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Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

The supplied matrices are the chiral gamma-matrix representation. The Pauli matrix multiplication law implies $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication gives $({\gamma }^0{)}^2=I_4$, $\{{\gamma }^0,{\gamma }^i\}=0$, and $\{{\gamma }^i,{\gamma }^j\}=-2{\delta }^{ij}{I}_4$. Hence their Clifford algebra relation isSince partial derivatives commute, multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ givesThus every component of the Dirac spinor satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation with the same mass.

For a Lorentz transformation $x^{\prime }=\Lambda x$, set ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$. The derivative transforms as ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. With the given spinor-matrix identity,This proves the Lorentz covariance of the Dirac operator explicitly.

For parity, let $x_P=(x^0,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. Spatial differentiation brings a minus sign, which is canceled by ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$. ThereforeSince ${\gamma }^0$ is Hermitian and squares to one, the Dirac adjoint transforms asThe chirality matrix ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with ${\gamma }^0$. The parity transformation of a Dirac bilinear consequently givesThe first bilinear is a Lorentz scalar even under parity, and the second is a pseudoscalar.

In this representation, ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$. Its square is one, so the chiral projectors $P_{\pm }=(1\pm {\gamma }^5)/2$ are idempotent, mutually orthogonal and sum to the identity. They select ${\psi }_L=P_{-}\psi$ and ${\psi }_R=P_{+}\psi$, the two Weyl spinors. Because ${\gamma }^{\mu }{P}_{\pm }=P_{\mp }{\gamma }^{\mu }$, a solution of the massless Dirac equation has each chiral component solving that equation separately. In two-component form,The two chirality sectors are thus independent under massless propagation and proper Lorentz transformations. A nonzero Dirac mass term couples them, while parity exchanges them. This explains why the projectors are particularly useful in the massless theory.