 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/4/d/solution

Under the combined local transformation,

D_{μ}^{'} ϕ^{'} = (\partial_{μ} + i q A_{μ} + i q \partial_{μ} α) (e^{- i q α} ϕ) = e^{- i q α} D_{μ} ϕ .

The derivative of the phase cancels the added gauge potential. The conjugate derivative gains the opposite phase, so their product and

∣ ϕ ∣^{2}

are invariant. Also

F_{μν}^{'} = F_{μν} + \partial_{μ} \partial_{ν} α - \partial_{ν} \partial_{μ} α = F_{μν} .

Every term of the interacting Lagrangian is therefore invariant.

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