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If we're conducting a double-slit experiment using coherent monochrome light, we know that, wherever two light-waves fall on the same point, whether they destructively or constructively interfere with each other depends on how out of phase or in phase the two waves are. My question is this. Where such waves converging on the same point are in phase enough, and thus cause constructive interference, does this imply that the energy of both light-waves is delivered at the same point? And, if so, does this mean that an atom is absorbing both photons simultaneously?

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    $\begingroup$ Usually it doesn't, but it can. Two photon absorption is a completely different physical interaction. Since twice the energy is being absorbed, the atom will change its energy by twice as much, i.e. it will be in a different quantum state than after the absorption of just one photon. The angular momentum selection rules are also different. This is clearly distinguishable from twice the number of single photon absorptions per unit time, which is what would happen at twice the intensity. Having said that, constructive interference between two waves quadruples the intensity. $\endgroup$ – CuriousOne Apr 19 '16 at 14:00
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    $\begingroup$ @CuriousOne You are absolutely correct, but I think the OP means something else. The way I read it, he's wondering if the double slit implies that each slit must contribute a photon. That is, two photons are fundamentally necessary in double slit interference. The answer to that question (if it is indeed the question he or she meant to ask) is absolutely not. $\endgroup$ – garyp Apr 19 '16 at 14:33
  • $\begingroup$ @garyp: You are right... I didn't read it that way. Hmmm... and once again we are looking at the abyss of how we can teach a proper physical intuition for the relationship between photons and electromagnetic waves. I wish I had a good answer for the OP beyond "absolutely not", which would be my reaction to your reading, too. $\endgroup$ – CuriousOne Apr 19 '16 at 14:38
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    $\begingroup$ Let's try this: When waves interfere, it is the amplitudes of the waves that add in a linear fashion. The intensity is proportional to the square of the amplitudes. In places of constructive interference the amplitude is doubled and the intensity is quadrupled, in places of destructive interference both are reduced to zero. The average of the intensity is doubled. In a classical picture of photons (this is false but sometimes useful) one can replace the local intensity with "photon density", i.e. there will be four times as many photon absorptions where the waves interfere constructively. $\endgroup$ – CuriousOne Apr 19 '16 at 14:46
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    $\begingroup$ The problem with any reasoning about "the atom" and "the light wave" is that in reality they are the same thing: different types of partial solutions of one and the same quantum field. The physically relevant differences between the two are in the energy scales. The electronic state is a small perturbation on the nuclear state, which is a perturbation on the high energy background. The interaction with light is, again, a small perturbation of the electronic state (for high Z atoms). The interference of light is not even that (as there is no interaction): it's just the linear structure of QM. $\endgroup$ – CuriousOne Apr 20 '16 at 0:48

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