Past exam of the laser science and quantum information processing course of the University of Oxford / 2015 / 2 / b

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- Solution b

Solution

Orthogonality of the number states makes the coherent state norm an exponential series:

⟨ α ∣ α ⟩ = e^{- ∣ α ∣^{2}} \sum_{n = 0}^{\infty} \frac{∣ α ∣ ^{2 n}}{n !} = 1.

Using

\overset{a}{^} ∣ n ⟩ = n ∣ n - 1 ⟩

and reindexing

m = n - 1

,

\overset{a}{^} ∣ α ⟩ = e^{- ∣ α ∣^{2} /2} \sum_{n = 1}^{\infty} \frac{α ^{n}}{( n - 1 )!} ∣ n - 1 ⟩ = α ∣ α ⟩ .

It is not an eigenstate of the creation operator. Indeed its supposed eigenvalue would have to be

α^{*}

, from its expectation, whereas

(\overset{a}{^}^{†} - α^{*}) ∣ α ⟩^{2} = ⟨ \overset{a}{^} \overset{a}{^}^{†} ⟩ - ∣ α ∣^{2} = 1.

This nonzero residual holds even at

α = 0

.

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