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 outputAfter the binomial expansion, the coefficient of $|2j,2N-2j⟩$ isHence every possible occupancy has even particle number in each port:All other occupancies have zero probability. The normalization identityfollows by taking the coefficient of $z^N$ in $(1-4z{)}^{-1/2}(1-4z{)}^{-1/2}=(1-4z{)}^{-1}$. For $N=1$ it reduces to (c), and symmetry under $j\leftrightarrow N-j$ gives equal mean occupancies in both outputs.