I'm still struggling a bit with some ideas around double slit experiments. One that keeps cropping up for me is the role of observers.

Imagine a classic double slit experiment with a hidden observer who has arranged an apparatus to detect which slit the electrons are passing through. This person and their measurements are hidden to you and you have no interaction with them.

So the question is, do you see an interference pattern or not?

Additionally: And if the answer is Not, then is the reason because they "disturbed" the electron (by say firing photons at them) or is it for another reason? And if it is because they "disturbed" the electron, then how is it that unobserved electron's are not disturbed since they certainly interact with other objects, for example other atoms in the matter around the slit(s) will feel a slight gravitational tug as it passes through.

  • 3
    $\begingroup$ "and you have no interaction with them" - Yes, I do: Both the hidden observer and I interact with the electrons. The hidden observer by bombarding them with photons, and I by watching their pattern on the screen. Thus, there is an "effective" interaction between myself and the observer $\endgroup$ – Lagerbaer Nov 26 '11 at 18:44

10 Answers 10


I'm a little unsatisfied by all the other answers because they don't have any units in them. In order to measure which slit the electron goes through, you have to disturb it by at least the order of one unit of action (that's ħ). If you disturb it enough to measure it, you destroy the interference pattern. You can disturb it less than that, and get a small amount of statistical information on which slit it went through, and this will only blur the interference pattern slightly. So there's a tradeoff between how much information you gain, and how blurry the interference pattern gets. I'm not going to work this out in detail.

  • $\begingroup$ Dear @Peter, you seem to refer to Bohr's quantitative "complementarity" which allows a "partly wave-like" and "partly particle-like" behavior of a quantum of the field. That's OK and you're right but that's not really what the original question was about. The original question talked about the situation when some apparata are completely able to measure the which-slit information (so the action change is very large, in your language), but this measurement isn't accessible to "us". So your answer is arguably off-topic. $\endgroup$ – Luboš Motl Dec 12 '11 at 11:16
  • 3
    $\begingroup$ Dear @Lubos, I was addressing the last part of the question, that you seem to have overlooked ... why does the gravitational tug of the electron on the apparatus (say) not count as a measurement? It's because it's too weak, and once you bring "too weak" into it, you need have some quantitative measure for it to make sense. $\endgroup$ – Peter Shor Dec 12 '11 at 11:59
  • $\begingroup$ Right. One should also explain why "too weak" won't destroy the interference pattern. A beginner may be tempted to think that any, arbitrarily weak influence will "measure", and that may really be the OP's case. Well, the interference pattern is only broken if the interaction evolves the two possibilities into (nearly) mutually orthogonal, and therefore mutually exclusive, states. A small which-slit-dependent modification of the state will only modify the states so that their inner products are still essentially the same, and they (mostly) interfere. $\endgroup$ – Luboš Motl May 17 '15 at 8:17

Of course that the interference pattern disappears whether or not you are aware of the experimenter who has bombarded the electrons with other particles.

The quantum mechanical predictions are surely independent of the "consciousness" of the other objects, which is what may lead some people to the same question as yours. Quantum mechanics is valid for predictions of any observed phenomena, whether they incorporate macroscopic objects and humans or not.

The electron is being entangled with some additional particles (photons?) that someone uses to bombard the electrons. These photons won't be detected again. We will only observe electrons, so it's enough to describe them by the density matrix for electrons only. Mathematically: $$ |\psi\rangle = a |{\rm left}\rangle + b |{\rm right}\rangle \to a |{\rm left}\rangle \otimes |{\rm left\,\,photons}\rangle + b |{\rm right}\rangle \otimes |{\rm right\,\,photons}\rangle $$ and $$ |\psi\rangle \to \rho ={\rm Tr}_{\rm partial\,\,over\,\,photons} |\psi\rangle \langle \psi| = |a|^2 |{\rm left}\rangle \langle {\rm left}| + |b|^2 |{\rm right}\rangle \langle {\rm right}| $$

That's why we're allowed to trace the density matrix over the photons' Hilbert space, and by doing so, the information about the relative phase of the left-slit and right-slit portions of the electron's wave function disappears (because these two portions are entangled with different, orthogonal wave functions of the photons) which is why interference is not possible anymore.

So the interference pattern disappears even if no one else is observing the reflected photons at all.

Where quantum mechanics "requires" consciousness or active knowledge is when you ask whom the predictions of QM are made for. They're not made for an objective world: at the fundamental level, none exists. Predictions of QM are meant to be used by a "conscious observer" who may observe the actual outcomes of experiments – whose probabilities are calculated as expectation values of the projection operators corresponding to the Yes/No questions.

But once you are such an observer, you may treat all objects in the world on par – as blind systems of particles that universally obey the laws of quantum mechanics. Their "humanity" or "knowledge" or "plan to exploit an observation" or "consciousness" is totally irrelevant for your predictions and their verification.

The real "paradox" about an observer knowing about another observer is that observer A may observe observer B who observes system S. According to B, the outcomes of measurements are known as long as B "perceives" them. However, A may evolve B+S into Schrödinger cat-like superpositions and only "collapse them" i.e. interpret them once A perceives his observations. So A,B may disagree when "facts became facts". But this question "when a fact became a fact" isn't measurable: any observer may "delay" this moment up to the moment when he actually perceives the outcomes, and there won't be any contradictions in the final perceptions of A,B. (Of course, A may also uniquely calculate the earlier moment when B says "now I know the result": this moment is before A observes the situation, and it is earlier because it's independent on the actual outcome that B perceives. However, B is still just a part of the physical dull world for A.)

  • 7
    $\begingroup$ I downvoted because if the OP doesn't get the double slit experiment, then talking about the partial trace of a density matrix will be pretty worthless to him. $\endgroup$ – Colin K Nov 26 '11 at 18:47
  • 5
    $\begingroup$ Dear Colin K, I am writing the answer not only for the OP but for everyone who asks the same question which may be asked, believe me, even if one has been exposed to double-slit experiment. Fraggle, nothing "causes" probabilities to shrink. Probabilities by definition always describe "shrunk" outcomes. For example, if you throw a dice, the probability distribution is spread over the numbers 1,2,3,4,5,6. It is completely spread. But it doesn't prevent "3" from being the result. In fact, it is guaranteed that one sharp number will be the result if you throw dice. $\endgroup$ – Luboš Motl Nov 27 '11 at 8:07
  • 1
    $\begingroup$ @Fraggle: Any interaction which is stronger if the photon goes through the first slit than if it went through the second slit, or vice versa, will cause the electron to shirt. The interaction has to be stronger by at least one unit of action to destroy the interference pattern. $\endgroup$ – Peter Shor Nov 28 '11 at 2:54
  • 1
    $\begingroup$ @Lubos: the issue with quantum mechanics is that only probabilities are allowed to shrink this way, because only probabilities have a consistent ignorance interpretation. But quantum amplitudes are not probabilities, and they are shrunk in this way after a measurement. The idea then is that probability is emergent from quantum mechanics, and this is philosophically difficult, because probability is so fundamental looking. $\endgroup$ – Ron Maimon Nov 28 '11 at 2:56
  • 1
    $\begingroup$ "But quantum amplitudes are not probabilities, and they are shrunk in this way after a measurement." Dear @Ron, quantum amplitudes and probabilities are related in a straightforward way: probabilities are squared amplitudes (in absolute value), or sums of such squared terms. Because this is a purely mathematical operation, it's clear that one must assign the same qualitative interpretation to both. So quantum amplitudes are (quantum-completed) probabilities. They surely can't be "more tangible" than probabilities - you couldn't get an "untangible" thing by squaring a tangible one. $\endgroup$ – Luboš Motl Dec 15 '11 at 6:57

In response to Luboš's answer, Fraggle writes

The issue for me is this. What causes an electron to "shift" from having position probabilities which are somewhat spread out in space to having position probabilities which are much more localized and not as spread out (localized to one slit or the other)? The answer, since it does not depend on consciousness, must only depend on the kinds of interactions the electron encounters. What kinds of interactions will cause its position to become more definite (less spread out) and what kind won't?

The position probabilities are localized by the electron's collision with the barrier that contains the slits. It can only get through the slits, so the wavefunction coming out the other side will start out looking like it arises from two point sources, one slit and the other slit.

But this isn't the issue. The issue is, what happens to that wavefunction as the two wavefronts from the slits spread out and combine? If electrons go through the slits unobserved, you will see interference effects in the impact pattern that builds up on the other side; but if they are being observed, there won't be interference effects. The "wave" nature will appear to have vanished, and you will just have "particle" behavior, a spray of bullet-like impacts.

The explanation for this has nothing to do with the existence of a hidden observer. All that is required is that there is some physical trace of which slit the electron went through. For example, there could be a microscopic magnetized object near each slit, which flips its polarity when an electron passes by.

The reason this removes the interference is that ultimately, quantum probabilities are joint probabilities. A quantum probability is associated with a total physical configuration, and interference of quantum probabilities occurs when two or more histories converge on the same total configuration. In the scenario I just described, which way the little magnets are pointing is an extra degree of freedom, and you don't just have a "wavefront from slit 1" and "wavefront from slit 2" which will then overlap and interfere on their way to the impact screen. You actually have one set of probabilities for "electron passed through slit 1, and the magnet at slit 1 flipped", and another set of probabilities for "electron passed through slit 2, and the magnet at slit 2 flipped". This is why, when it seems that the wavefronts from the slits should be combining and interfering, they don't: because they are actually probability waves for different configurations, when you look at the whole picture, including the state of the magnets, and so they never arrive at the same "point" in "configuration space": one wavefront is confined to the space of configurations in which magnet 1 flipped, the other wavefront is confined to the space of configurations in which magnet 2 flipped.

This is why some people end up believing in parallel worlds or in nonlocality: quantum probabilities look like they keep track of possible total states of the physical world, and allow for probability waves from "different histories" to converge and interfere. So they reason that either there are parallel worlds and they interact somehow, or there's a nonlocal coordination of probabilities within a single world.

  • $\begingroup$ +1 because it is a simple explanation in words. The only reservation I have is in "For example, there could be a microscopic magnetized object near each slit," : microscopic should be orders of magnitude larger than hbar measures. If one reaches hbar measures then the quantum mechanical setup for a solution changes. $\endgroup$ – anna v Nov 27 '11 at 6:12
  • $\begingroup$ I have found that there exists an experiment with minimal interference where the slit the electron has passed through is known and still there is an interference pattern after statistical accumulation. en.wikipedia.org/wiki/… $\endgroup$ – anna v Jan 4 '13 at 9:20
  • $\begingroup$ In those works, they are measuring neither which way the particle went, nor what the interference pattern on the screen was, but some third observable which is correlated with both of those properties, but so weakly that some quantum coherence is preserved. They call it an "unsharp observable" and it has some similarity to the idea of a "weak-valued measurement", a pointer variable which has the right expectation value to track the property it represents, but whose variance is enormous... $\endgroup$ – Mitchell Porter Jan 5 '13 at 6:06
  • $\begingroup$ In dx.doi.org/10.1007/BF00734319 they say that "transparency of the mirror" or "visibility of the interference pattern" can serve as the third observable, but I do not understand what concrete measurement is involved. $\endgroup$ – Mitchell Porter Jan 5 '13 at 6:07
  • $\begingroup$ In my opinion if they can tag which slit an electron went through it clears that what one is observing is an interference pattern in the probability /wavefunction even if it is a complicated wavefunction. The electron is in a probability wave, not a mass/energy wave which people imagine; the effect: if I observe it it disappears is refuted. $\endgroup$ – anna v Jan 5 '13 at 6:52

Leonard Susskind explains this well in lectures 6 and 7 of quantum entanglement. These lectures can be viewed online (see Stanford continuing education lectures; Leonard Susskind).

There he explains how any record of which way the particle went destroys the interference pattern no matter whether you as observer are aware of the record or not.


students should be aware of the many semantic problems caused by trying to describe QM behaviors with words that pre-existed the study of QM. 1) there is no such "thing" as a "wave". The word "wave" is a description, or name, for a pattern that is periodic - it repeats at regular intervals. "The wave (at the seashore) knocked me down". That burst of water that periodically surges on the beach is called a wave by convention, but it is not a wave, and neither is that single burst of water periodic, but observing and measuring and then graphing the ocean swells does result in a graph with a periodic attribute.

2) electrons, photons, etc are not particles. The word particle was defined well before QM and means a discrete little thing with finite dimensions AND was once part of something larger. The early QM explorers would have done all of us a favor by making up a new word.

3) there are no "packets" of energy. Packet was already defined as a small enclosure, fully sealed, if only by a drawstring, and that which was inside the packet was not outside the packet. With energy, the "field" (semantics again!) it is said extends forever, getting weaker according to the distance from the concentrated center. "spot blur" would be better than "packet". things can be delivered in packets, and so by that part of the the definition of packet the use of the word packet to describe a bit of energy is somewhat accurate.

On to double slit experiments: whatever an electron (or photon) is, it (may not be an it) can be isolated, toyed with, put to use, turned into profit. That profit can be had at least gives it cash value. When the electron is fired from a so-called "gun" at a double slit, or at a fine wire dividing a space in two (Hitachi double slit), the target is missed as often as it is hit. Crappy gun. You want to be asking why such bad markmanship? And when the misses pass on to the target, apparently they don't travel in straight lines. You want to ask why not straight? The complete trajectory of the electron is not known. Some may be curve balls, other sinkers or even knuckleballs, and a few seem to be hardballs. why? After each electron hits the target, (in a drunken state, it seems), that's it for that electron. The next electron hits somewhere else on the target, and that's it for that electron. It IS curious that after a lot of pitches have been thrown that the hit points do look like what we call a wave, but by no means does that mean photons or electrons are waves, as wave is not a thing but a named pattern. Apparently the electrons or photons have favorite trajectories. It is that which requires explanation.

As for "observation" which is just a general word for "measurement". Once we know why the electrons or photons or etc have favorite trajectories, it may be easier to explain the clumping-spray pattern seen when measurement apparatus are brought into play. It seems that once the measurement apparatus (pardon my French) "f*cks" with the electron, etc, it goes into a tumble, like when a spinning top which is on some trajectory is knocked over.

For whatever reason, in the physical sciences leading up to the time when the little ones were first found and named, an assumption was in place that all was "particle" or "wave", and so that expectation was foisted upon the little ones. That was a blunder. The blunder was never fixed, but words like "wavicle" were an attempt at a fix.

And note: the "wave function" does not "collapse". Bridges collapse, functions sometimes cease to be useful. And the measurement apparatus does not "destroy" the interference pattern, rather it changes the pattern of electrons on the target. Such really bad choice of words!

And I want to repeat what others have said: none of this has to do with human consciousness becoming aware of the measurement report. Once the measurement apparatus is turned on, the wave-like pattern goes away, and the spray-clumping pattern begins to manifest whether anyone is there watching or not. I liken it all to a good who-done-it story, and not a fantasy.

  • $\begingroup$ I agree that wavefunction is a bad name (I would prefer $\langle x \vert \psi \rangle$, but it not true that waves have to be periodic, unless you consider that solutions to the wave equation are not waves, like a square pulse. $\endgroup$ – jinawee Jun 26 '14 at 10:13
  • $\begingroup$ " Apparently the electrons or photons have favorite trajectories. It is that which requires explanation. " The explanation will come by solving the quantum mechanical equations describing the system and applying the boundary conditions, so that the state function ( not the trajectories) is known. Once the SF is known the probability distribution of the particles on the screen will be known. We have solved simpler problems and there has not been a falsification on the use of the state function for predicting measurements. It just is much harder for two slits etc. $\endgroup$ – anna v Jun 26 '14 at 10:37

What Luboš Motl said. But I want to address the second part:

then how is it that unobserved electron's are not disturbed since they certainly interact with other objects, for example other atoms in the matter around the slit(s) will feel a slight gravitational tug as it passes through.

When one is thinking of a double slit experiment, one is in the quantum mechanical region, i.e. the energies and wavelengths within the sizes of hbar. We do not have a double slit of macroscopic size with respect to the particles and expect to see interference.

The "slight gravitational tug as it passes through" is inconsistent with the quantum mechanical framework. One would have to solve the total quantum mechanical problem, including the gravitational tug from matter, which would affect infinitesimally the interference pattern, but would still work as a QM probability amplitude for passing through either slit.

  • $\begingroup$ Surely you don't mean wavelengths within the sizes of hbar; they're not even measured in the same units. $\endgroup$ – Peter Shor Nov 28 '11 at 13:50
  • $\begingroup$ That is why I used the word sizes, meaning measured in appropriate units, depending on what one measures: energy hnu, or space deltapdeltax~hbar, etc. $\endgroup$ – anna v Nov 28 '11 at 15:15

You say:

I'm still struggling a bit with some ideas around double slit experiments. One that keeps cropping up for me is the role of observers.

Observers don't have a special role in quantum mechanics. An observation is just a kind of interaction between two systems: the measurement apparatus and the system to be measured. This interaction need not be direct. For example, you can measure where an object is by reflecting light off it and looking at the light rather than looking at the object directly. A measurement takes information from the measured system and copies it into the measuring instrument and possibly into other systems as well.

When you do an interference experiment you put a system into a state in which some of its observables are not sharp: they don't have a single value. For example, in the case of an electron going through two slits the electron does not have a single value of position. You then change the system in such a way that it ends up in a state that reflects changes in phase between different instances of the electron along different paths and measure that state. For example, if you have a magnetic field between the slits and the screen that may change the interference pattern by virtue of making the different instances of the electron pick up different phase between the slits and the screen.

A measurement destroys the interference pattern while other interactions may just shift it. What's the difference? The difference is the information that spreads from the electron to the measurement apparatus. The interference pattern depends on the phase relationships between the different instances of the electron. The measurement spreads some of that information into the measurement apparatus and anything it interacts with and that prevents the interference. Other interactions such as the interaction with the magnetic field, don't produce that transfer of information and so don't stop the interference.

What about the slits? The screen with the slits is a relatively large object and is in a mixed state in which it has a probability distribution for being in states from say -1,000,000 to +1,000,000 electron momenta. (I don't know the exact numbers and all that matters is that they are large.) If the electron interacts with the screen then it shifts the momenta to -999,999 to +1,000,001, say. The probability of there being a detectable difference is the probability of finding it in a state it could not be in without the interaction: in this instance the probability of it being in the state +1,000,001, which is very small. So the probability that the information necessary for the interference will spread is very small. (Likewise for gravitational interactions between the slits and the electron.) By contrast, if you set up a system that is optimised to detect the electron then the probability that it will take some of that information is large because it is specifically designed to do that.

You don't see the interference pattern because of the interaction between the electron and their measurement apparatus. Whether you know they are there has nothing to do with it.

If you want to read more about this then you can read papers like these:






I'd like to add to user31182's answer and Peter Shor's anser.

From these and other answers you will glean that if there is a disturbance to the experiment through an interaction with the electrons, it will show up as a disturbance in the outcome, whether or not the "main" experimenter knows about the "hidden" experimenter.

So, now another way to look at what you've learnt is this: you now have a good grasp of the gist of how quantum cryptography works, or at least a good idea of how the interception problem is dealt with by it. That is, you can arrange things so that if someone eavesdrops your data, then you will know about it because this eavesdropping needfully involves a strong enough interaction with the quantum system to shift the statistics of the outcome. In Peter Shor's words, you arrange the coding scheme so that enough interaction to eavesdrop is enough interaction for the receiver's experimental outcome to shift enough to detect eavesdropping.

Generally such thought experiments involving "conscious observers" are not tackled much anymore as a conscious observer is a hugely complicated, uncharacterised system. It becomes meaningless to try to model such complexity. Instead we replace the experimenter and his or her tools of trade and measurements by quantum observables. Now I sense from your question that you may not quite have reached the level of dealing with the concept of an observable: an "operator" on a quantum state together with a special recipe for how we interpret the image of the quantum state under the action of this operator: otherwise put: how we decode the image of the quantum state to the statistics governing the outcomes of our experiment. If not, I'd suggest you try to learn something about this concept now, for it abstracts the notion of an "observation" or a "measurement" and if you're intelligent enough to ask the question you have, and if you have say freshman and second year mathematics behind you with linear algebra, then you're ready to go, IMO! The thorny question of "conscious observation" is neatly sidestepped: an interaction/ observation / measurement is simply replaced by these operators: they just are, whether or not they are hidden from the "main" experimenter. The "observation" in question has taken place if and only if the quantum observable modelling it has been imparted to the quantum state. End of story. The simple operator describing a measurement takes the quantum state as an input, return a real valued measurement and somehow (the answer to this somehow is the quantum measurement problem ) straight after the "observable's" application, the quantum system is in the eigenstate of the observable's operator that corresponds to the value measured. That's all there is to it.

A different, but related idea that you may wish to think about is the Wigner's Friend thought experiment. Here you know there is a second experimenter, but you don't know what they have observer. See my answer here for further details.


Actually it is not necessary to disturb the electron with photons or anything else. A measurement is required, but a measurement is just an observation.

Consider the case where you are watching one of the slits to see if the electron goes that way. If it doesn't, you know that it went the other way. This is an observation!

The question is, what is it that you have done? In the Copenhagen interpretation, your knowledge has collapsed the wave function. This interpretation leads to all sorts of questions like, "what is an observer"?

The Many Worlds interpretation deals with this in a different way, which is more natural for some people. Observing whether the electron went by creates and entanglement with the observer. That prevents the observer from seeing the other outcome. That is, he will never see the electron arrive in two places.

Elitzur-Vaidman bomb tester


Is it about dual state of the matter, interference, or collision with the interiors surfaces and the corners of the wall thickness??? The wall thickness is huge by comparison with the particles size.

Think about!!!

You say: "One that keeps cropping up for me is the role of observers". They say: "when we put sensors near slits", not " if we look at the experiment from a distance corner of the room"...

I've make a short movie explain this, hope you find it useful: http://www.youtube.com/watch?v=gBm6Y82Mz3g

  • $\begingroup$ Link broken.${}$ $\endgroup$ – jinawee Jun 26 '14 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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