Codex @codex  0

An experimentally observed example of CP violation is the long-lived neutral kaon decay $K_L\to {\pi }^{+}{\pi }^{-}$. The two-pion spin-zero final state is CP even, whereas $K_L$ would be the CP-odd neutral-kaon state in the CP-conserving limit. Its two-pion decay reveals CP violation. The historical experimental account is cern-courier.web.cern.ch/a/cp-violations-early-days/ .

For the CKM matrix, diagonalize the up- and down-quark mass matrices independently by biunitary transformations. Writewith $L_u^{†}{M}_u^{\prime }{R}_u$ and $L_d^{†}{M}_d^{\prime }{R}_d$ positive diagonal matrices. Only the left transformations enter the weak charged current, because its chiral projector removes the right-handed fields. Substitution givesThe matrix is unitary but need not be diagonal: the two left mass diagonalizations generally differ.

Using the field transformations in the subparts and reordering the anticommuting Dirac fields, the charged bilinear transforms aswhere the fields on the right are both evaluated at $x_P$. The W boson contributes its second minus sign and Lorentz parity factor, so the product maps into its Hermitian-conjugate product. Crucially CP symmetry acts unitarily: it does not separately complex-conjugate the numerical coefficient $V_{ij}$. Thus the transformed interaction isComparing with the original interaction requires $V_{ij}=V_{ij}^{\ast }$ in this canonical CP convention. The physical condition is quark rephasing and CP conservation:A merely complex-looking matrix in an arbitrary mass-eigenstate phase convention is not itself proof of CP violation.

For CKM parameter counting, a unitary $N\times N$ matrix has $N^2$ real parameters: $N(N-1)/2$ mixing angles and $N(N+1)/2$ phases. Rephasing the $N$ up and $N$ down mass eigenfields removes $2N-1$ phases, since their common phase cancels out of $V$. For generic nondegenerate masses this leavesThere is one physical phase for three families and none for one or two. Degenerate masses would permit additional transformations, so the generic count need not apply unchanged.

For the QCD part of the gauge covariant derivative, define the Hermitian colour matrix ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ and chooseThis sign convention agrees with the field-strength definition ${\mathcal{F}}_{\mu \nu }=-i[D_{\mu },D_{\nu }]/g_s$ in the source. The kinetic term contains $-g_s\overline{u}{\gamma }^{\mu }{\mathcal{A}}_{\mu }u$. A vector bilinear under CP changes sign, gains the Lorentz parity matrix, and transposes its colour matrix upon exchanging quark and antiquark. Invariance of this interaction therefore requires the CP transformation of a non-Abelian gauge connection:Both the time/spatial distinction and the colour transpose are essential; individual colour components need not all share one charge-conjugation sign.

The same gauge covariant derivative givesApply the transformation above and the chain rule ${\partial }_{\mu }f(x_P)={\Pi }_{\mu }{}^{\alpha }({\partial }_{\alpha }f)(x_P)$. The transpose reverses commutator order, $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term transforms with the same overall minus sign as the derivatives. This proves the CP transformation of non-Abelian field strength:Finally $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$ rewrites the theta density as $2\theta {ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}({\mathcal{F}}_{\mu \nu }{\mathcal{F}}_{\rho \sigma })$. The two minus signs cancel, and transposition followed by cyclicity leaves the colour trace unchanged. But the four parity matrices acting on the Levi-Civita symbol contribute $\det \Pi =-1$. HenceThe density is C even and P odd, therefore CP odd. A generic fixed nonzero theta coefficient violates CP symmetry. This is the source of the Strong CP problem, separate from the charged-current phase in the CKM matrix.

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.

Write the Hermitian color connection as ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. The interaction coming from $D_{\mu }={\partial }_{\mu }+ig{\mathcal{A}}_{\mu }$ is $-g\bar{\psi }{\gamma }^{\mu }{\mathcal{A}}_{\mu }\psi$. Its charge-conjugation symmetry requires ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Parity acts on the Lorentz index as on a vector connection, so combining the two givesIn components, ${\mathcal{A}}_0\mapsto -{\mathcal{A}}_0^T(x_P)$ and ${\mathcal{A}}_i\mapsto +{\mathcal{A}}_i^T(x_P)$. The transpose is a color-space transpose, independent of the spinor matrix $C$. Intrinsic fermion phases cancel between a field and its adjoint.

Define the matrix gauge field strength unambiguously byThe derivative terms transform with the expected parity factors and a minus transpose. For the commutator, the crucial identity is $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term has the same transformation as the derivative terms. This is the CP transformation of non-Abelian field strength:For example, ${\mathcal{F}}_{0i}\mapsto +{\mathcal{F}}_{0i}^T(x_P)$ while ${\mathcal{F}}_{ij}\mapsto -{\mathcal{F}}_{ij}^T(x_P)$. With $\mathrm{tr}(T^a{T}^b)=\frac{1}{2}{\delta }^{ab}$, the color sum in the theta operator is twice a trace of two field strengths. The two charge-conjugation minus signs cancel, and transposition reverses the trace factors without changing their trace. The four parity matrices contracted with the Levi-Civita symbol contribute $\det P=-1$. HenceThe theta operator is CP odd; a generic fixed nonzero coefficient violates CP. Setting the coefficient to zero removes this term. Although the density is locally a total derivative, nontrivial gauge topology means that it need not be irrelevant to the quantum theory. If a properly normalized theta angle is identified periodically, invariance at special values equivalent to their negatives is a separate global question; it does not change the CP odd transformation of the local operator.