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Quantum Entanglement - Equal Energy?

  1. Sep 9, 2011 #1

    referframe

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    In the spontaneous parametric down-conversion process, ultraviot photons split into two, entangled photons each with exactly 1/2 the energy of the original photon.

    What about other processes in which massive particles split into two or more entangled particles? Obviously total energy is conserved, but is the original total energy always EQUALLY DIVIDED between the new entangled particles?

    Thanks in advance.
     
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  3. Sep 12, 2011 #2

    DrChinese

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    The energy is NOT equally divided into 1/2 the energy of the original photon. Energy is conserved though, and the division is *very* close to equal. When entangled photons emerge from the BBo crystal, they come out slightly off axis. The actual variation in this angle is, to a small extent, a measure of the variation of the energy/wavelength of the photon stream. To say it another way: what is collected and used in experiments is extremely close to equal, but there is a dispersion of particles which are not collected which is less close to equal.

    Check out equations 15 and 16 from this:

    http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf
     
  4. Sep 12, 2011 #3

    Ken G

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    Indeed, isn't it true that the collected particles lie along paths that are consistent with membership in either of the separated populations? It would seem to be important that the particles be indistinguishable in every way, including which of the separated (but not separate) populations they belong to. (ETA: in other words, if they were not so indistinguishable, I would expect you'd only get the mundane variety of uncorrelated entanglement, like the left and right socks in my drawer. The word "entanglement" has come to mean more than "I can get information about A by looking at B", it has come to mean "that information has a phase relationship that is preserved and can be made to interfere later on.")
     
    Last edited: Sep 12, 2011
  5. Sep 12, 2011 #4

    DrChinese

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    Well, yes and no. :smile: (A typical DrChinese answer.)

    For them to be energy/frequency/wavelength entangled, they must be indistinguishable (as you say) within some kind of range of values. Often, that range is not enough to allow them to be entangled on that basis in any significant manner. But that does not stop them from being fully polarization entangled.
     
  6. Sep 12, 2011 #5

    Ken G

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    My question here is, why does the Wiki at http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion say that it is important that the two photons be culled from the regions of overlap of the two cones that emerge from the parametric down conversion? It suggests that the polarization entanglement does not occur unless it is ambiguous "which photon is which", if you like.
     
  7. Sep 12, 2011 #6

    DrChinese

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    This is Type I entanglement. It requires 2 BBo crystals, oriented 90 degrees apart. It must be ambiguous as to which is the source BBo crystal if you want polarization entanglement. That is because the output of each is known. Either VV only or HH only.

    So the crystals are placed in series, and the splitting occurs in one or the other. Their output cones are aligned so it is not possible to determine the source. The cones are actually like rings, hard to see from the diagram, but there is a very small spread that is picked up in the angle deflection from the forward axis. The output will be something like 10:00 + 4:00 or 8:30 + 2:30 or similar if you get the drift. The deflection will be something like 1.5 to 2.0 degrees off axis and that can be decided by the experimenter.
     
  8. Sep 12, 2011 #7

    Ken G

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    OK, that makes sense-- in this case the ambiguity I was referring to is required in order to achieve a superposition of two entangled states like VV+HH rather than the trivial entanglement VV. You were saying above that if the goal is instead a state like VH+HV, then there is no need for ambiguity in the source, you can still have interesting entanglement.
     
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