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In the Weyl representation of the gamma matrices, $C=i{\gamma }^2{\gamma }^0=\mathrm{diag}(\epsilon ,-\epsilon )$ with $\epsilon =i{\tau }_2$. The antisymmetry of $\epsilon$ gives $C^T=-C$, and direct multiplication gives $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$ and $C^{†}C=I$. Any two such intertwining matrices differ by a scalar: their ratio commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. Changing spinor basis sends $C$ to $SC{S}^T$, so transpose antisymmetry is basis-independent. Its sign is therefore fixed, not merely one of the two alternatives supplied by the general transpose symmetry of a charge-conjugation matrix argument.

After complexifying a nondegenerate metric in $D=2k$ dimensions, pair gamma matrices into $k$ fermionic creation and annihilation operators. Their binary occupation states span $2^k$ dimensions, and the gamma matrices generate all endomorphisms of this space. The Dirac module is therefore irreducible as a Clifford module. The even algebra preserves the chirality grading, splitting its Lorentz-spinor representation into two irreducible Weyl modules of dimension $2^{k-1}$. In $D=2k+1$, adjoining the old chirality matrix as the extra gamma gives an irreducible $2^k$-dimensional Spin group representation with no Weyl splitting. The two possible signs of that extra gamma distinguish Clifford modules but give equivalent Spin group modules.

An orthonormal coframe converts fixed flat gamma matrices to coordinate-index Clifford multiplication. Its metric identity gives the curved anticommutation relation. The spin connection is chosen so that these matrices are covariantly constant under the combined spin and tensor connection, which is the tetrad postulate in Clifford form.

The Dirac equation is the relativistic equation of motion for a spin-one-half Dirac field, $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, in units $\hslash =c=1$. The gamma matrices obey the Clifford algebra, ensuring that each component also obeys the free Klein-Gordon equation.

Being first order in time allows an initial spinor to determine its subsequent free evolution. The components encode spin as well as particle and antiparticle solutions; quantization interprets both kinds of excitation with positive physical energy. Coupling through the gauge covariant derivative describes an electron interacting with an electromagnetic field. The nonrelativistic limit contains the coupling of spin to a magnetic field, explaining why a scalar relativistic wave equation alone cannot describe an electron.

For Minkowski metric $\mathrm{diag}(1,-1)$, the displayed gamma matrices satisfy the Clifford algebra and ${\gamma }^5={\gamma }^0{\gamma }^1$ has chirality eigenvalues $+1,-1$. Ordering the spinor as $({\psi }_R,{\psi }_L{)}^{\mathsf{T}}$ gives $i({\partial }_t+{\partial }_x){\psi }_R=m{\psi }_L$ and $i({\partial }_t-{\partial }_x){\psi }_L=m{\psi }_R$. A plane wave has the Dirac Hamiltonian $p{\sigma }_3+m{\sigma }_1$, whose square is $(p^2+m^2)1$, proving $E^2=p^2+m^2$.

For $g=\mathrm{diag}(1,-1,-1,-1)$ and the block-diagonal standard representation of the gamma matrices, write a positive-energy Dirac spinor as $(\xi ,\eta {)}^T$. The Dirac equation gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. Choosing two orthonormal two-spinors ${\chi }_s$ yields the displayed positive-frequency solutions. Negative-frequency solutions are $v_s=\sqrt{E+m}((\mathbf{\sigma }\cdot \mathbf{p}){\eta }_s/(E+m),{\eta }_s{)}^T$ multiplying $e^{ip\cdot x}$. They obey $(/p+m)v_s=0$. Both have norm $2E$; their Dirac adjoints give ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$ and ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$. Spatial rotations act on each two-spinor with generators ${\sigma }_j/2$, explaining spin one-half.

With signature $(+,-,-,-)$, one standard representation is ${\tilde{\gamma }}^0=\mathrm{diag}(I_2,-I_2)$ and ${\tilde{\gamma }}^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$. Block multiplication and the Pauli matrix multiplication law give time-time anticommutator $2I_4$, mixed anticommutators zero, and spatial anticommutators $-2{\delta }^{ij}{I}_4$. Thus these gamma matrices represent the spacetime Clifford algebra.

For gamma matrices ${\gamma }^0={\sigma }_3$, ${\gamma }^1=i{\sigma }_1$, ${\gamma }^2=i{\sigma }_2$ and Minkowski metric $\mathrm{diag}(1,-1,-1)$, a one-axis spatial reflection acts with ${\gamma }^1$ on the two-component Dirac field. It preserves the kinetic term and reverses the Dirac mass term. Time reversal acts with ${\gamma }^2$ and includes conjugation of the operator coefficients; it has the same mass reversal. Their composition preserves a real mass. Charge conjugation can use ${\psi }_C={\gamma }^1{\psi }^{\ast }$, because ${\gamma }^{\mu }{\gamma }^1=-{\gamma }^1{\gamma }^{\mu \ast }$. The mass bilinear is charge-conjugation even when Grassmann field exchange signs are included, and the kinetic action is invariant up to a boundary term. These statements concern the classical free action.

For external electrons on shell, factor out signed electric charge in the parity-preserving interaction vertex:The $u$ are Dirac spinors; ${\gamma }^{\mu }$ are gamma matrices. With on-shell renormalization scheme normalization $F_1(0)=1$, the Gordon identity gives $g_s/2=F_1(0)+F_2(0)$ and $a_e=F_2(0)$.^[ref]

This explains a subtlety: the Dirac equation already predicts a spin magnetic dipole moment with $F_2=0$. Rewriting its Dirac electromagnetic current with the Gordon identity produces a spin-dependent term even before any loop order correction. $F_2$ measures the additional response after that baseline is included. The limit $q\to 0$ is taken after extracting its coefficient; setting $q=0$ in the interaction vertex at the outset would hide it. This electromagnetic anomalous magnetic moment is not a failure of gauge invariance of the kind called a quantum anomaly.

The arrow on a charged-fermion line in a Feynman diagram tracks fermion-number flow and fixes the order of the gamma matrices between external Dirac spinors. For an external particle it agrees with the particle direction; for an external antiparticle it is opposite to the antiparticle direction. It therefore need not point along the assigned four-momentum of every external leg.

Feynman slash notation contracts a four-vector with the gamma matrices. With mostly-minus signature, the Clifford algebra gives $/p^2=p^{2}I$, so $(/p-m{)}^{-1}=(/p+m)/(p^2-m^2)$ away from the mass shell.

Gamma matrices represent the spacetime Clifford algebra throughFor the Minkowski metric with signature $(+,-,-,-)$, ${\gamma }^0$ squares to $I$ and each spatial gamma matrix squares to $-I$. Distinct orthogonal generators anticommute. A four-dimensional complex representation is used for Dirac spinors in four-dimensional spacetime.

The particular matrix entries depend on the chosen basis; simultaneous similarity transformations preserve the defining relations. Their usefulness in the Dirac equation is that contracting them with derivatives produces a first-order operator whose square reduces to the second-order relativistic wave operator, apart from interactions or curvature contributions.

For gamma matrices with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, contracted Clifford algebra gives ${\gamma }_a{\gamma }^m{\gamma }^a=(2-D){\gamma }^m$. Expanding the two antisymmetrized factors gives ${\gamma }_a{\gamma }_b{\gamma }^m{\gamma }^a{\gamma }^b-D{\gamma }^m$. The five-gamma product equals $[2D-(D-2{)}^2]{\gamma }^m$, by moving the middle contracted gamma through and using the three-gamma contraction twice. Subtraction yields the displayed identity. A convention without the factor one-half in each bivector multiplies the result by four.

A hermitizing matrix for a set of gamma matrices is an invertible Hermitian matrix $H$ satisfying ${\gamma }^{a†}=H{\gamma }^a{H}^{-1}$. Equivalently, each $H{\gamma }^a$ is Hermitian. It defines a covariant adjoint $\bar{\psi }={\psi }^{†}H$. If ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$, then $H^{\prime }=M^{-†}H{M}^{-1}$ has the same property. In a standard Lorentzian Dirac basis, $H={\gamma }^0$ is a conventional choice.

For the QED proper Feynman vertex with both external four-momenta zero, a massive internal fermion and a massless internal photon in Feynman gauge, the even numerator reduces under Lorentz covariance to ${\gamma }^{\mu }[\alpha k^2+\beta m^2]$. The gamma matrices identity ${\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=(2-D){\gamma }^{\mu }$ and the integrated replacement $k_{\rho }{k}_{\sigma }\mapsto g_{\rho \sigma }{k}^2/D$ prove the coefficients. The term $m^2/[k^2(k^2-m^2+i0{)}^2]$ has neither an infrared divergence nor an ultraviolet divergence for $m>0$; its integral in four dimensions is $-i/(16{\pi }^2)$. Only the remaining squared massive denominator supplies the ultraviolet pole.

For action $-\int \bar{\psi }(\gamma \cdot \partial +m)\psi$, choose gamma matrices with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and signature $(-,+,\dots ,+)$. The Fourier kernel of $D=\gamma \cdot \partial +m$ is $i\gamma \cdot p+m$. Its algebraic inverse is $(-i\gamma \cdot p+m)/(p^2+m^2)$. The Grassmann Gaussian integral for $e^{-i\bar{\psi }D\psi }$ has covariance $-i{D}_F^{-1}$, giving the displayed Dirac propagator. Its denominator has positive-energy poles below and negative-energy poles above the real energy axis by the Feynman i-epsilon prescription. Both the metric and the factor of $i$ in the derivative kernel must be specified when comparing propagator formulas.

Set $s=(p_1+p_2{)}^2=m_h^2$, so $p_1^2=p_2^2=0$ and $p_1\cdot p_2=s/2$. The physical polarization vectors obey ${ϵ}_1\cdot p_1={ϵ}_2\cdot p_2=0$. Write $X_{\mu \nu }\simeq Y_{\mu \nu }$ when they agree after contraction with these polarization vectors; terms proportional to $p_{1,\mu }$ or $p_{2,\nu }$ may then be discarded.

The gamma matrix trace identities kill the terms with an odd number of gamma matrices. The remaining three two-momentum terms and the term with three factors of $m$ areIn the $g_{\mu \nu }$ coefficient the terms linear in $l\cdot p_i$ cancel, leaving $m^2-l^2-s/2$. Among the other terms, the surviving mixed products of $l$ and $p_i$ also cancel. ConsequentlyIn the second Feynman diagram, exchange the external labels and change the dummy loop momentum $l\mapsto -l$. Its projected numerator and denominator become identical to those of the first Feynman diagram. HenceWith the common factor $8m$ extracted, the primitive integer coefficients areEquivalently, extracting only $m$ gives $(A,B,C)=(32,8,8)$ and prefactor $i{N}_c{\lambda }_t{e}^2{e}_t^2{m}_t/\sqrt{2}$. A common rescaling of all three coefficients changes the prefactor inversely. The physical scattering amplitude is ${ϵ}_1^{\mu }{ϵ}_2^{\nu }{\mathcal{M}}_{\mu \nu }^{\prime }$; the projected integrand is not an identity for an arbitrary uncontracted off-shell tensor.

In pure QCD, different incoming and outgoing quark flavours allow only the tree-level $s$-channel gluon graph. A quark-gluon vertex preserves flavour, so no flavour-changing $t$-channel exchange is available. Label the incoming momenta $p_1,p_2$ and outgoing momenta $k_1,k_2$. For transparent spin traces, use the auxiliary metric $\eta =-g$ and its usual gamma matrices; the positive-energy Mandelstam variables areThis is a consistent change of metric notation, not a change of the physical invariants. Define the fixed-index colour tensor $C_{ji;kl}={\sum }_a(T^a{)}_{ji}(T^a{)}_{kl}$. Up to an irrelevant common phase,The indices $i,j$ label incoming colours and $k,l$ outgoing colours. Averaging over the two spins of each incoming particle and summing the outgoing spins givesThe gamma-matrix trace identity makes the contractionFor instance $p_1\cdot k_1=p_2\cdot k_2=-t/2$ and the crossed products are $-u/2$. Therefore, before the remaining colour sum and average,Here the bar averages spins only; it does not silently include the colour average requested next.

Because $({\gamma }_5{)}^2=1$, its even and odd powers are ${\gamma }_5^{2r}=1$ and ${\gamma }_5^{2r+1}={\gamma }_5$. Separate the matrix exponential into these two power series:No representation of the gamma matrices is needed. Hermiticity of ${\gamma }_5$ also makes this exponential unitary for real $\alpha$, while the projectors $(1\pm {\gamma }_5)/2$ acquire opposite phases.

First derive the contracted gamma matrices identities from the Clifford algebra in four dimensions. Moving the outer gamma matrix through two factors givesMoving it through three factors, and using this two-factor result, givesConsequently ${\gamma }^{\nu }/b{\gamma }_{\mu }/c{\gamma }_{\nu }=-2/c{\gamma }_{\mu }/b$ and ${\gamma }_{\mu }/b/d{\gamma }^{\mu }=4(b\cdot d)1$. Apply these in succession:The final step uses $\mathrm{tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. This is a genuinely four-dimensional instance of the gamma matrix trace identities; using it unchanged inside a $D$-dimensional loop calculation would require separate justification.