# Bose-Hubbard dimension arbitrary state

Physics Asked by HELP on September 3, 2020

Suppose we have the Bose-Hubbard model with $$N=2$$ particles and $$M=4$$ sites.
We can construct the Hamiltonian in the Fock basis $$|urangle =|n_1,n_1,…,n_Mrangle$$, where $$n_i$$ is number of particles on site $$i$$.
The vector $$|urangle$$ has $$M=4$$ elements in our case and there are $$D=(N+M-1)!/[N!(M-1)!]=10$$ state vectors.
My Hamiltonian has dimensions $$Dtimes D$$.

What is the dimension (elements) of any arbitrary vector state of the BH system?

I am asking, because I want to do a unitary time evolution of an initial state. However, the dimension of my unitary time evolution operator is $$Dtimes D$$, and if I take as my initial state $$|0rangle =[2, 0,0,0]$$ (all particles on same lattice), this has only 4 elements. So how can I apply the $$Dtimes D$$ operator on an $$Mtimes 1$$ vector?

## One Answer

The dimension of a vector is $$D$$, and the components represent the coefficients (weights) of the linear combination of basis Fock states. You arbitrarily chose an order for your Fock state basis set, so for example the state $$(1,0,0,0,0,0,0,0,0,0,0)$$ represents the Fock state arbitrarily chosen as first (Let's say $$|2,0,0,0rangle$$). Similarly, $$left(frac{1}{sqrt{2}},frac{1}{sqrt{2}},0,0,0,0,0,0,0,0right)$$ is the linear combination of the first two basis states, for example $$frac{1}{sqrt{2}}|2,0,0,0rangle + frac{1}{sqrt{2}}|0,2,0,0rangle$$.

Answered by Matteo on September 3, 2020

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