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In the double slit experiment, the observer doesn't know which slit the photon went through so the wavefunction is modelled as going through both slits at once and thus there's interference on the other side. But this seems too observer dependent because two different observers who have access to different information about the photon will not agree on the observation.

For ex: if one of the observers secretly sets up a measuring device near one of the slits without revealing it to anyone else, he'll know which slit the photon went through and thus not observe the interference pattern that all his colleagues still observe. This implies the interference is in the observer, not in the particle.

Is the observer's "many minds" interfering with itself?

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    $\begingroup$ Note that it's possible to stop the inteference pattern without knowing anything about the photon. E.g. you hook it up to a sensor and computer etc, but then you turn the screen off. Or you make it print the results directly into a trash can which then goes to the landfill. The interference pattern is either there or not; it doesn't disappear when you look at the paper in the bin. $\endgroup$
    – user253751
    Jul 13, 2020 at 11:04
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    $\begingroup$ The observer is the screen, not the scientist. $\endgroup$ Jul 13, 2020 at 13:27

8 Answers 8

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The term 'observe' does not mean watching the experiments from a camouflaged hideout so that no one notices you are there. 'Observe' here means 'making a measurement' and hence interacting with the system. The result of this interaction is the wavefunction collapse, a type of time-evolution that is not accounted for within the current quantum mechanical formalism (which is, otherwise, deterministic in that the initial state evolves according to the Schrödinger equation).

So it doesn't matter if the two observers are aware of each other. Even if they don't know what the other one is up to, they will realise someone must have made a measurement when they see no interference pattern.

To give a reductio ad absurdum example: if someone kicks a ball aiming at your head, but you don't see this happening, it doesn't mean that the ball 'might' not hit you. Whoever kicked the ball changed its state (trajectory), and it is now coming for you.

There's an interesting analogy to the two-slit experiments with photons in the answer to this question.

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    $\begingroup$ @PNS, thanks and sorry. Fixed it. $\endgroup$
    – SuperCiocia
    Jul 11, 2020 at 6:05
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    $\begingroup$ @Hierarchist the guy that kicked the ball in your direction has not disclosed this with you either. $\endgroup$
    – SuperCiocia
    Jul 11, 2020 at 6:30
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    $\begingroup$ @Hierarchist That doesn't matter. It doesn't even matter if you don't look at the detector result. But it does matter if the detector data isn't a permanent record. See en.wikipedia.org/wiki/Quantum_eraser_experiment $\endgroup$
    – PM 2Ring
    Jul 11, 2020 at 6:38
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    $\begingroup$ @Hierarchist Correct. As the quantum eraser experiments show, if the detector's "which way?" data is erased after the wave passes through the slits but before the wave reaches the screen, then the interference pattern will still form on the screen. Similarly, if the detector fails to function, interference will occur. That's dramatically illustrated in the Elitzur–Vaidman bomb-tester thought experiment. $\endgroup$
    – PM 2Ring
    Jul 12, 2020 at 4:47
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    $\begingroup$ Best opening sentence. $\endgroup$ Jul 12, 2020 at 8:46
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Neither the particle not the observer "interferes with itself" . The parts of the wave function passing through each slit interfere. The resulting interference pattern gives the probability of finding a particle. If the two parts of the wave function, left and right, are made distinguishable by a detector, they become mutually orthogonal or incoherent, resulting in disappearance of the pattern. No observer or observation is needed, just the experimental set up to enable observation.

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  • $\begingroup$ are you implying that it's not possible for one observer to "secretly set up a measuring device near one of the slits without revealing it to anyone else, he'll know which slit the photon went through and thus not observe the interference pattern that all his colleagues still observe"? $\endgroup$
    – Michael
    Jul 12, 2020 at 2:11
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    $\begingroup$ You said: the pattern disappears with "just the experimental set up to enable observation" (observer or observation not needed). --- I was under the impression that for the pattern to disappear, the observation had to (at least) be recorded, even if not actually viewed. So an experiment set up to enable observation (detectors and recorder(s) physically installed), but where, for example, the detectors were powered-off or not wired to the recorder(s), or the recorder(s) were powered-off, the pattern would NOT disappear. $\endgroup$ Jul 12, 2020 at 2:16
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    $\begingroup$ @Michael - Yes, it is not possible, specifically in the way you have asked. If anyone does anything to measure which slit the photon has passed through (even secretly), the pattern will disappear for all viewers. The "interference pattern", or "NO interference pattern" is seen by all viewers on the same screen. $\endgroup$ Jul 12, 2020 at 2:30
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The quantum mechanical description of the double slit experiment is as follows:

  • If the particle passes through the left slit it is described by the wave function $\psi_{l}$.
  • If the particle passes through the right slit it is described by the wave function $\psi_{r}$.
  • If it is impossible to know through which slit the particle passes, we have to account for all possible paths the particle could take. Therefore, the particle is described by $\psi_{lr} \propto \psi_{l} + \psi_{r}$.

The "weird thing" with QM is that the wave function is not the probability. Instead we have to square it, $|\psi|^2$, to obtain the probability distribution. This is similar to an electric field in optics, where $E$ is how we describe light, but (most often) we measure the intensity $I\propto |E|^2$. The "squaring" gives rise to a term which mixes the left wave function and the right wave function -- just like we learnt in school $(a+b)^2 = a^2 + 2ab + b^2$, but with a little twist, because the wave functions are usually complex. The mix term $2ab$ is responsible for the interference pattern, and it is only present, if the particle is described by $\psi_{lr}$.

The key point is that if we mess up the experimental setup and we (unknowingly) are able to determine through which slit the particle went, the interference pattern vanishes. We don't have to actively know though which slit the particle went. If it is in principle possible to determine the path of the particle, the wave function collapses. The observer merely measures the result of the collapsed wave function.

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To add to some other answers. Being able to tell that someone has tampered with the photon is a key aspect of secure quantum communications.

No communications system is totally secure from eavesdropping, but it is good to know whether a given communication has been intercepted. The most advanced quantum-encrypted communications links monitor the states of the photons carrying the message; if Alice sends a message to Bob and Charlie secretly intercepts it, Bob will know. This is particularly useful when sending a new decryption key, so that Alice and Bob will know whether it can be trusted.

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    $\begingroup$ I like the direction of this answer. But I think it would be better if you explicitly mentioned that the quantum nature of the system allows the observer who doesn't know about the tampering of the other observer to detect the tampering. That both observers won't see the interference pattern. And probably also the connection to the no-cloning theorem. The way you've worded the answer right now makes it feel like being a bit off-topic because you are detailing why detection of tampering is desirable, but not why quantum cryptography can actually detect any tampering. $\endgroup$ Jul 12, 2020 at 23:58
  • $\begingroup$ The phenomenon of wave collapse when the photon is detected mid-flight is implicitly understood in the question. It seems irrelevant to the answer to have to go through all that, or to explain any given communications technology in detail. $\endgroup$ Jul 13, 2020 at 7:22
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You are saying "if one of the observers secretly sets up a measuring device near one of the slits without revealing it to anyone else, he'll know which slit the photon went through and thus not observe the interference pattern that all his colleagues still observe.", but this is not correct. In reality, if anyone alters the boundary conditions, that is, puts a detector on one of the slits, the interference pattern disappears, because the photon's distribution on the screen will become random.

It is very important to understand what happens to a photon when it interacts with the detector. One of the things that can happen to the photon is to be absorbed. Think about it, if the photon gets absorbed at the detector, it cannot be absorbed on the screen. Each and every single photon can only be absorbed once. The other thing that can happen to the photon is to get inelastically/elastically scattered. In this case, the photon gives or gives not some of its energy to the interacting atom in the detector, and changes angle, and then lands on the screen. The ultimate answer to your question is the angle. It changes randomly.

But then why does the interference pattern disappear? Because of the interaction. The photon interacts with the detector. The boundary conditions change. The photon will be out of phase with the other photons. The scattering photons change angle randomly. The pattern disappears.

You are asking why does the pattern disappear for everybody? Because the detector changes the boundary conditions and that is observer independent. The photons will interact with the detector and that interaction is observer independent too.

You are saying the pattern is in the observer, and not in the particle. In reality, the pattern is in (caused by) the whole QM system, including the particle, the wave, the slits.

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In addition to the issue that others have mentioned of observing the photon near the slit, there is the issue of observing the interference pattern. What experimental set up would allow two people to see two different patterns? If both observers are watching the screen they will both see the same pattern.

I say "see", but actual vision would require a mechanism to absorb the photon and emit many photons isotropicly from the point of absorbtion. Both observers would see the photons emitted from the same point. In a practical experiment the location of the point of absorption would have to be recorded and then reviewed later, for example on a CCD imaging system. Both observers would see the same thing.

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I agree with the answer by m2cts, and here is an experiment that confirms the statements:

dblsslitelec

Electron buildup over time

Electrons are fired on the double-slit one at a time. In frame a) the individual footprints of the electrons are seen on the screen, and they look random and like footprints of particles, no fudge or spread in space. So a single electron has nothing to do with a wavefunction, i.e. one footprint can't tell you the form of the wavefunction reaching the screen.

As the number grows and the interference pattern appears, the interference is characteristic of wavefunctions. BUT what waves is the probability of detecting the electrons: it has high probability where a lot of spots appear, and low where few or none.

The wavefunction $Ψ$ of the experiment is the quantum mechanical solution of the boundary conditions: electrons with a given energy, scattering off double slits which each have a given width (in a standard double-slit these widths are the same), and the distance between the slits. $Ψ^*Ψ$ describes the plots.

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    $\begingroup$ “a single electron has nothing to do with a wave equation” — is this true? I thought the wave equation governed the probability distribution of of each individual electron's position on the screen? The pattern's only obvious after many electrons have hit the screen, isn't it there right from the start? $\endgroup$
    – gidds
    Jul 11, 2020 at 18:24
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    $\begingroup$ @gidds The probability functions are always there from the set up of the boundary conditions. The electron itself is not split,it is a quantum mechanical point paticlei in the elementary particle table of the standard model $\endgroup$
    – anna v
    Jul 11, 2020 at 18:43
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    $\begingroup$ @gidds no, it's wrong. The electron's state is described by the Schrödinger equation, which is indeed a wave equation for the wavefunction. $\endgroup$
    – fqq
    Jul 12, 2020 at 9:57
  • $\begingroup$ "The wavefunction $\psi$ of the experiment is the quantum mechanical solution of the boundary conditions" this does not make sense. It is the solution of a differential equation with some boundary conditions. $\endgroup$
    – fqq
    Jul 12, 2020 at 9:58
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Let me first state that measurements don't have to involve interaction with human beings or other lifeforms. So a double-slit pattern will emerge no matter if we observe it or not.

You write:

if one of the observers secretly sets up a measuring device near one of the slits without revealing it to anyone else, he'll know which slit the photon went through and thus not observe the interference pattern that all his colleagues still observe.

Although the observers who don't know of the secretly setting up of a measuring device on one of the slits (to measure if an electron or photon entered this slit) by one of the observers, the experimental set-up is the same for both groups of observers (even though the non-secret group isn't aware of this new set-up) and the one who put the device near one of the slits.
Although the non-secret observers don't know about the secretly added measuring device, they will nevertheless find out it has to be there if they see the same interference pattern appearing on the screen as the "secret" observer.
The fact that they thought there was no measuring device placed secretly doesn't have any influence on the pattern developing on the screen.
Suppose one of the observers found out about the secret and removed the secretly put measuring device, without the secret observer noticing. He/she will think the device is still there. He/she can't put it back just by thinking it's still there. So when they look on the screen while the experiment is done he/she will see (contrary to his/her expectation) a developing pattern (particles moving through both slits) that's the same as seen by the other observers.

Further, you write:

Is the observer's "many minds" interfering with itself?

I'm not sure what you mean with that exactly, but what the observers think about the experimental set-up has no influence on the developing patterns (as I said, a pattern will also arise in the absence of observers). It doesn't make any difference if the measuring device is put on one slit secretly, without the observers knowing it, or if it's put there while the observers are knowing it. The minds of all the observers are interfering if the secret observer tells his secret so everyone sees what is to be expected. If the secret observer hasn't been told that the device has been taken away then his mind interferes with itself. He/she thought a one slit pattern would appear but he/she sees a pattern compatible with a double slit. If the secret observer hadn't told the others about the device, the expectations of the observers will interfere with what they actually see.

You can try it at home (that is if you have the necessary equipment, or else ask a kind professor at a university if you can make use of a double-slit set-up). Now ask a friend if she can put a measuring device at one slit or the opposite i.e. not placing it. Without you knowing it. Your friend already knows what the pattern is gonna be. Your mind though doesn't know the outcome of the developed pattern. Your mind interferes with itself, which can't be said for your friend's mind who already knows the outcome (though a person in the night can always come by to make her action undone...).

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