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Is the direction of spin changed by measurement?

  1. Feb 24, 2016 #1
    I have been under the impression that application of a magnetic field will quantise the z component of an electrons spin in that direction. I see this as a physical process, but Is this incorrect as it is often described as a measurement of a random variable/ wavefunction collapse? I am familiar with the arguments of the Bell Inequality and appreciate how they would apply if the application of a magnetic field altered nothing, but am struggling to see how an application of a field would be assumed to have no effect on the initial state. Any insights would be greatly appreciated.
     
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  3. Feb 24, 2016 #2

    kith

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    Before the interaction with the magnet, you have a single beam. Afterwards, you have two beams. So certainly, the state of the particles is altered.

    Initially, the spin state and the spatial wavefunction are independent. After the interaction, you have entanglement: if a particle is detected in the upper part of the detector, its spin state is up. If it is detected in the lower part, its spin state is down.
     
  4. Feb 25, 2016 #3

    naima

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    After entanglement the state of the particle is no more a pure state. It is the eigenstate of no hermitian operator. There is one exception: when the SG is oriented along le direction of the initial spin state. the repeated measurement does not alter the first output.
     
  5. Feb 25, 2016 #4

    vanhees71

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    If your initial state of your atom is a pure state then just running through a magnetic field still keeps it in a pure state.

    A pure state is defined as a state, whose statistical operator is a projection operator
    $$\hat{\rho}=|\psi \rangle \langle \psi|$$
    with some normalized state vector. Unitary time evolution maps a projector into a projector, and thus a pure state stays a pure state.
     
  6. Feb 25, 2016 #5

    naima

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    I was talking about the entanglment with the SG apparatus with its screen
     
  7. Feb 25, 2016 #6

    vanhees71

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    How is the SG apparatus entangled with whatever screen? An SG apparatus essentially consists of a static inhomogeneous magnetic field with a large homgeneous component, leading to an entanglement between position and spin-projection state in direction of the magnetic field.
     
  8. Feb 25, 2016 #7

    naima

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    What you said is valid in classical physics. You can talk about static devices and of particles in an external field. In QM particles interact with the external field. and there is no static device.
    When a subsystem evolves while interacting with another subsystem, there is a unitary time evolution but it acts on the global system.
    Take the case of the "static" Young screen with two slits.
    It can be decribed with two orthogonal vector; Its ground state |Y0> and its excited state |Ye> when a photon is captured.
    The photon at the slit can be described with 3 orthogonal vectors |S1>, |S2> and |S3> (in front of slit 1 or 2 or somewhere else)
    At the beginning we have the tensor product of Y0 and of a superpositon S1,S2 andS3.
    A unitary interaction maps this to ##Y0> \otimes(aS1> +bS2>) + Ye>\otimes c S3>##
    After interaction, the photon is in a mixed state of## |S_0>< S_0|## and ##(|S_1> + |S2>)(<S_1|+<S2|)##
    We can also place the SG device behind the Heisenberg cut and have the same thing.
     
    Last edited: Feb 25, 2016
  9. Feb 25, 2016 #8
    I agree with that. So if the application of the field alters the state (and I am supposing also the spin directions) why are we surprised that the correlations in the SG experiment are greater than we would expect if the angles of spin were fixed in a given direction?
     
  10. Feb 25, 2016 #9

    kith

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    Ah, I didn't get that this is what you are interested in from your initial post.

    Ok, let's suppose we send a particle in state ##| \! \uparrow \rangle_x## through an SG-apparatus oriented in ##z##-direction. The probability to observe ##| \! \uparrow \rangle_z## is 50%. So the interaction with the SG-magnet at least doesn't simply align the spin magnetic moment with the magnetic field gradient deterministically.
     
    Last edited: Feb 25, 2016
  11. Feb 25, 2016 #10
    Thanks, but why would that be any different than to say suddenly applying a gravitational field to line of prepared coins all at slightly different angles to the vertical? Would not half fall as heads and half fall as tails?
     
  12. Feb 28, 2016 #11

    kith

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    How would you apply this logic to the state ##| \!\! \nearrow \rangle##?
     
  13. Feb 28, 2016 #12
    I would rotate the whole setup by 45 degrees.
     
  14. Feb 28, 2016 #13

    kith

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    Let me expand: if you prepare the state ##| \!\! \nearrow \rangle_z## and perform a measurement in the ##z##-direction, the probability to measure spin down is 25%. You can't just chose special measurement directions for your explanation, you need to explain what happens for all possible angles.
     
  15. Feb 28, 2016 #14
    I think you are saying that it wouldn't work for 30 degrees or 60 degrees. Is that correct?
     
  16. Feb 28, 2016 #15

    kith

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    Yes.

    On the other hand, it is a bit difficult for me to tell what exactly "it" is because you didn't elaborate much. I'm not interested in leading an endless discussion here. If you want to talk about why some classes of models don't work, please specify such a model in more detail so that there's something tangible to discuss.
     
  17. Feb 28, 2016 #16
    OK, thanks for your responses.
     
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