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?


1 Answer 1


I'm assuming that I can break down this system into three subspaces, each of just one particle

Fock space deals with identical particles, so no you can't break down the Hilbert space into a subspace for each particle, as you can never say which particle was which.

What you can do is break down the system into a space for each site, with the state being the number of particles on that site.

  • $\begingroup$ Does this mean that the entanglement entropy I would measure would be the entanglement between sites rather than entanglement between particles? $\endgroup$
    – Jlee523
    Aug 13, 2020 at 0:04
  • 1
    $\begingroup$ It would certainly be a natural entanglement entropy to measure. For a more complex system than a 3 site model there may be more than natural way to decompose the system into subspaces, in which case there will be multiple corresponding entanglement entropies $\endgroup$ Aug 13, 2020 at 0:10

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