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Is this mathematical proof that beamsplitters entangle photons

  1. Feb 4, 2016 #1
    Prof. S Lakshmi Bala from Department of Physics, Madras, India writes a blackboard of equations which show how beamsplitters used alone affect the wavefunctions of input photons. It seems that it depends on the number of photons you use and in which input port to get you a different entangled state. Using 1 in port A and 1 in port B gives a different entangled state than 2 photons on 1 input port and none in the other.

    The creation and anihillation operators a-dagger, b-dagger are used in association with the beam splitter alone, without use of other elements. Makes me think whether I could discard other elements from my quantum circuit scheme or not.

     
  2. jcsd
  3. Feb 6, 2016 #2

    Charles Link

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    This seems to be a rather advanced topic and I don't think I know the answer, but I think I can add something of use that may not be well known except perhaps by physics people with some expertise in optics. With a single beam incident on a beamsplitter, the energy reflection coefficient (R) can be used to compute the splitting or alternatively the Fresnel E-field coefficients (e.g. r where r^2=R) can be used to compute the resultant splittings, with intensity I=n*E^2 computed for each E-field that gets split. When two (mutually coherent) beams are incident from opposite directions, the R is no longer a good number, but the Fresnel E-field coefficients are used to determine what happens to each beam. (Maxwell's equations are linear, so the E-fields behave linearly, but the energy( e.g. from each beam) does not obey linear superposition principles.) One additional item worth mentioning is normally only one face of the beamsplitter actually does the splitting and the other face has an AR (anti-reflection coating) and has 100% transmission. For a 50-50 energy split, R=1/2 so that Fresnel r=+/-1/(sqrt(2). In general Fresnel r=(n1-n2)/(n1+n2). (and Fresnel t=2n1/(n1+n2)) Also note that a reflection off of a more dense (higher n) material picks up a "minus" sign which is essentially a "pi" phase change. (i.e. when n2>n1). Energy is completely conserved in the beamsplitter process, and the relative phase of the two beams will determine the distribution of the energy emerging from the beamsplitter.
     
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