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Furry's theorem states that a charge-conjugation-invariant vacuum has vanishing correlation functions containing an odd number of electromagnetic currents or photon fields. Consequently amplitudes with an odd number of external photons and no charged external state vanish.

A right-handed neutrino is a singlet of the entire Standard Model gauge group. Its same-chirality bilinear with the field obtained by charge conjugation therefore admits the displayed renormalizable Majorana mass term, in addition to a Higgs-induced Dirac mass term. It violates lepton number by two units. Analogous terms for electrically charged Standard Model fermions carry nonzero electric charge; quarks also have nontrivial color charge. The presence of a right-handed field permits mass but does not force a nonzero mass parameter.

The product of charge conjugation and a half-turn in isospin is G parity. Both conjugations act on the isospin generators by $(J_1,J_2,J_3)\mapsto (-J_1,J_2,-J_3)$, so their product commutes with every generator. The Schur lemma therefore gives a common G-parity value throughout each irreducible isospin multiplet on which the operator acts internally. It can classify a charged multiplet even though individual charged states are not eigenstates of charge conjugation.

For integer isospin $I$ and neutral-state charge conjugation value $\eta$, the entire isospin multiplet has $G=\eta (-1{)}^I$. Both charge conjugation and the second-axis half-turn reverse the first and third generators, so their product commutes with the full isospin algebra. Evaluate it on the $M=0$ state and use ladder connectivity to obtain the value for every $M$.

For an ordinary quark-antiquark meson, opposite intrinsic parities and orbital parity give $P=(-1{)}^{L+1}$. For a self-conjugate flavor state, charge conjugation exchanges the two constituents. Reversal of their relative momentum gives $(-1{)}^L$, spin exchange gives $(-1{)}^{S+1}$, and reordering the two fermionic creation operators gives one extra minus sign. Their product is $C=(-1{)}^{L+S}$. Electrical neutrality alone does not imply that a state is an eigenstate of charge conjugation; charged mesons have no individual C eigenvalue.

Use the kaon CP eigenstates convention $CP|K^0⟩=-|{\bar{K}}^0⟩$, matching kaon CP mixing parameter. Then $|K_1⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$ is CP even, while $|K_2⟩=(|K^0⟩+|{\bar{K}}^0⟩)/\sqrt{2}$ is CP odd. With CPT symmetry and a common small mixing parameter, the physical states can be writtenHere $S,L$ mean short and long mean lifetimes; the kaon CP mixing parameter $ϵ$ depends on the flavor-state phase convention. A positive flavor-exchange convention instead makes the symmetric flavor combination CP even, as explained in exchange-positive CP convention for neutral kaon mixing. Mixing these conventions would reverse the interpretation of the written sums.

Why does the two-pion final state test CP symmetry? A spin-zero kaon decaying into two spin-zero pions has relative orbital angular momentum $\ell =0$ by angular momentum conservation. Two negative intrinsic parities multiply to plus, and $(-1{)}^{\ell }=1$. Charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$; the $\ell =0$ state is even under this exchange. Hence this two-pion state has CP eigenvalue $+1$. If a decay transition operator $T$ preserves CP symmetry, inserting CP on both sides of $⟨\pi \pi |T|K_2⟩$ changes its sign without changing its value: that decay amplitude must vanish.

Assume for illustration that CP violation occurs only through the state admixture and the decay transition preserves CP symmetry. Put $A_1=⟨\pi \pi |T|K_1⟩$. ThenNeglecting the tiny difference in two-body Lorentz-invariant phase space between the physical masses gives ${\Gamma }_L^{\pi \pi }/{\Gamma }_S^{\pi \pi }\simeq |ϵ{|}^2$. A branching fraction, however, divides a partial decay width by the total decay width. With $\hslash =1$, ${\tau }_i=1/{\Gamma }_i$, soFor chosen values $|ϵ|=0.01$ and ${\tau }_L/{\tau }_S=100$, the branching fraction ratio is $0.01$, whereas the partial decay width ratio is $0.0001$. This is original arithmetic, not a historical fit. The longer observation time amplifies the integrated fraction relative to the instantaneous partial decay rate.

A neutral self-conjugate pseudoscalar with $C=+1$ cannot decay through charge-conjugation-conserving interactions into a neutral $C=-1$ vector and a neutral $C=+1$ pion. The two final particles are distinct and each maps to itself under charge conjugation, so their total C eigenvalue is the product, without an exchange factor $(-1{)}^L$. Their relative orbital angular momentum still enters their total parity. For $0^{-}\to 0^{-}{1}^{-}$, $L=1$ gives a CP-violating transition if a weak amplitude is retained.

The omega meson is an isospin-zero vector meson with charge conjugation value minus one and consequently G parity minus one. Three-pion decay is compatible with the isospin-symmetric strong interaction, whereas two-pion decay violates its G-parity rule.

A Feynman diagram with one independent closed cycle and hence one independent loop momentum. An $n$-vertex polygon of Dirac propagators in four dimensions has superficial degree of divergence $4-n$ before symmetry cancellations. For a fermion loop, the closed fermion loop sign is separate from any symmetry factor or pairwise charge conjugation cancellation.

In the Dirac representation of the gamma matrices, the parity symmetry in quantum field theory action ${\psi }_P(x)={\gamma }^0\psi (x_P)$ and the charge conjugation action ${\psi }_C=i{\gamma }^2{\psi }^{\ast }$ obey $({\psi }_P{)}_C(x)=-{\gamma }^0{\psi }_C(x_P)$, since ${\gamma }^0$ is real and anticommutes with ${\gamma }^2$. At rest, positive-frequency Dirac spinors have nonzero upper components and negative-frequency Dirac spinors have nonzero lower components; they are respectively $+1$ and $-1$ eigenvectors of ${\gamma }^0$. With the common conventional real intrinsic-parity phase, particle and antiparticle therefore have opposite intrinsic parity. The overall sign is conventional, whereas their relative sign is fixed.

Use the source's charge-conjugation matrix convention $C{\gamma }^{\mu }{C}^{-1}=-({\gamma }^{\mu }{)}^T$. It impliesA trace equals the trace of its matrix transpose, which reverses the order of all factors. Inserting the displayed identities givesThere is one minus sign from each of the three photon interaction vertices, and none from the transformed Dirac propagators. In the last line, change loop momentum to ${\ell }^{\prime }=-\ell -k_1$. Since $k_1+k_3=-k_2$, its three Dirac propagators become $K({\ell }^{\prime }+k_1)$, $K({\ell }^{\prime }-k_2)$, $K({\ell }^{\prime })$, exactly those of the reversed orientation. HenceThis is Furry's theorem. The regularization in quantum field theory matters: the individual Feynman integrals are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common dimensional regularization preserves charge conjugation and the vector Ward identity, making the change of variables legitimate. No chirality matrix occurs, so the axial triangle anomaly is not involved.

For an $n$-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by $p\mapsto -p$ and contributes $-1$ for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeysFor odd $n\ge 3$, the $(n-1)!$ cyclic orders pair into opposite orientations and cancel. For $n=1$, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor $(-1{)}^n$. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

All gamma matrices in this basis are real, and the Lorentz transformation $S(\theta )$ is real. Therefore complex conjugation commutes with the connected Spinor representation of the Lorentz group. With a suitable phase convention, charge conjugation is simply ${\psi }^c={\psi }^{\ast }$. The Majorana spinor condition ${\psi }^c=\psi$ becomesThe reality constraint is preserved by Lorentz transformations because $(S{\psi }_M{)}^{\ast }=S{\psi }_M$. For fermion fields, real components mean self-conjugate Grassmann variables or self-adjoint field components, rather than ordinary commuting real numbers.

Moreover, $\Gamma ={\sigma }_3$ is real, so charge conjugation commutes with the chiral projectors. The Majorana spinor and Weyl spinor conditions can therefore be imposed simultaneously. The Majorana-Weyl spinors areThe independent real component counts areThese are component counts before imposing an equation of motion. In particular the simultaneous reality and chirality constraints are compatible in this spacetime dimension, unlike the familiar four-dimensional case.

Use the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplying the block gamma matrices and using the Pauli matrix multiplication law givesThus these matrices satisfy the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$. Since partial derivatives commute, the antisymmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ drops out of its contraction with ${\partial }_{\mu }{\partial }_{\nu }$. Multiplying the Dirac equation on the left by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesTherefore every component of the Dirac spinor obeys the Klein-Gordon equation with mass $m$.

For Lorentz covariance, take $x^{\prime }=Lx$ and ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$, with constant $L$ and $S$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. The given intertwining relation then impliesThis proves invariance of the Dirac equation under the joint coordinate and spinor transformation.

The infinitesimal identity follows directly from the spinor Lorentz generator commutator. To derive it, first move ${\gamma }^{\rho }$ through a two-matrix product using the Clifford algebra:Subtract the expression with $\mu$ and $\nu$ interchanged and multiply by $i/2$. ThusWriting $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$ gives $S^{-1}=1+\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }+O({\omega }^2)$, and henceThe last step uses ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, including the metric signs in raising the first index.

For parity symmetry in quantum field theory, set $y=x_P=(t,-\mathbf{x})$. The time derivative of $\psi (y)$ is unchanged and each spatial derivative changes sign. Also ${\gamma }^0$ commutes with itself and anticommutes with every ${\gamma }^j$. ConsequentlyThis verifies the specified parity symmetry in quantum field theory action explicitly.

For charge conjugation, put $B=i{\gamma }^2$, so ${\psi }_C=B{\psi }^{\ast }$. In the Dirac representation of the gamma matrices, ${\gamma }^0,{\gamma }^1,{\gamma }^3$ are real and ${\gamma }^2$ is imaginary. The Clifford algebra therefore gives $B{\gamma }^{\mu \ast }{B}^{-1}=-{\gamma }^{\mu }$ for all four indices. Complex conjugation of the Dirac equation, followed by multiplication by $B$, yieldsThis is the antilinear charge-conjugation intertwiner property. No electromagnetic background is being held fixed in this free-field calculation.

Apply the specified parity symmetry in quantum field theory action before charge conjugation. Since ${\gamma }^0$ is real and $B{\gamma }^0=-{\gamma }^{0}B$,Thus the induced parity action on the charge-conjugate spinor has the opposite sign. In particular, if a spinor has intrinsic parity $\eta =\pm 1$, its charge conjugate has intrinsic parity $-\eta$. Equivalently, at rest the positive-frequency Dirac spinors occupy the upper two components and have ${\gamma }^0$ eigenvalue $+1$, whereas the negative-frequency Dirac spinors occupy the lower two components and have eigenvalue $-1$. The corresponding particle and antiparticle creation operators inherit these opposite signs under the specified parity convention. Hence electrons and positrons have opposite intrinsic parities; conventionally one assigns $+1$ to the electron and $-1$ to the positron. The common real overall parity sign may be reversed without changing this relative sign, as in opposite intrinsic parities of a Dirac particle and antiparticle.

Let $\epsilon =i{\sigma }_2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so ${\epsilon }^{-1}=-\epsilon$. Direct multiplication of the Pauli matrices gives $\epsilon {\sigma }_i^{\ast }\epsilon ={\sigma }_i$, equivalently $\epsilon {\sigma }_i^{\ast }{\epsilon }^{-1}=-{\sigma }_i$. For $U=\exp (i{\theta }_i{\sigma }_i/2)\in SU(2)$ this impliesTherefore ${\varphi }^c=\epsilon {\varphi }^{\ast }$ transforms as the same weak fundamental representation doublet as $\varphi$, while complex conjugation reverses its hypercharge to $-1/2$. This is the Higgs conjugate doublet. The Yukawa coupling product $\bar{L}{\varphi }^c{u}_R$ is an $SU(2)$ singlet, and its hypercharges add toIt is consequently gauge-invariant and can supply an up-quark mass.

In unitary gauge, ${\varphi }^c=((v+H)/\sqrt{2},0{)}^T$. Writing $y_u$ for the full arbitrary coefficient of the Yukawa coupling term, its physical form isA one-generation field rephasing makes the mass positive, and its Higgs boson coupling is $m_u/v$. The printed Yukawa expression has an additional trailing $m$ after $R^{+}$ which is never defined. If it denotes a constant, absorb it into $y_u=f^{+}m$; in the conventional dimensionless Yukawa notation without that factor, $y_u=f^{+}$. It does not affect the gauge-transformation argument.

For three families, use a singular value decomposition of the two mass matrices: $U_{uL}^{†}{M}_u{U}_{uR}=D_u$ and $U_{dL}^{†}{M}_d{U}_{dR}=D_d$, with positive diagonal entries. The original left-handed weak current then becomes a current between mass eigenstates withA complex $3\times 3$ matrix has eighteen real parameters, and $U^{†}U=I$ imposes nine independent real conditions, leaving nine. A generic unitary matrix can be built from three complex plane rotations and three diagonal phases: each plane rotation supplies one angle and one phase. Thus these are three rotation angles and six phases. Rephase each of the six Dirac quark mass eigenfields, applying the same phase to its two chiralities so the diagonal masses stay real. A common phase changes no entry of $V$, leaving five effective rephasings. Removing five of the six phases leaves three angles and one physical phase. This CKM parameter counting assumes generic nondegenerate masses; degeneracies give additional basis freedom.

Exactly massless neutrinos have an arbitrary unitary flavor rotation that changes no mass term. After diagonalizing the charged-lepton mass matrix, choose the neutrino basis with the same left-handed rotation. The apparent lepton charged-current mixing matrix becomes the identity. Thus mixing of exactly massless neutrinos is removable; it is not an analogue of physical CKM mixing.

For the requested CP symmetry calculation, write $j_{ij}^{\mu }={\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)d_j$. The supplied charge conjugation transformations, with the fermion-reordering sign, giveFor $\Gamma ={\gamma }^{\mu }(1-{\gamma }^5)$, use $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$ and the given gamma matrix relation to obtainUnder parity, the current acquires the Lorentz parity matrix and $1+{\gamma }^5$ becomes $1-{\gamma }^5$. Under charge conjugation the charged vector also changes sign and becomes its conjugate. The two minus signs cancel, and the parity matrices cancel in the contracted operator. ConsequentlyWith $k=g/(2\sqrt{2})$, the interaction is $-k\sum (V_{ij}{O}_{ij}+V_{ij}^{\ast }{O}_{ij}^{†})$. Thereforewith the transformed density evaluated at $x_P$. This is the CP transformation of a charged quark current.

Physical noninvariance requires a phase that cannot be removed by quark rephasings. The invariant quartetis the Jarlskog invariant, where $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$ in the standard three-angle parametrization. If $J\ne 0$, no allowed row/column phases can make the matrix real, and the charged-current interaction violates CP symmetry. For example all three angles $\pi /4$ with $\delta =\pi /2$ give $J=1/(8\sqrt{2})\ne 0$.

The printed noninvariance claim is conditional on a nonzero physical phase. Arbitrary Yukawa matrices do not force it: $V=I$, or any real CKM matrix, gives a CP-invariant charged-current interaction. The correct statement is generic CP violation when the irreducible phase is nontrivial, as explained by quark rephasing and CP conservation.

Finally a right-handed neutrino has gauge representation $(1,1{)}_0$. Its Lorentz-invariant same-chirality bilinear with the field obtained by charge conjugation is therefore a gauge singlet and permits a Majorana mass term, conventionallyThe factor $1/2$ is the usual convention for a field paired with itself; it may be absorbed into the printed coefficient. A Higgs Yukawa interaction can independently give a Dirac mass term. The Majorana term violates lepton number by two units but preserves the gauge symmetries. Corresponding terms for the other charged Standard Model fermions carry nonzero electric charge, and quarks also carry nontrivial color charge. Thus this gauge-invariant right-handed neutrino Majorana mass is allowed while those charged-fermion analogues are forbidden. Merely introducing a right-handed neutrino permits mass; vanishing mass coefficients would still leave it massless.

Choose the charge conjugation matrix so that $C^T=-C$ and $({\gamma }^m{)}^T=-C{\gamma }^m{C}^{-1}$, and write $\bar{\lambda }={\lambda }^{T}C$. Since the components are odd Grassmann variables,Thus a symmetric matrix gives zero, whereas a nonzero antisymmetric matrix can give a nonzero bilinear. The four-dimensional Majorana bilinear vanishing criterion follows by transposing an antisymmetrized product and reversing its $n$ factors:For $n=0,1,2,3$, the symmetry types are respectively antisymmetric, symmetric, symmetric and antisymmetric. These are the same-spinor selection rules of the Majorana spinor; they would be reversed for a commuting spinor. The vanishing choices are (a) and (c).

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.

Start with the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. The equation for the Dirac adjoint is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by the charge-conjugation matrix C to obtainFor this to be the original Dirac equation multiplied by minus one, the necessary matrix identity isUsing $C^{†}=C^{-1}$, the charge conjugation of the Dirac adjoint follows directly:The final step uses $({\gamma }^0{)}^T{C}^{-1}=-C^{-1}{\gamma }^0$ and $({\gamma }^0{)}^2=1$.

Let $j_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, $j_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, and $j_L=j_V-j_A$. For anticommuting fields, or normal-ordered operator bilinears, the minus sign in the transformed Dirac adjoint cancels the minus sign from interchanging the two fields. The charge conjugation of fermion bilinears is thereforeThe matrix identity just proved gives $j_V\mapsto -j_V$. Also $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$, obtained by transposing its four gamma matrices and reversing their order. Henceso $j_A\mapsto j_A$ and $j_L\mapsto -j_R$, where $j_R=j_V+j_A$. Since the external vector V is odd under charge conjugation, $j_L\cdot V$ becomes $j_R\cdot V$ and is generally not invariant.

For parity symmetry in quantum field theory, write $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. The given spinor transformation gives $j_V\mapsto P{j}_V$, $j_A\mapsto -P{j}_A$, and $V\mapsto PV$, with arguments evaluated at the reflected point. The contracted interaction again becomes $j_R\cdot V$. Each of parity and charge conjugation interchanges the two chiral contractions; applying both restores $j_L\cdot V$. This particular neutral current interaction is CP invariant, although neither P nor C is separately a symmetry. This conclusion uses the stipulated C-odd vector transformation; it is not a claim that all interactions of the Standard Model preserve CP symmetry.

For time reversal of a Dirac field, the gamma matrix adjoint and transpose identities giveWith $B={\gamma }^{5}C$, anticommutation of ${\gamma }^5$ with every gamma matrix yieldsEquivalently, $\boxed{B({\gamma }^{0\ast },-{\mathbf{\gamma }}^{\ast })B^{-1}=({\gamma }^0,\mathbf{\gamma })}$. Since time reversal is an antiunitary operator, the factor i in ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ must also be conjugated. ThusUsing the stated transformations of the field and its Dirac adjoint, the transformed bilinear consequently isThe external vector has the same component signs. Their contraction is unchanged, proving time reversal of a real neutral chiral vector interaction. A real coupling is understood; an antiunitary operator also conjugates coefficients, and relative complex weak phases need not satisfy this invariance.

A charged weak box contribution to kaon mixing is shown below. The incoming $d\bar{s}$ and outgoing $s\bar{d}$ connect through two charged W bosons and two internal up-type quarks. The black arrows are fermion flow, so the lower line's arrows oppose the momentum of the corresponding antiquark. The blue arrows specify the direction of positive charge carried by the charged propagators. Each vertex is a charged weak current vertex, with the appropriate CKM matrix element; the internal flavors are summed over $u,c,t$.

Use precisely the printed exchange-positive CP convention for neutral kaon mixing. In the flavor basis, CP symmetry acts by $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Write the restricted effective matrix asThen $SMS=\left(\begin{array}{cc}a& c\\ b& a\end{array}\right)$, so CP invariance of this mixing operator is exactly $b=c$. Therefore CP noninvariance in this neutral-kaon mixing block means $M_{12}\ne M_{21}$. Violation elsewhere in a larger Hamiltonian would not, by itself, prove an inequality for this particular block.

For neutral-kaon mass matrix diagonalization, choose correlated roots $r^2=b$, $s^2=c$, and eigenvalues ${\lambda }_{\pm }=a\pm rs$. The vectors $(r,s{)}^T$ and $(r,-s{)}^T$ are the corresponding eigenvectors, as direct multiplication verifies. Taking orthonormal flavor states and writing the printed CP eigenstates as $K_1=(K^0+{\bar{K}}^0)/\sqrt{2}$ and $K_2=(K^0-{\bar{K}}^0)/\sqrt{2}$, one findswhere the kaon CP mixing parameter isOverall phases have been absorbed. The kaon mixing square-root branch convention must be chosen continuously so that $r=s$ in the CP-conserving limit for the eigenstate labeled plus. These formulas are exact in the nondegenerate case with $r+s\ne 0$; the requested small admixtures follow if $|ϵ|\ll 1$. Smallness is an additional weak-CP-violation assumption, not a consequence of $b\ne c$ alone. A Hermitian mass operator has $c=b^{\ast }$; the same algebra also applies to the effective non-Hermitian operator that includes decays, where the two eigenstates need not be orthogonal. At a fully degenerate $b=c=0$, a CP eigenbasis can simply be chosen.

Use a finite-dimensional unitary representation of SU(2) and units $\hslash =1$, so the isospin generators are Hermitian and $I_{+}^{†}=I_{-}$. Their commutators areOn the normalized highest-weight vector, the given quadratic Casimir operator identity gives ${\mathbf{I}}^2|I,I⟩=I(I+1)|I,I⟩$. Since the quadratic Casimir operator commutes with every generator, this eigenvalue is unchanged down the ladder. Each application of $I_{-}$ lowers $M$ by one. For a normalized descendant,Positivity first gives $I\ge 0$. In a finite-dimensional representation the descending ladder must terminate; the norm vanishes only at $M=-I$. Thus the number of lowering steps is $2I$, a nonnegative integer, andFor $M>-I$, define the normalized descendant by dividing $I_{-}|I,M⟩$ by its positive norm. This fixes its relative phase and givesActing with $I_{+}$ and using $I_{+}{I}_{-}={\mathbf{I}}^2-I_3^2+I_3$ provesThe resulting isospin multiplet has $2I+1$ states. More explicitly,The factorial arguments are integers, even for half-integer $I$.

Put $R(\theta )=e^{-i\theta I_2}$. Differentiation and the commutators givewhile $I_2$ is unchanged. Hence at the half-turn,The half-turn preserves $I$ and reverses $M$, so $R|I,M⟩={\alpha }_M|I,-M⟩$. Since $R{I}_{+}{R}^{-1}=-I_{-}$, comparison of the two ways to rotate a raised state gives ${\alpha }_{M+1}=-{\alpha }_M$.

To fix the remaining phase, expand the supplied highest-state rotation identity as a finite sum. The $k$th term isAs $\theta \uparrow \pi$, all terms vanish except $k=2I$. The product of the lowering coefficients gives $I_{-}^{2I}|I,I⟩=(2I)!|I,-I⟩$. The surviving coefficient is therefore one, fixing ${\alpha }_I=1$. The half-turn phase of an SU(2) multiplet isIn particular the recurrence and highest-state phase are consistent with $R^2=(-1{)}^{2I}$.

Charge conjugation is a linear unitary symmetry. If $J_i=C{I}_i{C}^{-1}$, conjugation preserves the commutator and the scalar $i$, soThus the conjugated generators obey the same SU(2) Lie algebra. The charge conjugation action on $(I_1,I_2,I_3)$ is exactly the same pair of sign reversals as $R$. Their product $G=CR$ consequently satisfies $G{I}_i{G}^{-1}=I_i$ for all $i$. For integer $I$, its value on the neutral member isCommuting $G$ through the ladder operators transfers that value to every state in the same isospin multiplet. Therefore the G-parity of an isospin multiplet is