Commutation of bosonic operators on finite Hilbert space

Question

I'm checking this article of R. Somma about quantum simulations

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 commutator expression but obtained nothing useful. Can anyone help me?

Answer 1

The reason for this is because when you truncate the Hilbert space, applying the raising operator on the highest state raises you out of the Hilbert space, ie it gives a zero vector. Thus the commutator in matrix form is not the identity but a diagonal matrix with all ones expect for the last entry which is minus one. If your have any nonzero amplitude in that final state you will get a reduction in the expectation value of the commutator. When working with truncated Hilbert spaces, it is best to start from the definitions of the raising and lowering operators in this space and build everything up from there. Otherwise you can get inconsistencies, eg the displacement operator is not what you expect.