# Is wave function collapse non-local or local?

Imagine a particle in a very large box which takes years to travel from one end to the other. Alice and Bob are outside the box, on opposing ends. Each can remove their side of the box to check if the particle is on their side of the box.

Around the middle of the box, but outside the box, a star explodes. Alice and Bob agree to both check if the particle is on their side of the box when they see the star explode.

If Alice sees the particle on her side, we would expect Bob to not see it on his side, but the problem with this is that wave function would have to collapse instantly over all space in the box. In other words, it's non-local.

If the wave function were to collapse locally over time, it would start where Alice observed it, and information about the collapse would propagate at the speed of light to the other side of the box. But since it takes years to span the box, the probability distribution would still be non-zero probability on Bob's side.

Since it would break conserved quantities, a particle would not be able to be detected on both sides of the box, so it seems like the wave function collapses non-locally.

I think a non-local collapse looks something like the following.

I say it's non-local because the part of the probability distribution closer to Bob's side is changing by Alice's discovery of where the particle is or is not. A positive or negative observation by Alice affects Bob's chances of detecting the particle on his side. This is a non-local cause and effect.

Is this correct? This seems like an experiment that could be done. For example, a long tube has only an electron inside. The detectors Alice and Bob are photon detectors on each end of the tube. Turning on a very strong magnet on each end of the tube is equivalent to Alice and Bob opening the tube and looking inside. If the electron's spin flips in response to the magnetic field, then a photon is detected and may be detected. The time between turning on the magnet and detection of photon tells us where the electron was. If we run the experiment many times (throw away results where no photon is detected because electron spin did not flip or photon did not hit photon detector), will it produce a distribution of detections which is "uniform" (ignoring interference patterns), or will it produce a distribution which is more concentrated as you get further away from the magnetic field source (magnetic field is only on when we want to detect).

Further clarification: I understand the particle is everywhere in the box as some state before measurement. I understand that the particle IS the state. Measurement just forces the particle to be in a pure state. And, it seems like this transition from superposition to pure state happens instantly. No time passes. But my question is about what happens in the time between when Alice looks into the box and when she measures the position. As she sees that the space in front of her does not contain the particle, does this altar the particle's state? Does measurement of where the particle is NOT affect its state?

I have found this answer: https://physics.stackexchange.com/a/476738/159153 But just like the commenter on that answer, I am left unsatisfied with this story. The reason being, if negative measurement affects state, then I am still wondering how Bob's side of the box seems to be affected instantaneously.

Also, could this resolve the paradox seen in the delayed choice quantum eraser experiment? Since simultaneity can be broken by changing reference frames, it can also be fixed by changing reference frames. We can find a reference frame where the 2 events, detecting the particle on the main detected, and detecting the entangled particle in one of the other detectors, happens simultaneously. And if the wave function collapses instantly, then there's no paradox in this frame of reference.

• simultaneous detection at A and detection at B are mutually exclusive and perfectly correlated . maybe this means that the two events are "entangled"(don't know what that would mean in the context of events). Entanglement related changes occur instantaneously. Dec 6 '19 at 14:21
• "...the wave function would still have a non-zero probability of being detected..." Wave functions are not observables. If you don't think of them as physical objects, then your question just doesn't arise. Dec 6 '19 at 14:42
• Just want to express this in more concrete terms: Imagine a photon, that's been trapped since forever, in a resonant cavity, formed by two perfect mirrors. Bob and Alice each replace their mirrors with detectors at the same, agreed upon instant. You're asking, how does the photon know to come out from one end of the apparatus or the other but not both? Is that right? Dec 6 '19 at 14:43
• Yes, Solomon. D. Halsey, you are right, I am mis-speaking. I mean to say, "the particle has non-zero probability of being detected". However, I think my question still arises. Another way to ask my question is, how does the probability distribution change between when Alice detects the particle and when Bob detects it (He should detect it in the same place where Alice detected it). When Alice opens her box, the probability distribution is uniform over the entire box, but integrates up to 1, because it must be in the box. As her light cone spreads into the box, how does the distribution change? Dec 6 '19 at 19:43
• Hi, I updated my question to include a picture and possible experiment. Dec 6 '19 at 20:08

It's non-local in the sense that you said: Bob's wave function must be updated as soon as Alice makes a measurement, by setting $$\psi=0$$ in Alice's detection region and renormalizing it to 1 everywhere else. Up to changing the phase, this is the ONLY way to change the WF after measurement which doesn't allow faster-than-light communication between Alice and Bob. You can check by insisting that the probability that Bob finds the particle is unchanged whether Alice chooses to look for the particle or not that this is the case.

Physics (including QM) is local in the sense required by special relativity, basically that there is no FTL communication. There are other types of nonlocality like this one which are allowed. How you interpret this physically depends on interpretational questions. But there is an underlying nonlocality of a certain sort in QM.

Edit in response to Croolsby's comment: You said: if Bob's WF is changed when Alice measures the particle, won't he be able to tell by measuring the probability of finding the particle on his end?

Answer: If Alice DOES find the particle, the probability that Bob finds it in his measurement is $$0$$. This has a detectable effect on the probability on Bob's side. So in order to make up for this, the WF must be scaled up in the case that Alice doesn't find the particle. Balancing these out is the only way that Bob cannot tell what Alice has done. In particular, we need

$$P$$(Bob finds if Alice doesn't measure)$$= P$$(Bob finds if Alice does measure)

By splitting the right hand side of this equation into conditional probabilities, conditioned over whether Alice finds or doesn't find the particle, you can see that the only way for the RHS to equal the LHS is: If Alice doesn't find the particle, $$\psi$$ must be scaled to 0 in Alice's detector and renormalized (scaled up) everywhere else so that its norm is still 1. Up to a phase this is the only consistent way to make both sides of the equation equal.

• This might be correct but it is not important. What is important, is to understand that the two detectors at the two ends of the box are entangled, they have a common wavefunction, and that is why there is no need for anything instantaneous. Dec 7 '19 at 16:47
• Hi thanks for your answer. Since Bob's side of Probability Distribution (PB) is updated when Alice makes measurement, won't Bob notice a deviation from the PB he would have seen if Alice was not making measurements? In this case, Bob has some FTL information, which is, he can tell whether or not someone else is also making measurements on the other side of the universe. How is this problem resolved? This is different from the entanglement problem because in that one, someone measuring the other particles will not change the PB you see from your particles. Dec 10 '19 at 18:18

Croolsby,

There are different ways to interpret the wavefunction. You might consider it to be (1) a real, physical entity and in this case its collapse implies a non-local physical process, or (2) you might take it to be a representation of the available, incomplete knowledge about the system, in which case no nonlocality is required as the collapse represents a change of your knowledge about the system, not a change of the system itself.

We have solid evidence that the world is local, so, the most reasonable position is 2. In other words, we know that the particle is in the box, but we do not know where, so the wavefunction is spread inside the volume of the box. After the particle is detected you know where it is and you replace the old wavefunction with the new one, peaked around the detection locus. There is no paradox here.

"I understand the particle is everywhere in the box as some state before measurement.I understand that the particle IS the state."

This is not what QM says. Where did you get this information?

About the "negative" measurements. They change the state because they increase your knowledge about the system. There is also a physical interaction involved there because particles interact by long-range forces (like electric/magnetic fields). So, if you decrease the volume of the box by using a metal barrier and you do not find the particle in one of the two separated volumes you change the fields acting on the particle. The electrons and nuclei in the barrier will produce electric and magnetic fields that will exert a force upon the particle.

"We can find a reference frame where the 2 events, detecting the particle on the main detected, and detecting the entangled particle in one of the other detectors, happens simultaneously. And if the wave function collapses instantly, then there's no paradox in this frame of reference."

If you really want to go for option (1), a real wavefunction that undergoes an instantaneous collapse, you need to reject the modern interpretation of special relativity and go for an absolute reference frame. This is the only way you can avoid paradoxes.

• "This is not what QM says." The particle/system cannot be more than information which describes it. About negative measurements, you say, "There is also a physical interaction". This implies that negative measurement happens over time since the force carriers must be local. But this isn't right because measurement forces state into pure state. You say particle will experience force. A quantum system in the pesence of a potential field just changes the hamiltonian. The particle will continue to evolve according to that hamiltonian. It won't cause the particle to collapse to a pure state. Dec 7 '19 at 10:54
• "The particle/system cannot be more than information which describes it. " - do you have some evidence to back up this claim?; "negative measurement happens over time" - correct, it takes time to position the barrier and search for the particle. "this isn't right because measurement forces state into pure state" - what does the time required for a measurement has to do with the result of that measurement? I fail to see the argument here. Do you imply that a measurement that results in a pure state must be instantaneous? Why? Dec 7 '19 at 11:44
• "A quantum system in the presence of a potential field just changes the hamiltonian. The particle will continue to evolve according to that hamiltonian. It won't cause the particle to collapse to a pure state." - The collapse corresponds to a change in your knowledge, you know where the particle is. The presence of a barrier is required to limit the region where the particle can go, but in itself does not induce a collapse. If you only divide the box you still do not know where the particle is. Dec 7 '19 at 11:48
• What you are writing might be correct, but that does not help him to understand why there is no need for anything instantaneous. What is important, is to understand that the two detectors at the two ends of the box are entangled, they have a common wavefunction, and that is why there is no need for anything instantaneous. Dec 7 '19 at 16:48
• @Andrei Are you suggesting particles/systems have properties which are not described by the quantum state? Are you thinking about hidden variables? I would like to explore answers to this problem where measurement causes an instantaneous update to our knowledge of the system, and there are no hidden variables. Why? Because that's the standard Copenhagen interpretation. Entanglement alone does not answer the question because there are measurements happening. I've constructed a scenario where although measurement happens instantly, Alice is making many measurements over time via her light cone. Dec 10 '19 at 18:31

Let me reword your premise a bit.

There exists a system spanning some region of space. Two observers Alice and Bob, at rest wrt each other, simultaneously measure some observable quantity of this system at a previously agreed upon time. The observers are separated by space and do not affect each other or the system except for by their respective acts of observation.

Does the act of observation cause the wavefunction of the observable to collapse throughout the system instantaneously?

Let me reword it to say

Does the act of observation cause the entire system to acquire the observable's eigenfunction instantaneously?

In short, yes.*

## Consideration of the nature of system: role of entanglement

Lets say the system isn't what one would call an entangled quantum system. Its just a plain old quantum system. I now show that for a system that spans space to be called a single system, it must be an entangled quantum system.

Consider Bob's measurement called B.

1. Where does B occur?
• Over any finite part of the system but not including all of it. In effect, we are stating that Bob can be as large as he wants as long as he leaves enough room so that Alice can be called a separate observer.
1. Since B can at best only change those parts of the system that were directly involved with the measurement, there exists parts of the system which may not have acquired the new state - the eigenfunction.

2. This means an update wave of the wavefunction must sweep the system. This is local collapse*. The entire system collapses to the eigenstate causally.

3. But while this occurs, Alice could have performed her measurement called A. This would also induce a similar update wave to sweep towards Bob.

4. Where do these information-space waves meet? How do they interact? What would be the net result?

5. There is no reason why measurement of any one observer should be preferred over the others. Yet we must select one for the system can't be in two eigenstates.

6. Hence there should be no scope for other measurements while the wavefunction update is ongoing.

7. Since Alice is an independent observer, Bob's measuremnt can't restrict her.

8. So the update must occur instantaneously across the entire system -in other words the wavefunction collapse is non-local.

9. This only occurs if the entire system is a quantum mechanically entangled system.

• Even with instantaneous update, whose measurement takes precedence? Since true simultaneity doesn't exist, it doesn't matter. The system takes some state and that is the single source of truth for all observers.

The point of this reasoning is that in order to just define what one single system means, one has to bring in some sort of entanglement which introduces non-locality (in wavefunction collapse).

## Wavefunction measurement

Wavefunctions can't be measured. They are not observables.

1. Alice & Bob measure the observable's eigenvalues (millions of times with relaxation or over an ensemble). They can then approximately reconstruct the wavefunction by computing the PDF of the eigenvalues. They both should get the same result. One may term this reconstruction the "experimental" measurement of wavefunction though it really isn't.

2. Since this is all that measurements can do, their is no way to reconstruct the evolution of the pre-measurement wavefunction to the post-measurement delta function.

So then how does one say that the collapse is instantaneous or not if the evolution to the collapse itself can't be measured?

1. Bob can make a detection at some part of a quantum entangled system. Note that entanglement implies measuring a part is same as measuring the whole. So the entire system has indeed been observed.

2. Alice can now make her measurement arbitrarily close to the previously agreed upon time. She finds that no matter how close she gets, there is always only a single source of truth - the entire system is only in one eigenstate, that Bob measured - never in a flux or in the process of being updated - as causality would imply.

## Wavefunction collapse - necessary?

According to some (Everett, Coleman etc), not really. No collapse, no question of whether its local or not! Once the observer becomes part of the system (via entanglement), the evolution of the wavefunction from pre to post is fully defined by the Schrodinger eqn.

Also, there are no physically measurable non-local effects from non-local collapse.

## Application to the particle in a box

There a few problems with the premise you provided

1. Once the box is opened, the change in boundary condition stipulates an old and a new wavefunction even before any collapse related modifications could be made to it via measurement.

2. While the old wavefunction was a sinusoid, the new wavefunction is zero everywhere with unit norm. Since such a thing doesn't exist, it must be localized in some way in the form of a wave packet (most likely the sinusoid before the box was opened).

3. Till the time the particle's position is detected, the wave packet may have spread beyond the boundaries of the original box or moved. The light cones would therefore need to extend in both directions from both observers.

4. Even if we assume that the particle stays in the box, your middle picture of wavefunction collapse is IMHO incorrect

• After Alice's detection, due to unit norm, the indicated "blue step" on the right can't exist. Alice's detection implies a delta at the detected position which consumes all the norm.
• Why should Alice's light cone stop propagating once it detected the particle?
• Since the effects of detection propagate causally, why should the pregnant area of possible detection's on Bob's side not get its own detection spike too? After all, effects from Alice are still in transit. Note that you can't invoke "a single source of truth" argument to nullify the truth of the system's wavefunction - which from Bob's perspective is perfectly valid.

So what would really happen?

1. The wave function evolves as per a dynamic potential. This potential at $$t=0$$ restricts the particle to the box. At $$t>0$$, it restricts it to regions where its absence hasn't been detected. Such a wavefunction is evidently complicated.

2. One can circumvent all this by considering the box and the particle inside to be a giant entangled system.

• The opening of doors is an act of measurement. The system instantaneously collapses everywhere to its eigenstate. The particle is sitting duck at some point.
• Alice and Bob can read out the particle's position when their causal cones reach it. This has nothing to do with the collapse of the wavefunction. This way there is no ambiguity as to what is the collapse inducing actual measurement : the act of opening of doors or the light cones reaching the particle. (see On Clarifications below)

You make an important point in the following

A positive or negative observation by Alice affects Bob's chances of detecting the particle on his side. This is a non-local cause and effect.

What you have described is an entangled system and thereby non-local in its update of wave-function. The absence/presence of particle at Alice's location is perfectly (anti)correlated with that at Bob's.

## Consideration of the electron in a tube set up

Even though it would actually be difficult to trap a single electron in a tube, let alone measure its properties in a changing strong magnetic field and making it emit a photon on spin flip and so on... I get your point.

The act of turning on the magnetic field implies a dynamic hamiltonian. Quantum mechanically what the electron would do, I do not know.

If instead you had a single photon trapped in a very long thin tube, whose entire inner length was pixelated with photomultipliers, all initially off, turning them on should detect the photon somewhere instantaneously and nowhere else within the spatio-temporal resolution of the apparatus.

## On clarifications...

I understand the particle is everywhere in the box ...

• The particle is a theoretical particle: it exists only at a point in space. We just don't know where. What exists everywhere is the wavefucntion.

... as some state before measurement. I understand that the particle IS the state

• The particle and the state it is in are different things. An electron is an electron whether its trapped in the ground state of a $${}^1H$$ or free at the LHC.

And, it seems like this transition from superposition to pure state happens instantly. No time passes.

If you accept that, then you must accept non-local wavefunction collapse. Instantaneousness in time is non-locality in space

But my question is about what happens in the time between when Alice looks into the box and when she measures the position

• If opening the box (and looking within) is a separate act from the measurement of the particle's position, why include it in the discussion to begin with?

• To say that Alice and Bob measured the observable at some timestamp means they literally got the eigenvalue for the observable at that time - it doesn't mean that they initiated their measurements and are now in causal waiting.

• For e.g. in the classical quantum entanglement measurements, a measurement of spin is the actual measurement of spin - not the flicking on of the detector. Another way of saying this is that the moment of record is the moment of measurement.

• Why does this matter? For one, the act of opening the box to look within, if not considered to be a measurement, makes the Hamiltonian dynamic and the analysis complicated (as discussed in the sections above)

• But more importantly, the entire system is just one-single-big thing - an entangled quantum system. So observing any part i.e. any interaction anywhere not accounted for in the hamiltonian must induce a measurement on the enire system.

• So when we say that Alice and Bob made a measurement, the point is not so much as where the particle would be since their light cones haven't reached it but more like since the measurement was claimed to have been made, the particle was where the light cones reached it.

$$*$$ Note that this is opposite to your terminology. Local/non-instantaneous collapse respects causality and so is a wave at $$c$$. Instantaneous collapse would be called non-local.

You are asking in the comment "the probability distribution which tells you where the particle might be, and upon measurement, the particle's state collapses to one location."

Now it is very important to understand the difference between two things:

1. the probability distribution which tells you where the particle might be

2. the probability distribution which tells you where the particle is

Classically thinking, you would say, it must be 1. The particle might be at different places, with different probabilities, but not at the same time.

In QM, it is 2. The particle is actually everywhere in space, it is delocalized, when it is traveling in space (like a photon) as a wave. The probability distribution describes the probability of finding the particle everywhere in space.

Wavefunction collapse is a phrase that is confusing, it just means realizing one piece of the probability distribution.

It is a misunderstanding of this dubious word "collapse", which really means getting one instance from a probability distribution, in your question "wavefunction of the whole universe,", more complicated than the wavefunction for the scattering of two protons, but the principle is the same. One has to look for the effects of this particular point from the probability distributions describing it.

Spontaneous collapse of the universal wavefunction

You only realize this one piece of the probability distribution upon measurement. Until then, the particle traveling as a wave is delocalized.

You are basically asking if we have two detectors at two ends of the box, how will the detector at one end of the box know that the particle was measured on the other end, so it cannot be measured there too.

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.

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

It is basically the same as with two entangled particles. The information was already there, and no information needs to travel faster then light. In this case the measurement in one end of the box (finding a particle) means that the measurement at the other end will not measure (find) a particle, but this does not need information to travel faster then light from one end of the box to the other end.

The reason for that is that the two detectors at the two ends of the box are entangled. They have a common wavefunction. It describes the probability of finding the particle at one of the sides (exclusively, only at one side at the same time), that is why you cannot detect the particle at both sides of the box at the same time, and nothing instantaneous (no information) needs to travel between the two ends of the box.

Quantum field theory makes it easy to prove that the information cannot spread over spacelike separations - faster than light. An important fact in this reasoning is that the results of the correlated measurements are still random - we can't force the other particle to be measured "up" or "down" (and transmit information in this way) because we don't have this control even over our own particle (not even in principle: there are no hidden variables, the outcome is genuinely random according to the QM-predicted probabilities).

Why is quantum entanglement considered to be an active link between particles?

• I understand the particle is everywhere in the box as some state before measurement. I understand that the particle IS the state. Measurement just forces the particle to be in a pure state. And, it seems like this transition from superposition to pure state happens instantly. No time passes. But my question is about what happens in the time between when Alice looks into the box and when she measures the position. As she sees that the space in front of her does not contain the particle, does this altar the particle's state? Does measurement of where the particle is NOT affect its state? Dec 7 '19 at 7:06
• I have found this answer: physics.stackexchange.com/a/476738/159153 But just like the commenter on that answer, I am left unsatisfied with this story. The reason being, if negative measurement affects state, then I am still wondering how Bob's side of the box seems to be affected instantaneously. Dec 7 '19 at 7:41
• Árpád Szendrei: "the probability distribution which tells you where the particle is". This is false. QM tells you nothing about where the particle is, it only gives you the probability of finding a particle at that place if, and only if, a measurement is performed. Dec 7 '19 at 9:17
• Yes, more accurately, integrating the probability distribution over some area tells you the probability of finding the particle in that area. I used a short handed sentence because there's not much room in these comments to write out everything. Dec 7 '19 at 10:23
• @Croolsby "integrating the probability distribution over some area tells you the probability of finding the particle in that area." Yes this is correct. But it is not in contradiction with what I said. The photon as long as it travels as a wave is everywhere (delocalized) at the same time with different probabilities. If you measure it, you interact with the photon field at the location of the measurement. You are asking how the measurement on one side of the box affects the measurement on the other side. It does not. Dec 7 '19 at 16:32

This is the kind of thought experiment that leads to the Many Worlds view of QM, in which wave function collapse does not occur. But I think your question highlights an important point: that even the Many Worlds view appears to require nonlocality with respect to propagation of conditional probabilities. Somehow Bob's photon detector "knows" which branch of possible worlds it belongs to.

IMHO, the only really self-consistent interpretation may be a Many Worlds view in which the wave function contains all sets of mutually consistent possibilities. Alice's detection of the photon is inconsistent with Bob's detection of the photon, so the wave function does not contain the possibility that both Alice and Bob detect the photon. Trace all possible interactions throughout the universe back to the Big Bang, and it would turn out that the initial universal wavefunction, contained in an unimaginably tiny volume of spacetime, contains all possible subsequent configurations of the universe as "sets of mutually consistent possibilities". Perhaps simultaneity per se has a different meaning in that tiny context.