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Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

For $x^{\prime }=\Lambda x$, choose the corresponding spin liftThe spinor field transforms asThe infinitesimal Clifford algebra identity from part (b) exponentiates to $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, establishing Lorentz covariance of the Dirac operator. At fixed coordinates the infinitesimal variation also includes the orbital term $-{\Omega }^{\mu }{}_{\nu }{x}^{\nu }{\partial }_{\mu }\psi$.

A spatial rotation generator $M^{ij}$ is anti-Hermitian, but a Lorentz boost generator is Hermitian. For example,Its eigenvalues are $\pm \frac{1}{2}$. A nonzero rapidity $\zeta$ gives boost eigenvalues $e^{\pm \zeta /2}$, which are not all on the unit circle. This proves that the finite-dimensional component Spinor representation of the Lorentz group cannot be unitary for a positive-definite component inner product, even after a change of basis. It does not contradict a unitary action on the physical space of quantum states.

Consequently ${\psi }^{\prime †}(x^{\prime }){\psi }^{\prime }(x^{\prime })={\psi }^{†}{S}^{†}S\psi$ need not equal ${\psi }^{†}\psi$: it is the time component of the Dirac current, not a Lorentz scalar. The appropriate invariant form is indefinite. From $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ one obtainsThis Dirac spinor pseudo-unitarity gives ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$ and hence the Dirac scalar bilinear

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.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.