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I Spin measurements

  1. Dec 1, 2017 #1
    I’ve been watching the Susskind lectures on utube and already confused. Measurement of spin in up state gives up ( in sig z) while measurement of down gives minus down. What experimentally is the difference between down and minus down? Gf
     
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  3. Dec 1, 2017 #2

    PeterDonis

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    Do you understand why that is the case?

    There isn't one if you're just dealing with that one state. But if it is part of a larger quantum system, down and minus down can lead to different quantum interference effects between different parts of the system, which can be experimentally detectable.
     
  4. Dec 1, 2017 #3
    The reason for this line of question is to get to wether you can demonstrate “rotation of 720 brings it back to the original state”. with a Stern-Gerlach apparatus,can this be done?
     
  5. Dec 1, 2017 #4
    As to the first question, other than this is what the spin matrix gives, that is measurement of a spin down prepared electron in the z direction, gives minus down,just by the math. But no,I don’t really have a picture of it. To me it would just be down. Any help here?
     
  6. Dec 1, 2017 #5

    PeterDonis

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    The general rule for measurement in QM is that if you measure a system that is in an eigenstate of the measurement operator, the result is the eigenvalue times the state. The states "up" and "down" are eigenstates of the spin measurement operator. What are their eigenvalues?

    No, because "rotation of 720 brings it back to the original state" applies to a quantum system containing two fermions, not one: the idea is that you have to exchange the particles twice (each exchange corresponding to a 360 degree rotation) to get back the original state--one exchange gives you minus the state. Measurements on a single fermion won't give any information about this. The fact that measuring spin on the "down" state of a single fermion gives minus down is a different thing; it's not connected to the minus sign in a two-fermion state on particle exchange.
     
  7. Dec 2, 2017 #6
    Thanks very much, now I know where to start with this question. Regards,g
     
  8. Dec 2, 2017 #7

    vanhees71

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    I've no clue, what this debate is about. The confusion can only come from the not so accurate state that a pure quantum state is represented by a Hilbert-space vector. More precise and of great importance for quantum theory, particularly in making sense of half-integer spin (!), is the fact that as any general state, it's rather defined by a statistical operator. A pure state is distinguished by the fact that the statistical operator is a projection operator.

    Equivalently, but a bit more complicated to understand, you can say, a pure state is represented by a ray in Hilbert space.

    The reason, why it is so important in context of half-integer spin is that observables like angular momentum are formally defined and their algebra constructed by group-theoretical means. Angular momentum is defined, since the ground-breaking work of Emmy Noether, as the generator of rotations, which are a symmetry group of both Newtonian and special-relativistic spacetime. The symmetry groups are represented in quantum theory by unitary ray representations. That's why it makes perfect sense to consider the covering group of the rotation group SO(3), which is SU(2), and all of a sudden half-integer spin representations make perfect sense (as makes mass in the case of non-relativistic quantum theory where it is defined via a central charge of the Galileo Lie algebra).
     
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