Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 47 / 1 / Solution

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.