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Spin/angular momentum question

  1. May 9, 2005 #1
    For an electron in an arbitrary spin state, can an axis always be found along which the spin is 1/2 (as opposed to a superposition of 1/2 and -1/2 spins)? What about particles whose spin is 1 or greater? For example, for a spin 1 particle which is in an arbitrary spin state, can one always find an axis along which the spin is either 1 or 0. Or is it possible for a spin 1 particle to be in such a state that it's spin is a superposition of spin 1,0,-1 states along every axis? Moving on to angular momentum, if an electron in a hydrogen atom is in the n=2, l=1 state and in a superposition: a|1>+b|0>+c|-1> where |1>, |0>, |-1> are the eigenstates of L sub z, and a,b,c are arbitrary constants, can one always find an axis along which the angular momentum has a definite value (either 1, 0, or -1) and is not a superposition of states? (I assume that whatever is true for a spin 1 particle is also true for the n=2, l=1 state of the H atom).
     
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  3. May 9, 2005 #2

    James R

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    I think the answer to all your questions is "yes". You can always find an axis where the spin (whatever it is) is an eigenstate along that axis, and not a superposition. Mathematically, the process of finding the axis is equivalent to changing the basis of the spin space.
     
  4. May 9, 2005 #3

    dextercioby

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    Nope,what you're trying to describe is a collapse of an initially entagled state.According to the V-th postulate,the state after measurement will be an eigenstate of the measured observable.

    We can't know s*** about a quantum system,before any measurement...

    Daniel.

    EDIT:I still think he was trying to describe disentanglement. :confused:
     
  5. May 9, 2005 #4
    Dextercioby, my question had nothing to do with disentanglement and measurement.
    EDIT: How do you know that I'm a "he"? Well, I am, but you couldn't have known that.
     
    Last edited: May 9, 2005
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