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This seems an incredibly basic question, but one I've been unable to find an answer to on PSE; if this is a duplicate please point me in the right direction.

Concerning a simple Young's double-slit setup:

A sensor of some type is placed by one of the slits, such that if an electron were to pass through this slit, the sensor would register the passing and thus any possibility of seeing an interference pattern after many runs would be destroyed. The other slit has no such sensor.

Electrons are then fired one at a time. After each electron is detected at the downrange detection plate, a note is made whether the sensor positioned by the slit was triggered or not. In this way, two populations of detections may be built up: Marks on the downrange detection plate that were associated with the slit sensor being triggered $A$, and marks on the detection plate that had no associated triggering of the slit sensor $B$.

Now, if I observe the pattern of marks created by population $A$, I would expect to see no signs of interference as I have very clear which-way path information thanks to my sensor.

My question is this: If I choose to observe the pattern of marks created by population $B$ only, will I observe an interference pattern or not?

It seems my expectations could go both ways:

  1. I can argue that I should indeed observe an interference pattern since these electrons have not interacted with any other measuring device at all between the electron source and the detection plate, between which lie my double slits.

  2. I can argue that the very fact that my sensor at the one slit did not trigger a priori gives me which-way information, in that I now infer that my electron must have gone through the other slit thanks to the absence of which-way information through my sensor-equipped slit.

Which one of these assumptions aligns with reality would seem to have huge ramifications: the first implies that measurement is truly physical interaction of any kind, whereas the second implies that knowledge is measurement, even if that knowledge is obtained without physically interacting with the system (if my detector isn't triggered I cannot see how one could argue it interacted, so perhaps a more accurate statement would be there must be a different kind of interaction that may support non-epistemic views of the wavefunction).

Put another way more succinctly: It is one thing to understand that physical interaction destroys superposition. It is another to understand that a lack of interaction with a measuring device (generally pursued to preserve superposition) may also destroy it if it yields which-way information.

Given this I'm hoping the answer to my question will be #1, but expecting it to be #2.

JPattarini
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    You only get interference if the photon can go through either slit at the same footing without leaving traces. This is not the case here, so #2. – Norbert Schuch Jan 29 '16 at 12:30
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    BTW, your title is different than your question. You ask "Can the absence of information provide which-way knowledge?". There is no absense of information. The absence of seeing the photon in A (but seeing it on the screen) gives 100% information that it went through slit B. – Norbert Schuch Jan 29 '16 at 14:47
  • Behind every edge fringes appear. So fringes behind a slit or behind a double slit are the sum of two respectively four edges. So indeed you will see an intensity distribution behind a slit, even if you set a detector behind one of the slits. The pattern, of course, will be different from the pattern from a double slit. – HolgerFiedler Jan 29 '16 at 18:29
  • See the picture(s) here https://commons.m.wikimedia.org/wiki/File:Moellenstedt_biprisma_voltage_shadow.JPG about the influence of an electric potential on the intensity distribution of electrons. – HolgerFiedler Jan 29 '16 at 18:36
  • I asked here http://physics.stackexchange.com/questions/158105/can-the-intensity-distribution-behind-edges-and-slits-be-explaint-by-the-interac about the possible interaction between the particles and the surface electrons of edges. – HolgerFiedler Jan 29 '16 at 18:47
  • Photons don't go anywhere. While it's tempting to make inferences about what a classical object would have done, photons are not classical objects. @NorbertSchuch: Please don't suggest to the OP that one can infer classical information about quantum objects from the absence of any information. QM doesn't work that way. – CuriousOne Jan 29 '16 at 23:55
  • @CuriousOne I agree with the sentiment that photons don't have trajectories, and this is obviously key here. (Although the totally standard terminology of "which-way information" falls afoul of the same complaint.) However, you seem to be confused about continuous measurement. The absence of a detector click is information and leads to state reduction/update in the appropriate way. It might help to think about photon counting experiments. Every second the detector doesn't click you gain information: it becomes more and more likely the field is vacuum. – Mark Mitchison Jan 30 '16 at 12:46
  • @MarkMitchison: The absence of a detector signal is the absence of a detector signal. One can only infer a (as in one) path if the system is deterministic classical. The system isn't classical. Indeed, Feynman has given us a classical analogy of what such a particle would have to do to behave the way photons behave: it would have to scan all possible paths and calculate the complex exponential of the classical action. That's not what this "inferred from missing information" photon does. The argument that one can think about this classically is completely undermined by path integrals. – CuriousOne Jan 30 '16 at 19:51
  • @CuriousOne Doing away with paths altogether is the only interpretation that doesn't imply unnecessary strangeness, such as ontic realness of $psi$ interacting with the detector, as Mark suggests. It also seems nonsensical to speak of the electron "going through both slits" as if it were classically in 2 positions. I however admit having trouble understanding how, if truly no path exists, both strong and weak measurements consistently will report an electron at a single slit. The absence of a click somehow disturbing $psi$ seems problematic even for a non-classical object – JPattarini Jan 31 '16 at 09:21
  • @CuriousOne Put another way: the wavefunction of the system encompasses both slit conditions, and can be collapsed by interaction that disturbs this system. Forget about the physicist running the setup. An electron leaves the source and is registered on the detection plate with no click given by the one sensor on one slit. The other slit, of course, has no sensor at all. In what way is this system different from 2 slits with no detectors at all? No physical disturbance of the system occurs in either setup between emission and absorption at the downrange plate. – JPattarini Jan 31 '16 at 09:39
  • @CuriousOne I refuse to believe the universe truly cares whether there's a conscious observer present (which some of the answers seem to fall back on), so I'm trying to understand mechanistically why these setups would yield different outcomes. – JPattarini Jan 31 '16 at 09:41
  • @JamesPattarini: That only one electron can exist in the final state when one electron has existed in the initial state is simply the charge conservation property of the quantized em field in the low momentum limit. There is no magic here. Ramp up the energy to more than 1MeV and suddenly positrons will pop up in your double-slit experiment. Ramp it up to a couple hundred GeV and a Higgs will be found in your detectors once in a while. Schroedinger's $\psi$ is a kindergarten version of reality. Today we have accelerators and they make all of this look like rather trivial physics. – CuriousOne Jan 31 '16 at 10:03
  • @JamesPattarini: I certainly didn't ask you to believe in some conscious observer, either, but a quantum mechanical measurement has features that the free propagation does not have. That we are not teaching these sufficiently in QM I classes is understood. They didn't teach it to me, either, but it's rather trivial if you think about it: a measurement has to leave a lasting record, which is only possible if an irreversible change is made to the environment. Once this is properly expressed mathematically, the magic of quantum measurements leaves the room. – CuriousOne Jan 31 '16 at 10:05
  • @CuriousOne OK think I'm 90% of the way there, but hanging up on this: if a measurement must leave a lasting record in order to be considered a measurement, how in this setup does the lack of being detected satisfy this requirement? – JPattarini Jan 31 '16 at 11:49
  • The physics of the measurement doesn't happen when you make choices but when (in a classical picture to help you think about this) the electron "hits" the metal plate in the measurement device. With that an irreversible measurement is made, the rest is just philosophical guesswork. The proper way to express this would be to calculate the multi-electron wavefunction of plate electrons plus free electrons. That is, of course, total overkill, but it would give you the right physics. In reality you can learn to intuit these things and then you can get away with single particle calculations. – CuriousOne Jan 31 '16 at 19:11
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    @CuriousOne That does not answer the OP's comment, which is specifically about why, when the electron is not observed to hit the metal plate in the measuring device, one can still regard the interaction between field and detector as a measurement. (This should not be read as a pejorative comment, I'm genuinely curious to understand your viewpoint.) – Mark Mitchison Jan 31 '16 at 22:58
  • @MarkMitchison: When we don't observe, then we don't know. Is this different from the classical world? Yes, it is different because the classical world is always reduced to one possible outcome while the quantum world isn't. More problematic than that is that the future multitude of outcomes depends on the future measurement, so it's not even a known unknown, to go with an infamous American politician. All attempts to reduce the complexity of this situation to some sort of logical reasoning about classical paths have failed... so why not just let go of that idea entirely? – CuriousOne Feb 01 '16 at 00:34
  • @CuriousOne My opening statement of this conversation was that photons don't have trajectories. I don't know where you got the idea that I am talking about paths or any other classical property, I am not. I am talking about how to properly describe quantum measurements. – Mark Mitchison Feb 01 '16 at 00:49
  • (contd.) It is an experimental fact that every moment the detector doesn't click, the observer who wants to correctly predict the future outcomes needs to update their description of the quantum state. It is in precisely this sense that I claim that "no click" does correspond to information. Your insistence to the contrary is not only wrong, but probably confusing the OP. But I'm a bit tired of repeating myself so I'll leave it at that. – Mark Mitchison Feb 01 '16 at 00:50
  • @MarkMitchison: I was responding to "The absence of a detector click is information". You don't know if at any given moment there is an electron in your experiment, at all, or if that electron will arrive at a later time. Only when you register an electron do you know that there was an electron and when it was detected. Most importantly, a single electron doesn't tell you anything about the experimental outcome, which would be a frequency that is an estimator for an expectation value. QM predicts expectation values. The absence of a click is not an estimator for an expectation value. – CuriousOne Feb 01 '16 at 00:58
  • @Curious You say above "When we don't observe we don't know" - so how does the presence of a non-triggered sensor do anything to destroy an interference pattern? Sorry for being obtuse but nothing that's been said seems to answer this. If the wavefunction is something physical which is interacting with a sensor even when that sensor fails to register the presence of an electron, then this seems like a philosophical leap that has nothing to do with the QM I was taught, and I don't think that's what you're saying. – JPattarini Feb 01 '16 at 09:20
  • Nature doesn't know what a sensor is or that you are doing a measurement. You are simply arranging some matter in a certain configuration and that leads to certain physical phenomena. You would get exactly the same result if, instead of your sensor, there was a black piece of paper in an optical double slit. The wave function is not physical. The wave function is a human invention to describe a quantum field. The field is physical, but it permeates the entire universe, which part of it is your experiment is in your mind and in the paper of your theoretical description. Nature doesn't care. – CuriousOne Feb 01 '16 at 09:27
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    What I am saying is that people have to stop imagining things in QM that are simply not there, like paths and information from nothing. Let's say Sally goes into a room and nobody sees here coming out, even though they are watching the door... from that we can infer that Sally is still in the room. Why? Because Sally and the room are classical objects. If an electron is in a potential well and you are watching one side of the well but not the others you don't know anything because the electron could have escaped on one of the other sides by tunneling. QM is simply not the same as CM. – CuriousOne Feb 01 '16 at 09:32
  • @CuriousOne This makes perfect sense to me, and is how I'm trying to think of this setup. To me, the detector not clicking seems = to the detector not being there, and it seems you're saying that's not the case. As you say, the sensor could simply be a piece of paper at one of the slits, but in this case I'd think it more accurate to say that while a "clicked" detector could simply be a piece of paper, an "unclicked" detector should be indistinguishable from the other inert matter that makes up the slit. This is what I'm struggling to understand, if they do truly yield different outcomes. – JPattarini Feb 01 '16 at 11:06
  • @Curious Going one step further, your last comment seems to flatly imply that because the absence of a click in fact does not give you definite information about a path through the non-sensored slit (because this is a QM system and the electron could in fact be anywhere) it seems a priori that I should be able to observe an interference pattern with no issue at all. Why is this assumption incorrect? – JPattarini Feb 01 '16 at 11:09
  • The only things that make a difference to outcomes are the physical processes that are happening, not what you do with them in your mental process. An absorber does the same thing to the system as an absorbing detector, whether you get the information, or not. There is way too much quantum mysticism in the way people talk about quantum measurements. As soon as "the observer" enters the room, physics has left. OK, for the one thousands and first time: there are no paths in quantum mechanics other than those defined in path integrals. For your own sake, read up on path integrals. – CuriousOne Feb 01 '16 at 15:15
  • Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth Feb 04 '16 at 20:47

3 Answers3

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The OP's confusion seems to stem from the incorrect assumption that

if my detector isn't triggered I cannot see how one could argue it interacted [with the electron]

Just because the detector sometimes does not click, does not mean that there is no interaction at all.

A good way to think about this is in terms of continuous measurement. This and this are good (albeit quite involved) references for further reading on this topic.

You know that, uprange of the detector, the electron probability amplitude (or if you insist, the Dirac field) is delocalised in space. In particular, there is some amplitude for the electron to be found at the position of the detector. So in fact, the detector is always interacting with the electron (continuously measuring it). However, this interaction is weak because the detector doesn't cover all of space. Therefore the electron-detector interaction is not strong enough to cause the detector to "click" (i.e. trigger it) with 100% probability on a single run of the experiment.

More precisely, at the end of the experiment the detector and the electron (or if you insist, the Dirac field) are in the entangled state (roughly speaking) $$ \lvert \psi \rangle = \lvert A\rangle_e \lvert \mathrm{click}\rangle_d + \lvert B\rangle_e \lvert \mathrm{no~click}\rangle_d,$$ where $e,d$ label the states of the $e$lectron (or if you insist, the Dirac field) and $d$etector. You can see already that there is an interaction, because the presence of the electron changes the state of the detector (which was initialised in the pure state $\lvert \mathrm{no~click}\rangle$). You run into conceptual difficulty only if you believe that the state of the detector and the electron can be described independently of each other: in QM probability amplitudes refer to the state of the system as a whole. If you do not observe the detector to click on a given run of the experiment, the state of the electron is correctly described by $\lvert B\rangle_e$. However, in order to see interference, the electron (or if you insist, the Dirac field) must instead be in the state $\lvert A\rangle_e + \lvert B\rangle_e$. Therefore there is no interference.

Mark Mitchison
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  • while the latter half of your answer makes sense, the front half implies that the electron is physically delocalized and yet extant, as in De Broglie's matter waves - such an explanation may be intuitively pleasing, but if true would be indistinguishable from Bohmian mechanics and pilot waves, which complicate QM without adding explanatory power. I also fail to see how a matter-wave electron that somehow fails to trip a detector can still be disturbed by the presence of said detector, since the wavefunction describes both slits. – JPattarini Jan 31 '16 at 09:29
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    The electron doesn't have a state. The electron is a state of a quantum field. First quantization is a sure way to confuse the kids about what is really going on. To be honest, we should stop teaching it. Apart from a one time calculation of the hydrogen atom it has next to no useful properties that one has to concern oneself with. – CuriousOne Jan 31 '16 at 10:26
  • @JamesPattarini I am not talking about matter waves, which are not mentioned anywhere in the answer. Questions about ontic/epistemic "interpretations" have nothing to do with this problem. The wave function is a tool the physicist uses to predict stuff, and this answer is about that. The absence of a detector click gives you information because you understand what a detector is: it is a system continuously interacting with a quantum field. – Mark Mitchison Jan 31 '16 at 14:57
  • @CuriousOne I have to admit you are correct that the OP seems to be confused by the 1st quantisation set-up here. But "no useful properties"? You're kidding yourself mate. It's no secret that you despise anything that isn't high-energy physics, but plenty of us like to study other areas. Thinking about problems with a handful of bound states (which includes a whole bunch of interesting stuff) in terms of quantum fields is just a waste of effort. (Especially because QFT will just reduce to single-particle QM in the appropriate limits anyway.) – Mark Mitchison Jan 31 '16 at 15:01
  • @MarkMitchison: High energy physics gives you the correct intuitive picture of quantum mechanics, so why not use it? It is completely self-consistent and philosophy free, just as the scientific method needs it to be. The problem with thinking about all of this in simplified pictures is that one ends up mistaking the flaws of the simplified pictures for actual physics. This is all that is happening here. – CuriousOne Jan 31 '16 at 18:31
  • @CuriousOne Please don't be so naive, there is just as much or as little philosophy in QFT as in first quantisation. QFT is just a special case of QM with dynamical field variables. In QM there is always a wave function(al) whose ontic/epistemic status leads to all sorts of head-scratching by people who are that way inclined. I have attempted an answer in terms of "which-way information", which is the premise of the question. And thus one needs to understand that the detector does interact with the field and therefore its state does give info, even if that state is "no click". – Mark Mitchison Jan 31 '16 at 22:43
  • @JamesPattarini I updated the answer with some references if you want to read more about how to properly describe continuous measurements in quantum mechanics. – Mark Mitchison Jan 31 '16 at 22:55
  • @Mark my thanks for the references - I'm simply struggling with trying to interpret how a non-physical wavefunction can yield physical results in this setup. No click at the slits it seems should = a sum over many runs that contains zero path information and zero field interaction (hence no click), and thus a buildup of an interference pattern. If this isn't the case I still fail to see why. – JPattarini Feb 01 '16 at 09:26
  • @JamesPattarini I'm not sure I understand your comment. What do you mean by a "sum over many runs"? Do you mean in the sense of a Feynman path integral? You should understand that the path integral is not a sum over many different runs of the experiment. It is a way to calculate probability amplitudes by summing over all possible trajectories taken by the electron in each individual run of the experiment. – Mark Mitchison Feb 01 '16 at 13:41
  • The issue is that when you include a detector you have to sum over all possible trajectories (i.e. clicks or no clicks) of the detector as well. You can't just ignore the detector in the path integral. This is because the evolution of the detector is coupled to the evolution of the electron, or in other words there is an additional contribution to the energy depending on the state of the two subsystems: an interaction. If you consider an experimental run where the detector does not click, you are effectively left with a sum over paths going only through slit B, hence no interference. – Mark Mitchison Feb 01 '16 at 13:46
  • @Mark I think I've got it, thanks for the persistence with your comments. I'm not sure why the answer above from Asher suggests an interference pattern would be visible if only the NoClick subset was viewed, and just want to be sure the thought experiment I proposed actually aligns with what's been done in reality - if you know of any literature for similar setups you could steer me towards it'd be greatly appreciated. – JPattarini Feb 03 '16 at 13:39
  • Hi @JamesPattarini, the experiment works as you expect: there is no interference pattern when there is a detector placed at one slit – Mark Mitchison Feb 03 '16 at 16:43
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The problem is that you are treating quantum objects as both classical waves and classical particles simultaneously. More specifically, you talk about them passing through one slit or the other and sensing which slit an electron goes through. But in order for the interference pattern to emerge, the electrons have to pass through both slits at a time. We can expect one of two outcomes in your hypothetical scenario:

  1. The electrons pass through one slit at a time. Perhaps you can unintrusively detect them at one slit, but even without a detector you end up with two overlapping single-slit diffraction patterns, since we're only using one slit at a time.

  2. The electrons pass through both slits and we get an interference pattern, but consequently your sensor detects an electron at its slit every single time.

In neither case can you have both which-way information and an interference pattern, because either the electron takes both paths, or it doesn't self-interfere.

Asher
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  • This answer treats electrons as classical objects that can bilocate and literally "take both paths," and yet never do under observation. Quantum objects don't behave this way, and I'm looking for an explanation for how the wavefunction can be said to encompass both slits, and yet the absence of detection at one (which necessarily means no physical interaction with the system) can destroy interference. Saying "we know which path it took" relies on the observer - take the observer out of it and I'm trying to understand mechanistically why interference would be destroyed. – JPattarini Jan 31 '16 at 09:33
  • Without interaction at one slit the interference pattern is not destroyed, and I'm not certain why you assume it is. The only way for your sensor to be 100% sure that an electron has passed through its slit is for it to interact with the electron 100% of the time, i.e. if it is a shutter that completely blocks the slit, and then you're just doing a single slit experiment to begin with so there's no interference (as expected). If your sensor only sometimes detects an electron, then electrons can pass through without detection, which means you don't have any which-way information. – Asher Jan 31 '16 at 16:17
  • I also contest "...can bilocate and literally "take both paths," and yet never do under observation" since that is what the interference pattern is: an indirect observation that particles take both paths. You can't "detect" without "interacting" so there's no way to try to collect which-way information without destroying the interference pattern; the answer to "which way" is always either "we don't know" or "into the detector." – Asher Jan 31 '16 at 16:22
  • the other answers are both adamant that the mere presence of a sensor is enough to destroy interference even without a click, so please help me to understand where the disconnect is – JPattarini Feb 01 '16 at 09:16
  • @CuriousOne covers the same points I am in the comments on the question, and quite eloquently, so I'll leave the detailed explanation to him. For my part, here's a distillation of my point: there are either two slits, or one slit; if you have "two slits" but one is blocked off, you only have one slit; and it doesn't matter if the second "slit" is blocked by a "sensor" or by a brick wall. – Asher Feb 01 '16 at 18:34
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First, we need to define the interference pattern.
It is the pattern formed by the fundamental frequency of the wave properties of the electron, passing simultaneously through two slits with "suitable" width and separation distance.

When a "detector" is placed on one slit (A), it takes away some of the energy and allows only a higher harmonic (with lower energy) to pass through. This combination causes the pattern not only to change, but to "disappear" if the energy of the higher harmonic is too low to affect the wave that passes through the other slit (B).

It should be clear that placing the detector on one slit, destroys (changes) the pattern, and this is independent of the knowledge that one may obtain (or not) from the detector.

Guill
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