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Chiral symmetry allows independent flavor rotations of left-handed and right-handed fields. For two massless quark flavors, the non-Abelian chiral transformations form $SU(2{)}_L\times SU(2{)}_R$. The additional vector $U(1)$ symmetry of baryon number acts trivially on the pion field $U$. The low-energy pion field $U\in SU(2)$ transforms as $U\mapsto LU{R}^{†}$. Its derivative Skyrme model terms are invariant under constant $L,R$. A chosen vacuum $U=1$ preserves the diagonal subgroup $L=R$, which acts by conjugation and is the isospin symmetry. An ordinary common nonzero pion mass explicitly breaks the independent left/right symmetry to that vector subgroup; vacuum selection in the massless theory is instead spontaneous symmetry breaking. The singlet axial anomaly is a separate issue and is not part of this two-flavor non-anomalous group.

Couple the nucleon isospin $1/2$ to the pion isospin $1$. The highest state $|3/2,3/2⟩=|p{\pi }^{+}⟩$ is normalized. One application of the total isospin lowering operator gives $|3/2,1/2⟩=\sqrt{2/3}|p{\pi }^0⟩+\sqrt{1/3}|n{\pi }^{+}⟩$. For a common isospin-invariant two-body decay amplitude, the partial decay rates relative to the ${\Delta }^{++}$ channel are $2/3$ and $1/3$. The lowest state gives ${\Delta }^{-}\to n{\pi }^{-}$. Exact relations assume isospin-degenerate masses and common kinematic factors; electromagnetic and quark-mass differences cause corrections.

Each pion has G parity minus one, so an $n$-pion state has $G=(-1{)}^n$. In the isospin-symmetric strong interaction, a parent with positive G parity can decay only to an even pion multiplicity, and a negative parent only to an odd multiplicity. The rule does not forbid small isospin-breaking or electromagnetic amplitudes.

Suppose a transition operator obeys $[I_3,H]=H/2$ and $[I_{+},H]=0$, and acts on the lower component of an isospin-half kaon. Its output has third component zero, and $I_{+}^{2}H|K⟩=H{I}_{+}^2|K⟩=0$. This excludes an isospin-two output. A spinless parent decaying to two spinless pions has zero relative orbital angular momentum, so Bose-Einstein statistics excludes the antisymmetric isospin-one sector. The output is therefore the two-isovector singlet. In the equal-mass isospin limit, charged and neutral two-pion decay widths have ratio two, including the identical-particle counting.

An isorotation is a global transformation generated by isospin. In the pion Skyrme model it acts as $U\mapsto AU{A}^{†}$, with constant $A\in SU(2)$, rotating the three pion components while preserving the scalar component. The matrices $A$ and $-A$ induce the same classical rotation, so this action factors through $SO(3)$; quantum states may carry half-integer isospin on its $SU(2)$ cover. This global internal rotation is distinct from a spatial rotation or a local gauge transformation.

The pseudoscalar octet of flavor symmetry contains three pions, four kaons and the Eta octet state. In its weight diagram, the neutral pion and ${\eta }_8$ are separate states at the origin.

Approximate a hadronic matrix element of a product of two currents by a product of separate current matrix elements. For a decay to two mesons, one factor can create a pion from the vacuum while the other gives a pseudoscalar-to-pseudoscalar form factor. Nonfactorizable Quantum chromodynamics interactions are omitted; tree-level weak vertices do not alone justify this approximation.

The residual strong interaction between nucleons. Its long-distance part involves pion exchange and depends on spin and isospin. It is distinct from directly exchanging gluons between color-singlet nucleons.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.

Use $|m,m^{\prime }⟩$ as shorthand for the tensor product of two total-isospin-one states with third components $m,m^{\prime }$. A total-isospin-zero state must have total third component zero, so write $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$. Applying the total raising operator givesAn isospin singlet must be annihilated by this operator; hence $b=-a$ and $c=a$. Normalization gives the two-isovector singletIt is also annihilated by the total lowering operator. A raising or lowering action changes the third-component label by one; the unchanged output label in the PDF's preliminary ladder formula is a typographical omission, not the action used here.

Exchange commutes with the total isospin operators. The displayed singlet is symmetric. The $I=2$ multiplet has symmetric highest state $|1,1⟩$; applying the symmetric total lowering operator preserves that exchange symmetry. The symmetric square of a three-dimensional space has dimension six, exactly $1+5$, while its antisymmetric square has dimension three. The remaining $I=1$ multiplet is therefore antisymmetric. Thus the Clebsch-Gordan decomposition isPions have zero intrinsic spin and form an isospin triplet. Bose-Einstein statistics requires the complete two-pion state to be symmetric under exchange. Its relative orbital state changes by $(-1{)}^{\ell }$, so odd orbital angular momentum requires antisymmetric isospin: an odd-$\ell$ two-pion state has $I=1$. This statement treats the charge components as states of the same isospin multiplet, rather than ignoring exchange when their charges differ.

For the decay, let $|\chi ⟩=H_I|K^0⟩$. The stated commutator with $I_3$ gives $I_3|\chi ⟩=0$. More decisively, the commuting raising operator givesThe kaon doublet cannot be raised twice. An $I=2,M=0$ component would survive two raisings, so it is excluded. Meanwhile a spinless kaon decaying into two spinless pions has $\ell =0$ by conservation of total angular momentum, independently of whether parity is conserved by the decay. The exchange condition excludes $I=1$. This proves the highest-weight weak-isospin selection in kaon decay: the two-pion final state has $I=0$.

In a consistent pion phase convention, the singlet's charge components have the structureThe normalized symmetric charged channel has coefficient $\sqrt{2/3}$ and the neutral channel coefficient $-1/\sqrt{3}$. Thus, in the isospin limit with equal pion masses and matching two-body phase space,Equivalently, ordered charged and neutral amplitudes have equal magnitudes, but neutral identical particles contribute the phase-space factor $1/2!$. This is the same counting expressed in another basis, and must not be included a second time after using the normalized charge-channel coefficients. Small pion-mass and electromagnetic effects can modify the ideal ratio.

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

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.