 Solution

ID: past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2012/paper-1/1/c/solution

⟨ a ⟩ = α

⟨ a^{†} ⟩ = α^{*}

and

⟨ a^{†} a ⟩ = ∣ α ∣^{2}

⟨ a a^{†} ⟩ = ∣ α ∣^{2} + 1

. Therefore

⟨ \hat{E} ⟩ = E_{0} (α + α^{*}), ⟨ \hat{E}^{2} ⟩ = E_{0}^{2} [α^{2} + (α^{*})^{2} + 2∣ α ∣^{2} + 1],

(Δ E)^{2} = ⟨ \hat{E}^{2} ⟩ - ⟨ \hat{E} ⟩^{2} = E_{0}^{2}, Δ E = E_{0} .

Here

E_{0}

is taken positive as the field-amplitude scale. The coherent state displaces the mean field without changing this vacuum optical quadrature noise.

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