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I am reading Anil Ananthaswamy's book on double-slit interference and I realized that there's something which I thought I understood about this, but actually do not- as follows:

Photons propagate through space without interacting with each other. However, if one light beam made of photons in large number is phase-lagged by half a wavelength behind another light beam going in the same direction, the peaks of one beam are said to coincide with the troughs of the other and they then "cancel" each other out (according to Ananthaswamy), yielding a diffraction pattern with dark stripes in it where there are no photons anymore.

But since photons carry energy from one place to another, what happens to the energy when they "cancel each other out" in this way? Do the photons which would have otherwise ended up in a dark stripe get "steered" somehow into an adjacent bright band and add to the energy flux there?

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    $\begingroup$ More on energy conservation in interference. $\endgroup$
    – Qmechanic ♦
    Commented Nov 19 at 17:49
  • $\begingroup$ To clarify: you are asking about the process of interference in a photon-by-photon basis, correct? Not in the classical EM wave manner. $\endgroup$
    – agaminon
    Commented Nov 19 at 17:57
  • $\begingroup$ “ Photons have integer spin and therefore can pass right through one another without interacting as they propagate.” Regardless of spin state photons do not interact with each other, as electromagnetism is linear except for a very small effect known as Delbrück scattering. $\endgroup$
    – my2cts
    Commented Nov 19 at 19:05
  • $\begingroup$ Photons have no phase or wavelength. $\endgroup$
    – my2cts
    Commented Nov 19 at 19:07
  • $\begingroup$ Tip: Let's not have posts look like revision histories. $\endgroup$
    – Qmechanic ♦
    Commented Nov 20 at 8:37

3 Answers 3

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Photons do not cancel each other, yes that would be a violation of conservation of energy. Photons are only created by excited electrons in atoms and they are only and always destroyed by absorption by electrons in other atoms. The Huygen's/Fresnel cancelation explanation for the DSE is very convenient and it mathematically produces the "interference" pattern but it is an oversimplification from the 1800s/1930s and still used today.

Each photon acts on its own as concluded by great physicists like Dirac and Feynman, famously stated as "each photon interferes with itself". The pattern of interference has been shown to be created even if one photon is fired at a time. Interference can never be observed without an apparatus ... all apparati haves modes (like a laser cavity), the EM field of the apparatus interacts with the EM field of the photon, Feynman's path integral can calculate the modes, these are the paths for the photon(s).

The Feynman integral shows that photons like to travel path lengths full wavelengths multiples (resonance) and the calculated results are identical to Huygens/Fresnel.

Photons cancelling has nothing to do with interference ... because photons can't cancel ... they just get absorbed.

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    $\begingroup$ You could say that photons can "cancel", in the event that two photons with sufficient energy come together and create e.g. an electron-positron pair. This is not what the question is about, though, and to my knowledge has no relationship to phase at all. $\endgroup$
    – Hearth
    Commented Nov 20 at 13:55
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    $\begingroup$ An addition to your answer : The realization that „all apparati haves modes“ can also be consistently applied to slits. Then any explanation of the phenomenon of intensity distribution behind edges due to interference could be omitted. Especially since no wave property of light from a thermal source or from a laser can be measured - they do not exist. Except for Young's interpretation from the double-slit experiment. $\endgroup$
    – HolgerFiedler
    Commented Nov 21 at 4:41
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But since photons carry energy from one place to another, what happens to the energy when they "cancel each other out" in this way? Where did it go?

The energy goes somewhere else where the photons add constructively, rather than destructively. When the photons are crossing at a small angle, they cancel out in some regions, but add in others, so that overall they conserve energy.

In the double-slit interference case you mention, the energy in the bright stripes and the dark stripes averages out to what would be there if the sources were incoherent and there was just a smooth distribution.

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As I see it, the solution of the classical wave equation determines the probability of photon absorption. Assuming monochromaticity, the energy density divided by the photon energy gives the average number of photons available for absorption. For a low intensity the probability follows a Poisson distribution evolving into a Gaussian at high intensity. In this point of view interference effects occur in the wave solution and frequency, phase and wavelength are properties of the wave. The photon only has energy, momentum, orbital and spin angular momentum.

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