Commutation of bosonic operators on finite Hilbert space

Question

I'm checking this article of R. Somma

https://arxiv.org/abs/quant-ph/0512209

I understand the common commutation relations for creationd and annihilation operators, given by: $$[b_i, b_j^\dagger]|n_1 n_2\dots n_N\rangle = b_ib_j^\dagger|n_1 n_2\dots n_N\rangle-b_j^\dagger b_i|n_1 n_2\dots n_N\rangle\\ = \sqrt{n_i(n_j+1)}|n_1\dots n_i-1\ n_j+1\dots n_N\rangle\\ -\sqrt{n_i(n_j+1)}|n_1\dots n_i-1\ n_j+1\dots n_N\rangle=0$$ My question is about equation (2.57) on page 32. They restrict the Fock space to contain at most $N_P$ bosons per site, and then they affirm that the following relations hold: $$\left[\bar{b}_{i}, \bar{b}_{j}\right]=0,\left[\bar{b}_{i}, \bar{b}_{j}^{\dagger}\right]=\delta_{i j}\left[1-\frac{N_{P}+1}{N_{P} !}\left(\bar{b}_{i}^{\dagger}\right)^{N_{P}}\left(\bar{b}_{i}\right)^{N_{P}}\right]$$ Where does that come from? I started considering the general Fock state $|n_1 n_2\dots n_N\rangle=\prod_{k=1}^{N}\frac{1}{\sqrt{n_k!}}{b_k^\dagger}^{n_k}|0\rangle$ and trying to compute the commutaor expression but obtained nothing useful. Can anyone help me?

Answer 1

You can start with the exact matrix expression for the ladder operators, which we use for numerics, $$b^\dagger = \sum_{i=1}^N \sqrt{i} \mid i \rangle \langle i -1 \mid \quad b = \sum_{i=1}^N \sqrt{i} \mid i - 1 \rangle \langle i \mid,$$ where you will find that these two commute to $1$ for any state other than the final state $\mid N \rangle$, for which they yield $[b,b^\dagger]\mid N\rangle = -N \mid N \rangle$. The main point is that the second term is a kronecker delta function $$\delta_{\hat{n},N} = \frac{(b^\dagger)^N b^N}{N!}$$ because the ladder operators will destroy any state other than the highest allowed state, which is $\mid N \rangle$. In order to prove these things analytically it might help to know the following identity $$(b^\dagger)^m b^m \mid n \rangle = \frac{n!}{(n-m)!} \mid n \rangle, $$ and to remember that $(b^\dagger)^{N+1} = b^{N+1} = 0$.