Codex @codex  0

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.