How can a ground state of free particles have entanglement?


by nonequilibrium
Tags: entanglement, free fermions
nonequilibrium
nonequilibrium is offline
#1
May13-13, 03:18 AM
P: 1,406
Can somebody clear the fog in front of my eyes: how can a Hamiltonian describing free fermions have a ground state with non-trivial entanglement? My reasoning is: we can build the ground state in the independent electron approximation (which is now exact), hence it is a product state in the position base, hence trivial entanglement!

However, this seems far from true. (An extreme counter-example: the ground state of QHE even has long-range entanglement...)
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Jolb
Jolb is offline
#2
May13-13, 03:13 PM
P: 419
It's important in this context to define what we mean by entanglement--a box of a bunch of free fermions (a noninteracting electron gas, for example) aren't a typical example when we study "entanglement." A more common example would be something like entangled photons travelling away from one another, like in the Einstein-Podolsky-Rosen paradox or Bell's theorem. With those cases, we easily see the nonlocal effects of entanglement, but with an electron gas, what kind of measurement would probe entanglement? You see that it's a farther reach.

However, there IS a kind of nonlocal phenomenon in fermion gasses that you could describe as "entanglement." It's simpler than you would imagine: the exchange phase of e=-1 from swapping any two fermions. (This is just elementary spin-statistics.) It's still hard to imagine an experiment that would probe this nonlocal effect. It is actually easier to create an example using an "anyon" or a particle with an arbitrary exchange phase e. I see you've mentioned the QHE so you should be able to follow this argument.

If we started with two identical copies of an anyon gas, and without changing either one, somehow performed an Aharanov-Bohm type experiment to look at their relative phase, we'd obviously see no phase difference. However, if before performing the experiment, we first took one of the identical copies and somehow circled one anyon around another (recall that anyons are only observed to exist in 2-d systems), keeping all the other anyons in place, then when we perform the same aharanov-bohm experiment on the gasses, we would observe a phase difference equal to e2iθ. The difference in phases is between the multiparticle wavefunctions of the entire gas--not just the individual anyons that were in the looping process. Thus we see the entire multiparticle wavefunction changes due to the local process of circling one anyon around the other. This is a nonlocal effect we could describe as "entanglement."

For a fermion gas, you wouldn't see anything from that experiment because the exchange phase is e, so the phase accumulated when you loop one fermion around another is e2iπ=1. But the concept is still there. The reason is that the wavefunction is not just the direct product of individual electron states--it has to be antisymmetrized.
nonequilibrium
nonequilibrium is offline
#3
May13-13, 04:58 PM
P: 1,406
Thanks for the reply but I feel it misses the point. I'm not talking about anything special, just take for example entanglement entropy as a measure...

Jolb
Jolb is offline
#4
May13-13, 09:56 PM
P: 419

How can a ground state of free particles have entanglement?


I'm not exactly sure what "entanglement entropy" is, but why don't you take my suggestion--instead of the direct product of fermion states, form the correct wavefunction, the antisymmetrized version--and calculate it for yourself and see if you get what you want.

(Remember that even though the fermions are noninteracting, they still must obey Pauli exclusion.)


Edit: Here's a nice wiki article on how to antisymmetrize fermionic wavefunctions and how it relates to the exchange phase/pauli exclusion. http://en.wikipedia.org/wiki/Slater_determinant


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