Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 / 4 / b

Solution

Quantize the real massless scalar field using the normalized in-modes:

ϕ = \sum_{j} (a_{j} f_{j} + a_{j}^{†} \overset{ˉ}{f}_{j}), [a_{j}, a_{k}^{†}] = δ_{jk}, a_{j} ∣ 0_{in} ⟩ = 0.

Use the Klein-Gordon inner product convention that is antilinear in its first argument. Mode normalization gives

(f_{j}, f_{k}) = δ_{jk}, (\overset{ˉ}{f}_{j}, \overset{ˉ}{f}_{k}) = - δ_{jk}, (f_{j}, \overset{ˉ}{f}_{k}) = 0.

b_{i} = (p_{i}, ϕ)

. Applying the Bogoliubov transformation and the negative norm of conjugate modes gives

b_{i} = j \sum (\overset{ˉ}{A}_{ij} a_{j} - \overset{ˉ}{B}_{ij} a_{j}^{†}) .

Both the complex conjugation and the minus sign follow from the inner-product convention. These operators annihilate the out-vacuum, but need not annihilate the in-vacuum. The particle number operator in out-mode

i

N_{i}^{out} = b_{i}^{†} b_{i}

. In the in-vacuum, the only nonzero contraction is

⟨ 0_{in} ∣ a_{j} a_{k}^{†} ∣ 0_{in} ⟩ = δ_{jk}

⟨ 0_{in} ∣ N_{i}^{out} ∣ 0_{in} ⟩ = j, k \sum B_{ij} \overset{ˉ}{B}_{ik} δ_{jk} = j \sum ∣ B_{ij} ∣^{2} = (B B^{†})_{ii} .

This is particle number from Bogoliubov coefficients: mixing with negative-frequency in-modes produces out-particles even though no in-particles were present. For continuous mode labels the sum becomes an integral; normalized wave packets make an individual occupation number well-defined. No explicit collapse geometry or thermal spectrum is needed for this conclusion.

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