 Solution

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

H ∣0 ⟩ = 0, P ∣0 ⟩ = 0 .

In particular the stated Hamiltonian has already subtracted the free vacuum energy. For a one-particle state, evaluate the commutators using

[A B, C] = A [B, C] + [A, C] B

[H, a_{k}^{†}] = \int_{p} E_{p} a_{p}^{†} (2 π)^{3} δ^{3} (p - k) = E_{k} a_{k}^{†}, [P, a_{k}^{†}] = k a_{k}^{†} .

Acting on the Fock vacuum therefore gives

H ∣ k ⟩ = E_{k} ∣ k ⟩, P ∣ k ⟩ = k ∣ k ⟩ .

The extra factor

2 E_{k}

in the chosen relativistic normalization of a one-particle state is a scalar multiplier and does not change either eigenvalue. Applying the same argument to a multiparticle quantum state gives additive total energy and momentum.

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