 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2025/3/a/solution

For three odd fermionic operators, the identity

[A, BC] = {A, B} C - B {A, C}

i = (p, s)

, this yields

[a_{i}, a_{j}^{†} a_{j}] = δ_{ij} a_{j}, [b_{i}^{†}, b_{j}^{†} b_{j}] = - δ_{ij} b_{j}^{†} .

The opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels

d^{3} q / (2 π)^{3}

. Consequently

[a_{p}^{s}, H] = E_{p} a_{p}^{s}, [b_{p}^{s †}, H] = - E_{p} b_{p}^{s †} .

Substitute these into the mode expansion of a Dirac field. The result is

[ψ (x), H] = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} \sum_{s} E_{p} [e^{- i p \cdot x} a_{p}^{s} u^{s} (p) - e^{i p \cdot x} b_{p}^{s †} v^{s} (p)] .

Since

i \partial_{t} e^{- i p \cdot x} = E_{p} e^{- i p \cdot x}

and

i \partial_{t} e^{i p \cdot x} = - E_{p} e^{i p \cdot x}

, this is exactly

i \partial_{t} ψ (t, x) = [ψ (t, x), H] .

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