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The textbook argument against photons canceling each other draws upon the conservation of energy. Does this mean that energy conservation is a "stronger" principle than superposition? Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat). Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?

Deschele Schilder
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Tfovid
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  • Electromagnetic waves and sound/water waves resemble each other slightly in their behaviour but they are fundamentally different things and can't be compared in the way you're trying to compare them. – Charlie Aug 07 '20 at 15:27
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    Typically when two waves cancel each other by destructive interference, they only do that in a limited area or region of space. In some other location there will be constructive interference and the two waves will reinforce each other rather than cancelling. Can you show a specific scenario where waves cancel everywhere in acoustics or mechanics? – The Photon Aug 07 '20 at 15:31
  • @ThePhoton I may not have examples from classical mechanics, but from quantum optics, I'd use a beam splitter placed, say, at the origin of two orthogonal axes $\hat{x}$ and $\hat{y}$. Let's say I send in two photons, one from from the top and towards $-\hat{y}$ and one from the left towards $+\hat{x}$. With the appropriate phase difference, if the two photons are generated from the same source (e.g., a GHZ state), they should cancel out ½ of the time as they "emerge" from the other side of the beam splitter. – Tfovid Aug 07 '20 at 16:00
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    @Tfovid, But the beamsplitter has two "outputs". If there is destructive interference at one output ($+\hat{x}$, say), there will be constructive interference at the other ($-\hat{y}$). – The Photon Aug 07 '20 at 16:04
  • If a single photon can "cancel" itself (at least partially) during the two-slits experiment, then why two identical in all properties photons can not do so? – Andrey Lebedenko Sep 21 '21 at 09:48

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You need to careful when you talk about photons as waves because one photon is certainly not an EM wave, which is a superposition of many many photons. Indeed if you are talking about very high energy photons they will produce an electrons positron pair. In other words you can convert the energy of photons into the "mass energy" (think about the famous equation $E = mc^2 \gamma$) of an electron positron pair. This process is called pair production and is entirely quantum mechanical. That's why you don't learn about it in your standard EM course.

Gonenc
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    What does pair production have to do with destructive interference of photons, though? – Tfovid Aug 07 '20 at 21:45
  • @Tfovid I think his point is that talking about "interference of photons" is misleading and nonsensical. We can speak of interference for EM waves, but for photons we instead have processes like pair production. – electronpusher Aug 08 '20 at 16:58
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    @Tfovif the other comment is exactly right. We do not talk about interference of photons except to show the bizarreness of QM. Any real and honest to physics treatment will either talk about photons and pair production or talk about interference of EM waves. – Gonenc Aug 08 '20 at 19:00
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The textbook argument against photons cancelling each draws upon conservation of energy.

Our theory of particle physics is called the standard model, and photons are point particles of zero mass , in the axiomatic table of the model,

and that is what the textbook is using. Yes conservation of energy is a very strict law. (In addition, photon photon interactions, and that is how elementary particles behave, are rare, for low energy photons, see here )

Does this mean that energy conservation is a "stronger" principle than superposition?

The classical electromagnetic wave when mathematically broken down into photons of energy hν (ν the frequency of the classical light) emerges in a complicated way from the quantum mechanical "addition" of the complex wave functions of each photon, the photon "wave" is a probability wave for each photon. (see this answer of mine, individual photons behave exactly the same way)

Waves in other media than the EM field, e.g., sound or water, do cancel out---presumably by passing on their energy to some other degree of freedom (e.g., heat).

Classical electromagnetic waves do not need a medium to travel, as the michelson morley experiment showed. But they do interfere in bulk. See this instructive mit video.

Could this imply that EM waves don't have any alternative channel to pass on the destructed energy and thus can't cancel out?

see above.

Do not confuse classical EM waves with their component photons.A building may be made out of bricks, but a brick is not a building

anna v
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  • +1 because I really like the building analogy and I think the answer address the confusion of the OP – Gonenc Aug 08 '20 at 19:02
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We should be careful to distinguish interaction, correlation, annihilation and interference. Photons do not interfere. Any interference takes place at wave function level, so impacts the probability of finding a number of photons. Photons can annihilate but this requires two photons of a least 511 keV each in order to create an electron-positron pair. Photons can interact (scatter) via transient vacuum charge fluctuations. Finally photons can be correlated by Bose statistics.

my2cts
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All waves traveling through a medium don't cancel out. Sound waves, water waves, waves in a rope, etc. pass each other and travel further after they have passed. They don't (or almost not) exchange energy with the medium (like being converted to heat, though there is dampening). Two oppositely traveling waves may seem to vanish for a moment, but the medium contains the kinetic energy of the waves.
The same holds for classical em waves, although they don't travel in a medium. A circular em wave traveling outward from a center and a circular wave moving toward the same center will pass each other and continue their journey, without losing energy because they can't lose energy to the vacuum.
So they don't cancel out (they interact and travel along as if they didn't have encountered each other) because they don't have a channel to lose their energy.

Deschele Schilder
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It is a curious thing, because in the early days of quantum mechanics it was thought that wave interference effects (which is what I think you mean by cancelling) were only possible for a particle and itself (Dirac said exactly that, although I would have to put in a fair amount of work to find a reference).

This is indeed true for most particles in quantum mechanics. However, since the advent of quantum field theory, it has been recognised that the wave function for a photon does not describe the probability where where the particle may be found, but rather describes the probability for where it might be found that a photon has been annihilated. It actually does not have to be the same photon. Wave interference effects have actually been observed by astronomers for photons originating in different galaxies (sorry I don't have more in the way of details).

What I can say, is that your text book is out of date on this particular topic.

Charles Francis
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  • Is the takeaway that two photons can indeed interfere destructively? If so, what happens to the energy? – Tfovid Aug 07 '20 at 15:54
  • A region of destructive interference is balanced by a region of constructive interference, so there is no issue with conservation of energy. – Charles Francis Aug 07 '20 at 15:58
  • I just added a comment in response to @ThePhoton above where I have a hard time how such regions would arise in a single shot experiment. (If anything, these compensating constructive regions are more statistical than spatial.) – Tfovid Aug 07 '20 at 16:02
  • All the results of qm are statistical. Even the existence of space itself. That is why it is so hard to think about. – Charles Francis Aug 07 '20 at 16:13
  • I disagree because photons can and do cancel each other in pair production. This phenomena has absolutely nothing to do with the highly ambiguous words "wave-particle duality" for photons. – Gonenc Aug 07 '20 at 16:58
  • "it has been recognised that the wave function for a photon does not describe the probability where where the particle may be found, but rather describes the probability for where it might be found that a photon has been annihilated" Do you have a reference for that? The "modern viewpoint" I know is that photons don't have a wavefunction as a function of position because they're fully relativistic, cf. https://physics.stackexchange.com/a/463454/50583. – ACuriousMind Aug 10 '20 at 10:49
  • @ACuriousMind, digging up a reference would take time, but I am almost sure the basic idea was explained in either Bjorken & Drell or Peskin & Schroeder, which have been my two main text books. Otherwise my main source was lecture notes at Cambridge, given by a student of Dirac. Mathematically it is straightforward. Act on a photon state $|\psi\rangle$ with the photon field operator. The creation part of the photon field operator can be ignored (it is only relevant for states with an indeterminate number of photons, but here we are looking at a single photon state). – Charles Francis Aug 10 '20 at 16:27
  • The annihilation part of the photon field operator gives the photon wave function. You could be right that this is not the "modern view". My views are almost certainly closer to the pre-war views of Dirac, but Dirac was far more mathematically minded than modern physicists, and while I have found it possible to put his development into a rigorous mathematical framework, I cannot say the same for modern views of qft. – Charles Francis Aug 10 '20 at 16:36