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The vector current is odd and the axial current is even under charge conjugation. Under parity symmetry in quantum field theory, the first is a vector and the second a pseudovector. Thus their difference obeys $j_L^{\mu }(x)\mapsto -P^{\mu }{}_{\nu }{j}_L^{\nu }(x_P)$ under CP, with $P=\mathrm{diag}(1,-1,-1,-1)$. If a neutral vector has $V^{\mu }(x)\mapsto \xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$ under CP, the interaction transforms by the factor $-\xi$. It is invariant for $\xi =-1$. Being a real field restricts its intrinsic phases but does not alone fix this sign. The conventional C-odd neutral gauge-vector assignment makes real diagonal Z boson vector and axial couplings CP invariant, despite their separate C and P violations.

The combined operation of charge conjugation and parity symmetry in quantum field theory. A weak interaction can preserve this symmetry while violating each constituent separately.

The combined charge conjugation, parity symmetry in quantum field theory and time-reversal symmetry guaranteed under the hypotheses of the CPT theorem.

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 Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

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.

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 $\bar{\psi }={\psi }^{†}{\gamma }^0$, 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.

A symmetry of a classical field theory maps solutions to solutions; for the Noether theorem we require a continuous variational symmetry of the action, allowing its density to change by a total divergence. This is stronger than simply observing an accidental symmetry of one solution. Consider first-derivative fields ${\varphi }_a$ and defineThe Euler-Lagrange field equations are $E_a=0$. For a fixed-coordinate variation ${\delta }_0{\varphi }_a=\epsilon Q_a$, the product rule gives the off-shell identityThis identity is the proof's essential integration-by-parts step.

For a general spacetime symmetry write $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$. At a fixed point $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$, while the volume-element variation contributes ${\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })$. Suppose action invariance takes the formCombining with the preceding identity provesThis is the Noether current for a spacetime symmetry. It also covers internal symmetries, for which $\xi =0$. A constant parameter corresponds to one Noether current; several independent parameters give several currents. Integrating the continuity equation givesThus the Noether charge is conserved when the fields have suitable falloff or no flux through the spatial boundary. These are classical identities; quantum symmetries may additionally require absence of an quantum anomaly.

Translations conserve energy and momentum. If the density has no explicit coordinate dependence, take the active fixed-coordinate variation $Q_a={\partial }_{\nu }{\varphi }_a$ for translation in direction $\nu$. Then ${\delta }_0\mathcal{L}/\epsilon ={\partial }_{\nu }\mathcal{L}$, and the internal-form proof with $K^{\mu }={\delta }^{\mu }{}_{\nu }\mathcal{L}$ gives the canonical energy-momentum tensorIts four spatial integrals are the conserved energy and momentum. For a real scalar field with density $\frac{1}{2}(\partial \varphi {)}^2-V(\varphi )$ this becomes $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Direct differentiation gives ${\partial }_{\mu }{T}^{\mu \nu }=(\square \varphi +V^{\prime }(\varphi )){\partial }^{\nu }\varphi$, which vanishes by the field equation.

Lorentz symmetry conserves angular momentum and boost charges. For that scalar example, $T^{\mu \nu }$ is symmetric, so the Lorentz transformation currents can be writtenTheir divergence is $T^{\rho \sigma }-T^{\sigma \rho }=0$. Spatial pairs give angular momentum, and time-space pairs give the conserved boost charges. For a spinor field or a vector field, the intrinsic field variation supplies an additional spin current; equivalently the Belinfante-Rosenfeld stress-energy tensor combines spin and orbital pieces into a symmetric stress tensor. The nonunitarity of the finite component Spinor representation of the Lorentz group does not obstruct conservation of these physical spacetime charges.

Global phase symmetry conserves particle charge. For a complex scalar field with density ${\partial }_{\mu }{\Phi }^{\ast }{\partial }^{\mu }\Phi -V({\Phi }^{\ast }\Phi )$, the constant transformation $\Phi \mapsto e^{-i\alpha }\Phi$ leaves the density invariant. Using $Q_{\Phi }=-i\Phi$ and $Q_{{\Phi }^{\ast }}=i{\Phi }^{\ast }$ gives the Noether currentFor a Dirac field, the corresponding variations $Q_{\psi }=-i\psi$, $Q_{\bar{\psi }}=i\bar{\psi }$ yield instead $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, the Dirac current. Its divergence vanishes by the Dirac equation and adjoint Dirac equation. The Yukawa interaction of part 2 also preserves this phase symmetry, because each bilinear has one $\psi$ and one $\bar{\psi }$. Thus it preserves fermion number and forbids annihilation of two fermions into a single scalar. These currents express charge or particle-minus-antiparticle-number conservation, rather than conservation of total particle count in arbitrary interacting processes.

A global symmetry in field theory uses a spacetime-independent parameter and ordinarily relates physically distinct configurations. A gauge symmetry instead has arbitrary local parameters and represents redundancy in the description. For example, the Maxwell field is unchanged physically by $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$: its electromagnetic field tensor is unchanged. For charged matter use $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$ together with $\psi \mapsto e^{-iq\alpha (x)}\psi$. Differentiating shows $D_{\mu }^{\prime }{\psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\psi$, so the coupled Dirac action is locally gauge invariant. A local phase applied to free matter alone produces extra derivatives of $\alpha$ and is not an invariance; the compensating gauge field is necessary.

Local arbitrariness gives identities among field equations, rather than an independent physical charge for every function $\alpha$. In the pure Maxwell field, action invariance under a compactly supported $\alpha$ impliesThus ${\partial }_{\nu }{E}^{\nu }=0$ holds off shell, directly because $F^{\mu \nu }$ is antisymmetric. This exemplifies the Noether second theorem. Gauge fixing and the Gupta-Bleuler null-state quotient remove redundant photon components; they do not remove a physical global charge. Gauge transformations that are nontrivial at a boundary can carry boundary charges, so the redundancy statement concerns transformations satisfying the chosen boundary conditions. Finally, discrete symmetries such as the sign reversal of a real scalar or parity symmetry in quantum field theory can be important without furnishing a continuous-parameter Noether current.

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.