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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.

The Dirac adjoint of a chiral field has opposite projector placement because ${\gamma }_5$ anticommutes with ${\gamma }^0$: $\bar{\psi }{\gamma }_5=-\eta \bar{\psi }$. Since $C{\gamma }_5^T{C}^{-1}={\gamma }_5$, the algebraic charge conjugate satisfies ${\gamma }_{5}C{\bar{\psi }}^T=C(\bar{\psi }{\gamma }_5{)}^T=-\eta {\psi }^c$. This is conjugation of a projected spinor, distinguished from operator conjugation of a chiral field component, where the numerical projector is kept outside the conjugation operator.

The charge-conjugate field of a fermion field transforms in the complex-conjugate internal group representation. With a compatible Dirac adjoint and charge conjugation matrix convention it is ${\psi }^c=C{\bar{\psi }}^T$, up to a conventional constant phase. In four spacetime dimensions charge conjugation exchanges the chirality of the field; a right-handed quark can therefore be represented by a left-handed conjugate field when listing chiral unified matter. In two-dimensional Minkowski spacetime, charge conjugation can instead preserve chirality, permitting a Majorana-Weyl spinor.

For ${\psi }^c=C{\bar{\psi }}^T$, use a unitary charge-conjugation matrix with $C^{-1}{\gamma }^{0}C=-({\gamma }^0{)}^T$ and Hermitian ${\gamma }^0$ of square one. The Dirac adjoint then satisfiesIf the unitary charge conjugation operator sends $\psi$ to ${\eta }_C{\psi }^c$, it sends $\bar{\psi }$ to $-{\eta }_C^{\ast }{\psi }^T{C}^{-1}$. Antisymmetry $C^T=-C$ also gives $({\psi }^c{)}^c=\psi$. The unitary operator acting on fields is not an antiunitary operator acting on numerical coefficients.

The combined transformation $\psi \mapsto e^{-i\alpha {\gamma }_5/2}\psi$ and $({\varphi }_1+i{\varphi }_2)\mapsto e^{i\alpha }({\varphi }_1+i{\varphi }_2)$ preserves this interaction. The Dirac adjoint transforms with the same axial exponential as $\psi$. For potential $-{\mu }^2{r}^2/2+\lambda r^4/4$, $v=\mu /\sqrt{\lambda }$ gives a radial scalar of squared mass $2\lambda v^2$, a massless Goldstone boson and fermion physical mass $|g|v$. Cubic scalar vertices appear after the shift: $hhh$ has factor $-6i\lambda v$ and $h\pi \pi$ has factor $-2i\lambda v$.

The Dirac adjoint of a spinor is $\bar{\psi }={\psi }^{†}{\gamma }^0$. It transforms contragrediently, making $\bar{\psi }\psi$ a Lorentz scalar and $\bar{\psi }{\gamma }^{\mu }\psi$ a Lorentz vector.

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 generators in the Spinor representation of the Lorentz group satisfy $M^{†}{\gamma }^0+{\gamma }^{0}M=0$, so the finite transformation preserves the indefinite form $S^{†}{\gamma }^{0}S={\gamma }^0$. This makes $\bar{\psi }\psi$ a Lorentz scalar through the Dirac adjoint. Boost generators are Hermitian rather than anti-Hermitian: a boost has eigenvalues $e^{\pm \eta /2}$ and is not unitary in the ordinary positive-definite component norm. Rotations are unitary. This component representation is distinct from the unitary action on the physical state space.

For anticommuting components and an antisymmetric numerical matrix $A$, $\delta ({\psi }^{T}A\psi )=2\delta {\psi }^{T}A\psi$: move the second varied field past the first and use $A^T=-A$. Applying this to $C^{-1}$ and to $C$ cancels the factor $1/2$ in $\mathcal{L}=\bar{\psi }i/\partial \psi +\frac{1}{2}(m{\psi }^T{C}^{-1}\psi -m^{\ast }\bar{\psi }C{\bar{\psi }}^T)$. Varying the Dirac adjoint gives the first displayed equation. Integration by parts in the spinor variation gives $i{\gamma }^{\mu T}{\partial }_{\mu }{\bar{\psi }}^T+m{C}^{-1}\psi =0$; multiplication by $C$ gives the second. Squaring the coupled equations yields $(\square +|m{|}^2)\psi =0$, so the physical Majorana fermion mass is $|m|$.

With the usual Dirac adjoint, Hermitian ${\gamma }^5$ and $\{{\gamma }^0,{\gamma }^5\}=0$, taking the adjoint gives ${\psi }^{†}{\gamma }^5{\gamma }^0\psi =-{\psi }^{†}{\gamma }^0{\gamma }^5\psi$. Consequently a real pseudoscalar with coupling $-g\varphi \bar{\psi }{\gamma }^5\psi$ has a Hermitian interaction only if $g^{\ast }=-g$. Writing $g=i{g}_P$ with real $g_P$ gives the familiar interaction $-i{g}_P\varphi \bar{\psi }{\gamma }^5\psi$. Formal Euler-Lagrange field equations obtained with independent spinor and adjoint variables can still be computed for the original symbolic coupling; it must not silently be identified with a real physical coupling.

With $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }$, $\Omega ={\omega }^{\mu \nu }{\gamma }_{\mu \nu }/4$, and $\Lambda =e^{\omega }$, the Clifford algebra gives $[\Omega ,{\gamma }^{\rho }]=-{\omega }^{\rho }{}_{\mu }{\gamma }^{\mu }$. Exponentiation proves the inverse Lorentz action displayed above, equivalently $S^{-1}{\gamma }^{\rho }S={\Lambda }^{\rho }{}_{\mu }{\gamma }^{\mu }$. This second order is what combines with $\psi \mapsto S\psi$ and the Dirac adjoint action $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$ to make the gamma bilinears transform as ordinary contravariant tensors.

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.

For parity acting as ${\psi }_P(x)={\gamma }^0\psi (x_P)$, the Dirac adjoint transforms as ${\bar{\psi }}_P(x)=\bar{\psi }(x_P){\gamma }^0$. A bilinear with matrix $B$ therefore transforms by $B\mapsto {\gamma }^{0}B{\gamma }^0$, with its fields evaluated at $x_P$. Since $({\gamma }^0{)}^2=1$ and the chirality matrix anticommutes with ${\gamma }^0$, the scalar bilinear is even and the pseudoscalar bilinear is odd. The same calculation shows that parity exchanges the chiral projectors.

With the mostly-minus Minkowski metric, $|{\eta }_P|=1$ and the usual Dirac adjoint, this transformation makes $\bar{\psi }\psi$ even and $\bar{\psi }{\gamma }^5\psi$ odd under parity. The latter sign follows from anticommutation of ${\gamma }^0$ and ${\gamma }^5$ and does not depend on whether the convention includes an imaginary unit in ${\gamma }^5$.

Put $U=e^{-i\alpha {\gamma }_5/2}$. The transformed Dirac adjoint is ${\bar{\psi }}^{\prime }={\psi }^{†}{U}^{†}{\gamma }^0=\bar{\psi }U$, since ${\gamma }^0$ anticommutes with ${\gamma }_5$. It is therefore incorrect to transform the two sides of a Yukawa bilinear with mutually inverse matrices.

Write $M(\varphi )={\varphi }_1+i{\gamma }_5{\varphi }_2$. The transformed Yukawa interaction is invariant exactly whenSince $M$ commutes with ${\gamma }_5$, this is $M({\varphi }^{\prime })=e^{i\alpha {\gamma }_5}M(\varphi )$. Use i and equate the coefficients of $1$ and $i{\gamma }_5$:This is the global SO(2) rotation of a chiral scalar-doublet Yukawa model. Equivalently, $\Phi ={\varphi }_1+i{\varphi }_2$ transforms as ${\Phi }^{\prime }=e^{i\alpha }\Phi$, while the two chiral spinor components transform with opposite half-angle phases. The sign of the scalar rotation follows from the negative sign in the specified fermion transformation.

For a real local parameter $\alpha (x)$, the Dirac adjoint varies as $\delta \bar{\psi }=ie\alpha \bar{\psi }$. The matter kinetic term varies byThe QED interaction varies by $-e({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }\psi$, since the phases in $\bar{\psi }{\gamma }^{\mu }\psi$ cancel. These two variations cancel pointwise. The mass term is invariant for the same reason, and $\delta F_{\mu \nu }={\partial }_{\mu }{\partial }_{\nu }\alpha -{\partial }_{\nu }{\partial }_{\mu }\alpha =0$.

Only the gauge fixing term varies. Put $B={\partial }_{\mu }{A}^{\mu }$; then $\delta B={\partial }^2\alpha$, givingFor arbitrary local $\alpha$, the gauge-fixed action is therefore not invariant under this transformation of its dynamical fields. The restricted parameters satisfying ${\partial }^2\alpha =0$ give a residual invariance of the displayed density.

For a Dirac field, the path integral is a Berezin integral over independent Grassmann fields $\psi$ and $\bar{\psi }$:The vacuum boundary conditions and the Feynman i-epsilon prescription are understood; $\mathcal{N}$ fixes the vacuum normalization. This is the vacuum generating functional without sources, rather than a finite-temperature partition function.

The Dirac adjoint transforms as $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\alpha }$. For constant $\alpha$, ${\partial }_{\mu }(e^{i\alpha }\psi )=e^{i\alpha }{\partial }_{\mu }\psi$. Both the kinetic term and the Dirac mass term therefore retain their values. This proves the global phase symmetry, whose group is the unitary group $U(1)$.

The Standard Model is a chiral gauge theory: left-chiral quarks and leptons form SU(2) weak doublets, while the corresponding right-chiral charged fields are weak singlets. The covariant kinetic term contains $g_W{\bar{Q}}_L{\gamma }^{\mu }{t}^a{Q}_L{W}_{\mu }^a$. With $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$, its off-diagonal generators give the weak charged currentAn overall interaction sign depends on the convention for the covariant derivative and has no effect on this decay rate. The chiral projector $P_L=(1-{\gamma }^5)/2$ explains the vector-minus-axial combination. The Cabibbo-Kobayashi-Maskawa matrix comes from the different left-handed rotations that diagonalize up- and down-type fermion mass matrices. Set $V_{tb}=1$ and $m_b=0$ as stipulated.

Let $P_{\mu \nu }=-g_{\mu \nu }+p_{3\mu }{p}_{3\nu }/m_W^2$, the massive vector polarization sum. Sum over final spins and average over the two equally populated initial top quark spins. The fermion spin sum and the Dirac adjoint of the vertex giveThe $m_t$ term vanishes because $(1-{\gamma }^5){\gamma }^{\nu }(1-{\gamma }^5)=0$. Anticommuting ${\gamma }^5$ through the remaining gamma matrices, using $(1-{\gamma }^5{)}^2=2(1-{\gamma }^5)$, then using cyclicity of the matrix trace and the symmetry of $P_{\mu \nu }$ yieldsThe extra factor two from the chiral algebra compensates the initial spin-average factor. The term with ${\gamma }^5$ is antisymmetric in $\mu ,\nu$ and vanishes upon contraction with $P_{\mu \nu }$. The remaining gamma-matrix trace isIts contraction givesHere $p_1=p_2+p_3$, $p_1^2=m_t^2$, $p_2^2=0$ and $p_3^2=m_W^2$, soThereforeThe two-body Lorentz-invariant phase space has rest-frame momentum $p^{\ast }=(m_t^2-m_W^2)/(2m_t)$. Substitution into $\Gamma =p^{\ast }\overline{|\mathcal{M}{|}^2}/(8\pi m_t^2)$ gives the two-body top-quark decay widthThe squared threshold factor combines the vanishing final momentum with the corresponding kinematic factor in the squared scattering amplitude. The longitudinal W boson polarization contributes the high-mass enhancement proportional to $m_t^3/m_W^2$. The result is at tree level, for an on-shell W boson and $m_t>m_W$; finite bottom mass and radiative corrections change it. Restoring quark mixing multiplies the width by $|V_{tb}{|}^2$.

Choose the following Pauli matrix realization, with its spinor basis fixed for the later Dirac adjoint:The Pauli matrix multiplication rule verifiesExtending linearly therefore gives a two-dimensional Lie algebra representation. For an irreducible representation, it suffices to rule out an invariant line. Any such line would have to be an eigenline of $r(K_1)$, whose two eigenvalues are distinct. Those two coordinate lines are interchanged by $r(K_2)$, so neither is invariant. There is no proper nonzero invariant subspace.

This is a spinorial representation rather than a single-valued two-dimensional representation of the identity component of the specified orthogonal group:The mismatch explicitly exhibits the need for the spin double cover. It causes no obstruction to the requested infinitesimal construction.

The spinor basis in c is compatible with the supplied gamma matrices. The rotation generator $r(J)$ is anti-Hermitian and commutes with ${\gamma }_0=-i{\sigma }_1$. The two boost generators are Hermitian matrices and anticommute with ${\gamma }_0$. It follows separately for all three generators, and hence for every real $T\in \mathfrak{so}(1,2)$, thatFor the infinitesimal Lorentz transformation, $\delta \psi =r(T)\psi$ and $\delta {\psi }^{†}={\psi }^{†}r(T{)}^{†}$. Hence the Dirac adjoint transforms contragrediently:The two contributions to the scalar variation cancel:This proves the spinor invariant in three spacetime dimensions has infinitesimal Lorentz invariance. Real transformation parameters matter because the Hermitian adjoint conjugates coefficients; this is not an invariance under arbitrary complexified parameters. For ordinary commuting spinor components, the bilinear is purely imaginary in this mostly-plus convention, and multiplying it by $i$ gives a real invariant. Its reality is separate from its invariance.