Gamma-Matrix Conventions

Use these four-dimensional gamma matrices and identities with the conventions

$$\begin{aligned} \eta^{\mu\nu}&=\operatorname{diag}(1,-1,-1,-1),\\ \epsilon^{0123}&=+1,\\ \{\gamma^\mu,\gamma^\nu\}&=2\eta^{\mu\nu}I_4. \end{aligned}$$

The Dirac basis is the default for Hamiltonian and nonrelativistic-limit calculations; the chiral basis is given for comparison. The basis-independent Clifford algebra is derived in Gamma Matrices. The formulas below provide explicit matrices and translation rules.

Required background. Metric and Units fixes the spacetime and orientation signs; Gamma Matrices derives the representation-independent algebra used below.

Gamma matrices and the Dirac adjoint

The lowered matrices, slash notation, Hamiltonian matrices, and adjoint are

$$\gamma_\mu=\eta_{\mu\nu}\gamma^\nu, \qquad a\!\!\!/\equiv\gamma^\mu a_\mu,$$ $$\beta\equiv\gamma^0, \qquad \alpha^i\equiv\gamma^0\gamma^i, \qquad \bar\psi\equiv\psi^\dagger\gamma^0.$$

Hermitian conjugation obeys

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

so $\gamma^0$ and every $\alpha^i$ are Hermitian, while the spatial $\gamma^i$ are anti-Hermitian.

The chirality and antisymmetric sigma matrices are

$$\gamma^5 \equiv i\gamma^0\gamma^1\gamma^2\gamma^3, \qquad \sigma^{\mu\nu} \equiv \frac{i}{2}[\gamma^\mu,\gamma^\nu].$$

With these definitions,

$$\begin{aligned} (\gamma^5)^2&=I_4,\\ (\gamma^5)^\dagger&=\gamma^5,\\ \{\gamma^5,\gamma^\mu\}&=0. \end{aligned}$$

The chiral projectors are

$$P_L=\frac{I_4-\gamma^5}{2}, \qquad P_R=\frac{I_4+\gamma^5}{2}.$$

These are algebraic projectors in four spacetime dimensions. For a massive particle, chirality is not generally the same as helicity.

Dirac basis

Let $I_2$ and $0_2$ denote the two-dimensional identity and zero matrices, and let $\sigma^i$ be the Pauli matrices. The default Dirac basis is

$$\begin{aligned} \gamma^0_D&= \begin{pmatrix} I_2&0_2\\ 0_2&-I_2 \end{pmatrix},\\ \gamma^i_D&= \begin{pmatrix} 0_2&\sigma^i\\ -\sigma^i&0_2 \end{pmatrix}. \end{aligned}$$

Consequently,

$$\alpha^i_D = \begin{pmatrix} 0_2&\sigma^i\\ \sigma^i&0_2 \end{pmatrix}, \qquad \gamma^5_D = \begin{pmatrix} 0_2&I_2\\ I_2&0_2 \end{pmatrix}.$$

The rotation blocks are conveniently written with

$$\Sigma^i = \begin{pmatrix} \sigma^i&0_2\\ 0_2&\sigma^i \end{pmatrix}.$$

Then

$$\sigma^{0i}=i\alpha^i, \qquad \sigma^{ij}=\epsilon^{ijk}\Sigma^k.$$

This basis diagonalizes $\beta=\gamma^0$, which makes the rest-energy blocks and large/small component expansion transparent. It does not diagonalize $\gamma^5$.

Chiral basis

In the chiral, or Weyl, basis,

$$\gamma^0_C = \begin{pmatrix} 0_2&I_2\\ I_2&0_2 \end{pmatrix}, \qquad \gamma^i_C = \begin{pmatrix} 0_2&\sigma^i\\ -\sigma^i&0_2 \end{pmatrix},$$

and

$$\gamma^5_C = \begin{pmatrix} -I_2&0_2\\ 0_2&I_2 \end{pmatrix}.$$

Thus the upper two components are left-chiral and the lower two are right- chiral according to the projector definitions above. The Dirac and chiral bases are related by a constant unitary similarity transformation. A physical bilinear or trace gives the same result after every spinor and gamma matrix is transformed consistently.

Trace identities

In four dimensions,

$$\begin{aligned} \operatorname{tr}(I_4)&=4,\\ \operatorname{tr}(\gamma^\mu)&=0,\\ \operatorname{tr}(\gamma^5)&=0. \end{aligned}$$

The basic even traces are

$$\operatorname{tr}(\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu},$$

and

$$\operatorname{tr} (\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) =4\bigl( \eta^{\mu\nu}\eta^{\rho\sigma} -\eta^{\mu\rho}\eta^{\nu\sigma} +\eta^{\mu\sigma}\eta^{\nu\rho} \bigr).$$

With $\epsilon^{0123}=+1$ and the definition of $\gamma^5$ above,

$$\operatorname{tr} (\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) = -4i\epsilon^{\mu\nu\rho\sigma}.$$

The trace of an odd number of ordinary gamma matrices vanishes. The five- gamma identity is four-dimensional: its use in dimensional regularization requires an additional prescription for $\gamma^5$ and cannot be inferred by naive continuation of this table.

Useful contraction identities

The Clifford algebra gives

$$\gamma_\mu\gamma^\mu=4I_4,$$ $$\gamma_\mu\gamma^\nu\gamma^\mu=-2\gamma^\nu,$$

and

$$\gamma_\mu\gamma^\nu\gamma^\rho\gamma^\mu = 4\eta^{\nu\rho}I_4.$$

For slashed vectors,

$$a\!\!\!/\,b\!\!\!/ = a\cdot b\,I_4-i\sigma^{\mu\nu}a_\mu b_\nu,$$

and therefore

$$a\!\!\!/\,b\!\!\!/+b\!\!\!/\,a\!\!\!/ =2a\cdot b\,I_4.$$

The sign in the sigma term follows from this page’s definition $\sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/2$.

Translating another source

Before importing an identity, copy the source’s defining equations rather than its label “standard conventions.”

Source choice	Convention here	Required translation check
mostly-plus metric $\tilde\eta=-\eta$ with $\{\tilde\gamma^\mu,\tilde\gamma^\nu\}=2\tilde\eta^{\mu\nu}$	mostly-minus	one consistent map is $\tilde\gamma^\mu=i\gamma^\mu$; recompute adjoints and $\gamma^5$
$\epsilon^{0123}=-1$	$\epsilon^{0123}=+1$	reverse every epsilon-dependent identity
$\gamma^5=-i\gamma^0\gamma^1\gamma^2\gamma^3$	plus-$i$ definition	interchange $P_L$ and $P_R$ labels or translate them explicitly
$\sigma^{\mu\nu}=[\gamma^\mu,\gamma^\nu]/2$	$i/2$ times the commutator	supply the missing factor of $i$ in every sigma identity
slash formed with $a^\mu\gamma_\mu$	$a_\mu\gamma^\mu$	these agree only after the same metric is used for both factors

Some mostly-plus texts instead define the Clifford relation with an extra minus sign. The first row then does not apply. Always translate the metric and the defining anticommutator together.

Exercises

1. Chirality algebra

Use only the Clifford relation to prove $(\gamma^5)^2=I_4$ and $\{\gamma^5,\gamma^\mu\}=0$.

Solution

Moving the second ordered product $\gamma^0\gamma^1\gamma^2\gamma^3$ through the first requires six pair swaps. The product of the four squares is $1(-1)^3=-1$, while $i^2=-1$, so $(\gamma^5)^2=I_4$. Moving any one gamma matrix through the other three produces a minus sign, proving $\gamma^5\gamma^\mu=-\gamma^\mu\gamma^5$.

2. The epsilon-trace sign

Evaluate $\operatorname{tr}(\gamma^5\gamma^0\gamma^1\gamma^2\gamma^3)$ directly from the definition of $\gamma^5$.

Solution

Let $G=\gamma^0\gamma^1\gamma^2\gamma^3$. The Clifford algebra gives $G^2=-I_4$, so

$$\operatorname{tr}(\gamma^5G) = i\operatorname{tr}(G^2) = -4i.$$

Since $\epsilon^{0123}=+1$, this fixes the sign in the general identity.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill, 1964.
• Itzykson, Claude, and Jean-Bernard Zuber. Quantum Field Theory. McGraw–Hill, 1980.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995. doi:10.1017/CBO9781139644167.