10
$\begingroup$

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.

$\endgroup$
27
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 29 '16 at 12:30
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 29 '16 at 14:47
  • $\begingroup$ 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. $\endgroup$ Jan 29 '16 at 18:29
  • 1
    $\begingroup$ @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.) $\endgroup$ Jan 31 '16 at 22:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Feb 1 '16 at 9:32
4
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ 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. $\endgroup$
    – JPattarini
    Jan 31 '16 at 9:29
  • 1
    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Jan 31 '16 at 10:26
  • $\begingroup$ @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. $\endgroup$ Jan 31 '16 at 14:57
  • $\begingroup$ @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.) $\endgroup$ Jan 31 '16 at 15:01
  • $\begingroup$ @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. $\endgroup$
    – CuriousOne
    Jan 31 '16 at 18:31
2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – JPattarini
    Jan 31 '16 at 9:33
  • $\begingroup$ 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. $\endgroup$
    – Asher
    Jan 31 '16 at 16:17
  • $\begingroup$ 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." $\endgroup$
    – Asher
    Jan 31 '16 at 16:22
  • $\begingroup$ 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 $\endgroup$
    – JPattarini
    Feb 1 '16 at 9:16
  • $\begingroup$ @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. $\endgroup$
    – Asher
    Feb 1 '16 at 18:34
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.