge0$ . Why: a sum of squared moduli can never go negative, unlike the Klein–Gordon density $\propto\mathrm{Im}(\phi^*\partial_t\phi)$ — the first-order equation's key advantage, flowing directly from pairing $\psi^\dagger$ with $\gamma^0$ .

Limiting cases

Case	Result	Meaning
Square the operator	$(i\slashed\partial-m)(i\slashed\partial+m)=-(\Box+m^2)$	every component obeys Klein–Gordon $(\Box+m^2)\psi=0$ ; Dirac is its "square root"
Rest frame, $\mathbf p=0$	$(i\gamma^0\partial_0-m)\psi=0$	$\gamma^0=\mathrm{diag}(I,-I)$ splits the four components into energy $+m$ and $-m$ pairs
Momentum space	$(\slashed p-m)u(p)=0$ , $\slashed p\,\slashed p=p^2$	plane-wave spinors live on the mass shell $p^2=m^2$ ; the launch point for $u,v$ solutions
Massless limit $m\to0$	$i\slashed\partial\,\psi=0$ , decouples via $\gamma^5$	left/right chiral halves obey independent Weyl equations
Non-relativistic limit	reduces to the Pauli equation	recovers spin- $\tfrac12$ coupling with $g=2$
Change of representation	Dirac ↔ Weyl ↔ Majorana	same algebra (3), different $\gamma^\mu$ ; no physical result changes

Common trapThe gamma matrices are not a four-vector

The upstairs index on $\gamma^\mu$ is deliberately suggestive, and it misleads many students into thinking the $\gamma$ 's transform between frames like a four-vector. They do not: the four $\gamma^\mu$ are fixed, constant matrices, identical in every frame and never boosted. What transforms under a Lorentz transformation $\Lambda$ is the spinor $\psi\to S(\Lambda)\psi$ , through a $4\times4$ matrix $S(\Lambda)$ acting on the spinor index. The covariance of $(i\gamma^\mu\partial_\mu-m)\psi=0$ then follows from the compatibility identity $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ — a statement about how $\psi$ rotates, not about the $\gamma$ 's changing. The equation's Lorentz invariance is therefore genuine but not automatic: it is a theorem, and $\bar\psi\gamma^\mu\psi$ behaves as a four-vector only because the two factors of $S$ conspire with $\gamma^\mu$ to make it so.

Worked examples

1 · Verifying the Clifford algebra in the Dirac–Pauli representation (confidence builder)

Using the matrices of equation (5), confirm $(\gamma^1)^2=-I$ and $\{\gamma^1,\gamma^2\}=0$ , matching $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ .

Step 1. Square $\gamma^1$ in blocks. With $\gamma^1=\begin{pmatrix}0&\sigma^1\\-\sigma^1&0\end{pmatrix}$ , block multiplication gives diagonal blocks $-\sigma^1\sigma^1$ . With $(\sigma^1)^2=I$ this is $(\gamma^1)^2=\mathbf{-I_4}$ , matching $\{\gamma^1,\gamma^1\}=2(\gamma^1)^2=2g^{11}I$ since $g^{11}=-1$ .

Step 2. Compute the products: $\gamma^1\gamma^2$ has diagonal blocks $-\sigma^1\sigma^2$ , and $\gamma^2\gamma^1$ has $-\sigma^2\sigma^1$ .

Step 3. Add them: the diagonal blocks become $-\{\sigma^1,\sigma^2\}$ . The Pauli matrices anticommute, $\{\sigma^1,\sigma^2\}=0$ , so every block vanishes and $\{\gamma^1,\gamma^2\}=\mathbf{0}$ , matching $2g^{12}I=0$ . Both checks confirm $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ term by term, using only the Pauli algebra — so they hold in any Pauli-based representation.

2 · The mass shell from squaring the operator (a real-numbers calculation)

Show that $(i\slashed\partial-m)(i\slashed\partial+m)=-(\Box+m^2)$ , so every component of $\psi$ obeys Klein–Gordon; then take an electron ( $m=0.511$ MeV) with $p=1.000$ MeV along $z$ and verify the mass shell numerically.

Step 1. Multiply the operators. Since $m$ commutes with $i\slashed\partial$ , $(i\slashed\partial-m)(i\slashed\partial+m)=(i\slashed\partial)^2-m^2=-\slashed\partial\,\slashed\partial-m^2$ . Now $\slashed\partial\,\slashed\partial=\tfrac12\{\gamma^\mu,\gamma^\nu\}\partial_\mu\partial_\nu=g^{\mu\nu}\partial_\mu\partial_\nu=\Box$ . Hence $(i\slashed\partial-m)(i\slashed\partial+m)=\mathbf{-(\Box+m^2)}$ .

Step 2. Act on a solution: if $(i\slashed\partial-m)\psi=0$ , applying $(i\slashed\partial+m)$ gives $(\Box+m^2)\psi=0$ — Klein–Gordon, obeyed by each component separately. Relativistic consistency is automatic: the Dirac equation is engineered to be the square root of Klein–Gordon.

Step 3. Go to momentum space: a plane wave $\psi\sim u(p)e^{-ip\cdot x}$ turns $\Box\to-p^2$ , so the mass shell is $p^2=E^2-\mathbf p^2=m^2$ . For the electron, $E=\sqrt{p^2+m^2}=\sqrt{(1.000)^2+(0.511)^2}=\sqrt{1.261121}=\mathbf{1.1230\ MeV}$ .

Step 4. Verify: $E^2-p^2=1.261121-1.000000=0.261121\ \mathrm{MeV}^2$ , equal to $m^2=(0.511)^2=0.261121\ \mathrm{MeV}^2$ . They agree exactly, so the electron sits on its mass shell and $(\slashed p-m)u(p)=0$ is consistent precisely there. The same $u(p)$ seeds the plane-wave and, with energy-sign changes, the antiparticle spinors.

Physics of the portrait

The covariant Dirac equation is inseparable from the man who wrote it: not derived from deeper theory but conjured, by a demand for mathematical beauty, into predicting antimatter.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · the relativistic electron and the gamma matrices

Paul Adrien Maurice Dirac trained as an electrical engineer in Bristol before turning to mathematics and, at Cambridge, to the new quantum theory. Reserved to the point of legend — colleagues coined the unit "one dirac" for a rate of one word per hour — he worked by an aesthetic compulsion, holding that a physical law should above all be mathematically beautiful. In 1928, seeking a relativistic electron equation first order in time (and so admitting a positive probability density, unlike Klein–Gordon), he demanded the operator factor the energy–momentum relation. That forced the anticommuting coefficients now called the gamma matrices and, with them, a four-component wavefunction. Two triumphs fell out unbidden: the equation predicted the electron's spin and its gyromagnetic ratio $g=2$ — spin emerging from relativistic consistency, not put in by hand — and its negative-energy solutions, reinterpreted later as genuine antiparticles, predicted the positron, found by Carl Anderson in 1932. Dirac shared the 1933 Nobel Prize with Schrödinger. The covariant $\gamma^\mu$ notation, sharpened by Feynman's slash, became the alphabet of quantum electrodynamics, and his prediction of antimatter from a symmetry of an equation remains one of the purest cases of mathematics leading experiment. He died in Tallahassee in 1984.

History

1926

— Klein and Gordon publish the relativistic scalar wave equation, plagued by negative probability densities

1928

— Dirac factorises the energy–momentum relation, obtaining a first-order equation and the four gamma matrices

1928

— the Clifford algebra {γ^μ, γ^ν} = 2g^{μν} and the four-component spinor appear

1932

— Anderson observes the positron in cosmic rays, confirming Dirac's antimatter prediction

1948

— Feynman introduces the slash notation and covariant perturbation theory for QED

Going further

Texts. Thomson, Modern Particle Physics, §4.5–4.5.1 (covariant form, gamma matrices, adjoint and current) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.1, §7.3 (Eqs. 7.15–7.20, the factorisation route) · Halzen & Martin, Quarks and Leptons, §5.3–5.4 (gamma-matrix algebra, bilinear covariants) · Peskin & Schroeder, §3.2–3.4 (Dirac algebra, adjoint, Lorentz covariance) · Schwartz, QFT and the Standard Model, §10–11 (spinors and gamma matrices). The metric throughout is the particle-physics standard $g^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ ; sources using $(-,+,+,+)$ flip its sign and hence that of $(\gamma^k)^2$ , so check a text's signature before importing a formula. Natural units $c=\hbar=1$ throughout. Key results to keep in mind: (1) $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ follows from $\gamma^0=\beta$ , $\gamma^k=\beta\alpha_k$ ; (2) $(\gamma^1)^2=-I$ and $\{\gamma^1,\gamma^2\}=0$ in representation (5); (3) $(i\slashed\partial-m)(i\slashed\partial+m)=-(\Box+m^2)$ ; (4) the electron mass shell gives $E=1.1230$ MeV at $p=1.000$ MeV, with $E^2-p^2=m^2=0.261121\ \mathrm{MeV}^2$ ; (5) $\partial_\mu j^\mu=0$ with $j^0=\psi^\dagger\psi\ge0$ .

Prerequisites. The Dirac Equation (α,β form) · The Klein–Gordon Equation · The Dirac Probability Current. Leads to. Plane-Wave Dirac Solutions · Antiparticles · Spin and Helicity States.

Ask the tutor: Hand me any step — building the γ's from α and β, verifying the algebra in a chosen representation, squaring the operator back to Klein–Gordon, or proving the current is conserved — and I will walk it through line by line.

Why bother rewriting the Dirac equation in covariant form if it says the same thing?

Because notation shapes what you can do. The Hamiltonian form singles out time, so its Lorentz behaviour is obscured and a boost means re-deriving how each term transforms. The covariant form $(i\gamma^\mu\partial_\mu-m)\psi=0$ puts all four derivatives on equal footing, and once you prove it covariant you never re-check it. Every downstream calculation — plane-wave solutions, bilinears like $\bar\psi\gamma^\mu\psi$ , Feynman amplitudes — is far cleaner in $\gamma$ 's than in separate $\alpha$ 's and $\beta$ . The physics is identical; the covariant form is the language in which it is workable.

Are the gamma matrices really not a four-vector, despite the index?

Correct, and this is the most common misconception. The upstairs index on $\gamma^\mu$ labels which of four fixed matrices you mean; it does not mean they rotate under a boost. They are constants, the same in every frame. What carries the Lorentz transformation is the spinor, $\psi\to S(\Lambda)\psi$ , with $S(\Lambda)$ acting on the spinor components. The equation stays valid because $S$ and the $\gamma$ 's satisfy $S^{-1}\gamma^\mu S=\Lambda^\mu{}_\nu\gamma^\nu$ , letting the transformed $\gamma$ 's be reabsorbed. The notation correctly predicts that $\bar\psi\gamma^\mu\psi$ is a four-vector — but the $\gamma$ 's themselves never change.

What is the Feynman slash and why is it used everywhere?

The slash is a contraction shorthand: $\slashed a\equiv\gamma^\mu a_\mu=\gamma^0a^0-\boldsymbol\gamma\cdot\mathbf a$ for any four-vector. So $\slashed\partial=\gamma^\mu\partial_\mu$ and $\slashed p=\gamma^\mu p_\mu$ , letting the equation read $(i\slashed\partial-m)\psi=0$ , or $(\slashed p-m)u=0$ in momentum space. Its power is the identity $\slashed a\,\slashed a=\tfrac12\{\gamma_\mu,\gamma_\nu\}a^\mu a^\nu=a^2$ : a slashed vector squares to its invariant length, collapsing long $\gamma$ -strings in amplitudes. Feynman introduced it to make covariant perturbation theory tractable, and every QED computation is written in slashes.

Why does the smallest Dirac spinor have four components rather than two?

Because the algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ needs four mutually anticommuting matrices with the right squares, and a $2\times2$ space is too small: the three Pauli matrices anticommute but exhaust the traceless $2\times2$ matrices, leaving no room for a fourth. The next even dimension, $4\times4$ , is the first that works. Those four slots describe particle and antiparticle, each with two spin states, so the four-component spinor is the minimal object encoding both spin- $\tfrac12$ and antimatter — neither assumed, both forced by insisting on a first-order relativistic equation.

How exactly does the positive probability density arise, and why did it matter?

The density is $j^0=\bar\psi\gamma^0\psi$ . Since $\bar\psi=\psi^\dagger\gamma^0$ and $(\gamma^0)^2=I$ , this collapses to $j^0=\psi^\dagger\psi=\sum_i|\psi_i|^2$ , a sum of squared moduli that can never be negative. This mattered historically: the Klein–Gordon equation, second order in time, gives a density $\propto\mathrm{Im}(\phi^*\partial_t\phi)$ that can go negative, seeming to disqualify it as a probability. Dirac's first-order equation restored a positive probability — much of why it was accepted so quickly. (In modern field theory both are field equations and the density is reread as charge density, which may legitimately be negative — but the positivity was decisive then.)

Does choosing the Dirac–Pauli representation change any physics?

No. The Dirac–Pauli, Weyl (chiral), and Majorana representations are different $4\times4$ sets, all satisfying $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ and related by similarity transformations $\gamma^\mu\to U\gamma^\mu U^{-1}$ , $\psi\to U\psi$ . Any physical quantity is invariant under these, so the choice is pure convenience. Dirac–Pauli diagonalises $\gamma^0$ and suits rest-frame and non-relativistic problems; Weyl diagonalises $\gamma^5$ and suits high-energy, massless, or chiral problems where the handed halves decouple. Pick whichever is shortest.