Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 312 / 2 / d

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- Solution d

Solution

For constant

b_{i}

, the transformation is

δ A_{i}^{V} = b_{i}

. Its Noether charge is therefore

Q = \int d^{3} x b_{i} Π^{i} = q \to 0 lim b_{i} Π^{i} (q)

up to the Fourier-sign convention. Expand a transverse polarization as

A_{i}^{V} (q) = f_{q} a_{i} (q) + f_{q}^{*} a_{i}^{†} (- q), Π_{i} (q) = a^{3} (\dot{f}_{q} a_{i} (q) + \dot{f}_{q}^{*} a_{i}^{†} (- q)) .

Only the creation term survives on the vacuum, while

A_{i}^{V} (q) ∣0 ⟩ = f_{q}^{*} a_{i}^{†} (- q) ∣0 ⟩

. Hence

Q ∣0 ⟩ = q \to 0 lim [\frac{a ^{3} f ˙ _{q}^{*}}{f _{q}^{*}} b^{i} A_{i}^{V} (q) ∣0 ⟩] .

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