Skip to main content

Advanced Quantum Mechanics course sequence · Interactive lesson

Gamma matrices, the Clifford algebra and trace identities

Until now you have multiplied Dirac matrices out entry by entry. This topic retires that habit. Rewritten with γμ, Dirac's ten conditions collapse into one anticommutator with the metric inside it, and from that line alone come the slash identity and the traces that turn every spin sum later in the course into arithmetic.

UAQM-04.03 · Lesson 3 of 9 in this unit. Position is not a completion score.
Starting model: Gamma matrices, the Clifford algebra and trace identitiesLeft: a = (a⁰, 0, 0, a³) in units of mc, against the dashed light cone. Right: the eigenvalues λ of the 4 × 4 matrix a̸ in the complex plane, with an open circle of radius 1 mc. Since a̸a̸ = (a⋅a) × 1, a⋅a = 1.000 m²c², and Tr a̸ = 0, they are ±√(a⋅a), two of each: real parts ±1.000 mc, imaginary parts ±0.000 mc.a̸a̸ = (a⋅a) × 1, a⋅a = 1.00 m²c²a⋅a: + timelike, 0 null, − spacelikelight conea⁰a³Re λIm λλ = ±√(a⋅a), two of each, since Tr a̸ = 0
Start with a picture. Change the model in the lab below.
1

Build the model

One anticommutator holds all of Dirac's requirements, and every trace follows from it.

The key idea

The gamma matrices are the four fixed 4 × 4 matrices γ⁰ = β and γⁱ = βαᵢ, defined by the Clifford algebra (γμ, γν) = 2g(μν), in terms of which the free Dirac equation reads (iħγμ∂μ − mc)ψ = 0.

Example. For an electron at 0.8c, E = 0.8517 MeV and pc = 0.6813 MeV, so p̸p̸ = (E² − p²c²)/c² × 1 = (0.7253 − 0.4642) MeV²/c² × 1 = (0.511 MeV/c)² × 1: the measured mass returns as the square of a 4 × 4 matrix.

The complete physical model and assumptions

Dirac's requirements arrived as a list: four Hermitian matrices α₁, α₂, α₃ and β, each squaring to 1, every pair anticommuting. Multiply the equation by β/c, name γ⁰ = β and γⁱ = βαᵢ, and the list becomes one relation, (γμ, γν) = 2g(μν), with the Minkowski metric written into the algebra: the time matrix squares to +1 and each space matrix to −1. That sign is why the γⁱ must be anti-Hermitian, and why γμ† = γ⁰γμγ⁰ takes over from Hermiticity. The payoff is that calculation stops needing matrices. Contract with a four-vector and a̸ = γμaμ squares to (a⋅a) × 1, so p̸p̸ = m²c² for a free particle and the Klein-Gordon condition returns in one line. Traces need only the algebra, cyclicity and Tr 1 = 4: two gammas give 4g(μν), any odd number gives zero, four give three pairings signed +, −, +. None of this depends on the representation, which is why spin-summed rates never do. The tool has two edges. The algebra fixes the matrices only up to equivalence and says nothing about frames: the γμ are fixed matrices, not a four-vector, so the covariant-looking notation proves no covariance; Unit 5 has to. And traces evaluate nothing physical until topic 4.08's spin sums turn squared amplitudes into traces, while products containing Unit 5's γ⁵ bring in the antisymmetric ε(μνρσ).

1From α and β to γμ

Work in SI units with xμ = (ct, x), ∂μ = ∂/∂xμ = ((1/c)∂/∂t, ∇) and the metric g = diag(1, −1, −1, −1), kept throughout. Topic 4.01's equation is iħ ∂ψ/∂t = (−iħc αⁱ∂ᵢ + βmc²)ψ. Multiply on the left by β/c and use β² = 1: iħβ∂₀ψ + iħβαⁱ∂ᵢψ − mcψ = 0. Naming γ⁰ = β and γⁱ = βαᵢ gives (iħγμ∂μ − mc)ψ = 0, summed over μ = 0 to 3, derived. With pμ = iħ∂μ, so that p₀ = E/c, it reads (p̸ − mc)ψ = 0. Dimensions: the γμ are pure numbers, ħ∂μ has units J s m⁻¹ = kg m s⁻¹, and mc is a momentum, 2.731 × 10⁻²² kg m s⁻¹ for the electron, so the equation compares the spinor's variation with the inverse of λ̄C = ħ/mc = 3.862 × 10⁻¹³ m. Nothing new was assumed. This is topic 4.01's equation under new names, reached by multiplying by a matrix, not by changing frame.

2Ten conditions become one anticommutator

Compute each product from the α, β algebra. (γ⁰)² = β² = 1. (γⁱ)² = βαᵢβαᵢ = −β²αᵢ² = −1, because βαᵢ = −αᵢβ. For i ≠ j, γⁱγʲ + γʲγⁱ = −(αᵢαⱼ + αⱼαᵢ) = 0, and γ⁰γⁱ + γⁱγ⁰ = αᵢ + βαᵢβ = αᵢ − αᵢ = 0. All ten conditions now read (γμ, γν) = 2g(μν), the Clifford algebra, derived, with the metric's signs in the squares. Hermiticity changes character. γ⁰ is Hermitian, but γⁱ† = αᵢβ = −βαᵢ = −γⁱ, anti-Hermitian, and no representation can avoid it: a Hermitian matrix has real eigenvalues and cannot square to −1. One rule covers both, γμ† = γ⁰γμγ⁰, checked directly: γ⁰γ⁰γ⁰ = γ⁰ and γ⁰(βαᵢ)γ⁰ = αᵢβ = γⁱ†. In topic 4.02's Dirac representation, γ⁰ = diag(1, 1, −1, −1), and γⁱ has σᵢ in its upper-right block and −σᵢ in its lower-left, so each statement can be confirmed by block multiplication.

3Slash notation, and the square that gives back the mass

For any four-vector a, define a̸ = γμaμ = γ⁰a⁰ − γ⋅a. Because aμaν is symmetric, only the symmetric part of γμγν survives: a̸a̸ = ½(γμ, γν)aμaν = (a⋅a) × 1, derived. The same step gives a̸b̸ + b̸a̸ = 2(a⋅b) × 1. Now multiply (p̸ − mc)ψ = 0 by (p̸ + mc): (p⋅p − m²c²)ψ = 0, which for a plane wave is E² = p²c² + m²c⁴, Dirac's requirement recovered in one line rather than a page of matrices. For a plane wave p̸ is a numerical matrix with p̸p̸ = m²c² and, as the next section shows, Tr p̸ = 0, so its eigenvalues are +mc and −mc, two of each, and the equation puts the spinor in the +mc pair. The other two classes of vector behave differently. For a photon's k, k̸k̸ = k⋅k = 0 although k̸ is not zero: it is nilpotent, with no inverse. A spacelike vector's slash squares to a negative multiple of 1, as γ³ does, so its eigenvalues are imaginary. The figure shows all three.

4Traces from the algebra alone

Three inputs suffice: the Clifford algebra, cyclicity Tr(AB) = Tr(BA), and Tr 1 = 4, the one fact taken from the representation. Then Tr(γμγν) = ½Tr(γμ, γν) = 4g(μν), derived, so Tr(γ¹γ¹) = −4 and Tr(a̸b̸) = 4a⋅b. For odd products let Γ = γ⁰γ¹γ²γ³. Each γμ anticommutes with three of Γ's factors and commutes with the fourth, itself, so Γγμ = −γμΓ, and Γ is invertible because every factor is. For a product X of n gammas, ΓXΓ⁻¹ = (−1)ⁿX, while cyclicity gives Tr(ΓXΓ⁻¹) = Tr X. So Tr X = (−1)ⁿTr X, and every odd trace vanishes, derived; Unit 5 calls iΓ the matrix γ⁵. A change of representation, X → UXU⁻¹, leaves every trace unchanged, so no result can depend on which matrices were written down: Tr γ⁰ = 1 + 1 − 1 − 1 = 0 in the Dirac representation, and the Weyl γ⁰, with 1 in its off-diagonal blocks, has no nonzero diagonal entry at all.

5The four-γ trace: pairings and their signs

Move γμ to the right through the other three, each step costing a 2g term and a sign: γμγνγργσ = 2g(μν)γργσ − 2g(μρ)γνγσ + 2g(μσ)γνγρ − γνγργσγμ. The last term has the same trace as the left side, by cyclicity, so twice the trace is the sum of three two-γ traces: Tr(γμγνγργσ) = 4(g(μν)g(ρσ) − g(μρ)g(νσ) + g(μσ)g(νρ)), derived. Read it as a pairing rule, adjacent pairs with + and the crossed pair with −. A direct check: γ⁰γ¹γ⁰γ¹ = −γ⁰γ⁰γ¹γ¹ = −(1)(−1) = 1, of trace 4, and the formula gives 4(0 − (1)(−1) + 0) = 4. Contracting indices gives γμγμ = ½gμν(γμ, γν) = gμνg(μν) = 4 and γμa̸γμ = γμ(2aμ − γμa̸) = 2a̸ − 4a̸ = −2a̸, derived, shortcuts that keep longer traces short. Six gammas give fifteen pairings by the same reduction.

6What the notation does not prove

The index on γμ invites a reading the algebra does not support. The γμ are four fixed numerical matrices, the same in every frame; μ labels which matrix, as i labels σᵢ. Under a boost the coordinates change by Λ, but nobody replaces γμ by Λμνγν. Covariance therefore asks for something the notation cannot supply: a 4 × 4 matrix S acting on the spinor with S⁻¹γμS = Λμνγν, so that the equation keeps its form. Unit 5 proves that S exists and builds it; until then covariance is assumed. The algebra also leaves the matrices free up to equivalence, as Pauli's theorem, quoted in topic 4.02, says. What it fixes is every trace, and traces are what physics needs: topic 4.08 turns sums over unobserved spins into traces of p̸ ± mc, and topic 5.09 evaluates them into Mott's factor 1 − β²sin²(θ/2), quoted here, the baseline against which high-energy electron scattering from nuclei is measured.

2

Key equations

Read the units, assumptions and limits before substituting.

The Clifford algebra

(γμ, γν) = 2g(μν) 1

Ten conditions in one line: (γ⁰)² = 1, (γⁱ)² = −1, distinct gammas anticommute. Derived from topic 4.01's αᵢ² = β² = 1 and its anticommutators.

γμ: four dimensionless 4 × 4 matrices, γ⁰ = β, γⁱ = βαᵢ; g = diag(1, −1, −1, −1); 1 is the 4 × 4 identity

Hermitian conjugate through γ⁰

γμ† = γ⁰γμγ⁰

A Hermitian matrix cannot square to −1, so the γⁱ cannot be Hermitian in any representation. This one rule, derived, covers both cases and builds topic 4.04's ψ̄ = ψ†γ⁰.

dimensionless; γ⁰ is Hermitian and each γⁱ anti-Hermitian whenever β and the αᵢ are Hermitian

Slash notation squared

a̸a̸ = (aμaμ) 1, where a̸ = γμaμ = γ⁰a⁰ − γ⋅a

Derived from the symmetric part of γμγν. For a free particle p̸p̸ = m²c², so (p̸ + mc)(p̸ − mc)ψ = 0 is the Klein-Gordon condition; a null vector gives a nonzero a̸ squaring to zero.

aμ = (a⁰, a) any four-vector, γ⋅a = γⁱaⁱ summed; for pμ = (E/c, p) in kg m s⁻¹, p̸p̸ is in kg² m² s⁻²

Trace of two gammas

Tr(γμγν) = 4g(μν)

Half the trace of the anticommutator, by cyclicity Tr(AB) = Tr(BA). The metric's signs survive: Tr(γ¹γ¹) = −4, not +4.

dimensionless; the 4 is Tr 1, the only input taken from the representation; contracted, Tr(a̸b̸) = 4a⋅b

Odd products are traceless

Tr(γμ₁γμ₂ ⋯ γμₙ) = 0 for odd n

Every term with an odd number of gammas drops out of a spin-sum trace before any arithmetic, so the mc terms only ever survive in pairs.

dimensionless; proved with Γ = γ⁰γ¹γ²γ³, which anticommutes with every γμ (iΓ is Unit 5's γ⁵)

Trace of four gammas

Tr(γμγνγργσ) = 4(g(μν)g(ρσ) − g(μρ)g(νσ) + g(μσ)g(νρ))

Adjacent pairings enter with +, the crossed pairing with −. Derived by anticommuting γμ through the other three; six gammas give fifteen pairings the same way.

dimensionless; contracted, Tr(a̸b̸c̸d̸) = 4[(a⋅b)(c⋅d) − (a⋅c)(b⋅d) + (a⋅d)(b⋅c)]
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.25 mc
0 to 2 mc; steps of 0.05.
0.75 mc
−2 to 2 mc; steps of 0.05.

Try this. Start at a⁰ = 1.25, a³ = 0.75, an electron at 0.6c: the eigenvalues sit at ±1 mc, and the Dirac equation picks the +mc pair for u. Slide a³ to 1.25 and both collapse to 0 though a̸ is not zero; push outside the cone and they move onto the imaginary axis, like the ±i of γ³.

Live results

INVARIANT a⋅a
1.000 m²c²
REAL PARTS OF λ (±)
1.000 mc
IMAGINARY PARTS OF λ (±)
0.000 mc
Tr(a̸a̸) = 4a⋅a
4.000 m²c²

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelLeft: a = (a⁰, 0, 0, a³) in units of mc, against the dashed light cone. Right: the eigenvalues λ of the 4 × 4 matrix a̸ in the complex plane, with an open circle of radius 1 mc. Since a̸a̸ = (a⋅a) × 1, a⋅a = 1.000 m²c², and Tr a̸ = 0, they are ±√(a⋅a), two of each: real parts ±1.000 mc, imaginary parts ±0.000 mc.a̸a̸ = (a⋅a) × 1, a⋅a = 1.00 m²c²a⋅a: + timelike, 0 null, − spacelikelight conea⁰a³Re λIm λλ = ±√(a⋅a), two of each, since Tr a̸ = 0
Left: a = (a⁰, 0, 0, a³) in units of mc, against the dashed light cone. Right: the eigenvalues λ of the 4 × 4 matrix a̸ in the complex plane, with an open circle of radius 1 mc. Since a̸a̸ = (a⋅a) × 1, a⋅a = 1.000 m²c², and Tr a̸ = 0, they are ±√(a⋅a), two of each: real parts ±1.000 mc, imaginary parts ±0.000 mc.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Time component a⁰: 1.25 mc; Space component a³: 0.75 mc

Baseline 1.25; choose 0 to 2, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The upper index makes γμ look like a four-vector, and students draw two false conclusions: that in a boosted frame the matrices become Λμνγν, and that writing (iħγμ∂μ − mc)ψ = 0 with contracted indices proves the equation covariant. The γμ are four fixed numerical matrices, used unchanged in every frame. The index labels them, as the index on σᵢ labels the Pauli matrices, and a̸ = γμaμ is a matrix whose entries change from frame to frame although its square, a⋅a, does not. A contracted index shows invariance only when both objects transform, and here ∂μ does while γμ does not. The equation keeps its form only if the spinor transforms by some S with S⁻¹γμS = Λμνγν, a claim with content, proved in Unit 5. Until then the notation is a promise, not a proof.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

An electron moves along +z at v = 0.6c. Using the measured mₑc² = 0.511 MeV, write p̸ as an explicit 4 × 4 matrix in the Dirac representation, square it by blocks, and show that the result is m²c² times the identity. What are the eigenvalues of p̸?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

A 2.00 MeV photon travelling along +z meets an electron at rest, mₑc² = 0.511 MeV. Show that k̸k̸ = 0 although k̸ is not zero, find k̸p̸ + p̸k̸, and evaluate Tr(k̸p̸k̸p̸) twice: from the four-γ trace and by anticommuting.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

HardExample 3

An electron at v = 0.8c (E = 0.852 MeV, pc = 0.681 MeV) scatters elastically through angle θ, keeping its energy. Topic 4.08's spin sums will turn its summed squared Coulomb amplitude into T(θ) = Tr[(p̸′ + mc)γ⁰(p̸ + mc)γ⁰]. Evaluate T from the trace identities alone at θ = 0, 90° and 180°, in m²c² and in (MeV/c)², and find T(θ)/T(0).

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

Using only (γμ, γν) = 2g(μν) with g = diag(1, −1, −1, −1), what are (γ⁰γ¹)² and (γ¹γ²)²?
7

Review & continue

The gamma matrices are the four fixed 4 × 4 matrices γ⁰ = β and γⁱ = βαᵢ, defined by the Clifford algebra (γμ, γν) = 2g(μν), in terms of which the free Dirac equation reads (iħγμ∂μ − mc)ψ = 0.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.