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Varying the Dirac action with respect to the spinor, rather than its Dirac adjoint, gives the left-acting field equation after integration by parts. Together with the Dirac equation it proves conservation of the vector current.

In the complex-conjugate convention for charge conjugation of a Dirac field, a unitary matrix $B$ obeys the displayed gamma matrix identity. For $M^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, two minus signs cancel to give $M^{\rho \sigma }B=B{M}^{\rho \sigma \ast }$. Real Lorentz transformation parameters therefore give $SB=B{S}^{\ast }$, so $B{\psi }^{\ast }$ has the same Spinor representation of the Lorentz group transformation law as $\psi$. Complex conjugating the free Dirac equation and multiplying by $B$ proves that the conjugate also solves it for real mass. This $B$ is the matrix multiplying ${\psi }^{\ast }$; the convention multiplying ${\bar{\psi }}^T$ instead uses $B({\gamma }^0{)}^T$.

For ${\mathcal{L}}_{\mathrm{int}}=-g\varphi \bar{\psi }{\gamma }^5\psi$, the variational Dirac equation and adjoint Dirac equation are $(i/\partial -m-g\varphi {\gamma }^5)\psi =0$ and $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }+g\varphi \bar{\psi }{\gamma }^5=0$. Substitute these into the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. Anticommuting the chirality matrix past ${\gamma }^{\mu }$ gives two identical mass terms and two identical interaction terms, proving the displayed identity. These are classical field equations; the formula does not assert an unrenormalized quantum composite-operator identity. With Hermitian coupling $g=i{g}_P$, the second term is $-2g_P\varphi \bar{\psi }\psi$.

For $\mathcal{L}=\bar{\psi }(i/\partial -S+iP{\gamma }^5)\psi$, with real scalar backgrounds $S,P$, the Dirac equation is $(i/\partial -S+iP{\gamma }^5)\psi =0$ and the adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+S\bar{\psi }-iP\bar{\psi }{\gamma }^5=0$. Use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$ in the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ to obtain the displayed classical identity. Taking $S=m+g\varphi$ and $P=G\Phi$ includes both scalar and Hermitian pseudoscalar Yukawa couplings.

A free Dirac field has Lagrangian density $\mathcal{L}=i\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi -m\bar{\psi }\psi$ and obeys the Dirac equation.

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.

The historical interpretation fills the negative-energy electron states of the Dirac equation and uses the Pauli exclusion principle to prevent additional electrons from falling into them. Removing an electron of energy $-E$ and charge $-e$ gives a hole of excitation energy $E$ and charge $+e$, interpreted as a positron. A quantum field theory instead treats negative-frequency field coefficients as independent antiparticle creation operators in a Fock space, with positive excitation energies after normal ordering. No literal infinite material sea is needed. Bosonic antiparticles also exist, although a filled-sea exclusion argument cannot stabilize bosonic negative-energy levels.

The gamma matrix Clifford algebra and commuting partial derivatives give $(/\partial {)}^2=\square$. Thus every free Dirac equation solution also solves the Klein-Gordon equation componentwise. The converse is false without the first-order spinor constraint: the second-order equation alone allows twice as much independent initial data.

For the Dirac equation with two equal quark masses $m$, the equations for $q$ and $\bar{q}$ give the divergence of the nonsinglet axial current as $im\bar{q}{\gamma }_5{\tau }_{a}q=m{P}_a$. The color chiral anomaly is absent in this nonsinglet identity because the quark flavor generator is traceless. Equivalently, the mass term in the Hamiltonian is $m\int S{d}^{3}x$, and the nonsinglet axial charge algebra of quark densities gives ${\dot{Q}}_{5,a}=m\int P_a{d}^{3}x$.

The derivative of the local phase in $\Psi \mapsto e^{-iq\alpha }\Psi$ cancels the added gradient in the electromagnetic potential. Thus $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, and the charged Dirac equation maps solutions to solutions. The same cancellation makes its action gauge invariant.

The Gordon identity rewrites a vector-current matrix element between equal-mass on-shell Dirac spinors using a momentum term and an antisymmetric spin term. In the mostly-minus convention $\{{\Gamma }^a,{\Gamma }^b\}=2{\eta }^{ab}$, $(/p-m)u=0$, and ${\Gamma }^{ab}=[{\Gamma }^a,{\Gamma }^b]/2$, it reads $2m{\bar{u}}^{\prime }{\Gamma }^{a}u={\bar{u}}^{\prime }[(p^{\prime }+p{)}^a-{\Gamma }^{ab}(p_b^{\prime }-p_b)]u$. Derive it by adding the two external Dirac equation identities. The equivalent formula in the mostly-plus Dirac convention is ${\bar{u}}^{\prime }[{\gamma }^{ab}(p_b^{\prime }-p_b)+2m{\gamma }^a+p^a+p^{\prime a}]u=0$. The identity before division by $m$ remains valid for massless equal-mass states.

In the convention $⟨0|J^{\mu }|P(p)⟩=-i\sqrt{2}{F}_P{p}^{\mu }$, the Fermi interaction gives $\mathcal{M}=G_F{F}_P{p}_{\mu }{\bar{u}}_{\ell }{\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }$. With a massless antineutrino, the external Dirac equations reduce this to $G_F{F}_P{m}_{\ell }{\bar{u}}_{\ell }(1-{\gamma }_5)v_{\nu }$. The spin sum is $8G_F^2{F}_P^2{m}_{\ell }^{2}k\cdot q=4G_F^2{F}_P^2{m}_{\ell }^2(M_P^2-m_{\ell }^2)$. The two-body Lorentz-invariant phase space has integrated value $(1-m_{\ell }^2/M_P^2)/(8\pi )$, and the initial flux factor is $1/(2M_P)$, giving the displayed width. If the decay constant is instead defined without the square root of two, $f_P=\sqrt{2}{F}_P$ and the denominator is $8\pi$. A weak mixing coefficient multiplies the amplitude and its modulus squared multiplies the width.

With $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ and spin curvature $[{\nabla }_{\mu },{\nabla }_{\nu }]=R_{\mu \nu ab}{\gamma }^{ab}/4$, the Dirac equation operator squares to the rough Laplacian minus $R/4$. In the curvature contraction the four-gamma term vanishes by the first Bianchi identity, the two-gamma terms vanish by Ricci symmetry, and the scalar term is $-2R$. In Riemannian signature $i/\nabla$ is the self-adjoint convention and its square is $-{\nabla }^2+R/4$.

A spinor Lorentz matrix satisfies $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$. Combining this identity with the transformed derivative shows that $(i{\gamma }^{\mu }{\partial }_{\mu }^{\prime }-m){\Psi }^{\prime }=S(i{\gamma }^{\mu }{\partial }_{\mu }-m)\Psi$. Thus a Lorentz transformation carries every solution of the Dirac equation to another.

The massless Dirac equation is ${\gamma }^m{\partial }_m\Psi =0$. The Clifford algebra makes its square the massless wave equation. Quantization of a worldline spin variable as ${\hat{\psi }}^m={\gamma }^m/\sqrt{2}$ turns its fermionic constraint into this equation.

Gauge equivalence ${\psi }_{\mu }\sim {\psi }_{\mu }+{\partial }_{\mu }ϵ$ permits gamma-trace gauge. The Rarita-Schwinger field equation then gives transversality and a massless Dirac equation for each component. For momentum along the third axis, residual gauge removes the temporal and longitudinal components. The remaining constraint is ${\psi }_2={\gamma }^1{\gamma }^2{\psi }_1$, with ${\psi }_1$ in the two-dimensional kernel of $/p$. Spinor helicity $\pm 1/2$ is paired with vector helicity of the same sign, leaving only $\pm 3/2$. A Majorana spinor reality condition identifies antiparticles; a Dirac vector-spinor has separate antiparticle modes.

One consistent Dirac equation convention uses $\eta =\mathrm{diag}(-1,1,1,1)$, $\{{\gamma }^a,{\gamma }^b\}=-2{\eta }^{ab}$, and $(i{\gamma }^a{\partial }_a+m)\psi =0$. Taking the negatives of the standard mostly-minus gamma matrices, and defining the Dirac adjoint with the new ${\gamma }^0$, gives the same physical Dirac action. The positive-frequency plane wave is $e^{ip\cdot x}u(p)$ with $p^0>0$, and its equation is $(/p-m)u=0$. The mass shell is $p^2=-m^2$. These signs must be translated together when using a formula stated in another convention.

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.