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Entanglement and the Pauli exclusion principle

  1. Dec 4, 2012 #1
    Hey all,

    I have what I think (hope) is a relatively quick pair of questions regarding entanglement of fermions and bosons. First, am I right in saying that if two fermions are in the same position-state, they will necessarily be entangled? My reasoning here is just that if their position-state is the same, then some other aspect of their states (e.g. their spin) must be different (by the Pauli exclusion principle) - i.e. that that aspect of their states will be anti-correlated.

    Second, is there any such connection in the case of bosons? Obviously, the PEP doesn't apply; so I'm inclined to think that a pair of bosons could share a position-state and yet not be entangled. Is this right?
     
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  3. Dec 4, 2012 #2

    tom.stoer

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    Entanglement is not due to the exclusion principle but due to symmetrization for bosons (and antisymmetrization for fermions). That means that whenever we have a system consisting of two bosons (two fermions) the quantum state reads

    [tex]|a,b\rangle = |a\rangle_1 \otimes |b\rangle_2 \pm |b\rangle_1 \otimes |a\rangle_2[/tex]

    with + for bosons (- for fermions)

    '1' and '2' are 'labels' for the two particles. 'a' and 'b' represent all position or momentum space information, quantum numbers etc. Position space and spin are two specific examples. This entangled state simply says that it does not makes sense to say that "particle 1 is in state a and particle 2 is in state b". QM tells us that thet correct description is "there are two particles, one is in state a and the other one is in state b". So strictly speaking labelling particles is nonsense.

    And indeed one can formulate QM in a way where one does not refer to these labels.

    btw.: the PEP follows trivially for fermions with a = b:

    [tex]|a,a\rangle = |a\rangle_1 \otimes |a\rangle_2 \pm |a\rangle_1 \otimes |a\rangle_2 = 0[/tex]

    which means that this state does not exist (by construction i.e. antisymmetrization).
     
  4. Dec 4, 2012 #3
    No.
    1) The total wave function may not be stationary.
    -The entangled states are stationary states of the many-body Hamiltonian. However, the total wave function does not have to be a stationary state. The total wave function may be rapidly varying in time.
    2) There could be many other degrees of freedom other than position or linear momentum (which are conjugate properties).
    -There could be spin, energy, total angular momentum, isospin, etc.
    3) The ground state of a system of fermions has to be entangled.
    -You are probably thinking of the special case of the ground state (i.e., lowest possible energy) of a system of system of fermions.
    -The proof of this is along the general lines of what you are describing.
     
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