Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermionic Fock space superselected?

  1. Apr 24, 2012 #1
    If you construct a fermionic Fock state

    |psi> = a|psi1> + b|psi2>

    where |psi1> is a 1-particle state and |psi2> is a 2-particle state, and you apply a full 2pi spatial rotation to it, then each particle contributes a factor of -1 to the amplitude because of the half-integer spin SO(3) representation it comes with.

    R(2pi) |psi> = -a|psi1> + (-1)(-1)b|psi2> = -a|psi1> + b|psi2>

    Now this state is not equivalent to the one we started with because the relative phase has changed. But because the 2pi rotation is an exact symmetry for the physical state the usual argument is that this coherent superposition should not be possible. That's the argument used in fermion-boson (univalent) superselection. The usual solution is to say that the different spin states are in different superselection sectors, and there are no physically constructible observables that can determine the relative phase, so that we can as well mix them incoherently.

    So is it commonly accepted that the fermionic particle space separates into two superselection sectors of odd and even particle numbers? If not, why not?

    In case I've made a silly mistake in my argument please forgive me. I didn't get a lot of sleep recently.

    Thanks,
    Jazz
     
  2. jcsd
  3. Apr 24, 2012 #2

    DrDu

    User Avatar
    Science Advisor

  4. Apr 25, 2012 #3

    Demystifier

    User Avatar
    Science Advisor

    Yes, it's common.

    But I like to think of it in a different, perhaps more fundamental way. Is there a QFT interaction which can create such a superposition from an initial state which is not in such a superposition? If you have some experience with practical QFT, you can find out that interaction Hamiltonian leading to such a production would not be represented by a hermitian operator, while we need hermiticity for the sake of unitarity (probability conservation).
     
  5. Apr 25, 2012 #4

    DrDu

    User Avatar
    Science Advisor

    I fail to see this. E.g. [itex]1+\psi(x)+\psi^+(x)[/itex] would create a superposition of even and odd particle number states out of a state with only even or odd particle number. However, this operator, while being hermitian, is not invariant under a 360 degree rotation.
     
  6. Apr 25, 2012 #5

    Demystifier

    User Avatar
    Science Advisor

    Hm, but psi must have a spinor index (due to the spin-statistics theorem), while a Lagrangian should not have one. So you must contract it with something having another spinor index, which, I think, should contain another fermion field. But then you cannot create a single fermion, but only a pair of them.

    But then again, the spin-statistics theorem, as well as the claim that Lagrangian should not have a spinor index, rest on the assumption of relativistic invariance. On the other hand, in recent years it is quite popular to study field theories which break relativistic invariance. So after all, maybe you right. Maybe superpositions of even and odd numbers of fermions are not impossible.
     
    Last edited: Apr 25, 2012
  7. Apr 25, 2012 #6

    DrDu

    User Avatar
    Science Advisor

    That the lagrangian should not be spinorial is a good argument. However, the spin statistics theorem is not important here, as Jazzdudes argument will hold for any particle of half integer spin.

    Consider the following situation: Take a long box containing exactly one electron at low temperature so that only the symmetric ground state will be populated. Now insert a wall in the middle of the box. Each box will then contain a superposition of 0 and 1 electron if such a superposition were possible at all. Now rotate one of the boxes by 360 degrees and recombine the two halves: Tataa, the electron has disappeared!
     
  8. Apr 26, 2012 #7

    Demystifier

    User Avatar
    Science Advisor

    So far so good.

    That's wrong. Your trick with boxes is not different from a beam-splitter experiment with photons. In essence, there is no such thing as "state in the box". There is only a total state, which in this case has a form
    |1>=|1L>+|1R>
    which is a superposition of particle in the Left box and particle in the Right box. If you perform a measurement which determines in which box the particle is, the state will "collapse" to either |1L> or |1R>. But the vacuum state |0> plays no role here.
     
  9. Apr 26, 2012 #8

    DrDu

    User Avatar
    Science Advisor

    After writing the previous post I had gnarling doubts that this thought experiment is really due to my ingenuity and in fact I saw that it is discussed in the article by Wightman cited in a previous post and also I realized that the electron would not disappear. Rather, the rotation should transform a symmetric state into an antisymmetric state what has been observed experimentally.
    The discussion of Wightman I find somewhat dissatisfying:
    "As far as the relevance of these experiments to the
    univalence superselection rule is concerned, the standard response is that the
    experiments are beautiful, but they are not the ones involved in the univalence
    superselection rule where you must rotate the entire isolated system. "
     
  10. Apr 26, 2012 #9
    Thanks for the clarification! I know about univalent superselection, but I was just surprised that I've never seen any mentioning of superselection when constructing the fermionic multi particle state space. But I'm more a foundations of QT guy than a field theorist. So thanks for confirming my thoughts!

    Cheers,

    Jazz
     
  11. Apr 26, 2012 #10

    DrDu

    User Avatar
    Science Advisor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fermionic Fock space superselected?
  1. Fock spaces? (Replies: 3)

Loading...