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Spinor spreading

  1. Nov 2, 2008 #1
    Since a full description of a particle is the product \psi \chi, what's the relation between the spreading of the spatial factor \psi and of its spinor \chi?
     
    Last edited: Nov 2, 2008
  2. jcsd
  3. Nov 3, 2008 #2

    tiny-tim

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    Hi intervoxel! :smile:

    A spinor doesn't spread. :wink:
     
  4. Nov 5, 2008 #3
    Ok, tiny-tim, a spinor doesn't spread. Thank you. :)

    So, I suppose, we can visualize the spinor "cloud" as a conical surface (s=1/2, up state, say) made of up vectors with length hbar/2 centered around an axis in a certain direction in space so that its projection on that direction alone always returns the value hbar/2, while for others directions we might obtain any value, positive or negative. Is this picture correct?

    In the case when we apply a magnetic field in that direction the Lamour precession means that the cone is denser (greater probability) around a vector precessing at the Lamour frequency. Is it?
     
    Last edited: Nov 5, 2008
  5. Nov 5, 2008 #4

    tiny-tim

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    … on the road to Morocco …

    Hi intervoxel! :smile:

    Any "cloud" that you see in diagrams of electron distributions has nothing to do with the spinor.

    There is nothing conical about a spinor.

    The spinor just defines the direction of spin …

    if you want to visualise a spinor as a volume, then use a sphere, rotating about the axis defined by the spinor. :smile:
    Is that the Dorothy Lamour precession? :wink:

    She gets around, doesn't she?
     
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