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 Incoming links: Equal-time canonical anticommutator of a Dirac field

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 44 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-14

For the specified

1/ 2 E_{p}

mode normalization, the canonical anticommutation relations are

{a_{p}^{s}, a_{q}^{r †}} = {b_{p}^{s}, b_{q}^{r †}} = (2 π)^{3} δ^{sr} δ^{3} (p - q),

and every other anticommutator of the

a, b

operators vanishes. For the Dirac field, they imply the Equal-time canonical anticommutator of a Dirac field

{ψ_{α} (t, x), ψ_{β}^{†} (t, y)} = δ_{α β} δ^{3} (x - y), {ψ_{α} (t, x), ψ_{β} (t, y)} = 0.

To check the normalization, the two sectors contribute

u (p) u^{†} (p)

and

v (p) v^{†} (p)

with opposite spatial Fourier signs. Reverse

p

in the second contribution. The completeness relations then give

\sum_{s} [u_{s} (p) u_{s}^{†} (p) + v_{s} (- p) v_{s}^{†} (- p)] = [(\neq p + m) + (\neq \tilde{p} - m)] γ^{0} = 2 E_{p} I,

where

\tilde{p} = (E_{p}, - p)

. This cancels the mode denominator and leaves the Fourier representation of the delta function. Same-momentum

u^{†} v

orthogonality must not be used in place of this reversed-momentum identity.

For arbitrary spacetime points,

{ψ_{α} (x), \overset{ˉ}{ψ}_{β} (y)} = (i \neq \partial_{x} + m)_{α β} Δ (x - y), Δ (z) = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} (e^{- i p \cdot z} - e^{i p \cdot z}) .

For spacelike

z

, choose a Lorentz frame with

z^{0} = 0

; reversing spatial momentum shows

Δ (z) = 0

. Its derivatives vanish there as well, so the field anticommutators vanish at spacelike separation. This is fermionic microcausality. The anticommutation rules make the occupation of each normalized mode zero or one, implementing the Pauli exclusion principle, and make the normal-ordered free Hamiltonian positive for both particle sectors.

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Past exam of the quantum field theory course of the University of Oxford / 2021 / 2 / b / Solution 2026-10-25

The canonical anticommutation relations are

{a_{p}^{s}, a_{q}^{r †}} = {b_{p}^{s}, b_{q}^{r †}} = (2 π)^{3} δ^{sr} δ^{(3)} (p - q),

with the mixed and equal-type anticommutators zero. Insert the mode expansion of a Dirac field. The phases cancel at equal time, and a gives

{ψ_{α} (t, y), ψ_{β}^{†} (t, x)} = \int \frac{d ^{3} p}{( 2 π ) ^{3}} e^{i p \cdot (y - x)} \frac{\sum _{s} [ u _{α}^{s} ( p ) u _{β}^{s *} ( p ) + v _{α}^{s} ( - p ) v _{β}^{s *} ( - p )]}{2 E} = δ_{α β} δ^{(3)} (x - y) .

Since

\overset{ˉ}{ψ} γ^{0} = ψ^{†}

, the requested componentwise answer is

{(i \overset{ˉ}{ψ} γ^{0})_{α} (t, x), ψ_{β} (t, y)} = i δ_{α β} δ^{(3)} (x - y) .

This is the Equal-time canonical anticommutator of a Dirac field: the canonical momentum conjugate to

ψ_{α}

is

π_{α} = i ψ_{α}^{†}

. It is the fermionic counterpart of the canonical field canonical commutation relation, ensuring the correct local canonical quantization. For this first-order Lagrangian density, the conjugate variables are constrained; the displayed relation is the standard result after handling that constraint. It does not assert a nonzero anticommutator between the fields at separated spatial points.

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