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Epr spin and more

  1. Feb 7, 2014 #1
    The epr paradox is usually explained as something like:

    Suppose you have two electrons in the singlet state (+=spin up, -=spin down):

    lψ>= l+>l-> - l->l+>

    Now if you measure the spin on the first electron the explanation is (I think) that this collapses one electron onto l+> or l-> such that the wave function becomes:

    lψ>= l+>l-> (for measuring +)

    But I don't understand this. How is this form of wave function consistent with the antisymmetrization requirement for the wavefunction for electrons?
    Actually I'm also generally confused by how measurement in quantum mechanics is precisely defined. Suppose I want to measure the total spin. How does one do this? Is measurement of a quantity defined as the interaction of the quantity with an object large enough to exhibit classical behaviour.
  2. jcsd
  3. Feb 7, 2014 #2


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    ##\hat{P}_{12}|\psi \rangle = \frac{1}{\sqrt{2}}(|- \rangle |+\rangle - |+ \rangle |- \rangle ) = -|\psi \rangle##

    Yes so in other words a classically described measuring apparatus that can always take on definite measurement values. In the case of spin we could use a position measurement apparatus that records the position of the deflected electron on the screen. However if you allow for the measuring apparatus itself to be described by a quantum state, such as ##\{|z_0 \rangle , |z_{\uparrow} \rangle, |z_{\downarrow} \rangle \}## where in order this would represent the measuring apparatus in the pre-measurement state, measured spin up state (position corresponding to upwards deflection), and measured spin down state (position corresponding to dowwnards deflection), then the measurement interaction gets rather interesting when the deflected electron is not in an eigenstate of say ##\hat{S}_z## when we make a measurement of ##S_z##.
  4. Feb 8, 2014 #3
    Aaaa202, were you referring to the second expression rather than the first?
    Last edited: Feb 8, 2014
  5. Feb 8, 2014 #4
    Yes, I don't understand how the second expression fulfills the antisymmetrization requirement.
  6. Feb 8, 2014 #5
    I thought that this requirement only applied to a system of indistinguishable particles.
  7. Feb 8, 2014 #6
    But aren't electrons that?
  8. Feb 8, 2014 #7
    Not when you have measured them and made a distinction between them. (i.e. This one is spin up and the other one is spin down).
  9. Feb 8, 2014 #8


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    It is the total wave function that must be anti-symmetric, not just the spin part.

    If you have two experimenters, and one particle is localized near one experimenter, and the other particle is localized near the other experimenter, then the total wave function can be factored into a spatial part and a spin part. If the spin part is anti-symmetric, then the spatial part must be symmetric.

    The spin part of the joint wave function is anti-symmetric under particle exchange:
    [itex]| + \rangle | - \rangle \ -\ | - \rangle | + \rangle[/itex]

    The spatial part of the joint wave function is symmetric:
    [itex]| \psi_A \rangle | \psi_B \rangle \ +\ |\psi_B \rangle | \psi_A \rangle[/itex]

    where [itex]\psi_A [/itex] might be a spatial wave function localized near one experimenter (Alice), and [itex]\psi_B[/itex] might be a spatial wave function localized near the other experimenter (Bob).

    So the total wave function would look like this:

    [itex]|\psi_A, + \rangle |\psi_B, - \rangle + |\psi_B, + \rangle |\psi_A, - \rangle
    - |\psi_A, - \rangle |\psi_B, + \rangle - |\psi_B, - \rangle |\psi_A, + \rangle[/itex]

    Either particle can have spin-up or spin-down, and either particle can be localized at Bob, or at Alice.

    Now, if Alice measures a spin-up electron, then the "collapse" would reduce the total wave function to:

    [itex]|\psi_A, + \rangle |\psi_B, - \rangle
    - |\psi_B, - \rangle |\psi_A, + \rangle[/itex]

    Now, there is a correlation between the spatial part and the spin part. If the first particle is spin-up, then it is localized at Alice. If it is spin-down, it is localized at Bob, and similarly for the second particle. The total wave function is still anti-symmetric under particle exchange.
  10. Feb 8, 2014 #9


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    The rule of thumb, "if the particles are distinguishable, then the wave function does not need to be anti-symmetric" is sort of a short hand. If the particles are distinguishable through their spatial part, then you can always make a total wave function that is anti-symmetric. So there is no constraint that the spin parts alone be anti-symmetric. So it's as if they are distinguishable particles, and not subject to Fermi statistics. But that's a short cut. Fermi statistics still applies, but it no longer makes any observable constraints on spin.
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