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I sometimes hear statements like:

Quantum-mechanically, an interference pattern occurs due to quantum interference of the wavefunction of a photon. The wavefunction of a single photon only interferes with itself. Different photons (for example from different atoms) do not interfere.

First of all -- is this correct?

If it is correct -- how do we explain basic classical interference, when we don't care about where the plane waves came from?

I heard that there are experiments with interference of two different lasers -- is this considered as a refutation of the statement? If it is -- how should one formally describe such a process of interference of different photons?

Finally -- such statements are usually attributed to Dirac. Did Dirac really say something like that?

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    $\begingroup$ Phtons can only interfere with itself. But think again, if photons are part of an entangled whole, then what? $\endgroup$
    – user1355
    Commented Mar 2, 2011 at 11:36
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    $\begingroup$ Lasers generate coherent light, each pulse of which can have many millions of photons in it, nearly all of which come from different atoms. Can you say that these photons aren't interfering with each other? $\endgroup$
    – Peter Shor
    Commented Mar 17, 2011 at 21:41
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    $\begingroup$ This question was discussed in detail in Glauber's Nobel lecture: journals.aps.org/rmp/abstract/10.1103/RevModPhys.78.1267 .The short answer is that it is NOT photons that interfere. It is the complex probability amplitudes corresponding to source to detector paths that interfere. $\endgroup$
    – Girish Kulkarni
    Commented Oct 13, 2021 at 21:37

8 Answers 8

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The photon model of light may be the most frequently over-applied model in physics. Lamb expresses my opinion fairly clearly here:

The photon concepts as used by a high percentage of the laser community have no scientific justification.

In my experience, many physicists who answer simple questions about matter without unnecessary reference to quarks or gluons are incapable of answering simple questions about light without unnecessary reference to photons.

This is puzzling, because very few experiments are capable of distinguishing between the existence and nonexistence of photons.

If you are genuinely interested in the photon model of light, then be prepared to do an awful lot of math to predict even fairly simple experimental results. You will, of course, be using a more correct model, but one should use the right tool for the right job.

If, however, you're interested in the experimentally observable behavior of light, then Maxwell's equations will give you the right answer in the vast majority of cases. For example, you ask if two different lasers can interfere. They can! See this question: Is it possible to observe interference from 2 independent optical lasers?

I'm sure the photon model predicts this result, but I suspect not without a fairly strong grasp of the math. If you'd never heard of photons, and all you knew were Maxwell's equations, this result isn't very surprising.

I'll close my answer with a question: For what type of experimental prediction is the photon model actually relevant? For what types of predictions is the photon model confusing, misleading, or more effort than it's worth?

Examples so far:
The photon model is relevant to:

The photon model is not relevant to:

My last two claims are intentionally bold. Prove me wrong!

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  • $\begingroup$ More advanced theory are usually more complicated. But it is by now mean argument against them! There is a number of experiments that need photons to be explained (when you have only one - it is simple as long it is almost monochromatic, see physics.stackexchange.com/questions/437/…). For two photon interference (e.g. Hong-Ou-Mandel) there is no way to express it with only Maxwell equations. Ah - of course 'photon' is only short hand notation for 'elementary excitation of QED vacuum'. $\endgroup$
    – Piotr Migdal
    Commented Mar 4, 2011 at 9:44
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    $\begingroup$ Ok, thanks for addressing my final question. I'll add Hong-Ou-Mandel to the list of photon-relevant experiments. Regarding your first point, I believe complication is a fine argument against using more advanced theory, prematurely. Surely you don't use the quark model when predicting the spectrum of they hydrogen atom! $\endgroup$
    – Andrew
    Commented Mar 4, 2011 at 11:22
  • $\begingroup$ "'Conventional' interference fringes from two independent lasers" - true. However, if you interfere states with a fixed number of photons, you will need photons (anyway - it is not hard! only creation and annihilation operators and simple transformations on them). More generally - the most common state (i.e. coherent state) in fact can be described with classical optics (at least the cases I know). $\endgroup$
    – Piotr Migdal
    Commented Mar 4, 2011 at 12:37
  • $\begingroup$ This is totally wrong. Re Lamb, see physicsforums.com/showthread.php?t=372653 $\endgroup$
    – user4552
    Commented Jun 15, 2013 at 15:14
  • $\begingroup$ related: physics.stackexchange.com/q/68147/4552 $\endgroup$
    – user4552
    Commented Jun 15, 2013 at 16:35
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A single photon can only interfere with "itself". However, "itself" is ill-defined because all photons are identical in quantum mechanics. Because of their Bose-Einstein statistics, the wave function of all photons is symmetric - invariant under all permutations of the individual photons. So the states in which some photons are permuted actually do interfere with each other - the symmetry must be preserved.

Two independent atoms emit photons spontaneously and the process is "random", so there's no correlation between the phases of the two photons. That's really why they can't interfere with one another. Also, one-photon states can't interfere with two-photon states (that would be like adding apples and oranges - one can't define any meaningful "sum of two functions" if the two functions depend on different variables), and photon states with different (perpendicular) polarizations can't interfere with each other, either.

A classical electromagnetic wave is a condensate of a large number of photons - essentially all of them are in the same state. The wave function of all the photons is the tensor product of the wave functions of an individual photon - or a tensor power of the state of each photon is the same. So the probabilistic wave function may suddenly be given a classical interpretation. $$\psi_n(\vec x_1, \dots, \vec x_n) = \prod_{i=1}^n \psi_1(\vec x_i)$$ Each single-photon wave function $\psi_1$ essentially evolves independently, so the total state $\psi_n$ for all the photons keeps its factorized form. The average electric field and magnetic field from the many photons carries the same information as $\psi_1$, a single-photon wave function, and they evolve in the same way, too.

However, you may still say that the interference of the classical wave with itself is "due to the interference of each photon with itself".

Your question is non-quantitative - linguistic rather than mathematical - so it's hard to answer it sharply. However, the right description is in terms of mathematics. What you need to understand is that the wave function of the whole system is a function of a maximum set of commuting variables. If those variables include positions, the interference can only occur in between states whose all other quantum numbers take identical values.

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    $\begingroup$ Photon today does not interfere with a photon tomorrow so there are cases when photons are distinguishable. $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 3, 2011 at 9:35
  • $\begingroup$ "Two independent atoms emit photons spontaneously and the process is "random" [TRUE], so there's no correlation between the phases of the two photons [ILL-DEFINED]. That's really why they can't interfere with one another [FALSE]." There is a firm distinction between one photon in 2 places $\alpha |x_1\rangle + \beta |x_2 \rangle$ (here the phase is crucial) and two photons $|x_1\rangle \otimes |x_2\rangle$ (here the phase is totally irrelevant). Two photon interference can be measured if you have two detectors (see my answer). With only one detector you won't see it. $\endgroup$
    – Piotr Migdal
    Commented Mar 3, 2011 at 10:48
  • $\begingroup$ Two successive photons are described with a one-photon state, not with two-photon one. When you say "two-photon state" it means two photons now. $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 3, 2011 at 14:58
  • $\begingroup$ "A classical electromagnetic wave is a condensate of a large number of photons" It consists of large numbers of photons, but strictly speaking it is not a condensate. A condensate is a vacuum expectation value. $\endgroup$
    – flippiefanus
    Commented Nov 26, 2016 at 4:26
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    $\begingroup$ Great, @flippiefanus - I should have said a "coherent state" but it would be less comprehensible to the target audience. The coherent state is a simple generalization of the "condensate". $\endgroup$
    – Luboš Motl
    Commented Nov 28, 2016 at 17:28
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No. Even though such claim is in Dirac's classical work it is not true.

See e.g. Hong-Ou-Mandel interference, when exactly two photons interfere (they can be even from different sources). For quotation of Paul Dirac, and some more analysis, see:

However, there is some truth in Dirac's claim - photon interferes only with it itself if you have only one detector to measure.

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  • $\begingroup$ It is wrong to say that two different photons interfere. It is correct to say that HOM is an interference of two-photon state. It is analogous to interference of one-photon state. There is no destruction of energy in the destructive interference. $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 2, 2011 at 14:03
  • $\begingroup$ @Vladimir Well, 'different' in the sense of it's not interference of photon with itself. And they can be from different sources (different lasers or stars) if you like; anyway I erase 'different' as it is confusing. I am aware of all stuff with (in)distinguishability, etc. Of course there is no destruction of energy (never implied such). $\endgroup$
    – Piotr Migdal
    Commented Mar 2, 2011 at 14:10
  • $\begingroup$ Indistinguishableness is easy to understand in HOM if you consider the mirror as a true source of two photons. $\endgroup$
    – Vladimir Kalitvianski
    Commented Mar 2, 2011 at 14:23
  • $\begingroup$ So Dirac was wrong; can you recommend a better QM text book? $\endgroup$
    – Physiks lover
    Commented Jun 1, 2015 at 21:34
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There are cases when a photon cannot interfere with itself. Take an interferometer with two nearly equal paths and observe interference. As soon as a photon has a limited length itself, it can interfere with itself only if the path difference is inferior to its own length. Otherwise it cannot. The trick is to have a superposition of waves at the same time. If wave trains of photons arrive without overlapping, there is no interference because there is no superposition. Superposition is a local in time notion. It is often implied rather than pronounced.

Gradually changing the interferometer path difference leads finally to destructing the interference pattern due to finiteness of the photon wave packet.

Now, two distant atoms (or lasers) are considered as a single source and "their" photons do not interfere only if they do not overlap in time (see above). A single source means that one photon interferes with itself, not two different photons from different atoms.

Dirac in his book on QM describes interference of a photon with itself but his statement was not original. I read a H. Poincaré article (1912) where he concludes that if we accept quantum nature of the light, then each quantum (photon) interferes with itself, not with other photons. Poincaré arrived at this conclusion by considering a very low intensity beam (flux of one by one quanta). But I am not sure whether it was pronounced for the first time or he wrote what was a "quantum folklore" of that time.

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Lubos Motl's answer is correct, of course, but I would like to point out a physical system in which it appears that photons from different sources are interfering with each other.

Consider two external-cavity lasers, nominally identical but running independently. HeNe lasers will work just fine. On the first laser, mount one of the cavity mirrors on a piezoelectric stack so that the cavity length can be finely controlled. Combine beams from both lasers through a beamsplitter onto a detector, run the output of the detector to an amplifier, and run the output of the amplifier to the piezoelectric stack. Turn the system on, and the first laser will automatically phase-lock to the second laser if the two beams are well aligned. Beams split off from the two lasers will interfere, and can even be used to make holograms, using one beam for reference and the other for object illumination.

Any interference pattern formed using the two beams is comprised of single photon events, but it is not possible to determine which laser any particular photon is coming from without destroying the interference. IF we say that the interference is always due to a photon interfering with itself, then we are forced to accept that the wavefunction of each photon has its source in both lasers.

All this goes to say that a photon is not exactly a particle. Its definition is not simple!

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The existence of relativity large holograms (produced or reproduced by a single lasers) indicates to me that one photon can interfere with other photons if they all maintain a fixed phase relationship.

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  • $\begingroup$ One could say each photon is taking its own path thru the medium, the medium just offers many structured paths that generate many images the you observe based on your position. Peter Shor above said interference is occurring in a laser as evidenced by coherence .... but the laser action is photon --> electron --> stimulated photon .... i.e. the photons never interact (or interfere) with each other. $\endgroup$
    – PhysicsDave
    Commented Apr 21, 2022 at 1:58
  • $\begingroup$ To produce a hologram, a laser beam is split. A small part is spread by a lens and goes directly to the hologram. The rest is spread and illuminates a scene. The photons which reflect from the scene and arrive at the hologram interfere with with the direct beam to produce an interference pattern which is recorded on film. To my knowledge, this does not work unless all of the photons leave the light source in phase. $\endgroup$
    – R.W. Bird
    Commented Apr 21, 2022 at 12:56
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This statement is based on a misconception about photon. Based upon this misconception it was assumed that the interference between two spontaneous transitions namely from 2 to 1 and from 3 to 1 in a three-level atom consisting of energy levels 1, 2 an 3 could not be observed. However, they were observed in sixties. See the criticism of E.T. Jaynes regarding this prejudice.

https://bayes.wustl.edu/etj/articles/quantum.beats.pdf

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  • $\begingroup$ The photon interfering with itself statement came long before the sixties!? $\endgroup$
    – PhysicsDave
    Commented May 22 at 17:03
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A photon does not interfere with itself, or any other photons. The apparent interference pattern that results from photons, or even a single photon, passing though a double slit apparatus, is due to the quantum mechanical wave guide structure that determines the probabilities for photons to go from the emitter to the detection wall via various paths. See http://ps.missouri.edu/feynman for some animations of probability amplitudes.

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    $\begingroup$ quantum mechanical wave guide structure This phrase would seem to want some elaboration... $\endgroup$
    – McGarnagle
    Commented Nov 13, 2012 at 21:35

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