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About interchange phase of identical particles in Weinberg's QFT book

  1. Feb 24, 2013 #1
    In Weinberg's textbook on QFT(google book preview), he discussed the phase acquired after interchanging particle labels in the last paragraph of page 171 and the footnote of page 172. It seems he's suggesting interchanging particles of same species but different spin states will only bring a phase determined by convention, that is, the phase does not have to be ±1. I'm having a hard time understanding this, because I was taught that interchange of identical particles must give a phase of ±1 and took it for granted. Besides, for fermions antisymmetry seems to be an inevitable consequence of Pauli exclusion principle:
    Exclusion principle tells us the square of a fermion creation or annihilation operator must be zero, so [itex]a^2_m=0,\ a^2_n=0,\ (a_m+a_n)^2=0[/itex](assuming no superselection rule on m, n so that the 3rd operator is well defined), and we can easily see this implies [itex]a_ma_n+a_na_m=0[/itex]
    Is Weinberg actually treating particles of same species but different spin states as distinguishable particles? If so can I take it further and conclude particles of same species, same spin states but different momenta are also distinguishable, so that the interchange phase is also conventional?
    Cross-posted:http://physics.stackexchange.com/qu...-of-identical-particles-in-weinbergs-qft-book
     
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  3. Feb 24, 2013 #2

    tiny-tim

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    hi kof9595995! :smile:

    (weinberg's qtf is viewable online at http://books.google.co.uk/books?id=...a=X&ei=8A0qUam7CeqO0AWZnYGYDg&ved=0CDIQ6AEwAA )
    yes

    his argument (last two sentences of p171, and first two of p172) that phase2 = 1 (so phase = ±1) relies on interchange of two particles resulting in the same (ie "physically indistinguishable") state, so everything must be the same

    the argument does not work if eg the spins are different, and then it is only "convenient" (not necessary) to adopt the usual convention (because it conveniently fits in with rotational invariance :wink:)
    i think so … in that case, it is convenient to adopt the usual convention because it conveniently fits in with lorentz invariance
    are we allowed to add creation operators? :confused:
     
  4. Feb 24, 2013 #3
    Re: About interchange phase of identical particles in Weinberg's QFT b

    Why not, say if m means spin up and n means spin down, the sum is just another creation operator which creates a state of superposition of spin up and down.
    Then for fermions I doubt if this argument is still applicable since we have Pauli exclusion, which does not allow the existence of such states in the first place.
    I'd like to see why this is really nice for rotational invariance, my best guess is it has something to do with the phase acquired after rotation, but I don't see a concrete connection between these two types of phases(i.e. interchange and rotation)
     
  5. Feb 24, 2013 #4

    tiny-tim

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    but creation operators aren't even defined until p173 …

    and the formula aa = 0 (for fermions) in (4.2.7) comes from (4.1.6), which in turn assumes that we have normalised the phase factors as ±1

    if the phase factors are a general e, then (4.1.6) would have to be adjusted, and your proof would no longer work
     
  6. Feb 25, 2013 #5
    Re: About interchange phase of identical particles in Weinberg's QFT b

    Even if the phase is e, then interchanging two indistinguishable particles will give us [itex]a_n^2=e^{i\theta}a_n^2[/itex], which again implies [itex]a_n^2=0[/itex] as long as [itex]e^{i\theta}\neq1[/itex], and then by the argument in my original post, the phase has to be -1. Besides, I can't see another way to incorporate Pauli exclusion except [itex]a_n^2=0[/itex]
     
  7. Feb 25, 2013 #6

    tiny-tim

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    but the argument relies on (4.1.6) …

    what would (4.1.6) have to be changed to if the phase was e ?
     
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