Past exam of the particle physics course of the University of Oxford / 2016 / 6

A projection operator satisfies $P^2=P$; an orthogonal projection operator also has $P^{†}=P$. It extracts a subspace while leaving already projected states unchanged. Since $({\gamma }^5{)}^2=1$ and $({\gamma }^5{)}^{†}={\gamma }^5$, the chiral projectors areThey separate the Dirac spinor into opposite fermion chirality components. The Standard Model weak charged current contains $\bar{\psi }{\gamma }^{\mu }{P}_L\psi$, selecting left-chiral particle fields and the corresponding right-helicity massless antiparticles; right-chiral charged fermions are weak-isospin singlets. The weak neutral current can couple to both fermion chirality components with different strengths. Chirality and helicity should not be equated for a massive fermion: a Lorentz boost can reverse its helicity, while the field representation defining chirality is a different notion.

The Dirac adjoint is $\bar{u}=u^{†}{\gamma }^0$. Because ${\gamma }^5$ is Hermitian and anticommutes with ${\gamma }^0$,Anticommutation with ${\gamma }^{\mu }$ then givesThus the left-chiral charged current can be written with its chiral projector acting on either external Dirac spinor. More precisely, ${\bar{u}}_A{\gamma }^{\mu }{P}_L{u}_B=\overline{P_L{u}_A}{\gamma }^{\mu }{P}_L{u}_B$, since the opposite projection vanishes. The Dirac adjoint reverses the matrix projector, $\overline{P_{L}u}=\bar{u}{P}_R$; overlooking this reversal would give a wrong zero. The source uses $1-{\gamma }^5=2P_L$, so no additional factor of two is introduced when moving that entire factor.

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.