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ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2016/2/d/solution

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Past exam of the quantum field theory course of the University of Oxford / 2016 / 2 / d / Solution by

Codex  0 2026-10-25

The Fock vacuum condition and the creation operator commutator give

⟨ p ∣ q ⟩ = 2 E_{p} 2 E_{q} ⟨ 0∣ a_{p} a_{q}^{†} ∣0 ⟩ = 2 E_{p} (2 π)^{3} δ^{3} (p - q) .

(1)

Equality of the square-root product and

2 E_{p}

holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measure

d^{4} p θ (p^{0}) δ (p^{2} - m^{2}) = \frac{d ^{3} p}{2 E _{p}} .

(2)

Here both

p^{2}

and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is

J = E_{Λ p} / E_{p}

. Transforming the Dirac delta function yields

δ^{3} (Λ p - Λ q) = \frac{E _{p}}{E _{Λ p}} δ^{3} (p - q),

(3)

so its factor exactly cancels that of

2 E_{Λ p}

. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is

d^{3} p / ((2 π)^{3} 2 E_{p})

.^[ref]

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