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Topic 4 of 4

Dirac's Equation

In 1928 Paul Dirac sought a relativistic wave equation first-order in time — and discovered, to his own astonishment, that it required a four-component wave function, predicted spin-½ without assumption, and implied the existence of antiparticles. The Dirac equation remains one of the most beautiful and consequential equations in all of physics.

1. The Problem with the Klein-Gordon Equation

The first attempt at a relativistic wave equation was the Klein-Gordon equation, obtained by quantizing E² = p²c² + m²c⁴ directly. In natural units (ℏ = c = 1):

\Box\phi + m^2\phi = 0, \qquad \Box \equiv \partial_t^2 - \nabla^2 = \partial_\mu\partial^\mu

This equation looks fully covariant and its plane-wave solutions e^{−iEt+ip·x} correctly satisfy E² = p² + m². But it has a fatal problem: it is second-order in time. The conserved current associated with any Lorentz-covariant scalar wave equation is:

j^0 = i\bigl(\phi^*\partial_t\phi - \phi\,\partial_t\phi^*\bigr)

For a plane-wave solution φ ∝ Ae^{−iEt}, this gives j⁰ = 2|A|²E. When E < 0 — and the Klein-Gordon equation permits E = ±√(p² + m²) — the probability density becomes negative. There is no consistent probabilistic interpretation.

The relativistic energy spectrum E = ±√(p²+m²) has two branches separated by a gap of 2m. The Klein-Gordon equation admits both, with no way to restrict to positive energies while preserving the completeness of the solution set.

Dirac's Strategy: Factor the Dispersion Relation

Dirac's insight was to seek an equation first-order in both space and time derivatives. The Schrödinger equation iℏ∂ψ/∂t = Hψ is first-order in time — could we write a relativistic H that is first-order in spatial derivatives too?

i\frac{\partial\psi}{\partial t} = \bigl(\alpha^i p_i + \beta m\bigr)\psi

Squaring both sides must reproduce E² = p² + m². Expanding, the cross terms vanish only if αⁱ and β satisfy the anticommutation algebra:

\{\alpha^i,\alpha^j\} = 2\delta^{ij},\quad \{\alpha^i,\beta\} = 0,\quad \beta^2 = 1

These relations cannot be satisfied by numbers. They require matrices. The minimum dimension is 4×4, so ψ must have at least 4 complex components — the Dirac spinor.

2. Gamma Matrices and the Clifford Algebra

Define γ⁰ = β and γⁱ = βαⁱ. The anticommutation relations compress into the elegant Clifford algebra:

\{\gamma^\mu,\gamma^\nu\} \equiv \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2g^{\mu\nu}\,I_4

where g^{μν} = diag(+1,−1,−1,−1) is the Minkowski metric. This single equation contains everything about the gamma matrices. Key consequences follow immediately:

Consequences of the Clifford algebra

Property	Derivation	Result
(γ⁰)²	{γ⁰,γ⁰} = 2g⁰⁰I = 2I	+I₄
(γⁱ)² (i=1,2,3)	{γⁱ,γⁱ} = 2gⁱⁱI = −2I	−I₄
γμγν for μ ≠ ν	From {γμ,γν} = 0	= −γνγμ
tr(γμ)	tr(γμ) = tr(γν(γν)²) → cyclic + (γν)² = ±1	0
tr(γμγν)	Half of Clifford relation	4gμν
tr(odd product of γs)	Insert γ⁵(γ⁵)² = γ⁵, cyclic trace	0

The Three Standard Representations

The Clifford algebra does not uniquely fix the matrices — any unitarily equivalent set {UγμU†} also satisfies it. Physics is representation-independent.

Dirac (Pauli-Dirac) representation — natural for slowly-moving particles; upper 2 components are "large" (order 1), lower are "small" (order v/c):

γ⁰ = diag(I₂, −I₂) γⁱ = off-diagonal Pauli blocks

Weyl (chiral) representation — used in the Standard Model (Peskin & Schroeder, Zee); makes γ⁵ diagonal and left/right chiralities explicit in the upper/lower 2 components.

Majorana representation — all γμ are purely imaginary; admits real solutions ψ = ψ^c (self-conjugate fermions). Neutrinos may be Majorana particles.

Explicitly, in the Dirac representation (the most common for non-relativistic problems):

\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}, \qquad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}

In the Weyl representation (preferred in modern QFT):

\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}, \qquad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}

In the Weyl basis, γ⁵ = diag(−I₂, +I₂) — it is block-diagonal, revealing the chiral structure immediately. This is why Peskin & Schroeder and Zee prefer it.

3. The Dirac Equation — Multiple Forms

The Dirac equation has three equivalent forms, each revealing a different aspect of the physics:

Form 1: Hamiltonian (Schrödinger-like)

i\partial_t\psi = H_{\text{D}}\psi,\qquad H_{\text{D}} = \boldsymbol{\alpha}\cdot\hat{\mathbf{p}}+\beta m

This is first-order in time — Dirac's original goal, achieved. H_D is a 4×4 matrix operator acting on the 4-component spinor ψ.

Form 2: Covariant (Lorentz-symmetric)

(i\gamma^\mu\partial_\mu - m)\psi = 0

Manifestly Lorentz covariant: all four spacetime indices are contracted. γμ∂μ = γ⁰∂_t + γⁱ∂ᵢ. This is the form used in virtually all modern calculations.

Form 3: Feynman's Slash Notation

(i\not\!\partial - m)\psi = 0, \quad \text{where}\quad \not\!A \equiv \gamma^\mu A_\mu

The most compact notation. For any 4-vector Aμ, Ã = γμAμ (read "A-slash"). This notation is ubiquitous in QED Feynman diagram calculations.

The Adjoint Equation and Probability Current

Define the Dirac adjoint ψ̄ = ψ†γ⁰ (not simply the Hermitian conjugate). Taking the Hermitian conjugate of the Dirac equation and multiplying by γ⁰:

\bar{\psi}\bigl(-i\overleftarrow{\partial_\mu}\gamma^\mu - m\bigr) = 0

From this pair of equations, the current jμ = ψ̄γμψ satisfies ∂μjμ = 0. The crucial time component is:

j^0 = \bar{\psi}\gamma^0\psi = \psi^\dagger\psi = |\psi_1|^2 + |\psi_2|^2 + |\psi_3|^2 + |\psi_4|^2 \geq 0

The probability density is a sum of squares — always non-negative. The Klein-Gordon problem is solved.

The Dirac Lagrangian

\mathcal{L}_{\text{Dirac}} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi = \bar{\psi}(i\not\!\partial - m)\psi

Treating ψ and ψ̄ as independent variables in the Euler-Lagrange equations reproduces the Dirac equation and its adjoint. This Lagrangian is the starting point for quantum electrodynamics when coupled to the photon field: ℒ_QED = ψ̄(i∂̸ − m)ψ − eψ̄γμψ Aμ − ¼FμνFμν.

4. Spin-½ Emerges Automatically

One of the most remarkable features of the Dirac equation: it predicts spin-½ without assuming it. Spin falls out of the requirement that the equation be Lorentz-covariant.

Orbital Angular Momentum is Not Conserved Alone

Compute the commutator of the Dirac Hamiltonian H_D = α·p + βm with the orbital angular momentum Lᵢ = εᵢⱼₖxⱼpₖ:

[H_D, L_i] = [\alpha^j p_j,\,\varepsilon_{ijk}x_j p_k] = -i\varepsilon_{ijk}\,\alpha^j p_k \neq 0

Orbital angular momentum alone is not a constant of the motion for a free relativistic particle.

The Intrinsic Spin Operator

Define the spin matrix (Dirac representation):

\boldsymbol{\Sigma} = \begin{pmatrix}\boldsymbol{\sigma} & 0 \\ 0 & \boldsymbol{\sigma}\end{pmatrix} = \frac{i}{2}\varepsilon_{ijk}[\gamma^j,\gamma^k]

Then [H_D, Σᵢ] = +iεᵢⱼₖ αⱼpₖ, which exactly cancels [H_D, Lᵢ]. The total angular momentum J = L + S commutes with H_D:

[H_D, J_i] = 0, \qquad \mathbf{J} = \mathbf{L} + \mathbf{S}, \quad \mathbf{S} = \frac{\hbar}{2}\boldsymbol{\Sigma}

S has eigenvalues ±ℏ/2. The Dirac equation demands that the electron carry spin-½. There is no choice in the matter — it is a mathematical consequence of requiring a first-order, Lorentz-covariant equation.

The g-Factor = 2

In an external magnetic field B, taking the non-relativistic limit of the Dirac equation (a procedure known as the Foldy-Wouthuysen reduction) gives the Hamiltonian:

H' = -\frac{e}{2m}\boldsymbol{\sigma}\cdot\mathbf{B} = -g\frac{e}{2m}\mathbf{S}\cdot\mathbf{B}, \qquad g = 2

Classical electrodynamics for an orbiting charge gives g = 1. The Dirac equation gives g = 2 automatically. This is why the electron's magnetic moment was originally called "anomalous" — it was twice the classical prediction. Dirac showed it was not anomalous at all; it was the prediction of relativity.

QED correction: One-loop quantum electrodynamics gives g = 2(1 + α/2π + ...) ≈ 2.00232. This has been confirmed experimentally to 12 significant figures — the most precisely tested prediction in all of science. The starting point for this triumph is the factor g = 2 from the Dirac equation.

5. Free-Particle Solutions: Dirac Spinors

For a free particle with definite 4-momentum p^μ = (E, p), try plane-wave solutions. Two types:

\psi(x) = u(p)\,e^{-ip\cdot x} \quad (\text{particle, positive frequency})

\psi(x) = v(p)\,e^{+ip\cdot x} \quad (\text{antiparticle, negative frequency})

Substituting into (i∂̸ − m)ψ = 0 gives algebraic equations for the spinors:

(\not\!p - m)\,u(p) = 0, \qquad (\not\!p + m)\,v(p) = 0

where p̸ = γμpμ = γ⁰E − γ·p. Each equation has two independent solutions (two spin states), giving four total: u^(1), u^(2) (particles) and v^(1), v^(2) (antiparticles).

Explicit Rest-Frame Spinors (Dirac Representation)

At p = 0 (E = m), the four basis spinors are:

u^{(1)} = \sqrt{2m}\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\; u^{(2)} = \sqrt{2m}\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\; v^{(1)} = \sqrt{2m}\begin{pmatrix}0\\0\\0\\1\end{pmatrix},\; v^{(2)} = \sqrt{2m}\begin{pmatrix}0\\0\\1\\0\end{pmatrix}

For general momentum p, the boosted spinors are (Dirac representation):

u^s(p) = \begin{pmatrix}\sqrt{E+m}\,\chi^s \\ \dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{\sqrt{E+m}}\chi^s\end{pmatrix}, \quad v^s(p) = \begin{pmatrix}\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{\sqrt{E+m}}\chi^s \\ \sqrt{E+m}\,\chi^s\end{pmatrix}

where χ¹ = (1,0)ᵀ (spin up) and χ² = (0,1)ᵀ (spin down) are 2-component Pauli spinors. Notice that as p→0, the lower components of u vanish (they are of order p/m — the "small component") and u reduces to the rest-frame form.

Normalization and Completeness (Spin Sums)

The Lorentz-invariant normalization convention is:

\bar{u}^r u^s = 2m\,\delta^{rs}, \qquad \bar{v}^r v^s = -2m\,\delta^{rs}, \qquad \bar{u}^r v^s = 0

The spin sum completeness relations are indispensable for calculating unpolarized Feynman diagrams (averaging/summing over unobserved spins):

\sum_{s} u^s(p)\bar{u}^s(p) = \not\!p + m, \qquad \sum_s v^s(p)\bar{v}^s(p) = \not\!p - m

Trace technology — the workhorse of QED calculations:
When summing over spins: |ū Γ u|² → tr[(p̸'+m) Γ (p̸+m) Γ̄], where Γ̄ = γ⁰Γ†γ⁰.
Key trace identities (natural units):
tr(I) = 4
tr(γμ) = 0
tr(γμγν) = 4gμν
tr(γμγνγργσ) = 4(gμνgρσ − gμρgνσ + gμσgνρ)
tr(odd number of γs) = 0

6. Negative Energy and the Prediction of Antimatter

The four free-particle solutions split into two positive-energy states (E = +√(p²+m²) > 0) and two negative-energy states (E = −√(p²+m²) < 0). The negative-energy states are troubling: a particle could emit photons and cascade to arbitrarily negative energies, with no stable ground state.

The Dirac Sea (1930)

Dirac proposed an audacious resolution: the vacuum is not empty. All negative-energy states are already occupied — an infinite "sea" of electrons. By the Pauli exclusion principle (electrons obey Fermi statistics), no additional electron can occupy an already-filled state, and the cascade is blocked.

The Dirac sea. All negative-energy states are filled (blue, below E = −m). A real electron (bright blue) sits in a positive-energy state. A photon γ with energy ≥ 2m promotes a sea electron to positive energy, leaving a hole (red empty circle) in the sea. The hole has charge +e, mass mₑ, spin ½, and positive energy — it is the positron.

Properties of the Hole

Why the hole (= positron) has the properties it has

Property	Hole has...	Reasoning
Energy	E > 0	Removing a state of energy −E leaves energy deficit = +E
Charge	+e	Removing charge −e from a charge-neutral sea leaves net +e
Mass	mₑ	Sea states are electron states — same mass
Spin	½	Removing a spin-½ state leaves a spin-½ hole

Dirac initially thought the hole might be the proton (the only known positively charged particle), but Hermann Weyl pointed out that the hole must have the same mass as the electron. In 1931 Dirac predicted the antielectron (positron) with charge +e and mass mₑ. In 1932, Carl Anderson observed precisely this particle in cloud chamber photographs of cosmic rays, winning the 1936 Nobel Prize.

The Modern QFT Interpretation

The Dirac sea picture is conceptually cumbersome (an unobservable infinite background of particles). The modern interpretation — due to Feynman and Stückelberg — is cleaner: a negative-energy particle propagating forward in time is equivalent to a positive-energy antiparticle propagating backward in time. This is implemented precisely in the quantum field expansion:

\psi(x) = \int\!\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s\bigl[\underbrace{a_{p,s}}_{\text{destroys }e^-}u^s(p)\,e^{-ip\cdot x}+\underbrace{b^\dagger_{p,s}}_{\text{creates }e^+}v^s(p)\,e^{+ip\cdot x}\bigr]

Here a†_{p,s} creates an electron and b†_{p,s} creates a positron — both are genuine positive-energy particles in their own right. The positron is not a hole in a sea; it is its own real particle, an antielectron.

Charge conjugation C: The operation that exchanges particle ↔ antiparticle. For the Dirac field, ψ → ψ^c = Cψ̄ᵀ, where C = iγ²γ⁰. QED is symmetric under C (the photon field also transforms: Aμ → −Aμ). The weak force explicitly violates C — left-handed neutrinos exist, but right-handed antineutrinos do not have the same interactions, and C maps one to the other.

7. Chirality: Left- and Right-Handed Electrons

The Dirac equation contains one more layer of structure, invisible until you look closely. The four-component Dirac spinor decomposes into two two-component pieces — Weyl spinors — that carry fundamentally different transformation properties under the Lorentz group.

The Fifth Gamma Matrix γ⁵

\gamma^5 \equiv i\gamma^0\gamma^1\gamma^2\gamma^3 = \frac{i}{4!}\varepsilon_{\mu\nu\rho\sigma}\,\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma

The key properties of γ⁵ follow from the Clifford algebra:

Properties of γ⁵

Property	Statement	Consequence
Idempotent-like	(γ⁵)² = I₄	Eigenvalues can only be ±1
Hermitian	(γ⁵)† = γ⁵	Observable (real eigenvalues)
Anticommutes	{γ⁵, γμ} = 0 for all μ	Chirality flips under C and P separately
Traceless	tr(γ⁵) = 0	2 eigenvalues +1 and 2 eigenvalues −1
Weyl basis	γ⁵ = diag(−I₂, +I₂)	Block-diagonal; makes chirality manifest

Chiral Projection Operators

Since (γ⁵)² = 1, we can form projection operators:

P_L = \frac{1-\gamma^5}{2},\qquad P_R = \frac{1+\gamma^5}{2}

These satisfy P_L + P_R = I, P_L² = P_L, P_R² = P_R, P_L P_R = 0. Any Dirac spinor decomposes uniquely:

\psi = \psi_L + \psi_R, \qquad \psi_L = P_L\psi\;(\text{left-handed}),\quad \psi_R = P_R\psi\;(\text{right-handed})

ψ_L has γ⁵ eigenvalue −1; ψ_R has γ⁵ eigenvalue +1. In the Weyl basis, ψ = (ψ_L, ψ_R)ᵀ, where ψ_L and ψ_R are each 2-component Weyl spinors.

The Dirac Equation in Chiral Components

In the Weyl basis, (i∂̸ − m)ψ = 0 splits into two coupled equations:

i\bar{\sigma}^\mu\partial_\mu\,\psi_L = m\,\psi_R, \qquad i\sigma^\mu\partial_\mu\,\psi_R = m\,\psi_L

where σμ = (I₂, σ¹, σ², σ³) and σ̄μ = (I₂, −σ¹, −σ², −σ³). The mass term m couples left and right chiralities. Set m = 0 and the equations decouple completely:

i\bar{\sigma}^\mu\partial_\mu\,\psi_L = 0, \qquad i\sigma^\mu\partial_\mu\,\psi_R = 0 \qquad (\text{Weyl equations, } m=0)

A massless fermion is described by a single 2-component Weyl spinor and has an exact chirality. The neutrino was long treated as a massless left-handed Weyl fermion — and though we now know neutrinos have tiny masses, the approximation is excellent at energies much above those masses.

Chirality vs. Helicity

Chirality eigenstates for massless fermions. Left-handed (ψ_L): spin antiparallel to momentum (helicity h = −½). Right-handed (ψ_R): spin parallel to momentum (h = +½). For massive particles, a chirality eigenstate is a superposition of both helicity states.

Helicity vs. chirality

Property	Helicity h = S·p̂	Chirality (γ⁵ eigenvalue/2)
Values	±½	±½ (i.e., P_L gives −½, P_R gives +½)
Lorentz invariant?	No — a boost can flip h for massive particles	Yes — a Lorentz scalar quantity
Massless limit	Chirality = helicity exactly	Same
Massive particle	h-eigenstate ≠ chirality-eigenstate	Different in general
Parity	P: h → −h (reverses momentum, not spin)	P: chirality ↔ flips (P_L ↔ P_R)

Parity Violation and the Standard Model

Parity (spatial inversion P: x → −x) reverses momentum p → −p but leaves spin S unchanged (spin is an axial vector). Therefore P swaps helicity and also swaps chirality: P(ψ_L) = ψ_R. If a force couples to ψ_L but not ψ_R, it violates parity.

The charged-current weak interaction in the Standard Model couples exclusively to left-handed fermions:

\mathcal{L}_W = \frac{g}{\sqrt{2}}\,\bar{\psi}_L\gamma^\mu\nu_L W^-_\mu + \text{h.c.}

The W⁻ boson does not interact with right-handed electrons at all. This is maximal parity violation — the maximum allowed. It was discovered experimentally in 1957 by Chien-Shiung Wu in the angular distribution of electrons from Co-60 beta decay, winning the prediction's authors (Lee and Yang) the 1957 Nobel Prize.

Mass breaks chiral symmetry: In the massless limit, ψ_L and ψ_R transform independently under separate U(1) phase rotations — this is called chiral symmetry. The Dirac mass term ℒ_m = −m(ψ̄_L ψ_R + ψ̄_R ψ_L) breaks this symmetry explicitly, coupling the two chiralities. In the Standard Model, the gauge symmetry forbids a bare mass term. Fermion masses must arise from Yukawa couplings to the Higgs field: y ψ̄_L Φ ψ_R + h.c. When the Higgs acquires its vacuum expectation value ⟨Φ⟩ = v/√2, this becomes a mass term m = yv/√2. The Higgs boson is therefore the mechanism by which fermions acquire both mass and chiral mixing — the deepest consequence of the chiral structure first glimpsed in Dirac's equation.

Key Concepts

The Dirac spinor ψ has 4 components: 2 spin states × 2 energy signs (particle/antiparticle)
Gamma matrices γμ satisfy the Clifford algebra {γμ, γν} = 2gμν I₄
The covariant Dirac equation (iγμ∂μ − m)ψ = 0 is first-order in all spacetime derivatives
Positive-energy solutions (u spinors) describe electrons; negative-energy solutions (v spinors) describe positrons
Spin-½ is derived, not assumed — it emerges automatically from Lorentz covariance
γ⁵ = iγ⁰γ¹γ²γ³ defines chirality; P_L and P_R project onto left- and right-handed Weyl spinors
In the Standard Model, only left-handed fermions ψ_L feel the weak force (maximal parity violation)

Key Equations

Dirac equation — covariant form

(i\gamma^\mu\partial_\mu - m)\psi = 0

Clifford algebra

\{\gamma^\mu,\,\gamma^\nu\} = 2g^{\mu\nu}\,I_4

Dirac Hamiltonian

i\partial_t\psi = (\boldsymbol{\alpha}\cdot\hat{\mathbf{p}}+\beta m)\psi

Conserved probability current

j^\mu = \bar{\psi}\gamma^\mu\psi,\quad \partial_\mu j^\mu = 0

Fifth gamma matrix

\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3,\quad (\gamma^5)^2 = 1

Chiral projectors

P_L = \tfrac{1-\gamma^5}{2},\quad P_R = \tfrac{1+\gamma^5}{2}

Quantum Dirac field

\psi(x) = \int\!\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s\bigl[a_{p,s}\,u^s e^{-ipx}+b^\dagger_{p,s}\,v^s e^{+ipx}\bigr]

Spin sums (completeness)

\textstyle\sum_s u^s\bar{u}^s = \not\!p+m,\quad \sum_s v^s\bar{v}^s = \not\!p-m

Worked Example

Example Problem

Problem

An electron moves with momentum $|\mathbf{p}| = 2m$ (natural units). (a) Find its energy $E$ . (b) Evaluate $\text{tr}\bigl[\sum_s u^s(p)\bar{u}^s(p)\bigr]$ . (c) In the limit $m \to 0$ , what happens to the coupling between $\psi_L$ and $\psi_R$ in the Dirac equation?

Solution

(a) E = √(p²+m²) = √(4m²+m²) = m√5 ≈ 2.236m. (b) The spin sum gives p̸+m. Using tr(p̸) = tr(γμpμ) = pμ tr(γμ) = 0 (since tr γμ = 0), and tr(mI₄) = 4m: the total trace is 0 + 4m = 4m. (c) In the Weyl basis, the Dirac equation reads iσ̄μ∂μψ_L = mψ_R and iσμ∂μψ_R = mψ_L. As m→0, the right-hand sides vanish. ψ_L and ψ_R decouple entirely, each obeying a separate 2-component Weyl equation. A massless electron would be a pure chirality eigenstate, unable to convert between left- and right-handed components.

Practice

Exercises

8 problems

1 of 8

In natural units, an electron has momentum $|\mathbf{p}| = 3m$ . Find $E/m$ .

Your answer

(dimensionless)

2 of 8

From $\{\gamma^0,\gamma^0\} = 2g^{00}I_4$ , find the coefficient of $I_4$ in $(\gamma^0)^2$ .

Your answer

3 of 8

For a spin- $s$ particle, $\mathbf{S}^2 = \hbar^2 s(s+1)$ . For $s = 1/2$ , find $S^2/\hbar^2$ .

Unlock Exercise 3