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Spontaneous parametric down-conversion and energy-momentum cons.

  1. Apr 28, 2013 #1
    Is this graphic wrong, see,

    http://en.wikipedia.org/wiki/File:Spontaneous_Parametric_Downconversion.png

    Shouldn't k_s + k_i be less than k_pump in the top graphic because |k_s| + |k_i| = |k_pump|, as energy is proportional to momentum?

    If so is momentum transferred to the crystal after the signal and idler photon leave the crystal?

    Thanks for any help!
     
  2. jcsd
  3. Apr 28, 2013 #2
    Feynman diagram for spontaneous parametric down-conversion?

    What is, are, the Feynman diagrams for spontaneous parametric down-conversion? Does a single electron emit both photons?

    Thank for any help!
     
  4. Apr 28, 2013 #3

    Cthugha

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    No.

    You want to fulfill the phase matching conditions which correspond to conservation of energy and momentum/wave vector INSIDE the crystal. The wave vector depends on the refractive index of the material. The materials used for SPDC are birefringent. This means that you get a different refractive index for beams at different angles to the optical axis of the crystal. Therefore, the proportionality factor between wave vector and energy is different for the three beams as the refractive index changes depends on beam orientation.

    A thorough solution of these phase-matching condition depends on the exact pump conditions, including pump beam shape and angle. Often these equations need to be solved numerically.

    No! The important thing about parametric processes is, that they do not transfer anything to the crystal. No energy, no momentum, no nothing. Accordingly, there is also no absorption or emission in the crystal. You just create a polarization in the material like in any usual transmission process through a transparent crystalline material. SPDC arises from the fact that the polarization response of the material to the incoming light is nonlinear, so when you express the polarization response of the crystal in a Taylor series, terms beyond the first one are important, too. It is this nonlinear response, which gives rise to phenomena which include responses at energies which are not present in the incoming beam in processes like SPDC or four-wave-mixing, not any kind of absorption and emission.
     
  5. Apr 29, 2013 #4
    I don't see this but see the opposite, I will sketch it out, please if you would show me where I'm wrong.

    If I'm right and momentum is transferred there seems there might be a Feynman diagram representing the transfer of momentum?

    Thanks for your help!

    Edit, seems I did not upload the scan.
     
  6. Apr 29, 2013 #5
    This is the sketch, I hope.
     

    Attached Files:

  7. Apr 29, 2013 #6

    Cthugha

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    You have randomly drawn some pair of momenta which do not match up. This process will just not occur.

    The important point is to consider momentum and energy matching INSIDE the crystal which is what the image on the wikipedia page depicts. Inside the crystal photons having the same energy, but moving along different axes, will experience a different refractive index and will therefore not have the same wave vector. If they have the same magnitude of the wave vector, they will usually not have the same energy. This is why the wikipedia image also explicitly mentions one energy for the signal photon and one for the idler photon.

    You just assume a non-physical SPDC process - one which does not fulfill the phase-matching condition. As I said before, solving that is sometimes quite complicated. There are several papers discussing that, for example http://arxiv.org/abs/1010.1236. The original article appeared somewhere in Physics Reports. I do not have the exact citation right now, but I can look it up if you want to know the details.
     
  8. Apr 29, 2013 #7
    Ignore the crystal and only look at before and after and draw any final vectors you like, if the signal and idler photons go off at any angle to the pump beam axis I don't see how momentum can be conserved if the energy of the pump photon equals the energy of the idler and signal photons combined. Sorry if I'm being thick.
     
  9. Apr 29, 2013 #8

    Cthugha

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    I still do not really see what you mean. Are you aiming at the longitudinal momentum mismatch?

    Due to symmetry reasons, it is important to keep the transversal momentum mismatch at zero, but the longitudinal mismatch is not that critical. Assuming z as the longitudinal direction, the finite crystal length [itex]L_z[/itex] relaxes the requirement on the allowed longitudinal momentum mismatch [itex]\Delta k_z[/itex], so that you get the famous factor of [itex]sinc(\Delta k_z L_z/2)[/itex], so as long as the momentum mismatch is smaller than the order of [itex]1/L_z[/itex], we are fine.

    In a classical setting one would need the mismatch to be zero exactly. In qm it is enough to get below the uncertainties.
     
  10. Apr 30, 2013 #9
    If we can take as given,

    the pump photon energy equals the combined energy of the signal and idler photons, and,

    the signal and idler photons move away from each other, ie, at an angle to the beam axis, then,

    some of the momentum of the pump photon must be transferred to the crystal?
     
  11. Apr 30, 2013 #10

    DrChinese

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    I don't see why you are thinking momentum is transferred to the crystal. The sum of the momentum is conserved, that is what the original diagram was attempting to demonstrate.
     
  12. Apr 30, 2013 #11

    Cthugha

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    I just explained in my last post that the longitudinal momentum necessarily is ill defined due to the finite crystal size and this uncertainty is (under phase matching conditions) larger than the momentum mismatch if one assumed well defined momenta.

    This is NOT classical physics. Exact momenta in a spatially narrow system are unphysical. In practice you do not even know the exact emission position inside the crystal (and therefore the exact k) and have to integrate all possible probability amplitudes leading to detection at some point over the whole crystal. This gives you the aforementioned sinc dependence on the longitudinal momentum 'mismatch'. In reality even the pump energy is often badly defined as often femtosecond pulses are used which can be tens of nanometers broad. The quantity you want to conserve in qm is the expectation value.

    edit: As a comparison about the importance of uncertainty consider a mirror. If you reflect a photon at 90 degrees from a mirror, you also do not transfer momentum to the mirror (if you did things like Mach-Zehnder interferometers would not work as the transfer would yield information about which path the photon took).

    I think his point is simply that when the two downconverted photons are not collinear, their total magnitude of momentum should be larger than the initial momentum if energy is exactly conserved (however, energy usually also has large uncertainty, but it is often easier to assume it exact and shift all the mismatch to the longitudinal momentum). While that would be true in a classical system, the longitudinal momentum of the incoming and outgoing beams inside the crystal is so uncertain that this is no problem. However, this is also the reason why that kind of SPDC only works using not too thick crystals, limiting efficiency.
     
    Last edited: Apr 30, 2013
  13. Apr 30, 2013 #12
    Maybe it is the momentum inside the crystal?

    Again, ignor the crystal and look at the energy before and after the interaction with the crystal. Is it true that the energy of the pump photon is equal to the sum of the energy of the idler and signal photon?

    If yes then the fact that the signal and idler photon move away from each other, not on the beam axis, means that the sum of the momentum of the signal and idler photon in the direction of the beam can't equal the momentum of the pump photon. I would refer to my drawing which might clear things up.
     
  14. Apr 30, 2013 #13

    Cthugha

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    You still ignore that this effect requires a quantum treatment.

    The setup for SPDC is the same as for an optical parametric amplifier. You just do not have a seed beam, but rely on the vacuum as the seed beam which triggers the nonlinear process. Alternatively you can reinterpret that as including the uncertainty of the pump beam. Depending on which of these two treatments you prefer, the energy momentum of the process incoming pump+incoming vacuum state->signal/idler or incoming pump+fluctuations->signal/idler(+fluctuations) needs to be conserved. Classically (excluding the vacuum state or the fluctuations) SPDC does not exist, just like the simple Hong-Ou-Mandel effect does not exist classically without taking the vacuum state into account at a beam splitter. You can rarely neglect the role of the vacuum state in quantum optics.


    As another argument: SPDC - like all parametric processes - is instantaneous. No process involving some absorption process resulting in exchanging energy/momentum with the crystal could do that.
     
  15. Apr 30, 2013 #14
    Something still does not feel right but will read up on the subject and what you wrote before asking any more questions. Thanks for your help!
     
  16. May 1, 2013 #15

    Cthugha

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    Welcome to quantum optics. ;)

    It is hard to find a good treatment of SPDC. The easy treatments use a semiclassical approximation which is ok for few phenomena, but fails pretty quickly. A full treatment is not quite easy to follow. A reasonable treatment is given in the Mandel/Wolf, but I would not suggest this as a reference to someone wthout some already deepened knowledge in quantum optics. Learning QO from the Mandel/Wolf is like trying to learn mechanics from the Landau/Lifschitz. The Gerry/Knight has a lower-level treatment if I remember correctly, but I might be wrong on that.

    From the historical point of view, some early treatments can be found in B. R. Mollow, Phys. Rev. A 8, 2684–2694 (1973), W.H. Louisell, A. Yariv, A. E. Siegman, Phys. Rev. 124, 1646–1654 (1961) and B. R. Mollow and R. J. Glauber, Phys. Rev. 160, 1076–1096 (1967). The second paper gives a good explanation of how the parametric process should be treated as an amplification of the vacuum state in the output and the difference to a simple frequency converter. The math in these papers is nevertheless absolutely not for the faint at heart.

    edit: If you want to read the author's opinion on the figure you posted, you might want to read the beginning of Jeff Lundeen's PhD thesis, where the figure comes from. You can find it at http://www.photonicquantum.info/About/lundeenthesisv2.pdf.
     
    Last edited: May 1, 2013
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