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 Incoming links: Operator-valued distribution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 301 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-05

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Use the supplied oscillator expansion, with the boundary-term convention for momentum stated in part (b). Use a real polarization vector basis, as in the supplied unconjugated polarization completeness relation. Raise its second field index to obtain

\sum_{λ, λ^{'}} ϵ_{μ}^{λ} ϵ^{ν λ^{'}} η^{λ λ^{'}} = δ_{μ}^{ν} .

(1)

The mixed commutator contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the momentum expansion with the negative Minkowski metric oscillator commutator. Setting

r = x - y

and using the momentum Dirac delta distribution gives

[A_{μ} (x), π^{ν} (y)] = \frac{i}{2} δ_{μ}^{ν} \int \frac{d ^{3} p}{( 2 π ) ^{3}} (e^{i p \cdot r} + e^{- i p \cdot r}) = i δ_{μ}^{ν} δ^{3} (r) .

(2)

Here the last equality is the Fourier representation of the Dirac delta function. Similarly,

[A_{μ} (x), A_{ν} (y)] = - η_{μν} \int \frac{d ^{3} p}{( 2 π ) ^{3}} \frac{e ^{i p \cdot r} - e ^{- i p \cdot r}}{2∣ p ∣} = 0,

(3)

because the integrand is odd under

p \mapsto - p

. The momentum-momentum commutator is proportional to the same odd difference, now weighted by

∣ p ∣/2

, and also vanishes. Therefore the equal-time canonical commutation relations are

[A_{μ} (x), π^{ν} (y)] = i δ_{μ}^{ν} δ^{3} (x - y), [A_{μ}, A_{ν}] = [π^{μ}, π^{ν}] = 0.

(4)

These are identities of operator-valued distributions, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the canonical commutation relations after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.

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