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Stimulating water with two long straight bars one could produce linear waves. Having the right distance between the bars and the correct frequency of the steady system we get in the middle between the two bars complete destructive interference. In an ideal case the energy provided by the bars will be fully dissipated into chaotic heat energy of water molecules.

Thinking about to do a similar experiment with light, we get in trouble. First, sending crossed light beams normally do not led to a destructive interference, they are not interacting, or in the case of high energy photons it could lead to pair production. But production of a bigger amount of photons of lower energy (IR) is not imaginable. Dissipation does not take place.

Is it common to work in double slit argumentation with destructive interference arguments too? Young has done so. Do we too?

HolgerFiedler
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  • In general two light beams will not be coherent and the interference will wash out. If you take a coherent beam like a laser and make two beams with a splitter then they will indeed interfere just like water waves. – John Rennie May 12 '15 at 05:39
  • Incidentally, the energy is not dissipated into chaotic heat energy of water molecules. Interference just moves the energy around i.e. some bits of the water get less energy while others get more. – John Rennie May 12 '15 at 05:40
  • @John Rennie I agree that this process is more complicared then described. – HolgerFiedler May 12 '15 at 06:54
  • I have no idea why you think that "we" are in trouble. I can say for myself that I understand what is going on quite well (it seems to me John Rennie understands it even better). So that leaves only YOU in trouble. – CuriousOne May 12 '15 at 09:00
  • I've read this question three times. I still don't have any idea what you mean to say in either of the beginning paragraphs, nor what the questions are suppose to be. If you intend to imply that there is a deficiency in the understanding of interference phenomena by the scientific community then I believe you are simply wrong. And I have no idea what else this could be about. – dmckee --- ex-moderator kitten May 12 '15 at 14:54

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There are two ways to look at light, classical and quantum mechanical.

Electromagnetic waves given by the classical solutions of Maxwell's equations will have interference patterns as predicted mathematically from the sinusoid form of the solutions.

Are we working in the double slit argumentation with destructive interference arguments too? Young has done so. Do we too?

At the level of classical solutions there exists a corresponding formulation that does depend on the way the energy of the wave is deposited on the screen.

classical double slit

Two slits are illuminated by a plane wave.

The interference pattern is predicted and observed. It It is not destructive interference as in the case of sound waves, but the mathematics are the same. The wave deposits more or less energy as it hits the screen according to the sinusoid solution of the problem.

Questions may arise if one goes to a quantum mechanical formulation, where the plane electromagnetic wave emerges from an enormous number of photons who carry the energy of the wave in a quantized manner. The whole though is consistent as the way the classical wave is built up by the individual photons is known and consistent between the classical and quantum mechanical formulation.

In conclusion.

a) In the classical case the constructive and destructive interference observed is modeled with waves that transmit the energy of the beam and they interfere with each other similar to other classical waves.

b) When going to the quantum mechanical framework of photons the interference , single photon interference too, is , is in the probability distribution which describes the individual photon's probability to hit (x,y) on the screen. Thus the wave is a probability wave in this case, not an energy wave as classically. The two frameworks have been shown to be consistent when a large number of photons is involved, as can be seen in the link given above.

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anna v
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  • Anna, could you add a clear statement/ answer to my question. – HolgerFiedler May 12 '15 at 06:57
  • I would not encourage people to think in a picture where photons build up some sort of classical wave. A photon is a quantum of a quantum field, i.e. it's the outcome of a classical measurement on that quantum field. Photons do not have an independent reality of their own without that measurement any more than a quantum state of an atom has any reality to it without the atom. I know that particle physics easily misleads one to believe this, but that's an artifact of the way we measure particle states in detectors with series of weak measurements. – CuriousOne May 12 '15 at 09:05
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    @CuriousOne you will allow me to think also the same about field theory, that it is a very useful mathematical artifact that should not be confused with reality, which is the spots on the screen. In fact using such a framework to explain particle physics, and photons are particles, to people unaware of it is by far more confusing. That is why the shrodinger equation and simple quantum mechanics is taught, to build mathematical intuition, imo. – anna v May 12 '15 at 09:47
  • @CuriousOne . And certainly a probability wave is not a classical wave. Sines and cosines do not emerge naturally from field theory, one would have to go to great mathematical contortions. – anna v May 12 '15 at 09:49
  • I do agree with you on that completely. Field theory or any theory should never be confused with reality any more than a biography should be confused with a real life. The problem with the photon as a particle picture is that it is not contained in quantum field theory. Particle tracks are the result of weak measurements, they are NOT a fundamental property of every quantum system. – CuriousOne May 12 '15 at 09:49
  • What is a "probability wave"? Is that the kiddie formulation of quantum mechanics that we have left behind some 80 years ago? Do sines and cosines exist in nature? Of course not. If you EVER taught a student of yours otherwise then you have been lying to them. – CuriousOne May 12 '15 at 09:51
  • @CuriousOne Mathematics exists. it can be simple it can be complicated but it certainly can be used to model data and predict results of experiments. Wave is identified with solution of wave differential equations, called so because of their great success with classical physics data/obsrvations. It so happened that the differential equations modeling the micro frame of particles are wave equations. and their solutions squared identified with probability. – anna v May 12 '15 at 10:00
  • None of that has anything to do with my original remark that you are falsely suggesting that photons make up a classical wave. "Photons" are measurements that indicate the change of a quantum number of a quantum field. Nowhere does quantum field theory require a bunch of photons to march in lockstep to weave a classical wave. That's simply a false physical image of reality. Seriously, you need to stop putting this into the heads of the kiddies. They will believe you and then it will take them years to get rid of this old-thyme myth. – CuriousOne May 12 '15 at 10:05
  • @CuriousOne I think you are wrong or have not understood how the classical wave emerges from the photons. why do you not read Lubos Motl link I have given above that shows how this happens. http://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html – anna v May 12 '15 at 10:21
  • @CuriousOne Actually it would be terrible for physics if this did not happen, if there was no smooth transition from classical to quantum and vice verso at the limits of the validity of each view – anna v May 12 '15 at 10:24
  • I am not aware that Lubos Motl can or has changed the definition of "measurement" in QM? Anyway, you are welcome to hang on to your phantasy of "photons" zipping around in space weaving classical fields while I will simply stick to the basics of testable physics. I am quit happy with that. – CuriousOne May 12 '15 at 10:24
  • Anna, please, physics does not change just because YOU have some terrible fears of what might happen if YOUR personal interpretation of reality is wrong. NATURE IS, whether we think that thunder is a giant hammer coming down or not. The only question is whether there is a SCIENTIFIC reason to attach reality to photons. There isn't. A photon is a quantum number. That's all the theory says it is. Maybe you have seen too many Feynman diagrams? They are not showing photons, but terms in a mathematical perturbation series. – CuriousOne May 12 '15 at 10:28
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    @CuriousOne It is not an interpretation of reality, it is a modeling of reality. Both the simple quantum model and field theory model data, they are not reality. they are models. It is you who are confusing reality with mathematics dogmatically. There is consistency between the simple shrodinger picture and the field theoretical picture, and some modeling describes simply the data and other modelings describe other data. Calculating crossections for particle physics needs field theory. This does not mean that the simple solutions of the hydrogen atom of the potential model are wrong. – anna v May 12 '15 at 10:33
  • To me a "photon" is a number on my computer screen after a PMT has triggered. I am not attaching ANY physical reality in form a small ball with a complex phase to it that keeps zipping around in free space until it hits something. Only your interpretation does that. You are misinterpreting the math of perturbation expansions of quantum fields as actual physical objects, which is an eighty year old mistake. Even Feynman warned against doing that. No balls. Period. Not now, not eighty years ago, not since the beginning of this universe. – CuriousOne May 12 '15 at 10:37
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    @CuriousOne please do not interpret me and let us agree to disagree. I am an experimentalist and stick to measured numbers. The double slit single photon at a time shows a probability distribution, and sine and cosine functions fit it, that is all. It is a measurement. btw the way Lubos shows the classical wave to emerge from photons is starting from field theory. – anna v May 12 '15 at 10:40
  • When you talk about photons forming fields you are not sticking to numbers. Your numbers on the screen do not form the quantum field or a classical field or anything. They are merely ONE classical measurement on a quantum mechanical object. That measurement can not be repeated and only its expectation values are meaningful. Truthfully, as a particle physicist should know that what we keep measuring are black box responses. Even if the system under test was classical, it would still be a black box. But it isn't, so that removes anything but the expectation values from the non-mystics table. – CuriousOne May 12 '15 at 10:45