Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 5 / e

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- Solution e

Solution

Take unilateral Laplace transforms of the cavity equation, keeping the initial operator:

(s + γ /2) a (s) = a (0) - γ b_{in} (s) .

The input-output relation gives

b_{out} (s) = b_{in} (s) + γ a (s) = \frac{s - γ /2}{s + γ /2} b_{in} (s) + \frac{γ a ( 0 )}{s + γ /2} .

A transfer function describes the forced zero-state part of the response. Thus the passive cavity input-output transfer function is

G (s) = \frac{s - γ /2}{s + γ /2} .

The omitted initial-state term is a decaying transient

γ e^{- γ t /2} a (0)

, and is present for a general initial cavity state. This distinction avoids silently setting a physical annihilation operator equal to zero.

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