 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2023/3/c/solution

Put

x_{P} = (t, - x, y)

and

U_{P} = γ^{1}

. Since

U_{P}^{†} = - γ^{1}

and

γ^{0} γ^{1} γ^{0} = - γ^{1}

\overset{ˉ}{ψ}^{'} (x) = ψ^{†} (x_{P}) (- γ^{1}) γ^{0} = \overset{ˉ}{ψ} (x_{P}) γ^{1} .

The chain rule gives

\partial_{μ} ψ (x_{P}) = r_{μ} (\partial_{μ} ψ) (x_{P})

, with

(r_{0}, r_{1}, r_{2}) = (1, - 1, 1)

and no sum in this formula. The Clifford algebra gives

γ^{1} γ^{0} γ^{1} = γ^{0}, γ^{1} γ^{1} γ^{1} = - γ^{1}, γ^{1} γ^{2} γ^{1} = γ^{2} .

r_{μ}

therefore restores each kinetic term:

\overset{ˉ}{ψ}^{'} i γ^{μ} \partial_{μ} ψ^{'} = \overset{ˉ}{ψ} (x_{P}) i γ^{μ} (\partial_{μ} ψ) (x_{P}) .

However, the Dirac scalar bilinear changes sign because

(γ^{1})^{2} = - I_{2}

\overset{ˉ}{ψ}^{'} ψ^{'} = - \overset{ˉ}{ψ} (x_{P}) ψ (x_{P}) .

Changing variables in the action has unit absolute Jacobian determinant. Hence

S_{m} P S_{- m} : S_{kin} \to S_{kin}, - m \int \overset{ˉ}{ψ} ψ \to + m \int \overset{ˉ}{ψ} ψ .

A nonzero Dirac mass term for this single two-component field breaks parity symmetry in quantum field theory with the specified transformation.

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