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Commutation relations for Spin opertors

  1. Apr 17, 2012 #1
    Dear physicist,

    I designed an experiment for my undergraduate students. As we know, for spin operators, the commutation relation is

    [Si,Sj]=ihSk

    We also know, if we use two polarizers which are perpendicular each other, there is no light other side after polarizers. Namely apparatus is like,

    -------------------------------------------------------------------------------------
    Light source ------> 0 degree polarizer + 90 degree polarizer ------->Luxmeter (no light)
    -------------------------------------------------------------------------------------

    If we commute these two as,

    -------------------------------------------------------------------------------------
    Light source ------> 90 degree polarizer + 0 degree polarizer ------->Luxmeter (no light)
    -------------------------------------------------------------------------------------

    So we can say, 0 and 90 degree polarizers can commute. 0 degree polarizer may be named Sx operator and 90 degree polarizer may be named Sy operator.

    According to the result of this experiment,

    [Sx,Sy]=0.

    Whereas [Sx,Sy]=ihSz is the rule.

    Where is the problem in my mind?

    Best wishes.

    Serkan
     
  2. jcsd
  3. Apr 17, 2012 #2

    Bill_K

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    Science Advisor

    Serkan, You can't identify the E-field components Ex and Ey with spin components Sx and Sy. Light is made of photons, whose spin component is always parallel or antiparallel to the propagation vector. These correspond to left- and right-handed circularly polarized waves. Linearly polarized waves are linear superpositions of the circular polarizations.
     
  4. Apr 17, 2012 #3
    Is Serkan identifying the spin and E-field components? I don't think so.

    I think the important point is that, as you said, the polarization vector for light is orthogonal to the direction of propagation. If x and y are taken in the plane perpendicular to the light beam, then the z-axis points along the beam, so we have that Sz is zero, and hence there's no contradiction. (Serkan, note also the difference between mutliplying operators together, which makes no reference to states, and actually performing measurements of a real physical system!)
     
  5. Apr 18, 2012 #4
    Thank you very much for your interest. According to the results from you:

    1. SxSy-SySx=0. This equation dont tell us that these operators can be commuted each other. Actually, we can not measure the polarization of the light in the z direction via the apparatus given in the first post.

    2.SxSy or SySx multiplyings give pure imagine result, so we can not measure them. So we obtain SxSy=0 and SySx=0. Hence, SxSy-SySx=0-0=0.

    good works,

    Serkan
     
  6. Apr 18, 2012 #5

    Bill_K

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    Science Advisor

    As I thought, Serkan, you are still confused. A polarizer measures and filters the alignment of the E field. If you pass a beam of light through a polarizer aligned in the x direction, the only light that gets through will have its E field pointed in the x direction. You are confusing this with spin. The polarizer does not in any sense measure Sx.
     
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