 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2015/paper-1/5/d/solution

The input squared norm is

(N!)^{2}

, so divide its stated expression by

N!

. Commutativity and the inverse channel transformation give the normalized output

∣ ψ_{out} ⟩ = \frac{1}{2 ^{N} N !} [(b_{1}^{†})^{2} - (b_{2}^{†})^{2}]^{N} ∣0 ⟩ .

∣2 j, 2 N - 2 j ⟩

c_{j} = \frac{( - 1 ) ^{N - j}}{2 ^{N} N !} (j N) (2 j)! (2 N - 2 j)!, 0 \leq j \leq N .

Hence every possible occupancy has even particle number in each port:

P (2 j, 2 N - 2 j) = \frac{1}{4 ^{N}} (j 2 j) (N - j 2 N - 2 j), j = 0, \dots, N .

All other occupancies have zero probability. The normalization identity

\sum_{j = 0}^{N} (j 2 j) (N - j 2 N - 2 j) = 4^{N}

follows by taking the coefficient of

z^{N}

(1 - 4 z)^{- 1/2} (1 - 4 z)^{- 1/2} = (1 - 4 z)^{- 1}

. For

N = 1

it reduces to (c), and symmetry under

j \leftrightarrow N - j

gives equal mean occupancies in both outputs.

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