# Geometric measure of entanglement for fermions or bosons?

For a system consisting of multiple components, say, a spin chain consisting of $N\geq 3$ spins, people sometimes use the so-called geometric measure of entanglement. It is related to the inner product between the wave function and a simple tensor product wave function. But it seems that none used this idea on fermionic systems. Why? Is the reason that for the spin systems, the total hilbert space is a tensor product of the hilbert spaces of each spin, while for identical fermions, the total hilbert has not such a tensor product structure?

• What is a simple tensor product function? Something like $\psi^{\otimes N}$, where $\psi$ is the one-spin wavefunction? These type of function for fermions are not allowed because of the antisymmetrization; the "analogous" of a factorized state like that in fermion systems isoften assumed to be the so-called Slater determinant. However I do not know whether this is useful for you or not – yuggib Apr 11 '15 at 12:22
• After some search, I found a reference using this idea for fermions: journals.aps.org/pra/abstract/10.1103/PhysRevA.89.012504. Their idea is to use the slater wave function to approximate a given fermionic wave function. They mentioned that this will provide a geometric measure of entanglement for identical fermions, but they did not pursue this much further. – kaiser Apr 12 '15 at 7:55
• The problem is how to define entanglement for identical particles. See e.g. arxiv.org/abs/quant-ph/0610030 . However, in principle you can work in the so-called particle picture and define the geometric measure the usual way. – Karl Pilkington Jun 22 at 23:30