# Entanglement Entropy for indistinguishable Particles in a Fock Basis

Given some total system $$H = H_A \otimes H_B$$, we can find the entanglement entropy for a subspace by taking $$S = -tr(\rho_A ln \rho_A)$$ with $$\rho_A = tr_B \rho$$, where $$\rho$$ is the density matrix of the system. For a two particles in a two level system (i.e. a state can be in $$|0 \rangle$$ or $$|1 \rangle$$, this seems quite straightforward. For example, for the state:

$$\frac{1}{\sqrt{2}}(|00\rangle + |11 \rangle)$$

I could look at $$\rho_A = \frac{1}{2}(|0\rangle_A \langle 0|_A+|1\rangle_A \langle 1|_A)$$ and calculate the entanglement entropy from here.

I'm a little lost on how I would calculate entanglement entropy for identical particles in Fock space. For example, in the 3-Site Bose-Hubbard model with 3 particles, I can choose to label my states as $$|n_1, n_2, n_3 \rangle$$ where $$n_i$$ indicates the number of particles on site $$i$$. We thus have a 10 dimensional space: $$|300\rangle, |030\rangle, |003 \rangle, ...$$

I'm assuming that I can break down this system into three subspaces, each of just one particle. But how would I calculate the entanglement entropy from here?

Really my confusion comes from taking the partial traces in order to return a density matrix of just one particle. My assumption is that it can go something like this:

If our state is $$|300\rangle$$, I can rewrite this as $$|1 \rangle \otimes |1\rangle \otimes | 1\rangle$$, where now the number within the ket indicates what site the electron is on. Then, I can take the partial traces over two of the subspaces to return some $$\rho_{A} = tr_Btr_c\rho$$. This basically turns the 3-site Bose-Hubbard model into three, three level systems.

Is this approach correct?