 Solution

ID: past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2015/2/b/solution

⟨ 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

 New to topics? Read the docs here!