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Bohm Mechanics & Polarisation

  1. Dec 7, 2011 #1
    I was thinking: if you send light through a 45 degree filter (D filter), and then sent that light through another D filter, because it is now in superposition of V+H state, and according to 'Beyond Measure', light in that state (left circularly polarised) would go through the 2nd filter 100% of the time.

    But according to Bohm mechanics, the light be in a definite verticle or horizontal polarisation (even after going through the first D filter, thereby meaning that if it hit the 2nd D polariser, only half would go through, as expected if it were in a V or H polariser state (according to us, we expect it to be in a superposition, but its actually in a definite state).

    Now, Ken G responded and said things about additional instructions about what to do if the photon encounters filters in particular orders.

    But I see no difference in a photon with instructions and a V polarisation, and a photon with just a V polarisation. Both are the same physical realities, and the 'instructed photon' should behave just like a V polarised photon, only go through half the time if it encounters a 2nd D filter.
     
  2. jcsd
  3. Dec 7, 2011 #2

    Demystifier

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    No, this is not so according to Bohm mechanics. See e.g.
    http://xxx.lanl.gov/abs/quant-ph/9601013
    especially Sec. 4.
     
  4. Dec 11, 2011 #3
    Now I'm being told two different things: you say no, but two other Physics Professors tell me all physical properties of a quantum system have definite states at all times.
     
  5. Dec 12, 2011 #4

    Demystifier

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    I don't know who they are, but they are obviously not experts in Bohmian mechanics.
     
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