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Multiple Photon Absorption Question

  1. Aug 9, 2009 #1
    Question: Is it safe to say that if two photons were to simultaneously come in contact with a single atom, the total energy of the combined photons would be absorbed and then emitted as a single higher energy photon with a direction that is opposite that of the average angle of incidence of the two photons?

    This is just a thought I have been pondering the last day or so and I thought I would ask the helpful community here at PF.

    Here is the scenario I am thinking about: You have two high powered lasers; one red laser emitting photons with a wavelength of 645nm, and one green laser emitting photons with a wavelength of 535nm. Each laser is aimed at a tiny point on a reflective surface, say for example a highly polished front surface aluminum mirror. The red laser is mounted at an angle of [tex]\alpha[/tex], and the green laser is mounted at an angle of [tex]\theta[/tex]. By using the equation:

    E = hc / [tex]\lambda[/tex]

    you can calculate the energy of a single photon emitted by each laser. After plugging the values in, I ended up with 3.07x10-19J for the 645nm red, and 3.70x10-19J for the 535nm green. Now if my understanding is correct, both energies should be absorbed and then emitted as a single photon with a net energy of 6.77x10-19J, and when plugging this value back into the previous equation, a photon with a wavelength of 297nm and within the ultraviolet part of the EM spectrum.

    The emitted photon would also have a direction opposite to that of ([tex]\alpha[/tex] + [tex]\theta[/tex]) / 2. (by opposite, i dont mean like negative vectors, just think of opposite like what happens with light hitting a mirror)

    So...if you were to do this experiment in real life and decided to set up some kind of flat surface in line with the angle of incidence, you would detect one green dot, one red dot, and one dot in between the two in the ultraviolet. (Not sure how you could detect the ultraviolet really, maybe just have some fluorescent material which absorbs light in the 297nm range).

    I've heard somewhere that atoms can only absorb photons in certain energy levels which would mess up everything I've just said. Anyways, this is how I understand it, please clarify if my understanding is off and if this scenario would or wouldn't work. I am still very new to physics and have a lot to learn.
     
    Last edited: Aug 9, 2009
  2. jcsd
  3. Aug 9, 2009 #2
    Yes, this can happen in general but I have never heard of this effect at mirror surface. In bulk material the process is called four wave mixing, this page has good overview: http://www.rp-photonics.com/four_wave_mixing.html . At the atomic scale you have to take many different things into account. In your case you need an energy level that is close to 6.77x10-19J above the ground state, but even if such levels exists the process has to compete with other possibilities, for example the probability that the photons are absorbed individually, or that they are absorbed "together" but that the energy is not emitted as a single photon. The different transitions possible in a molecular system is usually shown in a Jablonski Diagram, http://en.wikipedia.org/wiki/Jablonski_diagram . You might want to look at http://www.rp-photonics.com/nonlinearities.html as well.
     
    Last edited by a moderator: Apr 24, 2017
  4. Aug 9, 2009 #3

    ZapperZ

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    In multiphoton photoemission, this is what actually happen. A photon with energy less than the work function of the material, causes an electron to be excited to a state that is still below the vacuum level. If another photon comes in before this electron can decay back to the ground state, then this electron will then have been excited an even higher energy state. In a 2-photon photoemission, this is what occurs, where photons with less energy than the work function can still produce photoelectrons.

    However, this will only work when there is a light source (typically high intensity laser) that has a very high photon density per unit area. The electron must undergo an excitation before it decays to its ground state, and this typically must take place within the order of femtoseconds to picoseconds.

    See, for example:

    K. Giesen et al., Phys. Rev. Lett. v.55, p.300 (1985).
    W.S. Fann et al., Phys. Rev. B v.44, p.10980 (1991).

    Zz.
     
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