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A flavor eigenstate is a state with a definite flavor label in the specified interaction or flavor-charge basis. For neutral kaons, $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$ have definite opposite strangeness. Weak interactions mix them, so these flavor states are not the mass eigenstates. The precise flavor basis must be specified when discussing mixing.

The additive flavor symmetry quantum number $Y=B+S$ combines baryon number with strangeness. For the $(u,d,s)$ quark triplet its values are $(1/3,1/3,-2/3)$. The Gell-Mann--Nishijima formula is $Q=I_3+Y/2$. This is distinct from the normalization of electroweak hypercharge.

Some formulas write $Q=I_3+Y$ rather than $Q=I_3+Y_{\mathrm{usual}}/2$. Then $Y=(B+S)/2$, where $B$ is baryon number and $S$ is strangeness. The light-quark weights are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$. Baryon weight rows therefore have half the usual vertical coordinates. This is a flavour convention, distinct from weak hypercharge.

A baryon containing one or more strange quarks and no heavier quark. The light hyperon multiplets include lambda, sigma, xi and omega baryons. Strong and electromagnetic interactions conserve strangeness, whereas a weak interaction can change it; a photon in the final state does not by itself make a hyperon decay electromagnetic.

The ${\Lambda }^0$ in the baryon octet is an isospin singlet with strangeness $-1$ and zero electric charge. Its ordinary valence quark content is $uds$.

Scattering with one meson and one baryon in the initial and final states. The strong interaction conserves electric charge, baryon number, net strangeness, four-momentum, total angular momentum, and parity. Approximate isospin symmetry imposes additional relations among charge-channel amplitudes.

The number of quarks of quark flavor $f$ minus the number of corresponding antiquarks. The strong interaction and electromagnetic interaction conserve each such difference even when they create or annihilate pairs. Weak charged currents can change these differences. For the strange quark, strangeness is $S=-F_s$.

The neutral strange meson with strangeness $-1$ and valence content $s\bar{d}$. It is distinct from the neutral kaon flavour state $K^0=d\bar{s}$, which has strangeness $+1$. These are flavour eigenstates, not the weak-interaction propagation eigenstates of neutral kaon mixing.

The light-quark $sss$ member of the spin-three-halves baryon decuplet has strangeness $S=-3$, baryon number $B=1$, flavor hypercharge $Y=-2$, and isospin zero. The Gell-Mann--Nishijima formula gives electric charge minus one. Its position at the bottom of the decuplet's triangular weight diagram follows by replacing all three up or down quarks with strange quarks.

A strong interaction decay must conserve quark flavour and have final rest energy below the initial mass. An excited light baryon can often decay to a lighter baryon plus a pion. A ground-state strange baryon may lack any such open flavour-conserving channel even when its spin permits one. In particular, the Omega baryon cannot decay strongly to a Xi baryon plus a pion because that changes strangeness; its flavour-conserving Xi-kaon channel is closed by the mass threshold. A spin-three-half assignment alone does not imply a strong decay.

For an A2 root system weight $w=(w_1,w_2{)}^T$, the specified electric charge and flavour hypercharge are the linear functionalsApply these to the meson octet weights in solution. The results, with electric charges measured in units of the positive elementary electric charge, areThe table illustrates the meson octet of pseudoscalar mesons; the same weight calculation applies to another meson species arranged in the same representation. Here flavour hypercharge is $Y=B+S$, with $B$ the baryon number and $S$ the strangeness, rather than weak hypercharge. If the full quark model tensor product $3\otimes \overline{3}$ is retained, its additional trivial representation is a ninth meson with $(q,y)=(0,0)$. This singlet is separate from the octet and need not share its mass.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by SU(3) leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate flavour symmetry, distinct from the exact colour gauge symmetry. The quark transforms as $3$ and the antiquark as $\overline{3}$.

For mesons, a quark-antiquark tensor is a $3\times 3$ matrix. Its trace and traceless parts are invariant subspaces, givingThe trace is the eta singlet state; the traceless part is the meson octet. For baryons, first separate a pair of quarks into symmetric and antisymmetric tensors:The antisymmetric pair transforms as $\overline{3}$ by contraction with the invariant epsilon tensor. In $6\otimes 3$, complete symmetrization gives ${\mathrm{Sym}}^{3}3$ of dimension $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{3}\otimes 3$, trace and traceless parts give a singlet and another octet. Thus the tensor cube of the defining SU(3) representation isThe two octets carry the two mixed permutation-symmetry components. Their weight diagrams are identical; the extra copy is a multiplicity, not a new arrangement of weights.

The printed charge relation fixes the half-hypercharge convention, rather than the usual Flavor hypercharge normalization:The quark weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the baryon weight diagrams in the half-hypercharge convention. The baryon octet has $(p,n)$ at $Y=1/2$, the isotriplet $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$ and the isosinglet ${\Lambda }^0$ at $Y=0$, and $({\Xi }^0,{\Xi }^{-})$ at $Y=-1/2$. The baryon decuplet has rows $\Delta$ at $Y=1/2$, ${\Sigma }^{\ast }$ at zero, ${\Xi }^{\ast }$ at $-1/2$, and ${\Omega }^{-}$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.

The fully symmetric flavour representation is precisely the baryon decuplet, $10={\mathrm{Sym}}^{3}3$. For a spatial ground state, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ quarks, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the Pauli exclusion principle for the three fermions.

An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state isIt is completely antisymmetric and an SU(3) colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third exterior power: ${\bigwedge }^3{\mathbb{C}}^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the quark model. This argument motivates three colours rather than deriving their exact number without further physical input.

The Pauli constraint on three-quark flavour multiplets also explains the absence of a flavour-singlet baryon in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. Butso three spin-$1/2$ quarks have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state baryon octet with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full Pauli constraint on three-quark flavour multiplets is applied. Excited orbital wavefunctions can permit three-quark flavour singlet; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every baryon at every energy.

A flavour-singlet meson is allowed: a quark and an antiquark are not three identical fermions, and the invariant flavour combination $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no meson exclusion.

Finally, ${\pi }^{+}p$ has baryon number one, strangeness zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ Delta baryon ${\Delta }^{++}$, the low-energy resonance with spin $3/2$. For $K^{+}p$, the kaon $u\bar{s}$ contributes strangeness plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ baryon can have positive strangeness, since every $s$ quark contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic quark content, for example $uuud\bar{s}$, and is not mandated by this spectrum. This is strangeness selection in pion-proton and kaon-proton resonances.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.

For the quark model, the relevant approximate flavour symmetry rotates the three light quarks $u,d,s$ in the fundamental $3$. The up quark, down quark, and strange quark basis states have the weight diagram of the defining SU(3) representation. In physical coordinates define isospin $I_3=H_1$ and flavor hypercharge $Y=2H_2/\sqrt{3}$. ThenThe last row follows from the Gell-Mann--Nishijima formula $Q=I_3+Y/2$, in units of the positive elementary electric charge. The antiquarks transform in $\overline{3}$ and have the negatives of these additive quantum numbers. Each quark has baryon number $B=1/3$; the strange quark has strangeness $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing flavor hypercharge from electroweak hypercharge. The Weyl group geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.

A conventional meson has valence content $q\bar{q}$, so its flavour symmetry gives a meson octet and a singlet. For the light pseudoscalar mesons, the octet contains the pions ${\pi }^{+},{\pi }^0,{\pi }^{-}$ at $Y=0$ with isospin $I=1$, the kaons $K^{+}=u\bar{s},K^0=d\bar{s}$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ Eta octet state. The central weight multiplicity two is realized byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The eta and eta prime mesons are mixtures of ${\eta }_8,{\eta }_1$, rather than exact octet and singlet mass eigenstates; this gives the pseudoscalar meson nonet. A different spin coupling of the same flavour product gives the vector mesons, again an octet and singlet, including the $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ mixtures. The lowest orbital states have $J^P=0^{-}$ and $1^{-}$ respectively.

A conventional baryon has three valence quarks, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest baryon octet has spin $1/2$ and positive parity. Its isospin rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the nucleons $p,n$, the three Sigma baryons, the Lambda baryon, and the two Xi baryons. The central states ${\Sigma }^0$ and $\Lambda$ have the same weight $(I_3,Y)=(0,0)$ but different total isospin; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity baryon decuplet is the triangle with rowsThe Delta baryons form its top row, and the Omega baryon is the bottom $sss$ state. All ten weights have multiplicity one. Their electric charges follow directly from the Gell-Mann--Nishijima formula; for example ${\Delta }^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and ${\Omega }^{-}$ has $I_3=0,Y=-2$, so $Q=-1$.

The two algebraic octets do not mean that two independent ground-state octets must occur. A three-quark colour singlet has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the Pauli exclusion principle therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the symmetric spin-flavour SU6 representation hasThe three-quark flavour singlet is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but ${\bigwedge }^3{\mathbb{C}}^2=0$. The Pauli constraint on three-quark flavour multiplets thus excludes it from this ground-state sector; orbital excitation permits other multiplets.

Finally, this flavour symmetry is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The weight diagrams classify flavour states, rather than asserting exact equality of their masses. Colour SU(3) is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the eightfold way. The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.

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.

Quark mixing arises because the up-type and down-type quark Yukawa matrices need not be diagonal in the same weak interaction basis. In the Standard Model the three left-handed quark doublets $Q_L^{\prime }$ transform as $(3,2,1/6)$ under $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$, whereas $u_R^{\prime }$ and $d_R^{\prime }$ transform as $(3,1,2/3)$ and $(3,1,-1/3)$. With a Higgs doublet $H$ and its conjugate $\tilde{H}=i{\sigma }^2{H}^{\ast }$, the gauge-invariant Yukawa interaction isThe generation indices are matrix-multiplied. Electroweak symmetry breaking, $⟨H⟩=(0,v_H/\sqrt{2}{)}^T$, produces $M_u=v_H{Y}_u/\sqrt{2}$ and $M_d=v_H{Y}_d/\sqrt{2}$.

Each complex fermion mass matrix admits a biunitary diagonalization,where the diagonal entries are real nonnegative quark masses. Define $u_L^{\prime }=U_{uL}{u}_L$ and $d_L^{\prime }=U_{dL}{d}_L$, and similarly for right-handed fields. The kinetic terms retain their canonical form because the transformations are unitary matrices. However, the weak charged current becomesThis unitary matrix is the Cabibbo-Kobayashi-Maskawa matrix. Its entries multiply transitions between up-type and down-type mass eigenstates. The minimal Standard Model has no analogous right-handed charged weak current.

Quark mixing does not introduce tree-level flavor-changing neutral currents. Photon, gluon and weak neutral current couplings are proportional to the identity within each fixed-charge quark sector, and $U^{†}IU=I$. The single Higgs doublet also has diagonal neutral scalar field couplings after the same transformations that diagonalize the masses. At higher loop orders, flavor changes are possible. For example, an $s\to d$ amplitude contains ${\sum }_i{V}_{is}^{\ast }{V}_{id}f(m_i^2)$. Unitarity of the CKM matrix sets the sum of coefficients to zero, cancelling any mass-independent part. This is the GIM mechanism; mass splittings leave a suppressed, generally nonzero amplitude.

For $N$ generations a unitary matrix has $N^2$ real parameters: $N(N-1)/2$ mixing angles and $N(N+1)/2$ phases. Rephasing the $N$ up-type and $N$ down-type quark fields removes $2N-1$ phases; a common baryon number phase leaves every matrix entry unchanged. Hence the physical numbers areTwo generations therefore give a real Cabibbo rotation,with strangeness-changing charged-current transitions suppressed by $\sin {\theta }_C$. Three generations have three mixing angles and one irreducible weak CP phase. This counting assumes nondegenerate masses; degeneracies allow additional basis rotations.

Writing $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$, a standard exact parametrization isComplex entries alone do not establish CP violation, because quark quantum field rephasings can change their phases. For nondegenerate quark masses the weak mixing sector conserves CP precisely when a basis makes $V$ real. A useful invariant test is the Jarlskog invariant,A nonzero $J$ requires all three mixing angles and a nontrivial phase. If two same-charge quark masses coincide, the additional basis freedom removes the corresponding physical CP effect. The possible QCD theta angle is a separate CP parameter, not part of this weak-mixing count.

The Wolfenstein parametrization displays the hierarchy of quark mixing economically. Put ${\lambda }_W=s_{12}$, $s_{23}=A{\lambda }_W^2$, and $s_{13}{e}^{i\delta }=A{\lambda }_W^3(\rho +i\eta )$. Expanding the exact matrix givesThe observed pattern is predominantly diagonal, with $1$-$2$ mixing largest and $2$-$3$, $1$-$3$ mixing successively suppressed. The expansion implies $J\simeq A^2{\lambda }_W^6\eta$. The truncated matrix must not be treated as an exactly unitary matrix.

Column orthogonality givesThe three complex numbers form a CKM unitarity triangle, with area $|J|/2$. Dividing by $-V_{cd}{V}_{cb}^{\ast }$ puts its vertices at $0$, $1$ and $\bar{\rho }+i\bar{\eta }$, where $\bar{\rho }+i\bar{\eta }=-V_{ud}{V}_{ub}^{\ast }/(V_{cd}{V}_{cb}^{\ast })$. The barred parameters include higher-order corrections to the Wolfenstein parameters. Comparing independent measurements of its sides and angles tests both quark mixing and the absence of additional flavor dynamics.

Semileptonic decay rates determine magnitudes of CKM entries through factors $|V_{ij}{|}^2$ once hadronic matrix elements are controlled. Neutral kaon and neutral B-meson mixing constrain products of entries through weak box scattering amplitudes. CP asymmetries probe their invariant relative phases, with kaon mixing providing indirect CP violation and decay amplitudes providing direct CP violation. These processes complement tests of the unitarity constraints on rows and columns. The minimal three-generation Standard Model describes weak quark mixing by a unitary matrix with three angles and one physical CP phase; it does not predict the Yukawa couplings or their hierarchy.

The flavor content is $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A charged-current box diagram changes strangeness by two units. The two internal quark lines can contain any up-type flavors $u_i,u_j\in \{u,c,t\}$, connected by two charged $W$ propagators. One allowed topology is shown below; its crossed counterpart also contributes to the full mixing amplitude.

Each charged-current vertex contains the appropriate CKM matrix element. The flavor sum has combinations ${\lambda }_i{\lambda }_j$, with ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, multiplying mass-dependent loop functions. CKM matrix unitarity gives ${\sum }_i{\lambda }_i=0$, illustrating the GIM cancellation of flavor-independent loop terms. The diagram is at order $g^4$ and second order in the weak interaction.

Let $\Theta$ be the antiunitary CPT operator. It interchanges the neutral-kaon flavor states up to phases. For a Hermitian Hamiltonian invariant under CPT,since the diagonal expectation is real. Therefore CPT gives $R_{11}=R_{22}$. More generally an effective decay Hamiltonian has $R=M-i\Gamma /2$, where both $M$ and $\Gamma$ are Hermitian. CPT gives $M_{11}=M_{22}$ and ${\Gamma }_{11}={\Gamma }_{22}$, hence the same equality of the complex diagonal entries. It does not require $R_{12}=R_{21}$; that is a CP condition. Nor is $R_{21}=R_{12}^{\ast }$ true for the full decay matrix in general.

Choose the CP convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Then CP acts as $-{\sigma }_x$ on the flavor basis. Invariance of $H^{\prime }$ means $R={\sigma }_{x}R{\sigma }_x$, and thereforeWith a different flavor-state phase convention this relation carries the corresponding phase factors; equality is the relation in the convention adopted here. For a Hermitian mass matrix it makes the off-diagonal element real.

For the requested production-time mass eigenstates, discard the absorptive part. From now on $R$ denotes the Hermitian part $(R_{\mathrm{original}}+R_{\mathrm{original}}^{†})/2$, so its entries have the formIf the original matrix already was Hermitian no replacement is needed. The neutral-kaon mass matrix diagonalization gives eigenvalues $r_0\pm |z|$, since $(r_0-\lambda {)}^2-|z{|}^2=0$. Write $z=r{e}^{i\phi }$, $r=|z|>0$. Use the kaon mixing square-root branch conventioncontinuously from the CP-conserving convention in which $z>0$. Independent unrelated root signs would interchange the eigenvalue labels. Direct multiplication gives $R(a,-b{)}^T=(r_0-r)(a,-b{)}^T$ and $R(a,b{)}^T=(r_0+r)(a,b{)}^T$. Assigning the larger mass to $K_L^0$ giveswith mass shifts $r_0-r$ and $r_0+r$ respectively. A common strong-interaction mass may simply be added to both. Their norms are one and their inner product is zero. A Hermitian mass-only calculation determines the lighter and heavier combinations; it does not determine their different lifetimes.

In the CP-conserving limit, the chosen lighter and heavier combinations areThey are CP eigenstates with eigenvalues $+1$ and $-1$ in the convention above. The sign of the real off-diagonal entry and the flavor-state convention are chosen together so that the stated $K_1^0=K_S^0$ limit is the lower-mass state.