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The Cabibbo-Kobayashi-Maskawa matrix $V=U_{uL}^{†}{U}_{dL}$ compares the left-handed unitary matrices that diagonalize the up-type and down-type Yukawa matrices. Its rows label up quarks, charm quarks and top quarks; its columns label down quarks, strange quarks and bottom quarks. In the weak charged current, the coefficient of ${\bar{u}}_{Li}{\gamma }^{\mu }{d}_{Lj}{W}_{\mu }^{+}$ is $V_{ij}$. A single CKM matrix therefore connects quark masses, flavour-changing decay amplitudes, and CP violation.

Write $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$. An exactly unitary matrix with three mixing angles and one CP-violating phase isThis convention agrees with Makoto Kobayashi's account of three-generation quark mixing.^[ref]

The Christenson-Cronin-Fitch-Turlay kaon CP violation experiment established the phenomenon that required an explanation. Makoto Kobayashi and Toshihide Maskawa showed that two-generation quark mixing cannot supply it, whereas adding a third quark generation permits an irreducible weak phase. Their 1973 paper also considered additional fields as alternative mechanisms; the three-generation conclusion concerns CP violation from the minimal unitary quark mixing mechanism.^[ref]

A common weak phase makes CP violation across mesons testable through the BABAR time-dependent CP violation experiment. It does not explain the cosmic baryon asymmetry.^[ref]

The approximate equality of proton and neutron parton distribution functions after interchanging up quarks and down quarks, including their antiquarks. It follows from the corresponding isospin symmetry when electromagnetic and light-quark mass-difference effects are neglected. Isospin-singlet sea distributions are unchanged.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

The exchanged particle is now a gluon, and the incoming up quark mass replaces $m_e$. DefineThe Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. Averaging over the $N_c^2$ initial color charge choices and summing the final color charge choices givesReplace the factor $N_c{e}^4{Q}_e^2{Q}_t^2$ from part (ii) by $g_s^4{C}_g$. ThereforeKeeping $m_u$ is allowed; if the usual negligible up quark mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial quark color charge states are averaged, whereas final ones are summed.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

Use units in which the proton electric charge is one. The up quark and down quark both have spin angular momentum $1/2$ and form a flavour isospin doublet:Their antiquarks have the opposite charges and third components of isospin, with the same total isospin and spin. This is strong-interaction flavour isospin, not the chiral weak isospin assignment.

The Clebsch-Gordan decomposition for SU2 says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and isospin. ThusBefore exchange symmetry is imposed, three light quarks have total spin $S=1/2$ or $3/2$, and total isospin $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero orbital angular momentum, $J=S$.

Physical baryons must also be colour singlets. The three-quark colour singlet is completely antisymmetric, so the Pauli exclusion principle requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the Pauli constraint on three-quark flavour multiplets; it leavesEquivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20=2\cdot 2+4\cdot 4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state baryons.

For three light quarks, charge addition gives $Q=I_3+1/2$. The observed nucleon doublet and Delta baryon quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:The proton and neutron have $J=1/2$; all four Delta baryons have $J=3/2$. These states have positive parity in the zero-orbital ground-state model.

For a light quark-antiquark meson, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states areEach row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero orbital angular momentum their parity is negative, following the meson parity and charge conjugation rule $P=(-1{)}^{L+1}$. The spin-zero isotriplet is the pion multiplet, and the spin-one isotriplet the rho meson multiplet. The nonstrange spin-one isosinglet is the idealized omega meson flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the eta and eta prime mesons; it is not automatically a physical mass eigenstate once other flavours and mixing are included.

Now introduce an isospin-zero quark $h$ of spin $1/2$ and charge $q_h$. For baryons with one isospin-singlet quark, the two light quarks have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt{2}$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence light-diquark spin-isospin correlation gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin producesThe three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half Lambda baryon, the spin-one-half Sigma baryon triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half Lambda c baryon of charge $1$, and the Sigma c baryon triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate antibaryons have opposite charges and unchanged $I,J$.

The meson doublets with one isospin-singlet quark are $h\overline{u},h\overline{d}$, withFor $q_h=-1/3$ their charges are $-1,0$, giving $K^{-},{\overline{K}}^0$-like kaons and their vector partners; the conjugate doublet has $u\overline{h},d\overline{h}$ and charges $1,0$. For $q_h=2/3$ the charges are $0,1$, giving $D^0,D^{+}$-like D mesons and their vector partners, with charge-conjugate partners as well. Finally a hidden-flavour pair $h\overline{h}$, if included, is neutral with $I=0$ and $J=0$ or $1$. These are quantum-number classifications; charge and isospin alone do not determine particle masses or mixing angles.

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.

The commuting Hermitian operators $H_1,H_2$ admit a weight-space decomposition. The six root vectors shift their simultaneous eigenvalues byor their negatives. Put $h_i=2{\alpha }_i\cdot H$, $e_i=\sqrt{2}{E}_{+}^i$ and $f_i=\sqrt{2}{E}_{-}^i$. The supplied relations giveThese are three sl2 triples, coming from the complexifications of the three SU(2) Lie algebra root-pair subalgebras. They act through homomorphisms into $\mathrm{End}(V)$. For a nontrivial representation, simplicity of the SU(3) Lie algebra makes its kernel zero, so these images are actual ${\mathfrak{sl}}_2(\mathbb{C})$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $i{h}_i$, $i(E_{+}^i+E_{-}^i)$ and $E_{+}^i-E_{-}^i$.

Decomposing under any one of these subalgebras gives finite weight strings in its root direction. The two simple-root highest weights must be nonnegative integers. In the SU(3) highest-weight coordinates used here, this condition isHere $(a,b)$ are the Dynkin labels; $(p,q)$ are the eigenvalues of $H_1,H_2$. An irreducible representation is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the orthogonal complement of an invariant subspace remains invariant.

The weight diagram has Weyl group symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary weight multiplicities are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min (a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the dominant weight multiplicity formula for sl3: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and setIts multiplicity is zero unless $r,s$ are integers, and otherwise is $\max \{0,1+\min (a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.

For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_{+}+E_{-})/\sqrt{2}$ and $(E_{+}-E_{-})/(i\sqrt{2})$. In the defining representation these eight generators satisfy $\mathrm{tr}(t_a{t}_b)={\delta }_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their structure constants of a Lie algebra antisymmetric, sothe antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a quadratic Casimir operator and is scalar on an irreducible representation by the Schur lemma.

Let $v$ be a highest-weight vector, annihilated by all $E_{+}^i$. On $v$, each $E_{-}^i{E}_{+}^i$ vanishes and each $E_{+}^i{E}_{-}^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt{3}{H}_2$. Henceand therefore the SU(3) quadratic Casimir eigenvalue isFor a reducible representation the same calculation applies separately to its irreducible blocks.

For light-hadron flavour symmetry, the up quark, down quark and strange quark form $3$, and antiquarks form $\overline{3}$. Conventional mesons have $3\otimes \overline{3}=1\oplus 8$. Conventional three-quark baryons haveobtained from $3\otimes 3=6\oplus \overline{3}$, $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$. Their Dynkin labels are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional meson octet, baryon octet, baryon decuplet and singlet multiplets give $\boxed{X=0,3,6}$, with the same values for conjugate antibaryon multiplets. More generally, the zero flavour triality of a light-quark hadron imposes $a+2b\equiv 0(\mathrm{mod}3)$ on multiplets constructed entirely from light quarks, antiquarks and gluons. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $27$, of labels $(2,2)$, has $X=8$.

Start with a nonzero highest-weight vector $v_0=|n⟩$ and set $v_r=E_{-}^r{v}_0$. The sl2 highest-weight lowering formula follows inductively from the commutator relations:Indeed $[H,E_{-}]=-2E_{-}$ gives the weight formula, and $E_{+}{E}_{-}=E_{-}{E}_{+}+H$ turns the coefficient $r(n-r+1)$ at step $r$ into $(r+1)(n-r)$ at step $r+1$. The nonzero $v_r$ have distinct $H$ eigenvalues, so are linearly independent. Finite dimensionality therefore forces a last nonzero vector $v_N$. Applying $E_{+}$ to $v_{N+1}=0$ gives $(N+1)(n-N)v_N=0$, hence $N=n$. This also proves that no earlier vector can vanish. Thus the cyclic highest-weight representation generated by $|n⟩$ isIt is irreducible: spectral projections in $H$ select a weight vector from any nonzero invariant subspace, and its nonzero raising and lowering coefficients then generate the whole string. This gives the classification of finite-dimensional sl2 representations used for each root direction below.

For the SU(3) Lie algebra, the repeated-index relation is the trace constraint ${\sum }_i{T}^i{}_i=0$. The displayed generators are a basis of its complexification, not nine independent generators or three individually vanishing diagonal elements. Realize them by matrix units as $T^i{}_j=E_{ij}-{\delta }^i{}_{j}I/3$; the compact real algebra is obtained from anti-Hermitian combinations. Thenand $E_{1+}=E_{12}$, $E_{1-}=E_{21}$, $E_{2+}=E_{23}$, $E_{2-}=E_{32}$. Substituting into the given commutators givesThus both root pairs obey the required sl2 Lie algebra relations. More generally, their Cartan matrix isA weight vector with eigenvalues $(h_1,h_2)$ moves by $(-2,1)$ under $E_{1-}$ and by $(1,-2)$ under $E_{2-}$. The third negative-root operator is $[E_{2-},E_{1-}]=E_{31}$ and moves it by $(-1,-1)$. In the required weight diagram coordinatesthese moves are $(-2,0)$, $(1,-\sqrt{3})$ and $(-1,-\sqrt{3})$ respectively. Successive lowerings from a highest-weight vector, with linear dependencies removed, construct the following bases.