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Compton-scattering = Entanglement ?

  1. Feb 12, 2005 #1
    Hi all,

    1) does the Compton-scattering produce an entangled state?
    That is, if I measure the energy of the photon, the energy of the
    electron is immediately known and vice versa.

    2) Can the photon after scattering be considered as a superposition
    of energy-states?

    -Edgardo
     
  2. jcsd
  3. Feb 12, 2005 #2

    dextercioby

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    Since the 4momentum is conserved,i would say that knowing one energy,automatically determines the other...
    As for second point,why...?The energy is known precisely from the energy-momentum conservation law...

    Daniel.
     
  4. Feb 19, 2005 #3
    My thought was the following:
    Before scattering, 'everything' is known, that is photon energy (wavelength).
    But right after scattering, we don't know the photon's energy. There's a certain probability that it will be scattered at [itex] \theta_{1}[/itex]
    with energy [itex] E_{1} [/itex], scattered at [itex] \theta_{2}[/itex]
    with energy [itex] E_{2} [/itex] etc.

    a) So could I describe the photon's state like the following?

    [itex] |\Psi \rangle = \sum c_{k} |k \rangle [/itex]

    b) When is the photon's energy determined?
    Right after scattering or when we measure the energy?
     
  5. Feb 19, 2005 #4

    dextercioby

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    But let's not forget the V-th principle of QM:we measure the energy of the outgoing photon and force his state vector to collapse.So even if,a priori,you can speak of a entangled state,one the measurement is performed,things are definitive...

    Daniel.
     
  6. Feb 22, 2005 #5
    Yes yes, I know that QM-axiom.
    But in your opinion, is the energy (for the Compton-scattering) determined before measurement?
     
  7. Feb 22, 2005 #6

    dextercioby

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    Of course not.

    Daniel.
     
  8. Feb 22, 2005 #7
    Would you then say that the photon's state can be written
    as a superposition of energy-states as I described above?
     
  9. Feb 22, 2005 #8

    dextercioby

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    Sure,before the measurement it is in an entangled state...I thought we had this cleared.

    Daniel.
     
  10. Feb 23, 2005 #9
    Hello dextercioby,

    do you know other examples where the photon's state
    can be written as a superposition of energy-states?
    I'm asking because we just recently had a discussion about white photons,
    and the question came up whether such a superposition for a single photon exists.
    https://www.physicsforums.com/showthread.php?t=62946&page=5&pp=15
     
  11. Feb 23, 2005 #10

    dextercioby

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    We have pure quantum states and mixed quantum states.A pure state wan be seen as an uniparticle state (photon with momentum "k" and helicity eigenvalue Delta),or a multiparticle state (a state in the tensor product of Hilbert spaces of uniparticle states).As for mixed states,they are (normalized) combinations of pure quantum states (for a photon a (normalized) combination of uniparticle states).This last part (statistical enesemble of photons) assumes entanglement and measurement of energy/helicity eigenvalue would collapse the state vector into the one corresponding to the measured eigenvalue/spectral value.

    Daniel.
     
  12. Feb 23, 2005 #11
    Argh! Please not so complicated :cry:
    I am not that expert in QM :rofl:

    I was rather asking for simple examples like Compton-effect.
     
  13. Feb 23, 2005 #12

    dextercioby

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    In told u,up until measurement,the energy of the photon could be any possible value allowed by the energy conservation law for the [itex] e^{-}\gamma [/itex] scattering.Once u know the final energy of the (outgoing) photon,then u can compute physical observables,as the differential (and then integral) cross-section.

    Daniel.
     
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