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I am not asking about whether the photon goes through both slits, or why. I am not asking whether the photon is delocalized as it travels in space, or why.

I have read this question:

Do we really know which slit the photon passed through in Afshar's experiment?

Which theory explains the path of a photon in Young's double-slit experiment?

Shooting a single photon through a double slit

Where John Rennie says:

The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through both slits. The photon position is only well defined when we interact with it and collapse its wave function. This interaction would normally be with the detector.

Lasers, Why doesn’t a photon go through the same slit every time?

Where ThePhoton says:

for example, if you put a detector after a two-slit aperture, the detector only tells you the photon got to the detector, it doesn't tell you which slit it went through to get there. And in fact there is no way to tell, nor does it even really make sense to say the photon went through one slit or the other.

In classical terms, this question might be obvious, because a classical billiard ball cannot be at two places in space at the same time. But this is not a billiard ball, this is a photon, a QM phenomenon. And this is not classical terms, but QM.

And if we truly accept that the photon travels through both slits, then it basically must exist in space at both places (both slits) at the same time.

But as soon as we interact with it (the wave function collapses), the photon becomes spatially localized, but only at a single location (at a certain time).

What is not obvious from QM, is how we can have these two things at the same time:

  1. the photon pass through both slits

  2. but we can only interact with it at one slit (not both)

What is that basic thing in QM, that will disallow for the photon to pass through both slits and be interacted with at both slits too? Somehow the QM world underneath will change to classical as soon as we measure, and interact with the photon. This change from QM to classical is where the possibility of the photon being at both places (both slits) at the same time gets disallowed somehow. This could be decoherence, as the QM entity gets information from the environment (because of the measurement), or just the fact that the wavefunction collapses and that has to have a single spatial location for the photon when measured.

So basically the photon goes through both slits, thus, it in some form exists at both slits at the same time. But when we try to interact with it, it will only be spatially localizable at one of the slits, not both at the same time.

Question:

  1. If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?
Árpád Szendrei
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    What's wrong with a good ol' EM wave interfering through the multiple slits? – user192234 Nov 16 '19 at 19:34
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    John is correct in saying there is not a defined trajectory. So why do you ask a question assuming there is a trajectory? – BioPhysicist Nov 16 '19 at 20:11
  • @AaronStevens I am not assuming a trajectory, just the opposite, I am assuming the photon goes through both slits. The question is, if it goes through both (thus in some form exists at both slits), then why can't we somehow measure this existence of the photon at both slits (at the same time)? What is the reason that when measured, from QM, we switch (maybe because of decoherence) to classical, where it is obvious that the photon cannot be (measured) at both slits at the same time? – Árpád Szendrei Nov 16 '19 at 20:39
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    This seems a question related to the measurement problem: https://en.wikipedia.org/wiki/Measurement_problem so I would say there is not a definitive answer to it – Claudio P Nov 17 '19 at 22:02
  • There is no proof or reason that a single photon goes through more than one slit at a time. Place a detector at both openings and only one will register. There is no reason it can’t be classical all the way and still produce QM predictions. – Bill Alsept Nov 18 '19 at 22:24
  • Why is this question protected? Weird. – user192234 Nov 27 '19 at 20:57
  • Note that you actually need true single photon sources to observe this fact (i.e. "no coincidence detections"). If you just attenuate a normal laser beam to have a "mean photon number less than one per some reasonably large time interval" (as is the most common suggestion in popular science discussions of single photon experiments) , you will in fact have a coincidence rate that matches the preductions of classical physics. – Adomas Baliuka May 28 '20 at 08:56

13 Answers13

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Think of it this way: A photon is the detection event. When there is only one photon, there is only one detection event. The probability distribution of detection events is associated with the photon's wavefunction.

S. McGrew
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    I really like this answer. The photon itself is the manifestation of the measurement that we make on the excitation of the field, and that can only be at one place. – Árpád Szendrei Nov 17 '19 at 00:40
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    While a deeper explanation might demonstrate truth to this statement, the current short version does not sit right with me. This answer feels very "hand wavy." – Aaron Nov 18 '19 at 15:13
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    Surely "photon" is the name we use to talk about a degree of excitation of a field, whereas a detection event is an avalanche of electrons in a detector, or a chemical reaction on a film, or a flicker of a needle. Furthermore, we can easily design detection events which indicate unambiguously that the field had non-zero excitation at two places at once. – Andrew Steane Nov 18 '19 at 19:08
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    Can that be done for single photons? If so, how? – S. McGrew Nov 18 '19 at 19:32
  • First map the photon's motional state onto some other degree of freedom (e.g. internal state of an atom) by a unitary process (the 'swap gate' in quantum computer language) then manipulate and measure that degree of freedom however you like. Alternatively, just shine light at the slits in Young's slits experiment performed with, say, atoms, and collect the scattered photons. If the interference pattern of the atoms is still visible, then the scattering events must be delocalized. – Andrew Steane Nov 19 '19 at 10:59
  • The 'swap' operation can be performed, for example, by passing a photon through an optical cavity containing an atom. The photon frequency should be far-off-resonant with any internal transition in the atom, to avoid absorption, but resonant enough for the atom to present a refractive index sufficient to shift the phase by $\pi$. Then the passage of a single photon rotates a suitably prepared atomic state to an orthogonal one. This has been done in the lab. – Andrew Steane Nov 19 '19 at 11:01
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    Please provide a reference. – S. McGrew Nov 19 '19 at 14:16
  • @AndrewSteane rotating to some orthogonal state is not sufficient for a swap gate.. Also, do we say the photon is a part? is it not? has it changed? – user192234 Nov 27 '19 at 21:11
  • Still hoping for a refrrence. – S. McGrew Nov 27 '19 at 21:26
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    Sorry to be slow on reference. My remark was prompted by a specific experiment, but I have forgotten the authors. Rempe perhaps? There have also been experiments in which photon number in a cavity was measured without absorbing any photons, by passing atoms through the cavity. e.g. Haroche group in Paris. – Andrew Steane Nov 27 '19 at 23:08
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If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

Alright, let's play some word games:

This isn't a well-defined question. "Detect a particle" doesn't mean anything in quantum mechanics. Quantum mechanical measurements are always measurements of specific observables. There is no holistic act of "observing all properties of a system at once" like there is in classical mechanics - a measurement is always specific to the one observable it measures, and the measurement irrevocably alters the state of the system being measured.

People often use "detect a particle" as shorthand for "perform a position measurement of a particle". By definition, a measurement of position has as its outcome a single position, and interacts with the state of the particle being measured such that it now really is in the state in which it is at that single position and nowhere else. So if you could perform position measurements that yielded both slits as the position of the particle, this would mean you have performed an impossible feat - there are now two particles, each in the state of being at one slit and that slit only. Quantum mechanics may be weird, but it is hopefully clear it is not this weird - we cannot duplicate a particle out of thin air just by measuring it.

If you don't insist on "detect" meaning "performing a position measurement", then of course the standard double slit setup is a "detection" of the photon at both slits - the pattern on the screen is only explainable by the particle's wavefunction passing through both slits and interfering with itself. This is of course just indirect reasoning - there simply is no observable whose eigenstates would naively correspond to "we have detected the photon at both slits at once".

Lastly, you seem to confuse "interacting" with "measuring" or "detecting". Of course we can interact with the particle at both slits - we just cannot perform position measurements (or other "which-way" measurements) at both slits and expect them to yield the impossible result of the particle split in two. But if you look at more sophisticated setups like the quantum erasers, there certainly is interaction with the particle at both slits - just carefully set up to not destroy the interference pattern, and hence no obtaining useable which-way information.

ACuriousMind
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  • Thank you i appreciate your answer. Do you mean that if there are detectors at both slits that do not collapse the wavefunction (I guess just elastic scattering) then it is possible to interact with the photon at both slits (this just does not qualify as measurement)? – Árpád Szendrei Nov 17 '19 at 02:10
  • "The measurement irrevocably alters the state of the system being measured" is false. Rather, the measurement constrains the possibilities of other measurements to ones consistent with it, as determined by the underlying probability model. – R.. GitHub STOP HELPING ICE Nov 17 '19 at 20:09
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    @R.. I fail to see the difference between changing the probabilities of future measurements and changing the state. A quantum state is essentially defined by the potential results of measurements on it, when you change these, you change the state. – ACuriousMind Nov 17 '19 at 21:29
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    @ÁrpádSzendrei: Yes! Put a vertical polarizer at one slit and a horizontal polarizer at the other. Then the particle still passes through both slits but no longer in a manner that results in the interference pattern. Changing the angle of one of the polarizers gradually results in a gradual reappearance of the intereference pattern. This clearly shows interaction can happen at both slits while not completely collapsing the wavefunction. More simply, just put a glass piece after one slit, which would shift the interference pattern due to the higher refractive index of glass than air. – user21820 Nov 19 '19 at 07:15
  • Despite my comment above, I may be closer to understanding the double-slit experiment with individual photons than I ever have been before, thanks to this answer. – T.J. Crowder Nov 19 '19 at 17:58
  • A two slit pattern can be derived from many individual photons passing through one slit at a time but still no explanation of a particle going through two slits at once. Also the best part of this question is if I and truly. – Bill Alsept Nov 20 '19 at 00:52
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We've had a lot of answers already (because this problem invites them), but let me offer one more way to think about it. (As best I can tell, this is the interpretation of quantum mechanics closest to the point I'll make. As @PedroA notes below, what follows is interpretation-dependent.)

If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

I think you're imagining we, as the scientists with our detector, are a classical system studying a separate quantum-mechanical one. But the entire experiment, including the detector and whoever inspects it, is also part of the quantum-mechanical setup. Our superposition isn't just of the photon passing through slit $1$ and its passing through slit $2$; it's of us detecting one and us detecting the other.

From a God's-eye point of view (if there is such a thing), we are superposed between announcing one result and announcing the other. We're not outside a quantum-mechanical system with such a God's-eye view, and therefore don't see the whole of the superposition. Hence we only see one result, not a bit of both.

J.G.
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  • This is an interesting answer after careful read, +1. However, even though you added fantastic links to wikipedia, I think you should emphasize that this is only one of the various interpretations, and this matter is mostly philosophical than physical. When I first read your answer, the tone seemed to imply that what you're stating is scientifically true. – Pedro A Nov 19 '19 at 23:00
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    This is the over-all approach of what is called Everettian Interpretation. It is focused on the Schrodinger equation, the dynamics of states and their time evolution. A lot of postulates can actually be derived if you obey the equation and study it. However you are not telling about some very surprising ontological facts that emerge if you pay close attention to this piece of work :) – user192234 Nov 27 '19 at 20:03
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Yes we can but the detectors should not completely destroy coherence. If not the interference pattern will be gone. For example two parallel polarisation filters should not destroy interference.

user21820
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my2cts
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  • Can you elaborate? (Really interested) – user192234 Nov 16 '19 at 20:31
  • Can you please tell me, are you saying that we can have two detectors on the two slits, and for a single photon that is shot, both detectors will interact with the photon, just this should be elastic scattering? The polarization filters should be set up how? – Árpád Szendrei Nov 16 '19 at 20:45
  • I somewhat remember from undergrad days, that the experiment has been done (with electrons). Turn on the detector for either slit, and the interference pattern vanishes because you know which slit each electron went through. (This for the regime where the electrons are microseconds apart, and the (non-)interference pattern is accumulated over seconds to minutes of "exposure"). – nigel222 Nov 19 '19 at 14:04
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You are asking for an answer that makes sense.

Quantum mechanics was not designed to make sense. It was designed to get correct answers. You can't expect it to make sense. That isn't what it's for.

If you want a story that makes sense (but might be wrong) here's one: Light traveling through space behaves exactly like a wave. There is no problem whatsoever about a wave going through two slits at the same time. That just vanishes.

Our methods to detect light are all quantized methods. Light changes a crystal on a photographic film. Or it sets off a photomultipler tube. Etc. They all give quantized detection. If you want a detector to tell you the amplitude of the wave, you need something that will take so many quantized measurements that they average out to something that seems continuous.

Since the measurements are quantized, of course QM will predict quantized results. That's what it ought to do if it is going to get correct answers. It will get answers that are compatible with the data.

There might be some weirdities in how light interacts with atoms. Those will affect the data. But there are no known weirdnesses about light traveling through space, it is all entirely compatible with light traveling as a wave.

QED is partly about describing light as quantum particles that behave exactly like waves. There's a lot of handwaving about probability functions etc. It's simpler and easier to just describe it as a wave, but QED gets the right measured answers too.

J Thomas
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    Do you mean "You can't expect it to make intuitive sense?" QM tends to make a lot of sense for people who accept that it is not intuitive in the sense that CM is... – Marius Ladegård Meyer Nov 17 '19 at 00:39
  • Yes, that's what I mean. – J Thomas Nov 17 '19 at 05:04
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    I dont think this answer is true. Its possible to place a detector at both slits, get no result and yet the diffraction pattern will be altered. – untreated_paramediensis_karnik Nov 18 '19 at 12:14
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    If light is acting as a wave, why would you expect that a detector at one or both slits would not affect the diffraction pattern? – J Thomas Nov 18 '19 at 14:04
  • @MariusLadegårdMeyer Do you mean "make intuitive sence to laymen"? QM tends to make a lot of intuitive sense for people who accept that intuition can be developed throughout your life, and make effort to learn intuition. As a comparison, NASA engineers in the 50s and 60s had very little intuition about orbital manoeuvring although it's purely CM, but nowadays people who clock thousands of hours on Kerbal Space Program find basic maneuvers very intuitive. – JiK Nov 19 '19 at 15:54
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    @JiK, touché. The point I was trying to make was that the phrase "does not make sense" implies that it is wrong. Of course intuition in most fields improves with experience and expertise, but I think I will stand by the statement that QM is not intuitive in the sense that CM is. At least the professors and postdocs I've worked with seem to feel that way. – Marius Ladegård Meyer Nov 19 '19 at 16:50
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    "Does not make sense" does not imply that it's wrong. "Common sense is what tells you the earth is flat." Not wrong, just hard to picture. – J Thomas Nov 19 '19 at 20:47
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Can the photon be detected at both slits, of course not, it can not even be detected at one slit ... it is only detected when the EM field energy collapses and excites an electron .... science today cannot detect when a photon passes close to an electron (in a slit) and maybe disturbs it somehow. So why do you even care whether a photon passes thru one slit or the other? ... you care because because you are trying to explain this mysterious pattern that appears on the screen and you have been told it is due to "interference". Historically it has been described as an "interference" pattern because the pattern looked much like water wave interference. (And of course this is the basis for the described wave nature of light.) You believe this explanation but it requires that energy passes in both slits in order to geometrically interfere and this is where things gets very confusing.

But there are 2 aspects you should be aware in the modern thinking, 1) Feynman allowed paths and 2) photon wave function. 1) Feynman attacked the same problem you are attacking, and his eventual proof was that photons needed to travel n times a multiple of their wavelength ... much like the length of a guitar string can only play one note (or frequency) and also much like a laser cavity where if the dimensions are not correct photons will fail to propagate in the desired path. (Note that the Feynman explanation also accounts for the observations in single photon experiments.) 2) as John Renee highlights the photon is delocalized and he even expresses that the photon as a fuzzy sphere, this is the photon wave function described in words. To take the description further we can say the sphere gets larger and larger at the speed of light until the "receiving" atom is found and decides (by probability and QM) that it will take all the energy. At his point the sphere collapses and all the energy proceeds to the "receiving" atom. Maybe one could argue that the fuzzy sphere was one big virtual photon with no energy and that the real photon is where all the energy goes and it takes the best path to the receiving atom, who knows.

Feynman has shown that the photon does not need to go thru 2 slits to have a wave like "interference" property, he has shown that light is a wave because it travels on paths that are harmonic, i.e. the path travelled is dependent on the photon energy/wavelength. The photon wave function (John Rennie) tells us that the photon looks everywhere for a path ... and eventually collapses to a single atom/electron. So in conclusion I would say both answers are correct ...it passes thru 1 slit and both slits!! ... but it is undetectable until the screen.

PhysicsDave
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  • I give you a +1 for considering the individual photon derivation of the Fringe pattern. billalsept.com – Bill Alsept Nov 20 '19 at 07:44
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    Yes the explanation works for double slit diffraction, single slit diffraction, thin film "interference" (especially with single photons), laser cavities and probably a few more. If the word interference is used then there is likely a "better" explanation using the Feynman theory, i.e. one that keeps energy conserved! – PhysicsDave Nov 20 '19 at 23:49
  • Yes it’s all individual photons convoluted to create the fringe pattern. It can even be demonstrated how thousands of individual and coherent photons radiating from a common source resemble a spherical wave. – Bill Alsept Nov 21 '19 at 00:03
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First try: We all know that if we just block one slit then it would definitely go through only one of them.

The only thing you can do in order to know that in some way it can be postulated that a photon goes through different slits at a given time is if you unblock the 2nd slit.

If you Detect and gain knowledge about having it gone past the slits then you just caused decoherence and it is not superposed anymore.

You can get more crazy about it but it's not necessary. You will not get more evidence then just the appearance of the interference pattern, it is quite a lot in favor of what you want to prove is occurring in reality :)

P.S:

You don't have to measure.. You can have an arbitrarily long period between every photon emitted, and come back years later to find a scatter plot that is converging to an interference pattern on the screen. Before enough time passes for the mod squared to actually reach the screen it will not interact (with thin air). Once it reaches the screen an interaction will happen in accordance with the running expectation of the mod squared..

user192234
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The photon goes through both slits

Keep in mind that this is really only the closest approximation of what happens that we have language for. Nothing can exist at two places at once, and QM doesn't change that.

However, it does some things with probability and uncertainty that we can only really describe as "totally weird".

In my very un-humble opinion, the best way to think of it is as a probability waveform itself travelling - much like any other wave travels. You could for example calculate that it's got 50-50 percent chance of going through either slit. Then after the slits, the probability waves interfere and create the known interference pattern.

If you instead measure the photon, then it's as if you emitted it from that specific location (since you know where the photon is and you have no waveform anymore), therefore creating a simple normal distribution.

Do note that this is NOT more accurate than stating it's at two places at once (as far as I'm aware, at least.). But it's a way to think about it that produces a mental image which is basically as accurate.

Gloweye
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If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

If you take seriously Feynman's many paths theory (path integral) of QM used in Quantum Field Theory, there is evidence that the particle doesn't just go through the two slits, but it actually takes every possible path. That means part of its "path" includes going around the sun, then Jupiter, and make its way back to earth to make a blip on a screen. Take all other possible paths you can think of, add them up, and that is the particles "path". In that sense, a defined path is a very classical idea.

Instead think of probabilities of being at various locations.

Let's modify your thought experiment and think about what happens if we put 1 detector behind the left slit. As soon as that one detector is added, the interference patterns disappear. We get classical results (perhaps single slit, I am not sure) both in our detector and our original canvas that captures the right side. As soon as we even attempt to detect which slit it goes through, QM reverts to classical results which give concrete answers to which side the particle went through. At that point it didn't go through both, it went through one as a classical particle.

As for why, we don't really know. See here for a recent experiment that tries to answer this. https://phys.org/news/2011-01-which-way-detector-mystery-double-slit.html

brian_ds
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    Why do you say we don't really know? We do! A detector (by definition) is simply a device that causes wavefunction collapse, so that we can obtain a classically measurable outcome. So obviously, if you put a detector near a slit, it will collapse the wavefunction. In layman terms, it forces the photon's wavefunction to rapidly evolve to be largely near the detector or largely far from the detector. – user21820 Nov 19 '19 at 07:27
  • Feynman's theory goes further to say the shortest path that is a multiple of the wavelength is the viable path with probability = 1. One thing you miss is that many of the paths have probability <1 and many close to zero. Also note that everybody forgets when you observe one slit you also get "interference" i.e. bright and dark areas. – PhysicsDave Nov 20 '19 at 23:45
  • @user21820 - what is waveform collapse? Its a word we came up with to describe something we don't fully understand using the best language we have (Quantum Mechanics). There are other interpretations such as quantum decoherence that can work too. – brian_ds Nov 22 '19 at 17:38
  • @PhysicsDave- True I should have added that most are close to 0, however my understanding is that you can have situations where more than 1 is not zero? Also, I was aware of interference in a single slit. However, again my understanding is that if you put a detector in front of 1 slit, it makes the other slit behave classically? – brian_ds Nov 22 '19 at 17:39
  • @brianh: I thought it is obvious. A detector forces the wavefunction of a particle near it to evolve to have either very high density or very low density near the detector. That is 'collapse'. – user21820 Nov 23 '19 at 19:54
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Quantum mechanics is a tool for answering questions. You ask it a question by setting up an experiment, and making a measurement. It answers that question, and that question only.

If you set up a light source, two slits, and a screen, and observe the flashes on the screen, then the question you're asking is 'how does the probability of a photon arrival depend on the position on the screen'. Repeat with enough photons and a pattern builds up on the screen.

This setup can tell you nothing about the path of the photon from light source to screen, or whether the photon even exists between them. If you want to investigate the path, then you create a different experiment with screens along the path you think might be involved, and if there are flashes, then you'll have an answer to 'was it here?'. What you won't have is an interference pattern on the screen, because that was a different experiment, without the intermediate screens, a different question.

Why doesn't QM answer where it is at all times? We don't know. We've only been clever enough so far to create a theory that tells you what happens at the measurement. It's a good theory, it works extraordinarily well, at what it works for.

Is there the likelyhood of any deeper theory which can tell you what happens prior to a measurement? I don't know. I'm rather intrigued by Lee Smolin's event based world where time is real, but distance is an emergent phenomenon, which explains entanglement in a rather mind-blowing way.

Neil_UK
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If the photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

The photon "goes through both slits" is not really a description I am comfortable with. A photon is a quantized potential for causing an effect. Its spatial existence is describable in the form/function of a wave subjected to the double slit setup. This wave function is spatially spread out, but it can only cause a single quantized effect in its domain.

So basically the nature of quantum particles can be described by wave functions but their interactions are discrete: the wave interacts as a whole or not at all.

Any "detection" will rely on an effect, and having an effect uses up the photon.

0

Detecting the photon at one slit would totally change the situation and destroy the "photon going through both slits" effect.

However, there is a notion of "weak measurement" that allows to obtain a small amount of information without changing too much the situation. Because the amount obtained is small, one has to repeat the "weak measurement" many times.

Now a paper just appeared that describes an experiment where each photon was "weakly measured" at the same time on both arms of an interferometer. Of course one needs to repeat the experiment many times and collect the results. The title of the paper is

Unambiguous joint detection of spatially separated properties of a single photon in the two arms of an interferometer

and you can find it at this address

https://www.nature.com/articles/s42005-023-01317-7

Alfred
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Photon goes through one slid, its wave function goes through both.

Benedetto
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