Quantum entanglement between fermions

  • #1
limarodessa
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Hi all

Can you help me?

Can the quantum entanglement exist between fermions which never interacted each other?

For example – if this states of fermions are described by Slater determinant

Does exist some papers from scientific journals about this theme?

Thank you in advance for answer
 

Answers and Replies

  • #2
ZapperZ
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Er.. back up a bit. When you already have a fermionic state described by Slater determinant, aren't there already 2 or more interacting fermions that have to have their spins aligned "just right" to produce the anti-symmetric state?

Zz.
 
  • #3
limarodessa
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When you already have a fermionic state described by Slater determinant, aren't there already 2 or more interacting fermions that have to have their spins aligned "just right" to produce the anti-symmetric state?

Zz.

So, if fermions had no interacted their states cannot be described by Slater determinant ?
 
  • #4
ZapperZ
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So, if fermions had no interacted their states cannot be described by Slater determinant ?

Is there any sense of using the Slater determinant for ONE fermion?

Zz.
 
  • #5
limarodessa
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Is there any sense of using the Slater determinant for ONE fermion?

Zz.

There may be several fermions. But can we use the Slater determinant for fermions which never interacted each other ?
 
  • #6
ZapperZ
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There may be several fermions. But can we use the Slater determinant for fermions which never interacted each other ?

I don't think you understand what the Slater determinant is for. Why do you think you use the Slater determinant?

If I have non-interacting fermions, I don't have to use the Slater determinant. How the total wavefunction really is for all these non-interacting particles is irrelevant, because they don't "sense" each other. But when they do, the fermionic statistics will kick in. By definition, they are now interacting with each other because now, the total wavefunction for all the particles involved must be antisymmetric.

Zz.
 
  • #7
limarodessa
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If I have non-interacting fermions, I don't have to use the Slater determinant.
Zz.

Thanks
 
  • #8
limarodessa
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How the total wavefunction really is for all these non-interacting particles is irrelevant, because they don't "sense" each other. But when they do, the fermionic statistics will kick in. By definition, they are now interacting with each other because now, the total wavefunction for all the particles involved must be antisymmetric.

Zz.

Well. But what you can say about this ? :

http://www.springerlink.com/content/mt368h1295458112/

'...It is then shown that a non-interacting collection of fermions at zero temperature can be entangled in spin...'
 
  • #10
tom.stoer
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If I have non-interacting fermions, I don't have to use the Slater determinant. How the total wavefunction really is for all these non-interacting particles is irrelevant, because they don't "sense" each other. But when they do, the fermionic statistics will kick in. By definition, they are now interacting with each other because now, the total wavefunction for all the particles involved must be antisymmetric.
Sorry, but I disagree.

Let's not discuss Slater determinants and wave functions but fermionic Fock states. A twp-particle state is described by

[tex]|1,1\rangle = b_\mu^\dagger b_\nu^\dagger|0\rangle[/tex]

Each fermionic creation operators create a single fermion; the two indices indicate that there are several quantum numbers, e.g. a spin; they must be different otherwise the resulting stet is exactly zero

[tex]b_\mu^\dagger b_\mu^\dagger|0\rangle = \left(b_\mu^\dagger\right)^2|0\rangle = 0[/tex]

Now if you interchange the two operators you get a minus sign

[tex]b_\nu^\dagger b_\mu^\dagger|0\rangle = -|1,1\rangle[/tex]

This is exactly what the we need in order to achieve a totally antisymmetric state; we do not have explicit antisymmetrization, the fermionic commutation relation already takes this into account.

It is important that we did not specify how the two fermions would interact. We didn't specify a Hamiltonian at all. That means that antisymmetrization comes prior to interaction; it must always be implemented for fermions - even for free = non-interacting fermions. Or let's say it the other way round: antisymmetrization and entanglement is not to be confused with interaction.
 
  • #11
ZapperZ
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Er... read again the logic of the question. ...if this states of fermions are described by Slater determinant..., meaning you START with states that have been regulated by it!

Zz.
 
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  • #12
dextercioby
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The definition of the Slater determinant (http://en.wikipedia.org/wiki/Slater_determinant) is independent on whether the identical fermions interact or not. A multi-electron state can be constructed with the Slater determinant, but the electrons could very well be free. The state of the 2 electrons in ionized helium atom for which the electrostatic interaction b/w the electrons can be neglected (thus one being kept in the atom and the other thrown to kilometers away) is still described by an antisymmetric 2-particle wavefunction, because the 2 electrons are identical fermions.

Anyways, the one of the articles quoted by the initiator of the thread (http://www.springerlink.com/content/mt368h1295458112/) pretty much answers his question.
 
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  • #13
tom.stoer
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Er... read again the logic of the question. ...if this states of fermions are described by Slater determinant..., meaning you START with states that have been regulated by it!

Zz.
If you want to talk about multi-fermion-states then they must always be constructed using antisymmetrization[/U]; whether you do that using a Slater determinant for wave functions or via Fock states doesn't matter. If you don't antisymmetrize you shouldn't call these babies "fermions".
And it has nothing to do with interaction.
 
  • #14
limarodessa
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