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This is a concept I don't fully understand. If I have two photons each with frequency $\nu$, then they each have an energy of $E = h\nu$. If they get matched with an inverted phase, then the summed wave will be null due to destructive interference. Then where does the energy go? It cannot radiate, since that would produce an extra E-M wave, right?

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    $\begingroup$ Are you wanting to consider exactly 2 photons and nothing else? Or are you wanting to consider full EM waves and look at two photons within that? $\endgroup$ Commented Dec 17, 2020 at 22:00
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    $\begingroup$ This question is not a duplicate of what it was closed as. This question specifically seems interested in photons, not just waves in general. $\endgroup$ Commented Dec 18, 2020 at 17:37
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    $\begingroup$ Photons are particles. Waves interfere. $\endgroup$
    – ProfRob
    Commented Dec 19, 2020 at 10:16
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    $\begingroup$ OP is applying classical reasoning to a quantum mechanical problem, this can’t possibly end good. Photons are waves due to wave particle duality. QM guarantees that the total energy of the photon-wave is preserved as expected. OP should read up on quantum mechanics $\endgroup$ Commented Dec 19, 2020 at 22:56
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    $\begingroup$ I think you are asking the right question. Too much attention is given to the "wave" instead of the particle. What is a light wave anyway if not billions of coherent photon particles? Photon frequencies in-phase or out of phase don't effect each others energy while propagating. But at the detection screen opposing phases add up and the contributions are apparent. What experimental proof determined it had to be so called "Light waves" instead of many individual oscillating photons getting diffracted across a detection screen that creates the fringe patterns we observe?? $\endgroup$ Commented Apr 7, 2021 at 5:24

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In the present day physics standard model photons are elementary particles , on par with the other particles in the table. This means they are point particles, of fixed (in this case zero) mass with spin 1 and $E=hν$ . The $ν$ is the frequency that the classical light will have, as it is composed of zillions of photons This can can be seen experimentally, how classical interference appears because the beam is composed out of a large number of same energy photons.

wavepart

camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

Single photons leave a point consistent with the particle nature. It is the accumulation of photons that shows the classical interference pattern.

So two photons will not interfere in any way, except if one is studying photon photon scattering, which is very improbable for low energy photons. For high energy photons, gamma rays, a lot of particle antiparticle pairs can be created and there are plans of gamma colliders.

So there is no problem with the individual photons, they do not interfere. It is the wavefunction of the set up ( in the case above "photon scattering through two given slits given distance apart") that carries the frequency information of the photon, and can thus appear in the probability distribution. This should not be surprising as it is a quantized maxwell equation that gives the photon wavefunctions.

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You cannot have complete destructive interference everywhere unless the photons have exactly the wave vector (that is, are propagating in the same direction with the same frequency*). At some places there is constructive interference, while in some places there is destructive interference. The total energy, including regions of constructive and destructive interference, is just the sum of the energies of the constituent waves.

Consider, as a simple example, two waves (of equal amplitude and polarization) traveling in opposite directions (and, for now, only worry about their electric fields), $$\vec{E}_{1}(\vec{r},t)=E_{0}\hat{\epsilon}\cos(kz-\omega t)\\ \vec{E}_{2}(\vec{r},t)=E_{0}\hat{\epsilon}\cos(-kz-\omega t)$$ When you add these together, you get a standing wave $$\vec{E}=\vec{E}_{1}+\vec{E}_{2}=2E_{0}{\hat\epsilon}\cos(kz)\cos(\omega t),$$ for which the time-average of the electric energy density** $u_{E}=\frac{\varepsilon_{0}}{2}\vec{E}^{2}$ is $$\langle u_{E}\rangle=\varepsilon_{0}E_{0}^{2}[1+\cos(2kz)].$$ There are places [nodes, where $2kz=n\pi$ for $n$ odd], where the electric energy density is zero because of destructive interference; and there are places [antinodes, with $2kz=n\pi$ for $n$ even] where it is four times that of each original wave, because of constructive interferences. Averaged over all space, the total energy is twice that of a single propagating wave—exactly what we expect for a system with two waves.

*If you really want to consider two waves with identical wave vectors, then you cannot emit a second photon that is $180^{\circ}$ out of phase with the first. Generating an electric field with that phase is actually absorbing the first photon, not emitting another one.

**The magnetic energy has as similar structure, but it is situated slightly differently spatially, because there is an additional relative minus sign between the magnetic fields $\vec{B}_{1}$ and $\vec{B}_{2}$ of the two waves.

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  • $\begingroup$ I think the key here is in your first "*". A complication is that you can't really create a wave with a single wave vector nevermind two with identical wave vectors. But apply the same question to point sources. Again, you can't put the two sources in exactly the same place. But in an imaginary world where you can (dangerous: your results are always meaningless if physics is applied to unreal situations) I think you have to conclude that each source does such work on the other that the energy remains in the system to be dissipated in whatever mechanism is driving the sources. $\endgroup$
    – garyp
    Commented Dec 17, 2020 at 23:05
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    $\begingroup$ I think the question is about photons, and this answer is about classical EM waves $\endgroup$
    – anna v
    Commented Dec 18, 2020 at 10:20
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Your question is correct, photons do not really interfere. The DSE taught at the high school level is a convenient theory and it also works well mathematically but 2 photons cancelling is a violation of conservation of energy. In university in quantum optics courses deeper explanations are provided.

Think of 2 tsunamis one from Japan and the other from USA, starting with opposite phase .... when they meet (at say Hawaii) they cancel and Hawaii is saved ... but a second later the waves emerge again and continue on their way to Japan and USA, the energy was only stored temporarily in the elasticity of the water! The energy will only be absorbed when the wave crashes on the land. For photons we can never really observe the field directly ... we can only see a photon when our eye or camera absorbs it. We assume the photons are interfering in the EM field .... it makes sense .... but every photon is created by an atom and eventually absorbed by an atom.

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    $\begingroup$ Dear PhysicsDave. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$
    – Qmechanic
    Commented Dec 22, 2020 at 20:37
  • $\begingroup$ @PhysicsDave what is "elasticity of the water"? $\endgroup$
    – Roger Wood
    Commented Apr 7, 2021 at 1:30
  • $\begingroup$ @RogerWood check out wikipedia, water elasticity and compressibility. $\endgroup$ Commented Apr 13, 2021 at 2:22
  • $\begingroup$ @PhysicsDave Tsunamis as surface gravity waves. Nothing to do with elasticity. The same earthquake may generate longitudinal (acoustic) waves which do involve a liquid's bulk modulus (elasticity). $\endgroup$
    – Roger Wood
    Commented Apr 13, 2021 at 5:07
  • $\begingroup$ @RogerWood Take 2 waves in a pond of opposite phase (a frog jumps in as a frog jumps out) when the waves meet they cancel, the water is calm, but only for a short time, the waves will reemerge and continue along their path, that's physics. Where was all the kinetic energy stored when the water was calm for that brief second? $\endgroup$ Commented Apr 14, 2021 at 2:56

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