Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 50 / 2 / a

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- Solution a

Solution

The normalized Fock vacuum obeys

a_{p} ∣0 ⟩ = 0

for every momentum and

⟨ 0∣0 ⟩ = 1

. Taking adjoints gives

⟨ 0∣ a_{p}^{†} = 0

. Of the four products in the mode expansion of

ϕ (x) ϕ (y)

, only

a_{p} a_{q}^{†}

has a nonzero vacuum expectation value. The commutator implies

⟨ 0∣ a_{p} a_{q}^{†} ∣0 ⟩ = (2 π)^{3} δ^{(3)} (p - q) .

Consequently

⟨ 0∣ ϕ (x) ϕ (y) ∣0 ⟩ = \int \frac{d ^{3} p d ^{3} q}{( 2 π ) ^{6} 2 E _{p} 2 E _{q}} (2 π)^{3} δ^{(3)} (p - q) e^{- i p x + i q y} = \int \frac{d ^{3} p}{( 2 π ) ^{3} 2 E _{p}} e^{- i p \cdot (x - y)} .

This unordered Wightman function differs from the time-ordered Feynman propagator when the temporal ordering is reversed.

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