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In the 2-slit experiment, is it possible to "account" for all of the energy in the incoming beam - i.e. does all of the incoming energy show up in the bright spots or is some of it "destroyed" when destructive interference takes place?

If it is destroyed, what form is it converted into?

If it's NOT destroyed, what would happen if you allowed the light to pass through a hole where the dark spot had been, then, exploiting the different angle that each path arrived from, organised things such that these waves now reinforced each other? We would then have a situation where more energy comes out than goes in. Which is impossible. Or does "spooky action at a distance" come into play and the overall brightness of the previous spots slightly dim to account for the new path?

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This question is really classical. If you model two-slit interference in Maxwell's electrodynamics, the same thing happens: opening the second slit causes the intensity at some points on the screen to decrease. It happens with water surface waves too, and any other kind of wave. Destructive interference at one point is always matched by constructive interference nearby, and energy is "diverted" from the former to the latter location through local processes. I find it counterintuitive myself, but you don't need any quantum weirdness to understand it.

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If you open up a new illuminated slit the interference pattern that results is different than it would be without the slit. If you open a new slit and keep the intensity of the light illuminating the screen the same energy that would formerly have been absorbed by the screen now arrives at the detectors. So the amount of energy that arrives at the detectors increases.

There is no "spooky action at a distance" in quantum mechanics. The equations of motion for observables and wave functions are local. The argument for quantum mechanics being non-local hinges entirely on the alleged collapse of the wave function. Before a measurement the state of a system may be $|A\rangle + |B\rangle$ but when you measure the observable whose eigenstates are $|A\rangle, |B\rangle$ you only see one of those states. People then leap to the conclusion that only one of them happens. But the theory states that the actual state is $|A\rangle|A_{obs}\rangle + |B\rangle|B_{obs}\rangle$, where $|A_{obs}\rangle, |B_{obs}\rangle$ stand for the states in which $A,B$ have been observed. It further predicts that the states $|A_{obs}\rangle, |B_{obs}\rangle$ can't undergo interference. So an observer in the post measurement state would exist in two versions, each of which would only have a record of one outcome. The people who believe in collapse are trying to model a system that is properly described in terms of wave functions and observables using a single number: the outcome of a measurement that some particular version of you happens to see. This leads to contradictions and to ideas about non-locality, see

http://xxx.lanl.gov/abs/quant-ph/9906007

http://arxiv.org/abs/1109.6223.

So what happens in the experiment you imagine? When you open a hole in the detector screen, each photon that arrives at the hole now has a new set of paths it can propagate through. Along each path the photon picks up phase and the way the phases add up at a particular determines whether constructive or destructive interference happens at a particular point beyond the screen.

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  • $\begingroup$ Thanks for the lengthy answer. Here are a couple of “observations” of my own: 1) I’m not talking about a new SLIT but a hole in the SCREEN allowing the out-of-phase waves to pass beyond the screen where (in some way I can’t quite envisage yet) they’re brought back into phase to reinforce. 2) The term “spooky action at a distance” is not my invention but John Gribbin in his 2 “Schrödinger’s Cat” popular accounts. If memory serves, he mainly uses the term in reference to entanglement. Being someone who struggles with the concepts of QM I quite like it. $\endgroup$
    – Lefty
    Commented Aug 18, 2014 at 13:21
  • $\begingroup$ Sorry, I’ve just spotted something in your answer that might answer the question and I missed before. You mention light that “would formerly have been absorbed by the screen”. Does this mean that when destructive interference takes place, the energy is lost as heat at the screen? $\endgroup$
    – Lefty
    Commented Aug 18, 2014 at 15:38
  • $\begingroup$ @Lefty: The phrase "spooky action at a distance" is actually Einstein's. He was referring to entanglement nonlocality, which is not relevant to this question (since it requires at least two particles, and in the double slit experiment there is only one at a time). $\endgroup$
    – benrg
    Commented Aug 19, 2014 at 7:42
  • $\begingroup$ When a photon arrives at a detector it would be absorbed by the detector as part of the detection process. Destructive interference just means that comparatively few photons would arrive at a particular detector. The photons that do arrive will not be treated any differently than the photons at places where the light constructively interferes. $\endgroup$
    – alanf
    Commented Aug 19, 2014 at 9:05

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