# Hilbert size of Bose-Hubbard system with maximum occupation

We have the Bose-Hubbard Model:

$$H=-t\sum_{} c_i^\dagger c_j+\frac{1}{2}U\sum c_i^\dagger c_i^\dagger c_i c_i -\mu\sum c_i^\dagger c_i$$

The Hilbert space dimension is:

$$D =\frac{(N+M-1)!}{N!(M-1)!}$$, with N number of Bosons, and M total number of lattice sites. This is true if the maximal occupation number of Bosons per site is equal to N.

However, let's say that we fix the maximal occupation number of Bosons to n (< N). How do we calculate the Hilbert space now?

• can't you subtract the number of disallowed states? there are $\sum_{j=1}^{N-N_{\max}} M (N_{\max} + j)!/(N! (N-N_{\max}-j)!) \times D_{N-N_{\max}-j, M-1}$ configurations with 1 site which is over-loaded, and the generalization to 2-sites etc. seems quite straight-forward, no?
– user275556
Commented Sep 28, 2021 at 15:17
• thank you for your comment! however, can you please explain what you mean by $D_{N-N_{\max}-j, M-1}$ ? Not sure I understand this notation Commented Sep 28, 2021 at 15:39
• just your formula for the number of configurations $D$ but for $N-N_{\max}-j$ particles and $M-1$ sites
– user275556
Commented Sep 28, 2021 at 16:00