Past exam of the particle physics course of the University of Oxford / 2017 / 1

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

Let $\chi =(\sqrt{5}/3,2/3{)}^{\mathsf{T}}$, so ${\chi }^{†}\chi =1$. The Pauli identity $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^{2}I$ givesusing $E^2=|\mathbf{p}{|}^2+m^2$. Thus unit spinor probability density is obtained withAn arbitrary constant phase is immaterial. For the alternative Lorentz normalization $u^{†}u=2E$, choose $N=\sqrt{E+m}$. Neither convention makes an infinite plane wave square-integrable; a box-normalized state includes an additional $V^{-1/2}$, or one forms a normalizable wave packet.

In the supplied Dirac representation, ${\gamma }^5$ exchanges the upper and lower blocks. ThereforeDefine $\eta =[I-\mathbf{\sigma }\cdot \mathbf{p}/(E+m)]\chi$. Acting with this projector givesFor completeness, its two independent components areThe equal-and-opposite block structure verifies ${\gamma }^5{\psi }_L=-{\psi }_L$. Projection isolates a chiral component of the massive Dirac solution; that component alone generally does not satisfy an independent massive Dirac equation, since the mass couples both chiralities.

Expand the original two-spinor as $\chi =c_{+}{\chi }_{+}+c_{-}{\chi }_{-}$. Orthogonality givesLet $r=|\mathbf{p}|/(E+m)$ and $w_{\pm }=({\chi }_{\pm },-{\chi }_{\pm }{)}^{\mathsf{T}}/\sqrt{2}$. These $w_{\pm }$ are normalized, left-chiral helicity eigenvectors. ThenThe two contributions have squared normsThese are weights relative to the original normalized state; divide by their sum if conditioning on a normalized left-chiral projection. In the ultrarelativistic limit $N\to 1/\sqrt{2}$, $r\to 1$, the positive-helicity amplitude is suppressed as $m/(2E)$, while the negative-helicity amplitude tends to $c_{-}$. HenceFor momentum specifically along $+z$, $c_{+}=\sqrt{5}/3$ and $c_{-}=2/3$: the original helicity probabilities are $5/9,4/9$, the left-chiral Norm is $1/2-\beta /18$, and its limiting weight is $4/9$. The source does not specify that momentum direction, so these numerical probabilities are a specialization, not the general answer.

The charged weak current contains $\bar{\psi }{\gamma }^{\mu }{P}_L\psi$. For an ultrarelativistic external fermion, it selects negative helicity; its antifermion partner has positive helicity. Wrong-helicity charged-current amplitudes vanish in the massless limit and are suppressed by powers of $m/E$ at finite mass. The active neutrino is consequently left-helicity and its antineutrino right-helicity in this limit.

This is a charged-current statement, not a prohibition on every right-helicity Standard Model fermion. Electromagnetic and strong vector interactions act on both chiralities and conserve helicity along a massless fermion line. A neutral-current $Z$ vertex also couples to both charged-fermion chiralities with unequal strengths. For example, a right-helicity electron can scatter electromagnetically: it is not excluded from the Standard Model. In vector-current pair annihilation the massless fermion and antifermion have opposite helicities; a purely charged-current vertex keeps only the left-fermion/right-antifermion choice. Yukawa interactions connect left and right fields and must be distinguished from these vector-current selection rules.