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Use the kaon CP eigenstates convention $CP|K^0⟩=-|{\bar{K}}^0⟩$, matching kaon CP mixing parameter. Then $|K_1⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$ is CP even, while $|K_2⟩=(|K^0⟩+|{\bar{K}}^0⟩)/\sqrt{2}$ is CP odd. With CPT symmetry and a common small mixing parameter, the physical states can be writtenHere $S,L$ mean short and long mean lifetimes; the kaon CP mixing parameter $ϵ$ depends on the flavor-state phase convention. A positive flavor-exchange convention instead makes the symmetric flavor combination CP even, as explained in exchange-positive CP convention for neutral kaon mixing. Mixing these conventions would reverse the interpretation of the written sums.

Why does the two-pion final state test CP symmetry? A spin-zero kaon decaying into two spin-zero pions has relative orbital angular momentum $\ell =0$ by angular momentum conservation. Two negative intrinsic parities multiply to plus, and $(-1{)}^{\ell }=1$. Charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$; the $\ell =0$ state is even under this exchange. Hence this two-pion state has CP eigenvalue $+1$. If a decay transition operator $T$ preserves CP symmetry, inserting CP on both sides of $⟨\pi \pi |T|K_2⟩$ changes its sign without changing its value: that decay amplitude must vanish.

Assume for illustration that CP violation occurs only through the state admixture and the decay transition preserves CP symmetry. Put $A_1=⟨\pi \pi |T|K_1⟩$. ThenNeglecting the tiny difference in two-body Lorentz-invariant phase space between the physical masses gives ${\Gamma }_L^{\pi \pi }/{\Gamma }_S^{\pi \pi }\simeq |ϵ{|}^2$. A branching fraction, however, divides a partial decay width by the total decay width. With $\hslash =1$, ${\tau }_i=1/{\Gamma }_i$, soFor chosen values $|ϵ|=0.01$ and ${\tau }_L/{\tau }_S=100$, the branching fraction ratio is $0.01$, whereas the partial decay width ratio is $0.0001$. This is original arithmetic, not a historical fit. The longer observation time amplifies the integrated fraction relative to the instantaneous partial decay rate.

In the exchange-positive CP convention for neutral kaon mixing, a coherent source with decay amplitudes $A$ and $B$ for $K^0$ and ${\bar{K}}^0$ has projected amplitudes $(A+B)/\sqrt{2}$ and $(A-B)/\sqrt{2}$. For equal phase space, their decay-rate asymmetry is $\mathcal{R}=2\mathrm{Re}r/(1+|r{|}^2)$, $r=B/A$. Expanding the squared moduli proves the identity. Knowledge of $|r|$ alone cannot fix the asymmetry: $r=i|r|$ gives zero, whereas $r=|r|$ gives constructive interference in the symmetric channel.

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.

A kaon is a pseudoscalar meson, with intrinsic parity $-1$. At rest, $P|\pm ⟩=-|\pm ⟩$. Combining this with the specified CP eigenvalues givesEquivalently, in the exchange-positive CP convention for neutral kaon mixing fixed by the question, $C|K^0⟩=-|{\bar{K}}^0⟩$ and $C|{\bar{K}}^0⟩=-|K^0⟩$. This sign is a convention for the flavor states; the specified CP eigenvalues and physical predictions must be kept consistent with it.

For two spin-zero pions, conservation of total angular momentum fixes the relative orbital angular momentum to zero. The resulting ${\pi }^{+}{\pi }^{-}$ or ${\pi }^0{\pi }^0$ state has CP $+1$. The lowest partial waves of ${\pi }^{+}{\pi }^{-}{\pi }^0$ and $3{\pi }^0$ have CP $-1$. Thus the dominant nonleptonic assignments in the CP symmetry approximation areThe two-pion mode has substantially more available kinetic energy and larger Lorentz-invariant phase space; the three-pion mode has an additional final particle and lies closer to its decay threshold. This helps explain why the CP-odd state has the longer mean lifetime. The actual mean lifetimes also depend on the decay amplitudes and on semileptonic channels, so counting final particles alone is not a quantitative prediction.

The middle panel of the figure in part (a) shows $\bar{s}\to \bar{u}{W}^{+}$, $W^{+}\to u\bar{d}$, with the $d$ as a spectator quark. The outgoing $d\bar{u}$ and $u\bar{d}$ form ${\pi }^{-}$ and ${\pi }^{+}$. In the bottom panel, an additional gluon produces $d\bar{d}$, supplying the neutral pion component for ${\pi }^{+}{\pi }^{-}{\pi }^0$. The charge-conjugate diagrams supply the other component of each of the kaon CP eigenstates. The drawn gluon splitting represents nonperturbative hadronization, rather than a quantitatively reliable perturbative prediction at the kaon scale.

These multiplicities describe the dominant low-orbital angular momentum modes, not an absolute rule for every three-pion configuration. In ${\pi }^{+}{\pi }^{-}{\pi }^0$, odd orbital angular momentum values of the charged pair and of its motion relative to ${\pi }^0$ can couple to total zero and give CP $+1$, but are strongly suppressed near decay threshold. Physical $K_L\to 2\pi$ also occurs through CP violation.

Choose the exchange-positive CP convention for neutral kaon mixing, $CP|K^0⟩=|{\bar{K}}^0⟩$. Neglecting CP violation, the kaon CP eigenstates can then be writtenWriting $A=A({\bar{D}}^0\to K^0{\pi }^0)$ and $B=A({\bar{D}}^0\to {\bar{K}}^0{\pi }^0)$, projection on these coherent final states givesThis is the convention implicit in the printed sum. Changing the sign convention for the neutral antikaon changes both its coefficient in the kaon CP eigenstates and $B$; it cannot change either physical decay rate. The extremely small $K_S$–$K_L$ mass difference can be ignored in the two-body phase space.