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Probability amplitudes and photons

  1. Jan 23, 2012 #1
    Suppose a Fock state contains 2 photons, both in the same spacetime mode and having the save (vertical) polarization. So we can write this state as |2>, or, if we want to emphacize its vertical polarization, we may write |2v> or |2v,0h>. Suppose now we want to measure polarization in the circular (R,L) basis.

    If we had only a single photon, the question would be trivial. It would end up R or L polarized with the same probability 1/2.

    What happens when there are 2 photons? If each photon again makes an independent decision, then we will have 1/4 chance to find both photons R-polarized, 1/4 chance to find them both L-polarized, and a 1/2 chance to find them in different polarization states. In other words, we will get the analogue of Malus's law for the 2-photon case.

    Is this picture correct? And if yes, then how come the Fock state |2v> is NOT identical to the product state |1v>|1v>, when it behaves exactly the same way?
  2. jcsd
  3. Jan 24, 2012 #2

    Ken G

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    Are they not identical? It seems to me the only difference would be if the polarizations were entangled, but I don't think they are. For example, if we have a laser polarized vertically, all the photons are supposed to be in the same state, but if the beam encounters a polarizer tilted at 45 degrees, half the beam comes through-- not all or none of it. So the polarization state must not be a "Bell state" that would entangle the polarizations, which also sounds like the |2v> Fock state is the same as |1v>|1v>. However, this also says that the |2v> state cannot be a general description of two photons that are vertically polarized, because you could imagine having entanglements in the polarizations that are not expressed in that notation. I'm no expert in Fock states, but does this not say that Fock states assume no entanglements, and is that not also the kind of states that lasers prepare?
  4. Jan 24, 2012 #3
    I must admit I did not quite understand this argument. If it were an entangled state (e.g. if one photon is |v>, then the other is also |v> but if one photon is |h> the other is also |h>), then we wouldn't denote such a state as |2v> in the first, right? Because we have as many reasons to denote it as |2h> as |2v>. In my view, the correct way to denote such a state would be
    (1 / sqrt 2) (|2v>±|2h>), which would be the same as (1 / sqrt 2) (|1v>|1v> ± |1h>|1h>).
  5. Jan 24, 2012 #4

    Ken G

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    I was referring to entanglements among the observations at 45 degrees to h and v.

    Those are the eigenstates of the 45 degree tilt, but again they would seem have no entanglements between the photons. The Fock states would appear to be equivalent to that, unless someone knows otherwise. That's why I think those are indeed the same.
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