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Understanding energy transfer during the process of entanglement

  1. Oct 28, 2012 #1
    Understanding energy transfer during the creation of entanglement

    Below is an attempt to understand the energy transfer process during entanglement, please correct where required.

    Broadly/conceptually speaking:

    There are two kinds of 'energies’ associated with a photon.

    1. The photon itself is 1 quanta of energy
    2. The energy in the form of momentum (frequency, velocity etc)

    During entanglement via SPDC

    1. The photon is made to strike a crystal
    2. Sometimes (about less than 1 in a billion strikes) two photons emerge from the crystal. The combined momentum of the two photons is equal to the momentum of the original photon. These photons are entangled.

    (However in case of carom/billiards......the two balls that emerge......have combined momentum less than that of the original striker.....the transfer of energy is not friction-less. Some energy is lost to the table, air molecules, etc)

    3. The transfer of energy is friction-less i.e. the original photon transfers all of its momentum to the two (entangled) photons that emerge from the crystal (once in a, say, billion times)
     
    Last edited: Oct 28, 2012
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  3. Oct 28, 2012 #2

    mfb

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    Where is the difference?

    A perfect splitting would violate energy-momentum conservation, unless both photons would travel exactly in the direction of the original photon (they do not, at least not in general). The photon has to transfer some momentum (and probably some energy) to the material, but that fraction can be very small.
     
  4. Oct 28, 2012 #3
    Seems like a contradiction in the above.

    For a moment lets assume that there is no difference.i.e there is just one quanta.

    This one quanta is now spit into, say, three things......the two photons plus the material.

    however quanta is the smallest unit, so how does it get split into smaller (energy)?
     
  5. Oct 28, 2012 #4

    mfb

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    ??

    Quantum theory does not say that energy (of a photon for example) cannot get used to produce several other particles.
    It just prevents you from detecting the same photon with 1/10 of its energy at 10 different locations.

    The photon energy is the smallest unit of energy of the incoming radiation. It is not some fundamental value.
     
  6. Oct 28, 2012 #5
    I agree with you, mfb

    however, I am not sure if the current Quantum theory (or rather hypothesis) is in support of the above, see below:
     
    Last edited: Oct 28, 2012
  7. Oct 29, 2012 #6

    mfb

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    That is in agreement.
    And quantum theory is a theory, not a hypothesis.

    Note that the wikipedia article (which is a bad source anyway) says "a quantum" (its energy depends on the setup, here: the radiation frequency), not "the quantum" (if that would be some fundamental value).

    An analog situation in everyday life: You can buy apples in a store. You can buy bigger or smaller apples - but you cannot buy 1/2 apple.
    You can split the apple after you bought it, however, then you have 2 "apple-like things" (here the analogy begins to break down :p).
     
  8. Oct 29, 2012 #7
    How about the below analogy:

    the apples/photons are same size, but they vibrate at different frequencies
    or apples are same size but moving at different velocities
     
  9. Oct 29, 2012 #8

    mfb

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    Vibrating apples, hmm...
    That model is problematic, as can (not) use one quickly vibrating apple to get two slowly vibrating apples ;).
     
  10. Oct 29, 2012 #9
    :)......so we have the 'two energies':

    1. the apple/photon (which can be converted to energy) i.e. the one quanta
    2. the 'vibration energy' of the apple

    Now to the question that I was getting at:

    Can the 'vibration energy', if you will, be (in increments of) less than a quanta/quantum?
     
  11. Oct 30, 2012 #10

    kith

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    A 'quantum' as the smallest unit of a physical quantity always refers to a single system. If you add another system, you cannot excite this system by less energy. But once excited, the second system can redistribute the energy according to its internal structure where smaller energy steps may occur.

    In the case of a photon, the first system is a single mode of the electromagnetic field. If you excite an atom with the energy of one photon, the excited atomic state may decay to the ground state via various intermediate states. At every transition, the atom emits a photon of a smaller energy than the initial photon. These photons occupy different modes of the electromagnetic field, each of which has to be considered as an own physical system with it's own 'quantum'.

    Also note that the possible energy steps usually depend on the state your physical system is in. So in most cases, there is no such thing as a 'quantum of energy' which is characteristic for the whole system. The harmonic oscillator (which is what elctromagnetic field modes essentially are) is a very special system wrt to this.
     
    Last edited: Oct 30, 2012
  12. Oct 30, 2012 #11

    mfb

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    That is the same.
     
  13. Oct 30, 2012 #12
    well answered, informative (I see what mfb might be referring to)....thanks for the post Kith

    I don't get the above fully yet, but I sense you are right.

    so do we have quanta of different energies here?

    Will the sum total of the emitted photons (at various intermediate states) equal to the "original/striker" photon?
     
  14. Oct 30, 2012 #13

    kith

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    Yes

    The sum of the energies? In a simple system, yes. But often there are also radiationless ways to redistribute/dissipate energy (for example vibrations). The important point is that energy is conserved.

    To illustrate the last paragraph of my former post, here are the energy diagrams of the harmonic oscillator (which corresponds to one photon field mode with fixed energy/frequency/wavelength) and the hydrogen atom. In the first case, you have a ladder with equal spacing, so the notion of a 'quantum' of the system makes sense. In the second case, the minimal transition energy strongly depends on where on the ladder you are.
     
  15. Nov 4, 2012 #14
    kith, interesting, do you mean like the example you gave:

    "the the energy of one photon, the excited atomic state may decay to the ground state via various intermediate states"

    could a similar thing be occurring during quantum entanglement?
     
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