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I understand how interference can arise when we start with a coherent light source where all the photons emitted are in phase. As they traverse the double slit, the distances to the screen from each of the double slits is different. The relative phases of the incident waves then sums to give the intensity.

And I have read various threads here

Young's Double Slit Experiment : What would happen if the "first slit" was too wide?

What makes the radiation behind slits coherent?

Why aren't interference patterns wiped out by random phase shifts?

and it seems in these answers, they are explaining the out of phase photons as being out of phase because the emanate from different locations on the source.

One commenter said "The first slit is an attempt to make the source look like a point. If the slit has a finite width the double slits produce displaced fringe patterns depending on where the light emanating from the single slit has come from".

Another commenter said if "you have an incoherent light-source, i.e. a non-point or extended source, and you place in its path a small enough opening, you're isolating light that was emitted, relatively speaking, from a single point on that non-point source, and hence that is already relatively spatially coherent."

But even from a given point on a filament, there is thermionic emission occurring concurrently at all depths into the filament(y coordinate) and heights(z coordinate) on the filament. This y and z coordinate spatial variability cannot be eliminated if your slits only diffract waves based on their x coordinate location.

There is also the fact that atoms are much smaller than the light wavelength they are emitting. So atoms next to each other or even a few atoms apart can emit much larger light waves slightly out of phase in time but they will not be resolvably different in their direction or their diffraction.

If we consider light emanating in a very specific direction and could measure the phase of light at some distance away from the source, then graph the phase of that light vs time we would see a band(the sum of all the phases of the unsynchronized photons arriving), not a sine wave as we we would with laser light.

So, at any given instant, the photons traversing the slits are not in the same phase even though they have an identical source direction. There is a population of photons (or waves). This population would only be 'coherent in direction' but not phase. As the waves diffract through the slits, then at every specific horizontal location on the screen we should be observing the interference pattern of two populations. Shouldn't this give a smear?

Yet we clearly see an interference pattern. I read the above referenced threads and their comments several times, but remain confused. One commenter even said 'The interference is washed out by random phase shifts! This is why the experiment is done with a laser,' but Thomas Young did this experiment in May 1801, centuries before lasers were around. His actual apparatus had a single slit far behind the double slit.

https://micro.magnet.fsu.edu/primer/java/interference/doubleslit/

I suspect such a single slit would 'filter out' all spatial phase variation of light arriving at the double slit. In other words, the light impinging on the double slit would be 'the same wave'. Yet temporal phase variation is not affected by this. Neither is phase variation that is spatial and smaller than the wavelength.

As I am relatively simple, I would appreciate a simple answer.

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  • $\begingroup$ The double slit is a perfectly classical wave experiment. The only places where you can have photons are the light source and the detector. Everything between tells us absolutely nothing about quantum mechanics. The semi-classical model of moving photons also doesn't do anything for you. It won't even let you calculate the classical double slit result. For that you would have to solve a path integral in the quantum mechanical description... but the path integral does not represent the motion of individual photons. $\endgroup$
    – FlatterMann
    Commented Oct 3, 2022 at 15:31
  • $\begingroup$ youtube.com/watch?v=WIyTZDHuarQ - This Veritasium video provides an interesting new view using the concept of 'pilot' waves. $\endgroup$
    – Janaaaa
    Commented Oct 4, 2022 at 6:48

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Historically photons are said to interfere ever since the days of Huygens but we physicists know that photons do not truly interfere as that is a violation of conservation of energy. But interference is still taught classically and does a good job of approximately modelling the concept of the slit and the thin film mathematically. So what is really happening? In the 1960s physicists experimented with sending one photon at a time for the double slit and amazingly the pattern still emerged. An explanation is offered by Feynman that said in summary every photon finds its own path that has a path length of an integer multiple of its wavelength. Thus every photon is constrained to travel paths that only satisfy this requirement. QM plays a role in this explanation as well, the photon has the highest probability of completing its path when the EM field is at a maximum at absorption, this occurs once every wavelength.

When we have many similar photons (i.e. laser or highly coherent) the paths are very similar and the interference pattern is highly visible. Young added a color filter and extra slit to make the photons more similar (or coherent) to aid in viewing.

A photon is an exchange of energy between 2 atoms, the 2 atoms are n wavelengths apart. If we consider a 2nd pair of atoms in close proximity to the 1st pair we could certainly say its photon is out of phase with the first one .... but by adding slits and lasers we are beginning to reduce/restrict the allowed paths so much so that we start to see patterns.

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    $\begingroup$ Interference certainly does not violate conservation of energy. $\endgroup$
    – garyp
    Commented May 16, 2020 at 23:24
  • $\begingroup$ Thank you. I appreciate the clarity of your sincere explanation. Especially I am enchanted with the Feynman interpretation of which I was unaware. However, this explanation seems to simply be the single slit interference (diffraction) pattern. khanacademy.org/science/physics/light-waves/… $\endgroup$
    – aquagremlin
    Commented May 17, 2020 at 3:15
  • $\begingroup$ So I’m back to the same point-if I have an excited atom and it emits a photon because of a rearrangement of one of its electrons, and another atom very near (in space and time) does the same thing- then you get two photons that are almost identical. Their integral path length will differ much less than a wavelength (because their sources differ by this very little amount) so they should land close together- closer than a wavelength. How do these photons group themselves into discrete packets a wavelength apart if their sources are packed only angstroms apart? $\endgroup$
    – aquagremlin
    Commented May 17, 2020 at 3:19
  • $\begingroup$ In practice I don't know if all the atoms in a filament are exicited, comparing 60 watts of photons to the size (#atoms) in the filament would be interesting. But based on Feynman paths, and also the steps Young took (color filter, single slit before double slit) we would conclude that bright spots on the screen do correspond to certain areas on the filament, i.e a group of proximal atoms, due to the tight geometry of the entire setup. These atoms can radiate at any time, phase (temporal?) not important. Dark areas correspond to filament areas where the photons radiate elsewhere into space. $\endgroup$
    – PhysicsDave
    Commented May 17, 2020 at 4:28
  • $\begingroup$ Khanacademy is at the high school or 1st year college level. Quantum optics is well beyond that. Feynman spent significant time on the double slit. The path approach works for single photon experiments like the YDSE and thin films. A laser is an example where if we upset the path length (mirror distance) then lasing stops. The word "interference" is practical but historical. Photons do not cancel each other out, they interact statistically with electrons mostly in atoms. $\endgroup$
    – PhysicsDave
    Commented May 17, 2020 at 4:39
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There is a very simple answer. Interference (in this case) occurs between each individual photon and itself-- not between different photons.

To calculate the interference pattern, you simply add the patterns due to the individual photons. When this is done, you obtain the pattern that is actually seen.

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  • $\begingroup$ Tho photons may interfere with themselves, that does not preclude the existence of other photons at the same locations with different phases. That is what my whole question is about. That we are making a macroscopic observation of many quantum events and resolving those events into groups (positively and negatively interfering events) implies that the collective behavior of the photons is grouped. Yet the source of the photons is a random array of very small particles. $\endgroup$
    – aquagremlin
    Commented May 17, 2020 at 2:54
  • $\begingroup$ It does not matter what the phases of the other photons are. Since the intensities (not the amplitudes) of the individual patterns are added, all that matters is where the peaks of the individual patterns are. The geometry of the slits ensures that the peaks mostly coincide. $\endgroup$
    – S. McGrew
    Commented May 17, 2020 at 3:19
  • $\begingroup$ "...existence of other photons at the same locations with different phases" - those other photons also interfere with themselves. For each single photon, the phase at one slit is always the same as the phase at the other slit. That is what gives rise the interference pattern. And that pattern is the same for all photons that have the same frequency. $\endgroup$
    – fishinear
    Commented Jan 11, 2022 at 16:36
  • $\begingroup$ The problem with the term "interference" is that it actually denotes the absence of interaction. It is a property of linear systems in which individual excitations (waves, photons) are passing each other without any change of their energy, momentum etc.. $\endgroup$
    – FlatterMann
    Commented Oct 3, 2022 at 15:24
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I woke up today with the realization that the populations of photons hitting each slit are identical. This in fact renders unnecessary any concern about what frequencies or phases the photons are. Of course this presumes a wave interpretation. No matter what crazy population is passing through a slit at a given instant, the SAME crazy population is hitting the other slit.

The 'particulate' behavior of photons was obscuring my thinking and allowing me to suppose that the populations of photons at the two different slits had differences, which is wrong. I need to remember that 'particulate' behavior is only the result of the fact that we can only detect photons with matter, which is particles. I think that statement 'photons interfere with themselves' is what S. McGrew was expressing. (so I marked his as the answer) But PhysicsDave's comment was actually more illuminating for me. (so I upvoted his)

My understanding leads me to predict two events: 1) if I could selectively remove certain phases after one slit, I would remove their interference with the 'full population' coming from the other slit and this would add a smeary character to the interference pattern. 2) If I added photons from another source to one of the double slits, this would also upset the symmetry of the two populations and also add a smeary character.

I wish there were a way for me to send venmo or paypal or some other REAL token of appreciation. When you take the time to sincerely explain and shed light, I appreciate such efforts.

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  • $\begingroup$ If 1000 scientists each made an identical apparatus and fired only one photon each and then got together for a big meeting to combine results (using a common reference/center point from each screen) they would indeed observe the pattern! Each photon determines its own path but has less freedom/choices given the setup; more historically and confusingly stated as "each photon interferes with itself". You can completely forget about classical "interference" (photons don't cancel) and similarly you don't need phase either! Dark areas in the pattern have no photons, bright areas all. $\endgroup$
    – PhysicsDave
    Commented May 18, 2020 at 1:46
  • $\begingroup$ @PhysicsDave Photons don't have a path. That a false semi-classical model that can't be used to explain the quantum mechanics of light. If anything, it is confusing. What is correct, however, is that wherever Maxwell equations predict a zero in the electromagnetic field, no photons can be detected. The "confusing" part is how the field "knows" where those zeros are without any self-interaction. The answer lies in a general property of linear wave equations: their solutions are determined by nothing else than the boundary conditions (because nothing happens in the volume). $\endgroup$
    – FlatterMann
    Commented Oct 5, 2022 at 9:23

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