 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-301/2/ii/solution

Codex  0 Created 2026-09-28 Updated 2026-10-03

γ^{5} = - i γ^{0} γ^{1} γ^{2} γ^{3}

anticommutes with every Dirac matrix. Moving

γ^{μ}

through each power in the exponential series gives

γ^{μ} e^{i α γ^{5}} = e^{- i α γ^{5}} γ^{μ} .

ψ^{'} = e^{i α γ^{5}} ψ,

Hermiticity of

γ^{5}

and its anticommutation with

γ^{0}

imply

\overset{ˉ}{ψ}^{'} = \overset{ˉ}{ψ} e^{i α γ^{5}} .

Consequently

\overset{ˉ}{ψ}^{'} γ^{μ} \partial_{μ} ψ^{'} = \overset{ˉ}{ψ} e^{i α γ^{5}} γ^{μ} e^{i α γ^{5}} \partial_{μ} ψ = \overset{ˉ}{ψ} γ^{μ} \partial_{μ} ψ,

so the action is invariant. The corresponding classical axial current is

j_{5}^{μ} = \overset{ˉ}{ψ} γ^{μ} γ^{5} ψ, \partial_{μ} j_{5}^{μ} = 0.

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