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Start with the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. The equation for the Dirac adjoint is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by the charge-conjugation matrix C to obtainFor this to be the original Dirac equation multiplied by minus one, the necessary matrix identity isUsing $C^{†}=C^{-1}$, the charge conjugation of the Dirac adjoint follows directly:The final step uses $({\gamma }^0{)}^T{C}^{-1}=-C^{-1}{\gamma }^0$ and $({\gamma }^0{)}^2=1$.

Let $j_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, $j_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, and $j_L=j_V-j_A$. For anticommuting fields, or normal-ordered operator bilinears, the minus sign in the transformed Dirac adjoint cancels the minus sign from interchanging the two fields. The charge conjugation of fermion bilinears is thereforeThe matrix identity just proved gives $j_V\mapsto -j_V$. Also $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$, obtained by transposing its four gamma matrices and reversing their order. Henceso $j_A\mapsto j_A$ and $j_L\mapsto -j_R$, where $j_R=j_V+j_A$. Since the external vector V is odd under charge conjugation, $j_L\cdot V$ becomes $j_R\cdot V$ and is generally not invariant.

For parity symmetry in quantum field theory, write $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. The given spinor transformation gives $j_V\mapsto P{j}_V$, $j_A\mapsto -P{j}_A$, and $V\mapsto PV$, with arguments evaluated at the reflected point. The contracted interaction again becomes $j_R\cdot V$. Each of parity and charge conjugation interchanges the two chiral contractions; applying both restores $j_L\cdot V$. This particular neutral current interaction is CP invariant, although neither P nor C is separately a symmetry. This conclusion uses the stipulated C-odd vector transformation; it is not a claim that all interactions of the Standard Model preserve CP symmetry.

For time reversal of a Dirac field, the gamma matrix adjoint and transpose identities giveWith $B={\gamma }^{5}C$, anticommutation of ${\gamma }^5$ with every gamma matrix yieldsEquivalently, $\boxed{B({\gamma }^{0\ast },-{\mathbf{\gamma }}^{\ast })B^{-1}=({\gamma }^0,\mathbf{\gamma })}$. Since time reversal is an antiunitary operator, the factor i in ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ must also be conjugated. ThusUsing the stated transformations of the field and its Dirac adjoint, the transformed bilinear consequently isThe external vector has the same component signs. Their contraction is unchanged, proving time reversal of a real neutral chiral vector interaction. A real coupling is understood; an antiunitary operator also conjugates coefficients, and relative complex weak phases need not satisfy this invariance.

A charged weak box contribution to kaon mixing is shown below. The incoming $d\bar{s}$ and outgoing $s\bar{d}$ connect through two charged W bosons and two internal up-type quarks. The black arrows are fermion flow, so the lower line's arrows oppose the momentum of the corresponding antiquark. The blue arrows specify the direction of positive charge carried by the charged propagators. Each vertex is a charged weak current vertex, with the appropriate CKM matrix element; the internal flavors are summed over $u,c,t$.

Use precisely the printed exchange-positive CP convention for neutral kaon mixing. In the flavor basis, CP symmetry acts by $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Write the restricted effective matrix asThen $SMS=\left(\begin{array}{cc}a& c\\ b& a\end{array}\right)$, so CP invariance of this mixing operator is exactly $b=c$. Therefore CP noninvariance in this neutral-kaon mixing block means $M_{12}\ne M_{21}$. Violation elsewhere in a larger Hamiltonian would not, by itself, prove an inequality for this particular block.

For neutral-kaon mass matrix diagonalization, choose correlated roots $r^2=b$, $s^2=c$, and eigenvalues ${\lambda }_{\pm }=a\pm rs$. The vectors $(r,s{)}^T$ and $(r,-s{)}^T$ are the corresponding eigenvectors, as direct multiplication verifies. Taking orthonormal flavor states and writing the printed CP eigenstates as $K_1=(K^0+{\bar{K}}^0)/\sqrt{2}$ and $K_2=(K^0-{\bar{K}}^0)/\sqrt{2}$, one findswhere the kaon CP mixing parameter isOverall phases have been absorbed. The kaon mixing square-root branch convention must be chosen continuously so that $r=s$ in the CP-conserving limit for the eigenstate labeled plus. These formulas are exact in the nondegenerate case with $r+s\ne 0$; the requested small admixtures follow if $|ϵ|\ll 1$. Smallness is an additional weak-CP-violation assumption, not a consequence of $b\ne c$ alone. A Hermitian mass operator has $c=b^{\ast }$; the same algebra also applies to the effective non-Hermitian operator that includes decays, where the two eigenstates need not be orthogonal. At a fully degenerate $b=c=0$, a CP eigenbasis can simply be chosen.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.