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Compared with the complete three-flavour meson tensor product, the additional flavour states are exactlyThus there are seven new flavour states. For the pseudoscalar mesons, the six open-charm states are ${\bar{D}}^0,D^{-},D_s^{-}$ and $D^0,D^{+},D_s^{+}$, respectively, using the D meson and Ds meson conventions. A pure $c\bar{c}$ pseudoscalar meson is the idealized flavour state associated with the eta c meson. The quark model flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other meson species.

If comparing the traceless SU(4) $15$ with the traceless SU(3) meson octet, the extra seven states are instead the same two triplets and the normalized SU(3) singletIts coefficient matrix is traceless and has squared norm one. In particular, $c\bar{c}$ by itself is not a vector in the traceless $15$; its coefficient matrix has trace one.

To relate these two descriptions, let $S_4=(u\bar{u}+d\bar{d}+s\bar{s}+c\bar{c})/2$ be the normalized SU(4) singlet and $S_3=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ the old SU(3) singlet. ThenThis orthonormal change of basis shows explicitly why the complete tensor product contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.

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.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger representations containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large representation.

A dual vector must be pulled back by $P^T$, since $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus a G2 root system weight $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet meson retains its values from solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor instance the first three come from the G2 root system weights $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The weight diagram displays the preserved octet and the new states.

The fractional electric charges make these additional states qualitatively different from ordinary quark-antiquark mesons, whose electric charges are integers. This is a consequence of the proposed symmetry extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

A meson has quark model flavour content $q_i{\bar{q}}^j$, so four quark flavours give sixteen flavour states inThis counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states $3\otimes \bar{3}=8\oplus 1$.

Taking the fourth quark to be the charm quark, the first nine combinations $q_{\alpha }{\bar{q}}^{\beta }$, with $\alpha ,\beta \in \{u,d,s\}$, are the light sector. The extra seven are $q_{\alpha }\bar{c}$ in $3$, $c{\bar{q}}^{\alpha }$ in $\bar{3}$, and $c\bar{c}$ in $1$ under light-flavour $SU(3)$. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from $SU(4)$ invariant. Exact flavour $SU(4)$ would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour $SU(4)$ is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour $SU(3)$.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.

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.

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

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.