Gamma Matrices

Gamma matrices arise because a relativistic wave equation first order in time and space must still reproduce the quadratic mass shell. Writing

$$H=c\boldsymbol\alpha\cdot\mathbf p+\beta mc^2$$

and demanding $H^2=\mathbf p^2c^2+m^2c^4$ for every momentum forces a matrix algebra. In covariant form it is

$$\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}I_4.$$

This Clifford algebra is representation independent. Particular Dirac or chiral matrices are coordinate choices on the spinor space and live on the Gamma-Matrix Conventions lookup page. Here the objective is to derive the algebra and the conclusions that do not depend on a chosen basis.

Required background. Metric and Units, Four-Vectors, and the Energy–Momentum Relation supply the sign and unit conventions, Lorentz invariants, and dispersion relation. This page also assumes complex linear algebra and elementary Pauli-matrix identities.

Linearizing the relativistic Hamiltonian

The positive square-root Hamiltonian

$$H=\sqrt{\mathbf p^2c^2+m^2c^4}$$

is nonlocal in position space and separates one energy branch before a local relativistic wave equation has been constructed. Dirac’s algebraic strategy is to seek a matrix-valued expression linear in momentum,

$$H=c\sum_{i=1}^3\alpha_i p_i+\beta mc^2,$$

acting on a multicomponent amplitude. The coefficients are independent of $\mathbf p$. Squaring gives

$$\begin{aligned} H^2 &= c^2\sum_{i,j}\alpha_i\alpha_jp_ip_j +mc^3\sum_i(\alpha_i\beta+\beta\alpha_i)p_i +\beta^2m^2c^4. \end{aligned}$$

Because $p_ip_j$ is symmetric in $i,j$, the antisymmetric part of $\alpha_i\alpha_j$ drops out. Equality with the scalar dispersion relation for every $\mathbf p$ and $m$ requires

$$\{\alpha_i,\alpha_j\}=2\delta_{ij}I, \qquad \{\alpha_i,\beta\}=0, \qquad \beta^2=I.$$

These conditions cancel every mixed term and leave

$$H^2=(\mathbf p^2c^2+m^2c^4)I.$$

The square-root relation has not disappeared: it has been factored into a first-order matrix operator whose eigenvalues occupy both energy branches.

From alpha and beta to gamma matrices

Define

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

Then $\alpha_i=\gamma^0\gamma^i$ and the relations above imply

$$(\gamma^0)^2=I, \qquad (\gamma^i)^2=-I, \qquad \gamma^\mu\gamma^\nu=-\gamma^\nu\gamma^\mu \quad(\mu\ne\nu).$$

Combining the cases gives

$$\boxed{ \{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}I }.$$

Lowered gamma matrices are obtained with the spacetime metric:

$$\gamma_\mu=\eta_{\mu\nu}\gamma^\nu, \qquad \gamma_0=\gamma^0, \qquad \gamma_i=-\gamma^i.$$

Lowering a Lorentz label is unrelated to Hermitian conjugation.

For any covector $a_\mu$, define slash notation by

$$a\!\!\!/\equiv\gamma^\mu a_\mu.$$

The Clifford algebra immediately yields

$$(a\!\!\!/)^{2} = \frac12\{\gamma^\mu,\gamma^\nu\}a_\mu a_\nu = a_\mu a^\mu I.$$

The commutator part vanishes because $a_\mu a_\nu$ is symmetric. More generally,

$$a\!\!\!/\,b\!\!\!/+b\!\!\!/\,a\!\!\!/ = 2a\cdot b\,I.$$

This is the representation-independent factorization behind the free Dirac operator.

Why four components are minimal in 3+1 dimensions

The four Hermitian matrices $\alpha_1,\alpha_2,\alpha_3,\beta$ square to $I$ and anticommute pairwise. Conjugating any one by another changes its sign, so its trace vanishes. Since its eigenvalues are $+1$ and $-1$, both must occur equally often: the matrix dimension is even. This already rules out dimensions one and three.

Dimension two also fails. The traceless $2\times2$ complex matrices form a three-dimensional space spanned by the Pauli matrices. Four nonzero Hermitian matrices with the required relations would be mutually orthogonal under $\operatorname{tr}(AB)$, hence linearly independent, which is impossible in that space. The familiar $4\times4$ Dirac matrices provide a construction at the next allowed dimension.

The complexified Clifford algebra in four spacetime dimensions is

$$\mathrm{Cl}_{1,3}(\mathbb C)\cong M_4(\mathbb C).$$

Its irreducible complex representation therefore has dimension four. A minimal Dirac amplitude has four complex components. This statement concerns the representation dimension of the Clifford algebra; it does not say that a spinor is a four-vector or that its entries transform with a Lorentz matrix.

Reducible representations of dimension $8,12,\ldots$ are possible by taking direct sums, but they repeat the minimal algebra rather than describe the ordinary irreducible Dirac spinor.

Hermiticity and the Dirac adjoint

For $H=c\boldsymbol\alpha\cdot\mathbf p+\beta mc^2$ to be Hermitian in the standard spinor inner product, one may choose

$$\alpha_i^\dagger=\alpha_i, \qquad \beta^\dagger=\beta.$$

It follows that

$$(\gamma^0)^\dagger=\gamma^0, \qquad (\gamma^i)^\dagger=-\gamma^i.$$

Both cases are summarized by the identity

$$(\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0.$$

This identity motivates the Dirac adjoint of a spinor,

$$\bar\psi=\psi^\dagger\gamma^0.$$

Under a spinor boost, $\psi^\dagger$ transforms with $S^\dagger$, which differs from $S^{-1}$. The inserted $\gamma^0$ makes $\bar\psi$ transform with $S^{-1}$, so that it can form Lorentz-covariant bilinears. The current and its positive time component are derived on the covariant Dirac page rather than assumed here.

Similarity transformations and representation independence

Let $S$ be any invertible constant matrix and define

$$\gamma'^\mu=S\gamma^\mu S^{-1}.$$

Then

$$\begin{aligned} \{\gamma'^\mu,\gamma'^\nu\} &= S\{\gamma^\mu,\gamma^\nu\}S^{-1} \\ &= 2\eta^{\mu\nu}I. \end{aligned}$$

If spinor components are changed simultaneously by $\psi'=S\psi$, the abstract equation and all consistently transformed bilinears describe the same physics. A trace, determinant, characteristic polynomial, or completely contracted gamma identity is similarity invariant.

Not every arbitrary similarity matrix preserves the simple Hermiticity conditions with the same Euclidean component inner product. Unitarily related standard representations do. For a general similarity transformation, the spinor metric used in the adjoint must be transformed consistently.

Lorentz generators on spinors

The antisymmetric products

$$\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$$

generate the spinor representation of the Lorentz algebra, conventionally through $\sigma^{\mu\nu}/2$. Their commutator with a gamma matrix is

$$[\sigma^{\mu\nu},\gamma^\rho] = 2i\left( \eta^{\nu\rho}\gamma^\mu -\eta^{\mu\rho}\gamma^\nu \right).$$

This identity ensures that conjugating the gamma matrices by the spinor transformation reproduces the vector transformation of their Lorentz label. Rotations are represented unitarily, while finite-dimensional boost matrices are not unitary with respect to the ordinary positive-definite component inner product. Lorentz covariance instead uses the Dirac adjoint.

The precise exponential sign depends on whether one writes active or passive Lorentz transformations and how the antisymmetric parameters are defined. The basis-independent condition is the intertwining relation

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

which will be used to prove covariance of the Dirac equation.

Common pitfalls

Treating the Lorentz index as a matrix index. In $(\gamma^\mu)^a{}_b$, $\mu$ labels a spacetime vector component while $a,b$ are spinor-space matrix indices. They transform in different representations.

Equating $\gamma_\mu$ with $(\gamma^\mu)^\dagger$. The first lowers a Lorentz index with $\eta_{\mu\nu}$. The second takes a matrix adjoint. Their component formulas happen to share signs in this convention but express different operations.

Checking a basis and declaring the result basis dependent. Matrix entries change under similarity transformations; the Clifford relation and fully contracted identities do not.

Importing Euclidean formulas unchanged. Euclidean gamma matrices square to $+I$ in every direction. Here spatial gamma matrices square to $-I$ because the metric has signature $(+---)$.

Exercises

1. Traces from anticommutation

Without choosing explicit matrices, show that $\operatorname{tr}\gamma^\mu=0$ and $\operatorname{tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}$.

Solution

Choose a different, invertible gamma matrix. Conjugation by it sends $\gamma^\mu$ to $-\gamma^\mu$ but preserves the trace, so that trace is zero. Cyclicity makes the two terms in the trace of the Clifford relation equal: $2\operatorname{tr}(\gamma^\mu\gamma^\nu)=2\eta^{\mu\nu}\operatorname{tr}I_4$.

2. Slash factorization

Prove $(p\!\!\!/-mc)(p\!\!\!/+mc)=(p^2-m^2c^2)I$.

Solution

The scalar mass commutes with every gamma matrix, so the mixed terms cancel. The symmetric product $p_\mu p_\nu$ selects the anticommutator:

$$(p\!\!\!/)^{2} =\frac12\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu=p^2I.$$

Subtracting $m^2c^2I$ gives the factorization.

3. A nonunitary change of components

Let $\psi'=S\psi$ with $S=\operatorname{diag}(2,1,1,1)$ and transform the gamma matrices by similarity. Find the matrix $G$ such that $\psi'^\dagger G\psi'=\psi^\dagger\psi$. What matrix replaces $\gamma^0$ in the Dirac adjoint if the original gamma matrices use the Dirac basis?

Solution

The positive inner-product matrix is

$$G=(S^{-1})^\dagger S^{-1}=\operatorname{diag}(1/4,1,1,1).$$

For example, the original unit vector $(1,0,0,0)^T$ becomes $(2,0,0,0)^T$. Its Euclidean squared length is four, while its length measured with $G$ remains one. The transformed adjoint uses

$$A=(S^{-1})^\dagger\gamma^0S^{-1} =\operatorname{diag}(1/4,1,-1,-1), \qquad \bar\psi'=\psi'^\dagger A=\bar\psi S^{-1}.$$

Here $A$ differs from $\gamma'^0=\gamma^0$. The Clifford algebra survives every similarity transformation, but the simple adjoint formula with $\gamma'^0$ requires the compatible inner-product convention.

References

• P. A. M. Dirac, “The Quantum Theory of the Electron,” Proceedings of the Royal Society A 117, 610–624, 1928, doi:10.1098/rspa.1928.0023.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995.