---
title: The Dirac Equation and Spinors
module: Relativistic Wave Equations
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  Dirac demanded a wave equation first order in time to fix the Klein-Gordon
  density problem. Factorizing $E^2 = p^2 + m^2$ into a linear form forces the
  coefficients to be anticommuting matrices — the gamma matrices of the Clifford
  algebra — so the wavefunction becomes a four-component spinor. The plane-wave
  solutions split into two particle and two antiparticle states, spin appears
  automatically with the correct $g = 2$ magnetic moment, and the chirality
  projectors that the weak interaction later needs fall straight out of the
  fifth gamma matrix.
topics: [Relativistic Wave Equations]
draft: false
sources:
  - book: Griffiths
    ref: "Ch. 7 — Quantum Electrodynamics, §7.1–7.2 (the Dirac equation, its solutions, bilinear covariants)"
  - book: Thomson
    ref: "Ch. 4 — The Dirac equation"
  - book: Halzen & Martin
    ref: "Ch. 5 (the Dirac equation and its solutions)"
  - book: Tong
    ref: "The Standard Model (Cambridge Part III), §1.2 Spinors and the Clifford algebra"
---

The Klein-Gordon equation carried a negative-definite density because it is second
order in time. Dirac's idea, in 1928, was to insist on an equation _first_ order in
time, which admits a positive density $\rho = \psi^\dagger\psi$, and then to make it
relativistic by requiring first order in space as well. That single demand, pursued
honestly, forces almost everything about the electron: the wavefunction must have
four components, it must transform as a spinor rather than a scalar, spin
$\tfrac12$ appears without being put in, the magnetic moment comes out with $g = 2$,
and the equation predicts antiparticles. The Dirac equation is the most productive
guess in the history of the subject. This lesson follows the demand to its
consequences.

We keep $\hbar = c = 1$ and $\eta^{\mu\nu} = \mathrm{diag}(+,-,-,-)$, with the Pauli
matrices $\sigma^i$ and the $2\times 2$ identity as building blocks.

## Factorizing the energy relation

Dirac wanted a Hamiltonian linear in the momentum,

$$
i\,\frac{\partial \psi}{\partial t}
  = H\psi,
\qquad
H = \vec\alpha \cdot \vec p + \beta m,
$$

with constant coefficients $\alpha^1, \alpha^2, \alpha^3, \beta$. Any solution of a
sensible relativistic equation must also satisfy $E^2 = \vec p^{\,2} + m^2$, so
applying $H$ twice must reproduce $\vec p^{\,2} + m^2$. Square it:

$$
H^2 = \left(\alpha^i p_i + \beta m\right)\left(\alpha^j p_j + \beta m\right)
    = \alpha^i \alpha^j p_i p_j
      + \left(\alpha^i \beta + \beta \alpha^i\right) p_i\, m
      + \beta^2 m^2 .
$$

Symmetrizing the first term over $i,j$, this equals $\vec p^{\,2} + m^2$ for
**all** momenta only if the coefficients obey

$$
\alpha^i \alpha^j + \alpha^j \alpha^i = 2\,\delta^{ij},
\qquad
\alpha^i \beta + \beta \alpha^i = 0,
\qquad
\beta^2 = 1 .
$$

No ordinary numbers satisfy these: two numbers cannot anticommute. The
coefficients must be **matrices**. The smallest matrices that work are $4\times 4$,
so $\psi$ must be a four-component column, a **Dirac spinor**. A common explicit
choice is the Dirac representation,

$$
\vec\alpha = \begin{pmatrix} 0 & \vec\sigma \\ \vec\sigma & 0 \end{pmatrix},
\qquad
\beta = \begin{pmatrix} \mathbb 1 & 0 \\ 0 & -\mathbb 1 \end{pmatrix},
$$

each block $2\times 2$. That the factorization of a scalar equation demands
matrices, and matrices of size four, is the origin of every structural feature that
follows.

$$
% caption: The logic of the Dirac equation. Demanding a Hamiltonian linear in
% momentum whose square returns E^2 = p^2 + m^2 forces the coefficients to
% anticommute; anticommuting objects must be matrices; the smallest that close the
% algebra are 4x4, so the wavefunction has four components.
\begin{tikzpicture}[>=stealth, font=\footnotesize, node distance=6mm, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[draw=black, thick, rounded corners=0pt, align=center,
        inner sep=5pt] (a) at (0,3.0) {linear Hamiltonian\\p times alpha plus m times beta};
  \node[draw=black, thick, align=center, inner sep=5pt] (b) at (0,1.5)
        {square must give\\p squared plus m squared};
  \node[draw=black, thick, align=center, inner sep=5pt] (c) at (0,0.0)
        {alpha and beta anticommute};
  \node[draw=black, thick, align=center, inner sep=5pt] (d) at (0,-1.5)
        {must be 4 by 4 matrices};
  \node[draw=acc, very thick, fill=acc!8, align=center, inner sep=5pt] (e) at (0,-3.0)
        {four-component spinor};
  \draw[->, thick] (a) -- (b);
  \draw[->, thick] (b) -- (c);
  \draw[->, black, thick] (c) -- (d);
  \draw[->, acc, thick] (d) -- (e);
\end{tikzpicture}
$$

## Gamma matrices and the Clifford algebra

The equation looks cleaner in covariant form. Multiply the Dirac equation by
$\beta$ and define the **gamma matrices**

$$
\gamma^0 = \beta,
\qquad
\gamma^i = \beta\,\alpha^i .
$$

Then $i\gamma^0\partial_t\psi = (\dots)$ tidies into the covariant **Dirac
equation**

$$
\left(i\gamma^\mu \partial_\mu - m\right)\psi = 0,
$$

often abbreviated with the Feynman slash notation $\gamma^\mu \partial_\mu$. The
anticommutation relations
above collapse into the single defining relation of the **Clifford algebra**,

$$
\left\{\gamma^\mu, \gamma^\nu\right\}
  = \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu
  = 2\,\eta^{\mu\nu}\,\mathbb 1 .
$$

This one line contains all the earlier conditions: $\mu = \nu = 0$ gives
$(\gamma^0)^2 = 1$; $\mu = \nu = i$ gives $(\gamma^i)^2 = -1$; and $\mu \ne \nu$
gives anticommutation. The Clifford algebra has a unique irreducible
representation up to change of basis, so different textbooks' explicit gamma
matrices (Dirac basis, chiral/Weyl basis) are physically equivalent; only the
algebra is fundamental.[^tong-clifford] It is the algebra, not any particular
matrix, that makes the equation work.

$$
% caption: The gamma-matrix multiplication pattern. On the diagonal the squares
% are fixed by the metric — the time gamma squares to plus one, each space gamma
% to minus one. Off the diagonal distinct gammas anticommute, so their symmetric
% product vanishes. This single table is the Clifford algebra.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % grid 4x4
  \foreach \i in {0,1,2,3,4} \draw[black] (\i,0) -- (\i,4);
  \foreach \j in {0,1,2,3,4} \draw[black] (0,\j) -- (4,\j);
  % headers
  \foreach \i/\lab in {0/{g0}, 1/{g1}, 2/{g2}, 3/{g3}} {
    \node[black, font=\scriptsize] at (\i+0.5, 4.35) {\lab};
    \node[black, font=\scriptsize] at (-0.35, 3.5-\i) {\lab};
  }
  % diagonal entries: g0 -> +1, gi -> -1  (row 0 top)
  \node[font=\scriptsize] at (0.5,3.5) {plus 1};
  \node[font=\scriptsize] at (1.5,2.5) {minus 1};
  \node[font=\scriptsize] at (2.5,1.5) {minus 1};
  \node[font=\scriptsize] at (3.5,0.5) {minus 1};
  % off-diagonal: anticommute -> 0
  \foreach \r/\c in {0/1,0/2,0/3,1/0,1/2,1/3,2/0,2/1,2/3,3/0,3/1,3/2}
    \node[black, font=\scriptsize] at (\c+0.5, 3.5-\r) {0};
\end{tikzpicture}
$$

## The plane-wave solutions

Look for plane-wave solutions of definite four-momentum,

$$
\psi(x) = u(p)\, e^{-i p\cdot x}
\quad\text{(positive energy)},
\qquad
\psi(x) = v(p)\, e^{+i p\cdot x}
\quad\text{(negative energy)},
$$

where $u$ and $v$ are four-component constant spinors. Substituting the first form
into the Dirac equation gives the momentum-space condition

$$
\left(\gamma^\mu p_\mu - m\right) u(p) = 0,
$$

and the second gives $(\gamma^\mu p_\mu + m)\,v(p) = 0$. Each is a set of four linear
equations; each has two independent solutions. So for every momentum there are
**four** solutions in all:

- $u^{(1)}, u^{(2)}$ — two positive-energy spinors, the two spin states of the
  particle;
- $v^{(1)}, v^{(2)}$ — two "negative-energy" spinors, which the next lesson
  reinterprets as the two spin states of the antiparticle.

In the Dirac representation, writing the four-spinor as two two-component blocks
$\psi = \binom{\phi}{\chi}$, the positive-energy solutions have a large upper
block $\phi$ and a small lower block $\chi \sim (\vec\sigma\cdot\vec p/2m)\,\phi$
in the nonrelativistic limit. The two-valuedness of $\phi$ is spin: the electron's
spin is not added to the theory, it is the residual freedom in the plane-wave
spinor once the Dirac equation is imposed.

$$
% caption: The four plane-wave solutions of the Dirac equation for each momentum:
% two positive-energy spinors u carrying spin up and spin down (the particle), and
% two negative-energy spinors v (reinterpreted in the next lesson as the
% antiparticle, again spin up and spin down).
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % particle box
  \draw[very thick] (-3.6,0.4) rectangle (-0.4,2.6);
  \node[font=\scriptsize] at (-2.0,2.3) {positive energy (particle)};
  \node[align=center, font=\scriptsize] at (-2.75,1.3) {spin up};
  \node[align=center, font=\scriptsize] at (-1.25,1.3) {spin down};
  \draw[->] (-2.75,0.8) -- (-2.75,1.9);
  \draw[->] (-1.25,1.9) -- (-1.25,0.8);
  % antiparticle box
  \draw[very thick] (0.4,0.4) rectangle (3.6,2.6);
  \node[font=\scriptsize] at (2.0,2.3) {negative energy (antiparticle)};
  \node[align=center, font=\scriptsize] at (1.25,1.3) {spin up};
  \node[align=center, font=\scriptsize] at (2.75,1.3) {spin down};
  \draw[->] (1.25,0.8) -- (1.25,1.9);
  \draw[->] (2.75,1.9) -- (2.75,0.8);
\end{tikzpicture}
$$

## Spin and the magnetic moment $g = 2$

That spin comes free is confirmed by coupling the equation to electromagnetism.
The minimal substitution $p_\mu \to p_\mu - qA_\mu$ inserts the electromagnetic
potential, and taking the nonrelativistic limit of the resulting equation for the
upper block $\phi$ yields the Pauli equation

$$
i\,\frac{\partial \phi}{\partial t}
  = \left[\frac{(\vec p - q\vec A)^2}{2m}
    + q\Phi
    - \frac{q}{2m}\,\vec\sigma\cdot\vec B\right]\phi .
$$

The final term is a magnetic moment interacting with the field, and it was _not_
put in by hand — it emerges from the cross terms in $(\vec\sigma\cdot(\vec p -
q\vec A))^2$ via the identity $(\vec\sigma\cdot\vec a)(\vec\sigma\cdot\vec b) =
\vec a\cdot\vec b + i\vec\sigma\cdot(\vec a\times\vec b)$. Reading off the moment,

$$
\vec\mu = g\,\frac{q}{2m}\,\vec S,
\qquad
\vec S = \tfrac12\vec\sigma,
\qquad
g = 2 .
$$

The Dirac equation predicts the electron's **gyromagnetic ratio** $g = 2$, twice
the classical value for orbital motion, with no free parameter. Experiment
confirms $g \approx 2$ to leading order; the tiny deviation $g - 2$ is a quantum
correction computed in the QED module and is one of the most precise tests in all
of physics. A spinless Klein-Gordon particle has no such term. Getting $g = 2$
right was the decisive early triumph of the equation.

## Chirality, helicity, and the fifth gamma matrix

Beyond the four $\gamma^\mu$ there is a fifth matrix built from their product,

$$
\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3,
\qquad
(\gamma^5)^2 = 1,
\qquad
\{\gamma^5, \gamma^\mu\} = 0 .
$$

Because $\gamma^5$ squares to the identity, its eigenvalues are $\pm 1$, and the
operators

$$
P_L = \tfrac12\left(1 - \gamma^5\right),
\qquad
P_R = \tfrac12\left(1 + \gamma^5\right)
$$

are projectors ($P_L^2 = P_L$, $P_R^2 = P_R$, $P_L + P_R = 1$, $P_L P_R = 0$) that
split any Dirac spinor into **left-handed** and **right-handed chiral** parts,

$$
\psi = \psi_L + \psi_R,
\qquad
\psi_L = P_L\psi,
\qquad
\psi_R = P_R\psi .
$$

Chirality is the eigenvalue of $\gamma^5$; it is Lorentz invariant. **Helicity**,
the projection of spin along the momentum $h = \vec S\cdot\hat p$, is a distinct but
related quantity: for a massless particle chirality and helicity coincide, while
for a massive particle they differ by terms of order $m/E$ because one can always
boost past a massive particle and reverse its helicity, but not its chirality.
This distinction is not idle bookkeeping — the charged weak interaction couples
_only_ to $\psi_L$, the left-chiral projection, which is why parity is violated and
why the weak force treats the two handednesses so differently. The electroweak
module builds its doublets out of the projectors introduced here for free spinors.

$$
% caption: The chiral projectors split a Dirac spinor into left- and right-handed
% parts using the fifth gamma matrix. For a massless particle chirality equals
% helicity (spin aligned against or along momentum); a mass mixes the two because a
% massive particle can be overtaken in a boost, flipping its helicity but not its
% chirality.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % central spinor
  \node[draw=black, thick, align=center, inner sep=5pt] (psi) at (0,0) {Dirac spinor};
  % left projection
  \node[draw=acc, very thick, fill=acc!8, align=center, inner sep=5pt] (L) at (-3.0,1.6)
        {left-handed part};
  % right projection
  \node[draw=black, thick, align=center, inner sep=5pt] (R) at (3.0,1.6)
        {right-handed part};
  \draw[->, acc, thick] (psi) -- (L) node[midway, above, font=\scriptsize, acc] {half of one minus g5};
  \draw[->, black, thick] (psi) -- (R) node[midway, above, font=\scriptsize, black] {half of one plus g5};
  % helicity glyphs below
  \node[font=\scriptsize, align=center] at (-3.0,-1.4) {spin against momentum};
  \node[font=\scriptsize, align=center] at (3.0,-1.4) {spin along momentum};
  \draw[->] (-3.9,-0.7) -- (-2.1,-0.7);
  \draw[->] (-3.0,-0.35) -- (-3.0,-1.0); % spin arrow
  \draw[->] (2.1,-0.7) -- (3.9,-0.7);
  \draw[black, ->] (3.0,-1.0) -- (3.0,-0.35);
\end{tikzpicture}
$$

## Summary

Demanding a first-order, positive-density relativistic equation forces the
coefficients $\gamma^\mu$ to obey the Clifford algebra $\{\gamma^\mu, \gamma^\nu\}
= 2\eta^{\mu\nu}$, which pushes the wavefunction up to four components — a Dirac
spinor. The covariant equation $(i\gamma^\mu\partial_\mu - m)\psi = 0$ has four
plane-wave solutions per momentum: two spin states of a particle and two of an
antiparticle. Spin $\tfrac12$ is not inserted; it is the residual freedom in the
spinor, and coupling to electromagnetism reproduces the electron's magnetic moment
with $g = 2$ automatically. The fifth gamma matrix $\gamma^5$ supplies the chiral
projectors $\tfrac12(1 \mp \gamma^5)$ that the weak interaction later needs. The
one loose end — the negative-energy solutions $v(p)$ — is the subject of the next
lesson.

[^tong-clifford]: The Clifford relation $\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}$ and the equivalence of the Dirac and chiral bases are stated in Tong, _The Standard Model_ (Cambridge Part III), §1.2, [damtp.cam.ac.uk/user/tong/standardmodel.html](http://www.damtp.cam.ac.uk/user/tong/standardmodel.html); Griffiths, §7.1–7.2, and Thomson, Ch. 4, give the physical derivation from the linear Hamiltonian, the plane-wave spinors, and the $g = 2$ result; Halzen & Martin, Ch. 5, works the solutions and bilinear covariants. Numerical values of $g$ follow the Particle Data Group, [pdg.lbl.gov](https://pdg.lbl.gov).