# Geometric measure of entanglement for fermions or bosons?

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1. Apr 12, 2015

### zweiling

For a system consisting of multiple components, say, a spin chain consisting ofN≥3spins, 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?

2. Apr 12, 2015

### kith

Yes, I think that's essentially the reason. See this entry of Arnold Neumaier's Physics FAQ for more detail.

3. Apr 12, 2015

### zweiling

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.

Essentially, their idea is that the slater wave function should be considered as un-entangled. Hence, if the wave function is close to a Slater determinant, then the fermions are weakly entangled. Quantitatively, the distance is measured by the inner product of the best Slater determinant and the given wave function.