 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2021/2/c/solution

ψ (t, p) = \int d^{3} x e^{- i p \cdot x} ψ (t, x) = \frac{1}{2 E} \sum_{r} [b_{- p}^{r †} v^{r} (- p) e^{i Et} + a_{p}^{r} u^{r} (p) e^{- i Et}] .

The orthogonality relations

u^{s †} (p) u^{r} (p) = 2 E δ^{sr}

and

u^{s †} (p) v^{r} (- p) = 0

v^{s} (- p)

a_{p}^{s} b_{- p}^{s †} = \frac{e ^{i Et}}{2 E} u^{s †} (p) \int d^{3} x e^{- i p \cdot x} ψ (t, x), = \frac{e ^{- i Et}}{2 E} v^{s †} (- p) \int d^{3} x e^{- i p \cdot x} ψ (t, x) .

The opposite time phases and reversed antiparticle momentum are essential. Both denominators are

2 E

, because the Dirac spinors have norm squared

2 E

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