 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2024/1/b/solution

Both a particle and its antiparticle carry the momentum of their mode label. Thus the normal-ordered total momentum operator is

P = \int_{p} p (a_{p}^{†} a_{p} + b_{p}^{†} b_{p}) .

[a_{p}, P] = p a_{p}, [b_{- p}^{†}, P] = - (- p) b_{- p}^{†} = p b_{- p}^{†} .

The minus sign in the antiparticle mode label is essential: both terms in the printed mode expansion of a free field contain

e^{i p \cdot x}

. Hence

[ϕ, P] = \int_{p} \frac{p}{2 E _{p}} (b_{- p}^{†} e^{i E_{p} t} + a_{p} e^{- i E_{p} t}) e^{i p \cdot x} = - i \nabla ϕ .

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