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The quantum electrodynamics interaction is ${\mathcal{L}}_{\mathrm{int}}=-e{j}^{\mu }{A}_{\mu }$, where the Dirac electromagnetic current is $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$. Under parity symmetry in quantum field theory,Invariance of the interaction therefore requires the electromagnetic four-potential to transform as the same Lorentz four-vector:so $A^0$ is parity even and $\mathbf{A}$ is parity odd. Under charge conjugation, the current is odd, $j^{\mu }\mapsto -j^{\mu }$, so invariance requires

The operator $ia\bar{\psi }{\sigma }^{\mu \nu }{\gamma }^5\psi F_{\mu \nu }$ is the relativistic fermion electric dipole moment operator. Its nonrelativistic limit contains $\mathbf{S}\cdot \mathbf{E}$: spin angular momentum is an axial vector, whereas the electric field is a polar vector. It is consequently odd under parity symmetry in quantum field theory. Both the pseudotensor fermion bilinear and the electromagnetic field tensor are odd under charge conjugation, so their product is even. ThereforeThe CPT theorem then makes it odd under time-reversal symmetry. Such an interaction can arise only from CP violation. The Cabibbo-Kobayashi-Maskawa matrix therefore induces a nonzero Standard Model electric dipole moment at sufficiently high loop order, but its flavor structure and loop suppressions make the result extraordinarily small.

Since $C{J}_3{C}^{-1}=-J_3$, the operator $C$ sends a weight-$m$ state to a weight-$-m$ state. AlsoFor the spin-one normalization,Write $C|1,1⟩=a|1,-1⟩$. Using $C|1,0⟩=|1,0⟩$ in the first relation gives $a=-1$, and applying $C$ to the second gives the other phase. Therefore, in the stated phase convention,A simultaneous phase redefinition of the charged pion states changes both displayed signs but not their physical interchange under charge conjugation.

Under charge conjugation, $\varphi \leftrightarrow {\varphi }^{\ast }$ must be accompanied byThen $D_{\mu }\varphi$ is exchanged with $(D_{\mu }\varphi {)}^{\ast }$, while $F_{\mu \nu }\mapsto -F_{\mu \nu }$. The scalar kinetic and mass terms are therefore exchanged with themselves and $F_{\mu \nu }{F}^{\mu \nu }$ is unchanged, proving invariance of scalar quantum electrodynamics.

Charge conjugation reverses the gauge charge, soUsing the antisymmetric spinor metric $ϵ=i{\sigma }^2$, a compatible action on the two Weyl fields isComplex conjugation reverses the sign of $i$ and of the gauge representation, while $A_{\mu }\mapsto -A_{\mu }$ restores the original gauge covariant derivative. The identities ${\sigma }^2({\sigma }^{\mu }{)}^{\ast }{\sigma }^2={\bar{\sigma }}^{\mu }$ and ${\sigma }^2({\bar{\sigma }}^{\mu }{)}^{\ast }{\sigma }^2={\sigma }^{\mu }$ then exchange the two equations of motion. Unit phases can be inserted in the two spinor transformations without changing the conclusion, subject to the mass-term relation.

With matrix-valued fields, charge conjugation acts asUnder parity, $A_0$ is even and $A_i$ is odd:Consequently $F_{0i}$ is odd and $F_{ij}$ is even. Under time-reversal symmetry, which is antiunitary, $A_0$ is even and $A_i$ is odd after $t\mapsto -t$; hence $F_{0i}$ is even and $F_{ij}$ is odd. Equivalently, the non-Abelian electric and magnetic fields transform as

For the neutral quark-antiquark mesons here, $C=(-1{)}^{L+S}$. Thus $C({\eta }^{\prime })=C({\pi }^0)=+1$, while $C({\rho }^0)=C(\gamma )=-1$. The radiative ${\rho }^0\gamma$ mode conserves charge conjugation and admits a quark-line photon-emission diagram. Both components of the supplied ${\rho }^0$ wavefunction contribute: the charge-weighted flavour overlap is proportional to $Q_u-Q_d=1$, so it does not cancel. Its photon energy is $(m_{{\eta }^{\prime }}^2-m_{\rho }^2)/(2m_{{\eta }^{\prime }})\simeq 165\mathrm{MeV}$.

The ${\pi }^0\gamma$ mode changes charge conjugation and, more decisively, a spin-zero particle cannot decay into another spin-zero particle and one real photon. In the parent rest frame the photon has helicity $\pm 1$, while the opposite-moving pion has helicity zero; the spin-zero two-body helicity selection rule cannot be satisfied. Its on-shell two-body amplitude vanishes even if a weak interaction violates C. A virtual photon subsequently making another pair is a different final state.

For ${\pi }^0{\rho }^0$, angular-momentum conservation requires relative orbital angular momentum $L=1$. Its parity is then negative, matching the parent, but its charge-conjugation eigenvalue is $C_{\pi }{C}_{\rho }=-1$: the two distinct self-conjugate particles are not interchanged by C, so there is no additional $(-1{)}^L$ factor in this C eigenvalue. The decay is forbidden by strong and electromagnetic charge-conjugation conservation despite being above threshold. A weak contribution would require CP violation as well. At the strong/electromagnetic accuracy of the diagram classification,If extremely small weak CP-violating amplitudes are retained, ${\pi }^0{\rho }^0$ need not be exactly zero; the one-real-photon spin prohibition remains exact for ${\pi }^0\gamma$.

Angular momentum and parity. Work in the rho meson rest frame. Since each pion has zero spin, their combined intrinsic spin is $S=0$ and their total angular momentum is entirely relative orbital angular momentum:The parity selection rule for a two-body decay givesThe supplied $P_{\rho }=-1$ requires odd $L$, so both conservation laws agree:Here the statement about two-pion decays is understood as the dominant strong interaction channel; the assertion that it occurs literally always is an idealization, since other interactions can produce other channels.

Exchange symmetry. Pions are spin-zero bosons, so the complete two-particle wavefunction must have bosonic exchange symmetry, including its spatial and isospin factors. The unique spin-zero product is symmetric. Exchanging the particle slots reverses the relative position, so an $L=1$ spatial state has orbital parity $(-1{)}^L=-1$. Its isospin state must therefore be antisymmetric to make the product symmetric.

For two neutral pions, the internal charge state $|{\pi }^0{\pi }^0⟩$ is symmetric under exchange. There is no antisymmetric internal factor to compensate odd $L$. Equivalently, the odd-$L$ projection of two identical spin-zero neutral particles vanishes. ThereforeThis argument does not depend on treating isospin as an exact symmetry of nature.

All nine isospin states. In the following two-pion isospin decomposition, let $|ab⟩=|{\pi }^a{⟩}_1\otimes |{\pi }^b{⟩}_2$ denote ordered particle slots; each of $a,b$ is $+,0,-$. The second label in $|I,q⟩$ is the total electric charge in units of the elementary charge. For this pion system it equals the total isospin projection $I_3$.

Addition of angular momentum for two isospins of one gives $1\otimes 1=2\oplus 1\oplus 0$. Start from the highest-weight quantum state $|2,2⟩=|++⟩$ and apply the sum of the two ladder operators. The ladder recurrence for Clebsch-Gordan coefficients gives the normalized symmetric quintet:The normalized quantum state orthogonal to $|2,1⟩$ in the two-dimensional charge-one subspace is antisymmetric. Lowering it gives the antisymmetric triplet:The remaining charge-zero quantum state must be orthogonal to both displayed charge-zero states and annihilated by the total ladder operators. This gives the symmetric singlet:Together the quintet, triplet and singlet contain $5+3+1=9$ orthonormal states, accounting for the full $3\times 3$ charge-product space. Multiplying an entire isospin multiplet by a common quantum phase changes none of these physical conclusions.

Rho isospin and strong decay modes. Only the $I=1$ two-pion isospin decomposition is antisymmetric under exchange; $I=0,2$ are symmetric. The required $L=1$ spatial state therefore selects $I_{\pi \pi }=1$. Conservation of isospin by the strong interaction, in its usual approximation, impliesThe three charge states have the corresponding strong two-pion channelsThe ordered antisymmetric combinations above describe the internal part; multiplying by the odd spatial state gives the required symmetric complete bosonic state. The neutral $I=1$ state contains no $|00⟩$ component, in agreement with the independent identical-particle argument.

Charge conjugation. Charge conjugation preserves mass and reverses electric charge. A charged rho meson is therefore changed into a distinct state of opposite charge, so neither ${\rho }^{+}$ nor ${\rho }^{-}$ is itself a charge-conjugation eigenstate. The neutral state can be an eigenstate. Acting on its two-pion internal state interchanges the charge labels:The charged-pion quantum phase conventions cancel between the two particles. Thus the two-pion charge-conjugation parity is $C=(-1{)}^L=-1$, and conservation by the strong interaction fixesOnly ${\rho }^0$ among the individual charge states is a charge conjugation eigenstate.

For a neutral $c\bar{c}$ bound state, meson parity and charge conjugation give $P=(-1{)}^{L+1}$ and $C=(-1{)}^{L+S}$. In spectroscopic notation $n{}^{2S+1}{L}_J$, the integer $n$ labels radial excitation, $L$ is orbital angular momentum, $S$ the total constituent spin, and $J$ their total angular momentum. The surviving level diagram contains eight states:

The spin-dependent splittings distinguish the singlet and triplet levels and the three triplet P levels. The radial excitation changes neither parity nor charge conjugation.

The leading subprocess is $u\bar{d}\to W^{+}$. There are two beam assignments, and charge conjugation relates antiproton PDFs to proton PDFs. ThereforeIn a narrow-width leading-order normalization the prefactor is $\pi g^2/(12s)$ when $|V_{ud}|=1$; only the PDF dependence is needed for the fractions below.

Write the common sea PDF as $S=\bar{u}=\bar{d}=\bar{s}=\bar{c}=\Sigma /8$. The Valence and sea decomposition of proton-antiproton W production gives the three parton luminositiesThe factor two in the sea-sea term includes both beam assignments. The antiproton's valence antiquarks make valence-valence Annihilation possible.

Use the combinations $a_{\pm }=a\pm b$ at each momentum. Part a gives $[Q,a_{+}]=0$, $[Q,a_{-}]=-2a_{-}$, while their creation operators obey $[Q,a_{+}^{†}]=0$, $[Q,a_{-}^{†}]=2a_{-}^{†}$. Applying b givesThe creation operators transform by the Hermitian adjoint formulas with $e^{+2i\lambda }$. At $\lambda =\pi /2$, both exponentials are $-1$, soSubstituting into the fixed-time mode expansion of a free field of the complex scalar field givesThus $C$ exchanges particles and antiparticles, as required for charge conjugation. To see the charge reversal explicitly, the conserved charge $Q_{\mathrm{em}}={\int }_{\mathbf{p}}(a_{\mathbf{p}}^{†}{a}_{\mathbf{p}}-b_{\mathbf{p}}^{†}{b}_{\mathbf{p}})$ obeys $C{Q}_{\mathrm{em}}{C}^{†}=-Q_{\mathrm{em}}$. The generator $Q$ in the question is distinct from this charge.

The two occupation terms in the Hamiltonian are exchanged, with the same relativistic energy $E_{\mathbf{p}}$, and thereforeIn fact $[Q,H]=0$ because its mixing terms exchange equal-energy modes. With normal ordering, $:Q:=2{\int }_{\mathbf{p}}{c}_{-,\mathbf{p}}^{†}{c}_{-,\mathbf{p}}$ for $c_{-}=(a-b)/\sqrt{2}$. Then $C=e^{i\pi N_{-}}$ fixes the vacuum state and satisfies $C^2=1$. The construction is a bosonic species-swap generator.

Combine the two equal-mass Dirac fields into $\Psi =({\psi }^1,{\psi }^2{)}^T$. A constant unitary matrix $U$ acting on this flavour index commutes with the common Dirac Hamiltonian, so $\Psi \mapsto U\Psi$ gives the unitary flavour symmetry of degenerate Dirac fields $U(2)$. Its real Lie algebra has dimension four. A convenient Hermitian generator basis is $T^0=1_2/2$, $T^a={\tau }^a/2$, $a=1,2,3$, where ${\tau }^a$ are the Pauli matrices.

For each generator the canonical anticommutation relations giveThe calculation is exactly the one in b, since $T^A$ commutes with both the Dirac gamma matrices and the common fermion mass matrix. ThusExplicitly their densities areThe two off-diagonal conserved currents are why counting only the individual fermion numbers misses two generators. The singlet generates $U(1)$; the other three generate special unitary group $SU(2)$. Unequal masses would break the off-diagonal conservation laws.

Scope of the count. Four is the count of currents of the displayed ${\bar{\psi }}^A{\gamma }^{\mu }{T}_{AB}{\psi }^B$ form. The completely free system also has an enhanced internal symmetry of free Dirac fields if mixing with charge-conjugate fields is admitted. For $m\ne 0$, write its two Dirac fields as four equal-mass Majorana spinors ${\chi }^I$. Their orthogonal group $O(4)$ rotations have six continuous generators, with conserved currents proportional to $i{\bar{\chi }}^I{\gamma }^{\mu }{\chi }^J$ for $I<J$; the four $U(2)$ currents form a subgroup. This larger count is relevant if “similar” means all local internal currents of the free system, without preserving a chosen fermion number. If $m=0$, the usual number-preserving chiral flavour symmetry of massless Dirac fields is $U(2{)}_L\times U(2{)}_R$ with eight currents; allowing charge conjugation mixing enlarges it to $U(4)$ with sixteen. These counts concern internal currents, excluding spacetime Noether currents.

First conjugate the Dirac equation: $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. A matrix $\Gamma$ taking ${\psi }^{\ast }$ into another solution of the same Dirac equation must obeyBecause ${\gamma }^1$ anticommutes with the real ${\gamma }^0,{\gamma }^2$ and commutes with the purely imaginary ${\gamma }^1$, a unitary choice isAny constant overall phase is equally valid. To verify invariance of the action, rather than only of its equation, keep the fermion exchange signs. Define the charge-conjugation matrix $C=\Gamma {\gamma }^0={\sigma }_2$. It obeysFor a classical Grassmann field, interchanging two components contributes a minus sign; equivalently use normal ordering for quantum fermionic operators. ThusandThe last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the kinetic term and the Dirac mass term therefore leave $S$ invariant under charge conjugation. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the fermion statistics is essential.

Independent global phase symmetries would rotate $\varphi \mapsto e^{-i\alpha }\varphi$ and $\Phi \mapsto e^{-i\beta }\Phi$. The cubic interaction vertex preserves only $2\alpha +\beta =0$ modulo $2\pi$. The continuous internal symmetry of a classical field theory is therefore the circle groupNormalize the Noether charge so that $q_{\varphi }=1$. Then $q_{\Phi }=-2$, $q_{\bar{\varphi }}=-1$ and $q_{\bar{\Phi }}=2$. The Noether current iswhere $A\overleftrightarrow{\partial }B=A\partial B-(\partial A)B$. For example, the Euler-Lagrange field equations givewhich cancel with the relative coefficient $-2$. The quartic interaction vertex separately preserves both original phase rotations.

There is also a discrete charge conjugation symmetry $(\varphi ,\Phi )\mapsto ({\varphi }^{†},{\Phi }^{†})$ for these real couplings. Including it, the internal symmetry is $U(1)⋊{\mathbb{Z}}_2\cong O(2)$, with charge conjugation reversing the Noether charge. The connected symmetry and its conserved current are the $U(1)$ above.

With the conventional particle labels, the allowed cubic decay channels areEach has one tree-level Feynman diagram, consisting of one cubic Feynman vertex. The final identical-leg contractions supply $2!$ and cancel the $1/2!$ in the Lagrangian density.

Taken literally with these labels, the printed $\Phi \to \varphi \varphi$ channel is forbidden: its initial Noether charge is $-2$ and its final Noether charge is $+2$. Its decay amplitude is zero, at every order preserving the global phase symmetry. If the source uses “$\Phi$” to name the heavy antiparticle instead, its intended answer is the nonzero decay amplitude $-h$ for $\bar{\Phi }\to \varphi \varphi$. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.

The neutral pion has positive charge conjugation eigenvalue and its spin-one isospin half-turn has eigenvalue minus one. Hence the G parity eigenvalue is minus one. Since G parity commutes with isospin, both charged members have the same value. Charged-state phase changes alter individual charge-conjugation and rotation formulas together, leaving this eigenvalue unchanged.

The rho meson is an isospin-one vector meson. Its neutral member has charge conjugation value minus one, so all three charge states have G parity plus one. The two-pion decay is allowed by G parity, whereas a purely strong isospin-conserving three-pion decay is excluded.

For a charged spin-zero complex scalar field, time reversal preserves charge and can include an intrinsic phase. In its free-field mode expansion of a free field, antiunitarity conjugates the plane-wave coefficients. Relabel spatial momentum to its negative and compare the annihilation and creation terms: $\hat{T}a(\mathbf{p}){\hat{T}}^{-1}={\eta }_{T}a(-\mathbf{p})$ and $\hat{T}{b}^{†}(\mathbf{p}){\hat{T}}^{-1}={\eta }_T{b}^{†}(-\mathbf{p})$. Particle annihilation remains annihilation; unlike charge conjugation, this operation does not exchange particle and antiparticle operators.

For a neutral pion pair in a state of definite relative orbital angular momentum $L$, charge conjugation interchanges ${\pi }^{+}$ and ${\pi }^{-}$. Relabelling the particle slots reverses the relative momentum and contributes $(-1{)}^L$. The individual charged-pion phase conventions cancel in the pair. Thus a neutral vector rho meson decaying strongly to a pion pair has $L=1$ and $C=-1$. A charged rho meson is taken to the state of opposite electric charge, so it is not itself a charge-conjugation eigenstate.

For charged current production $u\bar{d}\to W^{+}$, charge conjugation of the antiproton distributions gives $u(x_{+})d(x_{-})+\bar{d}(x_{+})\bar{u}(x_{-})$. With a common sea quark distribution $S$, the valence quark terms are $u_v{d}_v$, the mixed terms are $u_{v}S+S{d}_v$, and the two sea assignments give $2S(x_{+})S(x_{-})$. Omitting the second beam assignment undercounts the pure sea contribution by two.