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A box diagram has a one-loop cycle of four propagator edges and four vertices connecting external legs. Its internal particle types depend on the theory. In a charged weak box contribution to kaon mixing, two charged weak-boson propagators connect two up-type quark lines, with flavor-changing charged-current vertices.

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.

Adopt the phase convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Changing both phases changes the convention for the mixing parameter, not physical predictions. The flavor contents are $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A pair of charged weak currents changes strangeness by two units through the following box contractions, with $u_i,u_j=u,c,t$ summed over.

The drawing uses unitary gauge: the W boson gauge-boson propagator contains its longitudinal part. In a gauge of a renormalizable quantum field theory, charged Goldstone boson boxes must also be included to obtain the same gauge-independent result. The external lines represent the valence quarks; the meson matrix element also contains strong interaction binding effects.

Put ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, $x_i=m_i^2/m_W^2$ and $P_L=(1-{\gamma }^5)/2$. The charged weak box contribution to kaon mixing has the effective four-fermion interaction structureHere $G_F$ is the Fermi constant, $V$ is the CKM matrix, and $S$ is a dimensionless loop function from a Feynman integral; the estimate suppresses convention-dependent numerical factors. Four weak interaction Feynman vertices and a one-loop Feynman integral give the coupling and loop order suppression. Unitarity of the CKM matrix gives ${\sum }_i{\lambda }_i=0$, so the GIM mechanism cancels the mass-independent part. With a negligible up quark mass, the leading charm quark term has $S(x_c,x_c)\sim x_c$. Define $f_K$ by $⟨0|\bar{s}{\gamma }_{\mu }{\gamma }^{5}d|K^0(p)⟩=i{f}_K{p}_{\mu }$. Estimating the hadronic matrix element by $f_K^2{m}_K{B}_K$, with the dimensionless kaon bag parameter $B_K$ of order unity, givesThe hadronic normalization includes division by $2m_K$ when relativistically normalized states are converted to the effective Hamiltonian for neutral-meson mixing. Since $|V_{cs}^{\ast }{V}_{cd}|\simeq \sin {\theta }_C\cos {\theta }_C$, the Cabibbo suppression is explicit. Illustrative scales $G_F\sim 10^{-5}{\mathrm{GeV}}^{-2}$, $m_c\sim 1\mathrm{GeV}$, $f_K\sim 0.1\mathrm{GeV}$ and $m_K\sim 0.5\mathrm{GeV}$ give an order $10^{-16}$ to $10^{-15}\mathrm{GeV}$ estimate. The full sum includes charm quark-top quark and top quark-top quark terms, important for CP violation; long-distance strong interaction contributions prevent this dimensional estimate from being a precision prediction. A meson matrix element necessarily needs hadronic input in addition to electroweak parameters.

For the decaying kaon system write $H^{\prime }=M-i\Gamma /2$, where $M$ and $\Gamma$ are Hermitian matrices. The entries denoted $M_{ij}$ in the question are entries of this effective Hamiltonian for neutral-meson mixing; they need not themselves form a Hermitian matrix. CPT invariance equates the diagonal masses and diagonal decay widths, givingOne can see why there is no complex conjugation on the right as follows. In the flavor subspace let $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and represent the antiunitary CPT operation by $J$ followed by complex conjugation. CPT changes a projected forward propagator into its adjoint. Indeed, if $P$ projects onto the flavor subspace and $H_{\mathrm{full}}$ is the Hermitian microscopic generator, its invariance and the antiunitary reversal of $i$ give $\Theta P{e}^{-i{H}_{\mathrm{full}}t}P{\Theta }^{-1}=P{e}^{+i{H}_{\mathrm{full}}t}P=(P{e}^{-i{H}_{\mathrm{full}}t}P{)}^{†}$. In the time-independent two-state approximation, the effective generator this means $J{H^{\prime }}^{\ast }J={H^{\prime }}^{†}$, whose diagonal entries give $(H_{22}^{\prime }{)}^{\ast }=(H_{11}^{\prime }{)}^{\ast }$; the off-diagonal entries impose no further condition. Equivalently, apply CPT separately to the dispersive and absorptive Hermitian matrices. Treating the decaying effective Hamiltonian for neutral-meson mixing as a Hermitian matrix would incorrectly discard the absorptive part.

In the chosen flavor phase convention CP is represented by $-J$. Its invariance requires $J{H}^{\prime }J=H^{\prime }$. Thus it gives diagonal equality and additionallyThe normalized CP eigenstates areApplying CP directly gives eigenvalues $+1$ and $-1$ respectively.

Now use CPT to put $H^{\prime }=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ and take correlated square roots $\alpha =\sqrt{b}$, $\beta =\sqrt{c}$. Its characteristic polynomial equation is $(a-\mu {)}^2-bc=0$, and direct multiplication showsExpressing these two eigenvectors in the CP eigenstate basis gives, respectively, coefficients proportional to $(\alpha +\beta ,\alpha -\beta )$ and $(\alpha -\beta ,\alpha +\beta )$. Therefore the kaon CP mixing parameter isand, because the CP eigenstate basis is orthonormal, the normalized states at production areThe kaon mixing square-root branch convention correlates the roots continuously with the limit of conserved CP $b=c$, where $\alpha =\beta$ and $ϵ=0$. Reversing one root interchanges the two eigenstates. The labels short and long are fixed by their decay widths: their complex eigenvalues are ${\mu }_{S,L}=m_{S,L}-i{\Gamma }_{S,L}/2$, and subsequent evolution supplies $e^{-i{\mu }_{S,L}t}$. The displayed formula assumes nonzero mixing and $\alpha +\beta \ne 0$, as in the physical kaon system; degenerate or defective matrices require a separate limiting treatment.