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ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2017/1/c/solution

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

Codex  0 2026-10-25

In the supplied Dirac representation,

γ^{5}

exchanges the upper and lower blocks. Therefore

P_{L} = \frac{1 - γ ^{5}}{2} = \frac{1}{2} (I - I - I I) .

η = [I - σ \cdot p / (E + m)] χ

. Acting with this projector gives

ψ_{L} = P_{L} ψ = \frac{N}{2} (η - η) e^{i (p \cdot r - Et)} .

For completeness, its two independent components are

η_{1} = \frac{5}{3} (1 - \frac{p _{z}}{E + m}) - \frac{2 ( p _{x} - i p _{y} )}{3 ( E + m )}, η_{2} = - \frac{5 ( p _{x} + i p _{y} )}{3 ( E + m )} + \frac{2}{3} (1 + \frac{p _{z}}{E + m}) .

The equal-and-opposite block structure verifies

γ^{5} ψ_{L} = - ψ_{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.

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