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Entanglement @ Beam Splitter

  1. May 2, 2012 #1
    In regards to the attached image;

    If the input photons were, for input b = photon #2, and for input c = photon #3, where:
    photon #1 and #2, and #3 and #4 are entangled as |H>|V> + |V>|H>,

    for the outputs:|H>(b’)|H>(b’) - |V>(b’)|V>(b’) + |H>(c’)|H>(c’) - |V>(c’)|V>(c’) + H(b’)|V>(c’) - |V>(b’)|H>(c’) (Not sure if I got the + and – signs correct?)

    in the case of getting the outputs |H>(b’)|H>(b’) - |V>(b’)|V>(b’) + |H>(c’)|H>(c’) - |V>(c’)|V>(c’), would photons #2 and #3 be entangled with each other? Or would they remain entangled with their original partner?

    So up until the actual detection of H(b’)|V>(c’) - |V>(b’)|H>(c’) (one specific bell-state) - or once that basis state is created when the photons entangle with the final detectors at the end output - the photons remain entangled with the original partner?

    would the above be the same if photons #1 and #2, and #3 and #4 were entangled as |H>|V> - |V>|H>?
     

    Attached Files:

  2. jcsd
  3. May 3, 2012 #2

    mfb

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    You send one photon of each entangled pair through a (non-polarizing) beam splitter?
    Why don't you get parts like H(b’)>|V(b’)>?

    Why don't you look at each photon individually? If they are not entangled in some way, I would not expect anything special from using both at the same time.
     
  4. May 3, 2012 #3
    And they reach the beam splitter at zero delay.

    Well, I don't know. Perhaps you can shed some light on this?
     
  5. May 5, 2012 #4

    mfb

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    Well, you left them out in your state, and I ask why you did this.
     
  6. May 5, 2012 #5
    My guess is they get cancelled out.
     
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