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The antiproton is the antiparticle of the proton: it has the same rest mass and spin $1/2$, with opposite electric charge and baryon number. A negative electric charge alone cannot identify an antiproton, because a negative pion also bends in that direction in a magnetic field. Combining magnetic rigidity with velocity determines a rest mass; agreement with the proton rest mass then makes the identification much more discriminating.

The mode expansion of a Dirac field explains how independent particle and antiparticle excitations enter a relativistic field description. It supplies a theoretical language for the antiproton, while the Chamberlain-Segrè antiproton identification supplies an experimental test. The proton is a composite baryon, so a free Dirac field is not a complete account of its internal dynamics or of antiproton production in a target.

Choose the negative-frequency Dirac spinors as the charge conjugates of the positive-frequency spinors. Transposing $\bar{u}(/p-m)=0$ and using the charge-conjugation matrix identity gives $(/p+m)C{\bar{u}}^T=0$, so the displayed choice solves the correct antiparticle equation. In the mode expansion of a Dirac field, the unitary charge-conjugation operator then exchanges particle and antiparticle annihilation operators, with conjugate phases on their creation operators. A different spinor basis can insert conventional phases or a spin-index relabeling; the displayed equality specifies a compatible choice.

For mode normalization $1/\sqrt{2E_{\mathbf{p}}}$, particle and antiparticle operators satisfy $\{a_{\mathbf{p}}^s,a_{\mathbf{q}}^{r†}\}=\{b_{\mathbf{p}}^s,b_{\mathbf{q}}^{r†}\}=(2\pi {)}^3{\delta }^{sr}{\delta }^3(\mathbf{p}-\mathbf{q})$. The Dirac spinor completeness relations imply ${\sum }_s[u_s(\mathbf{p})u_s^{†}(\mathbf{p})+v_s(-\mathbf{p})v_s^{†}(-\mathbf{p})]=2E_{\mathbf{p}}I$. Inserting this into the mode expansion of a Dirac field gives the displayed local anticommutator. The reversed momentum in the antiparticle completeness term is essential.

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.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.

Use the metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$.

Field transformation. The mode expansion of a Dirac field isCharge conjugation exchanges particle and antiparticle operators. Choose their phases so thatThe conjugate relations for creation and annihilation operators preserve their canonical anticommutation relations. Thus the transformed expansion is ${\eta }_C$ times the same integral with $a_s{u}^s$ replaced by $b_s{u}^s$ and $b_s^{†}{v}^s$ replaced by $a_s^{†}{v}^s$. On the other hand, transposing the adjoint expansion gives $b_{s}C{\bar{v}}^{sT}{e}^{-ipx}+a_s^{†}C{\bar{u}}^{sT}{e}^{ipx}$. The given spinor identities identify these expressions, proving

Symmetry of the spin matrix. Transposing $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$ gives ${\gamma }^{\mu }{C}^T=-C^T{\gamma }^{\mu T}$. Substitute ${\gamma }^{\mu T}=-C^{-1}{\gamma }^{\mu }C$ and multiply on the right by $C^{-1}$:The complex four-dimensional Clifford algebra representation generated by the gamma matrices is irreducible; equivalently its sixteen independent gamma products span the full matrix algebra. The Schur lemma therefore gives $C^T{C}^{-1}=zI$. Hence $C^T=zC$. Transposing once more gives $C=z^{2}C$, so $z^2=1$ andThis is the transpose symmetry of a charge-conjugation matrix. The given antisymmetric choice is consistent with this conclusion.

Equation of motion. Taking the adjoint of the free Dirac equation yields $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Its transpose, multiplied by $C$, isThe defining matrix relation converts this intoThus the charge conjugation of a Dirac field maps free solutions to free solutions. With an external electromagnetic field, its charge would also reverse; no such background is present here.

Chirality. Define the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$. The operator $\hat{C}$ acts on the field operators and commutes with these numerical matrices:The transformed component is therefore left-chiral, as requested. There is a useful distinction: the algebraic charge conjugate of a left-chiral spinor is right-chiral. Indeed $C{\gamma }^{5T}{C}^{-1}={\gamma }^5$ and $\overline{P_L\psi }=\bar{\psi }{P}_R$ giveThis is the operator conjugation of a chiral field component; it avoids confusing projection after operator conjugation with charge conjugation of the already projected spinor.

Baryogenesis. Consider equal initial populations of a heavy particle $X$ and its antiparticle, with two baryon-number-violating decay channels carrying different final baryon numbers $b_1,b_2$. Let $r$ and $\bar{r}$ be the branching fractions into a channel and its antiparticle channel. The net baryon number produced per initial particle-antiparticle pair isExact charge conjugation symmetry equates the conjugate branching fractions and makes this vanish. Exact CP symmetry also equates the fully summed conjugate decay rates, because parity reverses momenta and helicities without changing baryon number. Thus both violation of charge conjugation and CP violation are necessary for generating an asymmetry from a symmetric initial state.

Violation of charge conjugation alone can generate a helicity asymmetry while leaving total baryon number zero. For example, with $L,R$ denoting helicities, CP symmetry may enforce $\Gamma (X\to q_L{q}_L)=\Gamma (\bar{X}\to {\bar{q}}_R{\bar{q}}_R)$ and the analogous relation with $L,R$ exchanged, even when same-helicity charge-conjugate rates differ. The CP rate cancellation in baryogenesis means that summing over helicities cancels the baryon asymmetry. These are the symmetry requirements in the Sakharov conditions; baryon-number violation and departure from thermal equilibrium are also needed in the conventional decay mechanism.

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.

Write $f_s=(-1{)}^{1/2-s}$ for $s=\pm 1/2$, so $f_{-s}=-f_s$. Under the antiunitary operator $\hat{T}$, the coefficients and exponentials in the mode expansion of a Dirac field are conjugated as well as the creation and annihilation operators being transformed. ThusRelabel $q=p_T$ and $r=-s$. The stated Dirac spinor identities implyAlso $q_T\cdot x=-q\cdot x_T$. Substitution reconstructs the original Dirac field at the reflected time:The minus sign comes from reversing the spin label, not from anticommuting field operators. This is the relation between the time-reversal matrix and the charge-conjugation matrix in the supplied spin phases and intrinsic phase convention. Rephasing the charge-conjugation matrix or the intrinsic time-reversal phase can change its displayed form. It proves time reversal of a Dirac field without identifying an antiunitary operator with its finite-dimensional spinor matrix.

In the mode expansion of a Dirac field, $b^s(p)$ is the fermionic annihilation operator for a particle of four-momentum $p$ and spin angular momentum label $s$, while $d^{s†}(p)$ is the fermionic creation operator for the corresponding antiparticle. The Dirac spinor $u^s(p)$ is the positive-frequency particle wavefunction and $v^s(p)$ is the negative-frequency antiparticle wavefunction; they solve $(/p-m)u^s=0$ and $(/p+m)v^s=0$, respectively.

Apply the parity symmetry in quantum field theory to the given mode expansion of a Dirac field:The stated Dirac spinor identities are equivalently $u^s(p)={\gamma }^0{u}^s(p_P)$ and $v^s(p)=-{\gamma }^0{v}^s(p_P)$. Relabel the momentum sum by $p\mapsto p_P$ and use $p_P\cdot x=p\cdot x_P$ to obtainThe phase $|{\eta }_P|=1$ is the intrinsic parity convention.

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.

The spinor matrix $B$ obeys $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$, with $t_0=1$ and $t_i=-1$. Conjugating the explicit $i$ and three spatial matrices also gives $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$. With spin phases $f_s=(-1{)}^{1/2-s}$ and identities $f_s{u}^{-s\ast }(p_T)={\gamma }^{5}C{u}^s(p)$, relabeling spin and momentum in the mode expansion of a Dirac field gives $B=-{\gamma }^{5}C$, because $f_{-s}=-f_s$. That last relation is phase-convention dependent; the quantum time-reversal operator itself is not just a spinor matrix.