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Wavefunction collapse and antisymmetry

  1. Nov 1, 2012 #1
    Hey guys,

    I was reading a book about the philosophy of science, and in the chapter about QM the author uses a well known example in order to explain quantum entaglement and illustrate the non-separability of individual system in QM. He describes a system composed of two spin-1/2 particles. Said system is in a singlet state:

    [tex] \psi_{12} = \frac{1}{\sqrt{2}} (\psi_{1}^{+} \otimes \psi_{2}^{-} - \psi_{2}^{+} \otimes \psi_{1}^{-})[/tex]

    The subscripts indicate particle 1 or 2, and the subscipts spin up or down. This is clearly an antysymmetric. Now suppose these two particles are emitted from the same source, but go in different directions. When they are spatially separate, we do an experiment to measure the spin of one electron, and it comes up as plus, and we'd automatically know the spin of the other electron. So far so good. The author then goes on to say that in this case the wavefunction would collapse into:

    (\psi_{1}^{+} \otimes \psi_{2}^{-})

    Which is really weird, because this wavefunction is not antisymmetric. How do you conciliate wavefunction collapse with the antisymmetric requirement of the w.f.?
  2. jcsd
  3. Nov 2, 2012 #2


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  4. Nov 2, 2012 #3
    Wow, took me quite a few minutes to understand that post, but that's great. So what you're saying is that we cannot really know whether it is electron one or two that is localized on the right side (or left). What we do know is that a certain orbital (spatial w.f.) is more localized on the right or the left. So we have to pair up the orbital on the right with with spin up, if we did indeed measure the rightmost electron to have spin up. So in my case, the initial w.f. is:

    [tex] (\varphi_R(r_1) \otimes \varphi_L(r_2) + \varphi_L(r_1) \otimes \varphi_R(r_2)) [\alpha(s_1) \otimes \beta(s_2)-\beta(s_1) \otimes \alpha(s_2)][/tex]


    [tex] \varphi_R(r_1)\alpha(s_1) \otimes \varphi_L(r_2)\beta(s_2) -
    \varphi_R(r_1)\beta(s_1) \otimes \varphi_L(r_2)\alpha(s_2) +
    \varphi_L(r_1)\alpha(s_1) \otimes \varphi_R(r_2)\beta(s_2)-
    \varphi_L(r_1)\beta(s_1) \otimes \varphi_R(r_2)\alpha(s_2)[/tex]

    So measuring that the electron on the right has up (alpha) spin, means the w.f. will collapse to:

    [tex] \varphi_R(r_1)\alpha(s_1) \otimes \varphi_L(r_2)\beta(s_2) -
    \varphi_L(r_1)\beta(s_1) \otimes \varphi_R(r_2)\alpha(s_2)[/tex]

    i.e. we keep only the first and last term, because the pair up R with alpha. This w.f. is still antisymmetric. I think the statement in the book is actually inaccurate and leads to confusion, but maybe it's just used and example to discuss quantum entaglement (without considering the exclusion principle; it's not a physics book after all).

    Thanks for replying :D
    Last edited: Nov 2, 2012
  5. Nov 3, 2012 #4


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    Before measurement the system was in an antisymmetric singlet | J M> = |0 0>.
    (the total spin is always null)

    Is it still in an antisymmetric singlet after? (entangled?)
    Cannot we have up up> result sometime after?
  6. Nov 3, 2012 #5
    This is most definitely NOT my area of expertise - but as I understand it, a singlet state is characterised by a complete lack of information about the spin of each particle, as opposed to a simply 'unmeasured' state where the spin has a value (which may already be known to another party).

    After the collapse, the latter condition prevails so the electrons are no longer entangled.
  7. Nov 3, 2012 #6

    I don't know if you can label it a singlet, but you'd still have S=0 and Sz = 0. You can't have and up-up result unless you flip the spin of one of the electrons (i.e. go to one of the triplet states).
  8. Nov 3, 2012 #7


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    look at a Clebsch Gordan table for 1/2 * 1/2 coupling: here
    we see that up down > is in a superposition of the initial antisymmetric singlet and of the |1 0> triplet with equal probabilities.
    I think this means that entanglement (|J M> =|0 0>) stops after measurement.
    So your book was right.
    I would be happy to read Demistifier's explanation
  9. Nov 5, 2012 #8


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    Yes. As I have already explained (see post #2), you are missing the particle-position part of the wave function.

    Entanglement may be destroyed by measurement.
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