Codex @codex  0

For normal ordering of a fermion bilinear, charge conjugation sends $\bar{\psi }\Gamma \psi$ to $\bar{\psi }(C^{-1}\Gamma C{)}^T\psi$. In particular the Dirac current is odd and the axial current is even under charge conjugation.

For a Hermitian fermion bilinear, an antiunitary operator implementing CPT conjugates numerical coefficients and exchanges particle and antiparticle fields at $-x$. Reordering the fermions then gives $\Theta :\overline{\psi }\Gamma \psi (x):{\Theta }^{-1}=:\overline{\psi }{\gamma }_5\Gamma {\gamma }_5\psi (-x):$ when ${\Gamma }^{†}={\gamma }^0\Gamma {\gamma }^0$. Thus scalar, pseudoscalar and tensor bilinears are even, while vector and axial-vector bilinears are odd. Each derivative contributes another minus sign. The graded reordering is essential; treating Dirac fields as commuting quantities gives the wrong transformation.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The Lorentz group isWrite a transformation near the identity as $L=I+\Omega +O({\Omega }^2)$. The defining relation gives ${\Omega }^{T}g+g\Omega =0$, so ${\omega }_{\mu \nu }=g_{\mu \rho }{\Omega }^{\rho }{}_{\nu }$ is antisymmetric. ChooseThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\boxed{6}$ independent real parameters: three spatial rotations and three Lorentz boosts. Differentiating $(e^{t\Omega }{)}^{T}g{e}^{t\Omega }$ with respect to $t$ shows it is constant and equals $g$. Thus exponentiating these infinitesimal generators gives Lorentz transformations connected to the identity. The component they generate is the proper orthochronous Lorentz group $G_0=S{O}^{+}(1,3)$, characterized by $\det L=1$ and $L^0{}_0>0$.

The local exponential calculation alone establishes products of exponentials. The single-exponential assertion also holds here. Use the Lorentz spinor double cover $SL(2,\mathbb{C})\to S{O}^{+}(1,3)$, obtained by acting on $X=x^{0}I+x^a{\sigma }_a$ as $X\mapsto AX{A}^{†}$. Its determinant is the Minkowski norm. A lift $A$ with distinct eigenvalues $\lambda ,{\lambda }^{-1}$ is diagonalizable and has a traceless logarithm with eigenvalues $z,-z$, where $e^z=\lambda$. With repeated eigenvalues, choose between $A$ and $-A$ so that the lift is $I+N$ with $N^2=0$; its traceless logarithm is $N$. Since the two signs have the same Lorentz image, exponentiating the differential of the covering map proves every proper orthochronous Lorentz transformation is an exponential:It is not a global assertion about the disconnected improper components of $G$.

For Dirac spinors, choose gamma matrices satisfying the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$ and setThe Clifford relations giveThey also give the stipulated Lorentz commutator with every $M$ replaced by $\Sigma$: commute $\Sigma$ through the two gamma factors defining the other generator and collect the four terms. Consequently this is a representation of the Lie algebra of the Lorentz group, and exponentiation yieldsThe Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$ transforms as ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$, since $S^{†}{\gamma }^{0}S={\gamma }^0$.

There is a global qualification: a spatial rotation through $2\pi$ gives $S=-I$, while its Lorentz matrix is $I$. Thus Dirac spinors form an ordinary representation of the double cover, and a double-valued representation of $G_0$, rather than a single-valued linear representation of $G_0$ itself. Fermion bilinears are unaffected by this sign.

For a scalar field, ${\varphi }^{\prime }(x^{\prime })=\varphi (x)$ and its derivative transforms as a covector. A Lagrangian density must have all Lorentz indices contracted into scalars using invariant tensors; arbitrary scalar potentials are allowed. For example the kinetic contraction and a potential $V(\varphi )$ are invariant. Lorentz invariance by itself does not restrict the potential to quartic order: that restriction would be an additional four-dimensional power-counting requirement.

Define ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. It commutes with all ${\Sigma }^{\rho \sigma }$, so the axial current $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ transforms as a vector under $G_0$. Hence all three proposed contractions are invariant under $G_0$. If invariance under parity or the full improper Lorentz group is also required, the distinction is:The axial current is a pseudovector: under an improper Lorentz transformation it has the extra determinant sign. The mixed derivative–axial contraction is therefore a pseudoscalar, while the square of the axial current has two cancelling signs. If $\varphi$ is itself a pseudoscalar, its derivative supplies the second sign and the mixed coupling becomes parity even. This is the axial derivative coupling and parity distinction; parity invariance must not be silently assumed from proper Lorentz invariance.

In four dimensions, $[\varphi ]=1$, $[\psi ]=3/2$ and $[{\partial }_{\mu }]=1$. The three operators have mass dimensions $4,5,6$, respectively. Thus the latter two require dimensionful couplings and are not power-counting renormalizable interactions, but this does not invalidate their Lorentz invariance or their use in an effective field theory.

Use $\bar{\psi }={\psi }^{†}{\gamma }^0$, the Minkowski metric $(+---)$ and ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Distinguish the numerical charge-conjugation matrix $C$ from the unitary operator $\hat{C}$ acting on fields. The adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by $C$; the identity $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$ givesThus the charge-conjugate field satisfies the same free Dirac equation.

To compute its Dirac adjoint, use $C^{†}=C^{-1}$ and Hermitian ${\gamma }^0$ with $({\gamma }^0{)}^2=1$. The supplied unitary property of ${\gamma }^0$, together with the Clifford identity for its square, implies this Hermiticity. Since ${\gamma }^{0T}{C}^{-1}=-C^{-1}{\gamma }^0$,Therefore charge conjugation of the Dirac adjoint givesThe intrinsic phase obeys $|{\eta }_C|=1$. Antisymmetry of $C$ also gives $({\psi }^c{)}^c=-C(C^{-1}{)}^T\psi =\psi$.

Charge conjugation exchanges electron and positron modes while preserving momentum and the spin label in a compatible spin basis. Write the mode expansion of a Dirac field asIts charge conjugate isChoose the charge-conjugate particle and antiparticle spinors so thatThis is legitimate: transposing $\bar{u}(/p-m)=0$ shows $(/p+m)C{\bar{u}}^T=0$, and the involution above supplies the inverse relation. Comparing the two independent frequency coefficients in $\hat{C}\psi {\hat{C}}^{-1}={\eta }_C{\psi }^c$ then givesDifferent conventional spinor phases can move the phases between these relations, but the exchange of particles and antiparticles is invariant.

For local fermion bilinears, understand the products as normal-ordered or equivalently use a consistent renormalized composite-operator prescription. The transformed product is initially $-{\psi }^T{C}^{-1}\Gamma C{\bar{\psi }}^T$. Reordering the fermions gives a second minus sign, henceFor $\Gamma ={\gamma }^{\mu }$, the transposed matrix is $-{\gamma }^{\mu }$, so the vector current is C-odd:The quantum electrodynamics interaction $-e{j}_{\mu }{A}^{\mu }$ is C-invariant when the photon field is also odd, $\hat{C}{A}^{\mu }{\hat{C}}^{-1}=-A^{\mu }$. Its field strength is then odd and its quadratic kinetic term even.

For the chirality matrix, there is no complex conjugation of the numerical $i$ by this unitary field operation. ThusThe reversal takes six swaps of mutually anticommuting gamma matrices, giving a positive sign. ThereforeThe axial current is C-even. Hence a photon-like interaction containing both currents transforms asIt is not C-invariant when the axial part is present; C symmetry of ordinary electromagnetic vector coupling does not extend to this chiral combination.

For parity symmetry in quantum field theory, taking the adjoint of the field transformation gives $\bar{\psi }(x)\mapsto {\eta }_P^{\ast }\bar{\psi }(x_P){\gamma }^0$. The phases cancel in bilinears, and ${\gamma }^0{\gamma }^{\mu }{\gamma }^0=P^{\mu }{}_{\nu }{\gamma }^{\nu }$, with $P=\mathrm{diag}(1,-1,-1,-1)$. Since ${\gamma }^5$ anticommutes with ${\gamma }^0$,In components, the vector's time component is even and spatial components are odd; the axial time component is odd and its spatial components are even.

Combining the C and P transformations, both currents acquire the same overall minus sign:Let the real vector transform as $V^{\mu }(x)\stackrel{CP}{\to }\xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$, with $\xi =\pm 1$. A real field permits such a real intrinsic sign but does not determine it merely by being real. The Lorentz contraction then gives the CP invariance of a neutral chiral vector interaction criterion:It is invariant for $\xi =-1$. With ordinary vector parity and the C-odd neutral-gauge-field assignment, this is precisely the usual transformation, $V^0\mapsto -V^0(x_P)$ and $\mathbf{V}\mapsto +\mathbf{V}(x_P)$.

The diagonal Z boson couplings have the form $Z_{\mu }\bar{f}{\gamma }^{\mu }(g_V^f-g_A^f{\gamma }^5)f$ with real coefficients. The same calculation applies to any real vector and axial coefficients. Thus the tree-level flavour-diagonal neutral-current Z interaction preserves CP although it violates C and P separately when both current structures occur. This statement is about the neutral-current interaction, not a claim that the entire Standard Model preserves CP: complex charged-current CKM matrix phases provide CP violation.

The CPT theorem is a structural consequence of relativistic quantum field theory. In flat spacetime, assume a positive physical state space, the positive-energy spectrum condition, Lorentz invariance, microcausality, the usual Spin-statistics theorem, and a Hermitian dynamics. For ordinary field theories these are realized by a local Hermitian Lorentz-scalar Lagrangian density. There is then an antiunitary operator $\Theta$ implementing the combined charge conjugation, parity symmetry in quantum field theory and time-reversal symmetry, with a CPT-invariant vacuum. The combined symmetry is required even when the three separate symmetries fail.

Under the combined spacetime transformation, $x\mapsto -x$. A complex scalar field can be taken to transform into its adjoint at $-x$; a Hermitian vector field transforms into minus itself at $-x$. A derivative contributes a minus sign through differentiation of its reflected argument. For a Dirac field the transformation supplied in the paper may be writtenField phases are conventional and do not affect the physical conclusions. Crucially, antiunitarity means $\Theta i{\Theta }^{-1}=-i$ and conjugates every numerical coefficient. The reversal of time is compatible with positive energies because this conjugation changes $e^{-iHt}$ to $e^{iHt}$ while leaving the Hamiltonian invariant. Treating time reversal as a unitary change of coordinates would instead give incorrect signs.

One can see explicitly how the CPT transformation of Dirac bilinears works. Let $B_{\Gamma }(x)=:{\psi }^{†}{\gamma }^0\Gamma \psi :$ be a Hermitian fermion bilinear, with ${\Gamma }^{†}={\gamma }^0\Gamma {\gamma }^0$. The colons indicate normal ordering so that the graded reordering below has no vacuum contact term. Using the reflected fields and conjugating ${\gamma }^0\Gamma$, then interchanging the two fermion factors, givesThe last step uses the Dirac-Hermiticity condition and $\{{\gamma }^0,{\gamma }_5\}=0$. The identities for the gamma matrices then giveThe pseudoscalar's sign is particularly instructive: its explicit $i$ must also be conjugated. The minus sign from exchanging fermions is equally essential. These signs depend on the number of free vector indices, rather than on a field's separate parity assignment.

This supplies a local Lagrangian argument for CPT invariance. Hermitian local field combinations with $r$ vector indices transform with sign $(-1{)}^r$, including the appropriate reordered derivative terms. A Lorentz scalar contracts those indices in pairs or with four-index invariant tensors, giving a positive total sign. For charged terms, the reflected conjugate fields and conjugated coefficients exchange a term with its Hermitian-conjugate partner. ThusThe integration variable $x\mapsto -x$ has unit absolute Jacobian. The general field-index argument is discussed in Srednicki's treatment of discrete symmetries, section 40.

For example, the scalar mass term $\overline{\psi }\psi$ is even. The Dirac electromagnetic current and vector potential are both odd, making their contracted interaction even. The Hermitian Dirac kinetic term $(i/2)\overline{\psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\psi$ is also even: reflection and reordering interchange the two derivative placements, while antiunitarity conjugates its $i$. A complex Yukawa coupling and its Hermitian conjugate are exchanged rather than requiring the coupling itself to be real. Hence CP violation in the CKM matrix is compatible with exact CPT. Renormalizability is not necessary: local Hermitian higher-dimensional Lorentz-scalar operators satisfy the same argument.

The axiomatic form of the CPT theorem does not start by assuming a polynomial Lagrangian density. The positive-energy spectrum condition provides analytic continuations of Wightman functions, and complexified Lorentz transformations relate configurations to their total spacetime inversions. At suitable spacelike configurations, microcausality permits reversal of the order of field operators with the signs dictated by the Spin-statistics theorem. Uniqueness of analytic continuation extends this relation to the correlators generally, from which the Hilbert-space antiunitary operator is obtained. This explains why locality and positive energy are genuine hypotheses, not incidental features of a particular interaction. The elementary argument above displays the mechanism for the scalar, spinor and vector fields relevant to the Standard Model; it is not a substitute for all the analytic steps of the axiomatic proof.

For one-particle states, CPT reverses additive charges and spin projections. The spatial momentum is unchanged by the combination: parity symmetry in quantum field theory and time reversal each reverse it. Thus, up to phases,Since the energy-momentum spectrum is preserved, the CPT constraints on particle properties give equal particle-antiparticle masses, equal spin magnitudes, and opposite conserved additive charges. Relating the corresponding two-point functions also fixes equal pole positions, so unstable conjugate particles have equal total decay widths, hence equal lifetimes. Magnetic moments at the same chosen spin orientation have opposite signs. These are precise experimental tests of CPT, not tests that every individual discrete symmetry is respected.

For scattering, CPT interchanges incoming and outgoing boundary conditions. The antiunitary operator identity $⟨a|b⟩=⟨\Theta b|\Theta a⟩$ relates $⟨f,\mathrm{out}|i,\mathrm{in}⟩$ to the amplitude with transformed $f$ incoming and transformed $i$ outgoing. Thus the reversed process with antiparticles and reversed spins has the same transition probability. This is not, in general, equality with the forward CP-conjugate process. In particular, CP violation can make corresponding partial decay widths different while their complete sums remain equal. With CPT preserved, genuine CP violation requires corresponding time-reversal symmetry violation.

Finally, a proposed CPT violation requires revisiting at least one assumption: for example, a nonlocal interaction, a preferred spacetime direction violating Lorentz invariance, or loss of unitary Hermitian dynamics. A generic curved background need not possess the global spacetime inversion used here. Those observations delimit the theorem; they do not weaken its prediction for the ordinary local Standard Model in Minkowski spacetime.

Use the free-boson operator product expansionWrite $E_p=:e^{ip\cdot X}:$ and $ϵ=z-w$. The needed Wick contractions are $\partial X^{\mu }(z)\partial X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }/(2{ϵ}^2)$ and $\partial X^{\mu }(z)E_p(w)\sim -i{\alpha }^{\prime }{p}^{\mu }{E}_p(w)/(2ϵ)$. For $V_{\zeta }={\zeta }_{\mu }:\partial X^{\mu }{E}_p:$, the cross double contraction in the worldsheet stress tensor has a cubic pole: either of its two derivatives can contract with the derivative in $V_{\zeta }$, while the other contracts with $E_p$. Including the factor $-1/{\alpha }^{\prime }$ givesA primary operator has no cubic pole. Hence the first vertex is primary precisely whenThe antiholomorphic conformal weight is $\bar{h}={\alpha }^{\prime }{p}^2/4$. Primarity alone puts no mass-shell condition on $p$.

For $W_{\zeta }={\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, the holomorphic cubic pole is $-i{\alpha }^{\prime }{p}^{\mu }{\zeta }_{\mu \nu }:\bar{\partial }{X}^{\nu }{E}_p:/(2{ϵ}^3)$. The antiholomorphic operator product expansion gives the analogous pole involving $p^{\nu }{\zeta }_{\mu \nu }$. Thus the transversality of derivative-exponential primary vertices requiresThere is no trace-free requirement for primarity. A physical integrated string vertex operator must additionally have conformal weights $(1,1)$; for this vertex that gives $p^2=0$.

Now keep the fermionic signs in the real chiral Majorana fermion calculation. Contracting the external $\psi (w)$ through $:\psi (z)\partial \psi (z):$ givesMultiplying by $-1/2$ and Taylor expanding $\psi (z)$ at $w$ producesThis proves that $\psi$ is a primary operator with weight $h_{\psi }=1/2$. To find the central charge, in $T(z)T(w)$ the two double-contraction pairings contribute, before the overall factor $1/4$,The single Wick contractions give $2T(w)/{ϵ}^2+\partial T(w)/ϵ$, soComparing the leading term with $c/(2{ϵ}^4)$ proves $c_{\psi }=1/2$, as in the free chiral Majorana fermion conformal field theory.

With four embedding bosons and the ordinary reparameterization bc ghost system, cancellation of the worldsheet Weyl anomaly in each chirality requiresThus there are 44 real left-moving and 44 real right-moving fermions, equivalently 44 two-dimensional nonchiral Majorana fermions. These are internal free fermions in a four-dimensional bosonic string. Adding internal matter to the bosonic string does not introduce local worldsheet supersymmetry or its superghosts.

For the requested polynomial string vertex operators, work in the unprojected Neveu–Schwarz sector. A full model also chooses spin structures and a consistent projection, which the central-charge condition alone does not specify. If the holomorphic and antiholomorphic oscillator contributions to the conformal weights are ${\ell }_L,{\ell }_R$, physicality requiresTherefore only equal levels at most one can contribute to the requested nonpositive bosonic string mass spectrum.

With zero fermion insertions, level zero gives the tachyon vertex operator $E_p$, a scalar with $M^2=-4/{\alpha }^{\prime }$. Level one gives the massless closed-string vertex operator ${\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, with the transverse conditions above and the string-state gauge redundancy that removes longitudinal polarizations. The physical transverse polarization tensor decomposes into a symmetric trace-free graviton, an antisymmetric Kalb–Ramond field, and a scalar dilaton. In four spacetime dimensions these have respectively two, one, and one physical polarizations; the two-form can be dualized to a scalar.

With two fermion insertions there are two possibilities. One fermion in each chirality givesThese are $44^2=1936$ scalar tachyons before any projection. There is no spacetime derivative or spacetime polarization to impose a vector transversality condition.

Two different fermions in the same chirality form an antisymmetric fermion bilinear, a weight-one current $J^{ij}=:{\psi }^i{\psi }^j:$; equal-species undifferentiated bilinears vanish. Closed-string level matching then requires a weight-one bosonic derivative in the opposite chirality. The resulting string vertex operators areThey have $M^2=0$ and are gauge vectors, with $p\cdot A^{ij}=0$ and $A_{\mu }^{ij}\sim A_{\mu }^{ij}+p_{\mu }{\lambda }^{ij}$, and likewise for $\tilde{A}$. Each chirality supplies $\left(\genfrac{}{}{0ex}{}{44}{2}\right)=946$ gauge vectors. The fermion bilinears generate the two $\mathfrak{so}(44)$ current algebras associated with the Special orthogonal Lie algebra, giving the unprojected gauge group $SO(44{)}_L\times SO(44{)}_R$. The tempting vertex $:{\psi }^i{\psi }^j{E}_p:$ alone fails closed-string level matching because its weights are $(1,0)$ before the common momentum contribution. Additional derivatives raise the levels above the nonpositive-mass range; a product of a left and a right bilinear has four fermion insertions and is outside this requested sector. This completes the zero- and two-fermion vertices of an internally fermionized bosonic string; a specified global projection can remove some of the listed states.

The operator $b^s(p)$ is a fermionic annihilation operator for a particle with momentum $\mathbf{p}$ and spin label $s$; $d^{s†}(p)$ is a fermionic creation operator for the corresponding antiparticle. The c-number Dirac spinors $u^s(p)$ and $v^s(p)$ solve the positive-energy particle and antiparticle equations respectively. They supply the spinor wavefunctions in the mode expansion of a Dirac field; they are not creation or annihilation operators.

Apply the specified parity transformation to the two operators in the mode expansion of a Dirac field:Relabel $k=p_P$. The momentum sum is unchanged, $p\cdot x=k\cdot x_P$, and the spinor identities give $u^s(p)={\gamma }^0{u}^s(k)$ and $v^s(p)=-{\gamma }^0{v}^s(k)$. The two minus signs in the antiparticle term cancel. ThereforeTaking the Hermitian conjugate and using $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^0{)}^2=1$ in the Dirac adjoint yieldsThe unit-modulus intrinsic-parity phase cancels from all neutral fermion bilinears.

Use the printed gamma matrix conventions throughout, in particular ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. The transpose identity for ${\gamma }^0$ gives $C{\gamma }^0=-{\gamma }^{0}C$, and its Hermitian conjugate gives the corresponding identity for $C^{†}$. Thus the Dirac adjoint of the charge-conjugated field isReordering the anticommuting components of the fermion bilinear supplies a minus sign:Invariance for arbitrary fields therefore requires $C^{†}C=I$, so the charge-conjugation matrix is a unitary matrix. This is the required normalization constraint; it is not a bosonic reordering calculation.

For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ giveWith $V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ and $A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, the supplied transpose identities implyConsequently the vector current is C-odd and the axial current is C-even: $V^{\mu }\mapsto -V^{\mu }$ and $A^{\mu }\mapsto A^{\mu }$.

For parity, put $\Lambda =\mathrm{diag}(1,-1,-1,-1)$. The gamma matrix identities ${\gamma }^0{\gamma }^{\mu }{\gamma }^0={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$ giveThus $V^0$ is P-even and $\mathbf{V}$ P-odd, whereas $A^0$ is P-odd and $\mathbf{A}$ P-even. The axial current transforms as a pseudovector.

For a normal ordering of a fermion bilinear, the result of part (a) and the fermionic sign from interchanging the fields giveFor the Dirac current, $(C^{-1}{\gamma }^{\mu }C{)}^T=-{\gamma }^{\mu }$. For the axial current, the definition of ${\gamma }^5$ and the gamma matrix identities implyThus the charge conjugation of fermion bilinears givesThe Dirac current is odd under charge conjugation, and the axial current is even under charge conjugation. The normal ordering specifies the usual treatment of coincident-field contact terms.

Use the Dirac representation of the gamma matrices and commuting external-wavefunction components, rather than anticommuting quantum fields. Let $A={\gamma }^0{\gamma }^2$, $B={\gamma }^2{\gamma }^0=-A$, and ${\Gamma }^{\alpha }={\gamma }^{\alpha }(g_V-g_A{\gamma }^5)$. Since $({\gamma }^2{)}^{†}=-{\gamma }^2$ and $v=i{\gamma }^2{u}^{\ast }$,Transpose this scalar and use $({\gamma }^2{)}^T={\gamma }^2$, $B^T=A$ and ${\bar{u}}_B=u_B^{†}{\gamma }^0$:Here $A^2=1$ and $A({\gamma }^5{)}^{T}A={\gamma }^5$. The supplied identity therefore yieldsHenceThe external labels and momenta are exchanged; the scalar constants are transposed, not complex-conjugated. At the level of these numerical fermion bilinears, charge conjugation reverses the relative sign of the vector and axial terms, relating particle and antiparticle weak amplitudes. For $g_V=g_A$, the two displayed spinor chains have opposite chiral projectors. A physical charge-conjugation transformation of an operator current also includes the minus sign from reordering Dirac fields. In particular this numerical identity does not imply equal electric charges for a particle and its antiparticle.

Writing $x_P=(t,-\mathbf{x})$, the standard parity transformation of a Dirac field isSince ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, the fermion bilinear $\bar{\psi }{\gamma }^5\psi$ changes sign. Invariance of the pseudoscalar Yukawa interaction therefore requiresThus $\varphi$ is a pseudoscalar, despite having the usual scalar-form kinetic term and mass term. In this convention $({\gamma }^5{)}^{†}=-{\gamma }^5$, but $\bar{\psi }{\gamma }^5\psi$ is Hermitian because of the additional ${\gamma }^0$ in the Dirac adjoint. A real $g$ is consequently consistent with a Hermitian Lagrangian density.

In perturbation theory about the parity-invariant vacuum state, renormalization can preserve this non-anomalous parity symmetry in quantum field theory. Every local counterterm must then be parity even, at every loop order. The allowed terms are precisely ${\mathcal{L}}_{\mathrm{ct}}^{(+)}$ in b; all coefficients in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ are zero. In particular, no $\varphi$ or ${\varphi }^3$ counterterm is generated, whereas ${\varphi }^4$ is allowed and required. The one-point and three-point pure-boson vertices also vanish to all orders: a parity-odd Lorentz scalar cannot be formed from their at most two independent external four-momenta. This last reasoning is specific to these low-point vertices; it does not assert that parity symmetry in quantum field theory is an independent $\varphi \mapsto -\varphi$ internal symmetry with the fermions held fixed.

The matrix in time reversal of a Dirac field depends on the convention for the spin-flip phases and on the sign of the spinor identity used when relabeling the mode expansion of a Dirac field. If ${ϵ}_s=(-1{)}^{1/2-s}$ and ${ϵ}_s{u}^{-s\ast }(p_T)=-{\gamma }^{5}C{u}^s(p)$, the relabelled coefficient is ${ϵ}_{-s}{u}^{-s\ast }(p_T)={\gamma }^{5}C{u}^s(p)$, giving $B={\gamma }^{5}C$. A common phase changes $B$ but leaves its gamma matrix conjugation property and the induced fermion bilinear transformation unchanged. The antiunitary operator acts on both numerical coefficients and creation or annihilation operators.

For ${\psi }^C={\eta }_{C}C{\bar{\psi }}^T$ and ${\gamma }^{0T}={\gamma }^0=-C{\gamma }^0{C}^{-1}$, the Dirac adjoint transforms as ${\bar{\psi }}^C=-{\eta }_C^{\ast }{\psi }^T{C}^{†}$. Anticommuting the components of the fermion bilinear gives $(\bar{\psi }\psi {)}^C=\bar{\psi }(C^{†}C{)}^T\psi$. Invariance of the scalar bilinear therefore fixes $C^{†}C=I$.

A fermion bilinear transforming as a Lorentz four-vector. For a single Dirac field this is the Dirac current; distinct flavours can give a charged or flavour-changing current.