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The information about a particle is contained in a vector of unit-norm called the wave function. One postulates says that this wave function is supposed to evolve with time as the particle interacts with other particles (by being in the potential field generated by other particles).

A measurement of a state variable, like position, is equivalent to an interaction of a particle, with a measurement apparatus specifically designed to change the wave function of the particle into one of the position eigenstates.

So there's no special phenomenon like a "wave function collapse". The measurement is just an example of an interaction, i.e. an interaction which evolves the wave function into an eigenstate of the variable being measured. The wave function doesn't collapse, or cease to exist after the measurement. The wave function just evolves in time, as usual.

I often see proposed explanations for the wave function collapse (decoherence is something I remember). What's all that about? The "collapse" seems to be just the usual evolution of the wave function. Can someone explain why the collapse is so mysterious?

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    $\begingroup$ maybe this thread helps? physics.stackexchange.com/q/27 $\endgroup$
    – Wihtedeka
    Sep 16 at 9:08
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    $\begingroup$ The difference between the pou describe an the "collapse" language is that you are describing a wavefunction in (x,y,z,t) and collapse is talking only of wavefunctions as (x,y,z), i,e, a complete solution for exact boundary conditions. Each new boundary condition is a "collapse" I agree it is a stupid term, the wavefunction is a mathematical formula, not a balloon in space. $\endgroup$
    – anna v
    Sep 16 at 9:19
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    $\begingroup$ The "collapse" seems to be just the usual evolution of the wave-function - no. The time evolution of the wave function in quantum mechanics is unitary, while the 'collapse' is not. $\endgroup$
    – Jakob
    Sep 16 at 10:00
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    $\begingroup$ Another way of stating Jakob´s comment is: the time evolution of the wave function is described by the Schrödinger Equation, the „collapse“ isn’t. There exist no mathematical equation to describe this. Example: what would you expect such an equation to look like to describe the collapse of an superposition state of spin up and down to one of the (completely random!) result of either state spin up or down. $\endgroup$ Sep 16 at 10:24
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    $\begingroup$ @Andrea We are talking of modelling data with mathematics, so the boundary conditions are determined by the specific system under observation for which we use a wave function to model it. The wave functions change according to the boundary conditions, no? $\endgroup$
    – anna v
    Sep 16 at 10:29

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The collapse becomes `mysterious' once you realise that:

  1. All things, including lab equipment is arguably composed of atoms that should satisfy quantum mechanics
  2. It is impossible to design an interaction between quantum systems that automatically results in sending "the wave-function of the particle into one of the position eigenstates". What you can get, at best, is that you particle and your apparatus get entangled. This is a result of unitary evolution.

You have (very, very broadly) two solutions. Either you decide that there are special physical systems called "measuring apparata" that obey different rules of evolution than the rest of quantum systems. Or you say "everything can go in a superposition." Both options are weird, and hence "the problem" and the endless debate on interpretations.

[Edit to include @MikeScott's comment]
There is a third option, namely, to say that "quantum mechanics is incomplete, there must be a better theory describing what is actually going on". This option also leads to weirdness, and so did not put a stop to the endless debate.

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    $\begingroup$ The second option forms the basis for the many worlds interpretation. If you're interested read P.37 and onward from users.ox.ac.uk/~lina0174/introduction.pdf $\endgroup$ Sep 16 at 11:15
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    $\begingroup$ @Andrea There’s a third option, which is broadly that there are hidden variables that don’t respect locality (but do respect causality). That’s the Bohmian pilot wave interpretation. $\endgroup$
    – Mike Scott
    Sep 16 at 18:32
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    $\begingroup$ The first option is obviously absurd. $\endgroup$ Sep 16 at 21:16
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    $\begingroup$ @R..GitHubSTOPHELPINGICE So obvious that the physics community has been arguing over it for about a hundred years :) $\endgroup$
    – d_b
    Sep 17 at 1:23
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    $\begingroup$ @R.. GitHub STOP HELPING ICE : It may seem absurd, but it is at least in theory, actually testable, which makes it solid science. $\endgroup$ Sep 17 at 4:35
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I think the mystery in part stems from the Copenhagen interpretation as subsequently codified by von Neumann, which held that a particle's wave function, upon measurement of some observable property of the particle, switched to be one of the eigenfunctions of that observable. There was no associated explanation for how that switch happened in a continuous way- it was simply assumed to happen.

As you have, I have often considered that a measurement is simply a type of interaction between particles, and that the so-called collapse is a result of modelling such interactions with over-simplified Hamiltonians in which some potential function is assumed that masks the underlying complexity of the interaction between the particle being observed and the particles that comprise the measuring device.

That said, the principle seems odd when you consider extreme examples. Take as a thought experiment a two-slits arrangement in deep outer space in which the detecting screen is five light-years from the slits. If an electron is fired through the slits its wave function, according to the standard interpretation, spreads out across an enormous region of space, yet when the electron eventually hits the detecting screen five years later, its wave function instantly becomes a highly localised wave packet. That, to me, still seems rather difficult to accept.

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  • $\begingroup$ The last paragraph doesn't seem hard to accept for me (probably because I've just started with the subject). Can we not define the detection interaction as the "collision" with a photon of a very high momentum? This interaction seems dramatic to me, so I can accept that it dramatically changes the wave-function. $\endgroup$
    – Egg Man
    Sep 16 at 11:46
  • $\begingroup$ The last paragraph is not a correct description of a 5 light-year long double slit experiment because it refers to "the electron's wavefunction". There is only a joint wavefunction for the electron and the detector and it never becomes more localized, just more correlated. $\endgroup$ Sep 16 at 12:00
  • $\begingroup$ @ConnorBehan. The point about the Copenhagen interpretation is that it denies what you've said, and treats the detecting screen as a 'measuring device' that somehow remains aloof from what happens at a quantum level. $\endgroup$ Sep 16 at 12:14
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    $\begingroup$ Replying to @EggMan, the unacceptable aspect to me is not that the wave function changes in the example I gave, but that the Copenhagen interpretation is that the wave function changes instantly from being evenly spread through a region of several cubic light years to being a sharp peak within a microscopic region within the detector. $\endgroup$ Sep 16 at 12:19
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    $\begingroup$ Oh, so this was meant as a reason why the Copenhagen interpretation is hard to accept. Then I completely agree. $\endgroup$ Sep 16 at 12:49
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It is not clear to me that the 'collapse of the wave function' even makes much sense in the 'Copenhagen Interpretation'.

This perspective is presented in the canonical reference [1] (which is sometimes referred to as the closest elucidation of Bohr's views, for example in [4]), which I will summarize below for convenience. Please see [1] for more details. It will hopefully become clear from my summary that, except in a very qualified (and in fact, trivial) sense discussed below, the 'collapse' is really referring to an idealized collapse that never really happens.

A famous article by Bell [4] specifically 'criticizes' some of the passages I will describe below, in the sense of saying they adequately represent the incompleteness of standard quantum mechanics he is not satisfied with (and he talks about "jumping" which simply does not seem to be correct, one can judge for themselves), so one can see how authoritative the perspective in [1] is, even to it's famous critics. One can see some further commentary in the literature on this approach for example in [2] page 12.

To see what the 'collapse' really means we should examine the theoretical description of what it means to actually do a measurement in quantum mechanics as described in [1], specifically section 7. (The justifications for why this is the approach to measurement that they take are given in the earlier sections 1, 2 and 6 - I would get side-tracked if I start justifying all this too, e.g. why it's unavoidable that we must assume the existence of classical objects, so please consult [1] for that, below we will just take it as a given that we must describe measurement in the following way).

A measuring device is a classical system with a quasi-classical wave function. Assuming the measurement process can 'completely describe' the quantum system (i.e. within the limits of quantum mechanics) it means the quasi-classical wave function is part of a complete basis of eigenfunctions characterizing a measurement process, i.e. eigenfunctions associated to the possible eigenvalues of the measuring apparatus. Assuming it's spectrum is discrete for simplicity it means a collection of $$\Phi_n(x)'s$$ represent the possible states of the measuring apparatus, and we can say with certainty that the classical apparatus is described by one, and only one, of these wave functions, if we know the value of a measurement.

In other words, the classical nature of the measuring apparatus means we can be absolutely certain that the measuring apparatus is in a given state $n$ and so has as it's wave function the stationary state $$\Phi_n(x).$$

Before a measurement of a system, the apparatus and system are independent subsystems of a total system, and so the total wave function is a product of their wave functions, $$\Psi(x,y) = \Phi_n(x) \psi(y)$$ for $\psi(y)$ the wave function of the system we want to measure (I will always refer to this as the 'system', and refer to the 'total system' when I want to include the measuring apparatus with it, referring to the measuring apparatus separately when I talk about it).

After the measurement, which involves an interaction between the apparatus and system we're measuring, the total wave function $\Psi'(x,y)$ is a complete mess in general, however because the systems are no longer interacting the apparatus is again independent and so we can Fourier expand the total wave function in terms of the $\Phi_n$ basis via $$\Psi'(x,y) = \sum_n A_n(y) \Phi_n(x).$$

Now again we invoke the classical nature of the classical measuring apparatus to say the following. If we measured a single eigenvalue from the discrete spectrum of the classical measuring apparatus with certainty, then the classical apparatus after the measurement in fact has a definite wave function, it is again a single eigenfunction from the spectrum of possible eigenfunctions, so in fact this sum 'collapses' down to a single term $$\Psi'(x,y) = A_m(y) \Phi_m(x).$$ It is important to note that, because of the classical nature of the measuring apparatus, and the fact we know the value of a measurement from the discrete spectrum with certainty, the initial sum that 'collapsed' was never really there, it was just a convenient tool to understand what's going on. Obviously if the measurement process was such that we can only be sure it was one from a set of possible measurements, the sum collapses to that set.

What we really care about is the case where we know the precise eigenvalue of the classical measuring apparatus after the measurement, but it's important to keep in mind the case where we don't know the precise value in what follows below (I will mention it explicitly when this case arises).

This immediately implies, since the systems are again independent after the measurement, that $A_m(y)$ is proportional to the wave function of the system we measured after the measurement. It is only proportional because $A_m(y)$ not only has to account for the state of the system after the measurement, it also has to account for the probability that we would find the $m$'th reading of the classical apparatus. We will see this explicitly below.

Therefore we can set it equal to a multiple of the true normalized wave function of the system, $\phi_m(y)$, after measurement $$A_m(y) = a_m \phi_m(y).$$ An obvious implicit assumption here is that $A_m(y)$ does not depend on the initial wave function $\phi(y)$. In other words, $A_m(y)$ (and as a consequence, $\phi_m(y)$,) is completely determined by the measuring process alone, assuming the measuring process can completely describe the state of the system as I said at the beginning of this post (otherwise the initial conditions could clearly affect things).

However, the linear nature of the equations of of quantum mechanics implies that there still should be a linear relation between the wave function before measurement, $\psi(y)$, and the wave function after measurement, $A_m(y)$. In other words, $\psi(y)$ should evolve into $A_m(y)$ under some evolution operator which we can write as $$A_m(y) = \int K_m(y,y') \psi(y') dy'.$$ Since $A_m(y)$ is completely determined by the measurement process, it means $K_m(y,y')$ is completely determined by the measurement process.

We now have two different interpretations of $A_m(y)$, the 'collapse' interpretation and the 'evolution' interpretation, i.e. $$A_m(y) = a_m \phi_m(y)= \int K_m(y,y') \psi(y') dy'.$$ They must clearly be the same thing so that we can set $$K_m(y,y') = \phi_m(y) \psi_m^*(y')$$ implying that $$a_m = \int \psi_m^*(y')\psi(y') dy'.$$ At this stage (ignoring the obvious implication that the notation suggests for now) all we can say about these $\psi_m^*(y')$'s is that they depend on the measurement process.

But this is all just saying that the abstract wave function of the total system after a measurement, allowing for situations where we could not measure the eigenvalue of the classical measuring apparatus even with any certainty (i.e. an extreme version of the special case I warned about above), is $$\Psi'(x,y) = \sum_n a_n \phi_n(y) \Phi_n(x)$$ where the $\phi_n(y)$ are normalized functions (which represent the total system we measured after a measurement, note counter-intuitively they are not actually 'eigenfunctions' of anything), the $\Phi_n(x)$ are normalized eigenfunctions of the measuring apparatus, so that the $a_n$'s are just the usual coefficients representing probabilities and satisfying $$\sum_n |a_n|^2 = 1.$$

The fact that this last relation should hold, coupled with the fact that $a_n$ is defined by $a_n = \int \psi_n^*(y')\psi(y') dy'$, means that the $\psi(y')$ should expand in a complete basis of $\psi_n(y')$'s, but the $\psi_n(y')$'s were determined by the measuring process. In other words, the wave function after measurement should expand in a basis of eigenfunctions of an operator characterizing the measurement process.

But again, invoking the classical nature of the measuring apparatus for a (at least theoretically) precisely known measurement from the discrete spectrum, the sum $\Psi'(x,y) = \sum_n a_n \phi_n(y) \Phi_n(x)$ thus 'collapses' to $\Psi'(x,y) = a_m \phi_m(y) \Phi_m(x)$ (i.e. it was always of this form for the specific interaction between the classical apparatus and system where we know the precise eigenvalue of the measuring device after measurement), meaning the measuring apparatus gave the $m$'th eigenvalue associated to $\Phi_m(x)$, but since $a_m = \int \psi_m^*(y')\psi(y') dy'$ this tells us that the wave function $\psi(y')$ was actually 'measured' to be in the state $\psi_m(y')$ when we did the measurement. In other words, this is the best we can say about the state of a quantum system when we do a measurement, meaning we cause a measuring apparatus to interact with the system. All we can infer about the quantum system is that in the process of interacting, the $\psi_m(y')$ wave function 'rubs off' onto the measuring apparatus under the time evolution of the total system, in the sense that the quasi-classical wave function evolves (through interacting with the system) from one 'stationary state' to another in the process of interacting. It's not "jumping", one is completely ignoring the fact the measuring apparatus is interacting during the measurement and so obviously can linearly evolve (not discontinuously jumping) to a new state.

We are completely shielded, in principle, from saying anything more about what the system was 'really' doing, all we can do is infer from the final measurement of the apparatus what the system was doing from how it caused the measuring apparatus wave function to register the measured eigenvalue. Further the wave function after this measurement process is also given by this discussion, it is this new wave function $\phi_n(y)$, which in general is completely different from the initial wave function $\psi(y)$. Everything is encoded in the above discussion.

Thus you see the idea of 'collapse of the wave function' is nonsensical if one means anything other than the trivial collapse of the Fourier expansion discussed above. It would completely contradict the linearity of the equations of quantum mechanics if there was some jarring 'collapse' of the wave function going on. The above process completely accounts for this properly. The initial wave function of the system, $\psi(y)$, just evolved into $A_m(y) = a_m \phi_m(y)$ via a linear evolution operator $A_m(y) = \int K(y,y') \psi(y') dy'$ where $\phi_m(y)$ is the normalized wave function of the system after measurement, and $a_m$ encodes the (experimental) fact that we measured a certain value due to the system somehow 'rubbing off' onto the measuring apparatus during the interaction. That's intrinsically the best we can say about the state of a quantum system within quantum mechanics. It would thus completely contradict the linearity of quantum mechanics to think the wave function of the system actually 'jumps', and it's always stated in a completely hand-waving unjustified manner, unlike the discussion above where everything fits into place.

The really non-trivial thing going on here, that the concerns about 'quantum jumping' are really expressing, is that the fact we can measure anything at all. This is intrinsically caused by the fact that quantum mechanics can only be defined in the first place by assuming the existence of classical mechanics, to which it must reduce in the 'classical limit'. That is the real 'jarring' thing about this. This assumption of the existence of classical mechanics means we must have the measuring process is such that the total wave function (expanded in a basis of eigenstates of the measuring apparatus) always (when we measure the precise eigenvalue of the measuring apparatus) 'collapses' to a single term, but the 'collapse' doesn't really happen what really happens is 'classical mechanics' says only one term was there the whole time.

The wave function just evolves from one wave function into another wave function via linearity and due to the interaction between the (classical) apparatus and the system (we're measuring), that something 'rubs off' onto the measuring apparatus during this is simply an experimental fact the theory is trying to capture.

So, through all this, there is no 'collapse' of the total wave function. It is just a mathematical tool allowing us to say that the total wave function 'jumps' from the complete Fourier sum down to a single individual term in the sum if we keep things general in the beginning. If that sum doesn't 'collapse' then we can never measure anything, or actually even say the system even had a wave function, indeed how can we even talk about a measurement. In other words, nothing makes sense without the 'classical limit'.

If you assume classical mechanics exists however, then there is never any 'jump', the abstract 'Fourier expansion' we proposed actually just contained one term the whole time. Again, the point is: without classical mechanics, that Fourier sum can never be said to really just be a 'single term', so we get completely stuck and just have no theory.

Not only is it the case that the initial wave function $\Psi(y)$ transitioning into a new wave function $\phi_m(y)$ is just a consequence of the fact that a classical measuring apparatus interacts with a quantum system, i.e. an interaction makes it evolve into a new wave function, it is absolutely vital to notice that it is precisely only because of the classical nature of the measuring apparatus that means we can even be aware of the fact that the system evolved from one wave function to a new wave function.

In other words, without the existence of classical mechanics, there is absolutely no theory of quantum mechanics at all. Without the quantum system "rubbing off" onto a classical apparatus through the interaction, we just have nothing. Any discussion of "collapse" that mentions it in some hand-waving fashion is either just a misunderstanding of the above description of the measurement process, or an 'alternative interpretation' of quantum mechanics which you can likely bet isn't even internally logically consistent (compared to Copenhagen).

There is a(n unbelievable) claim that 'decoherence' allows us to understand how the 'classical world' arises from quantum mechanics, e.g. via diagonal entries on a density matrix. At least from the above discussion, it is very likely completely circular and nonsensical from the above perspective. One has to trust that an 'alternative' perspective (e.g. those mentioned in [3]) to the Copenhagen interpretation given above (from reference [1] below) is as logically internally consistent as this while also somehow defining quantum mechanics without classical mechanics and properly accounting for the measurement process without a contradiction. It is hopefully clear from this discussion why people sometimes say there is no alternative to the 'Copenhagen interpretation'.

A side comment is to notice how absolutely vital it is that the continuous spectrum eigenfunctions in quantum mechanics are actually "wave functions" (a view actually commonly denied even on this site, see my answer here for how serious some of the the flaws are with this perspective, on top of the critical flaw in the whole theory that would occur due to the measurement process according to this post). Indeed re-reading the above discussion for this case, if they aren't, we can never even know what the wave function of a system after a measurement is, the measuring apparatus could never even have a definite value thus we can never even fix the wave function of the system after measurement.

References:

  1. Landau and Lifshitz, "Quantum Mechanics", 3rd Ed., Sections 1, 2, 6, 7.
  2. Zinkernagel, "Niels Bohr on the wave function and the classical/quantum divide".
  3. Weinberg, "Quantum Mechanics", 1st Ed., Sec. 3.7.
  4. Bell, "Against Measurement", 1990 Phys. World 3 (8) 33.
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  • $\begingroup$ After reading this answer 3 times I am still unsure what the point is, which is why down voted. I am under the impression that you simply avoid the conflict between unitary time evolution and the measurement process/projection by introducing an axiomatic "classical measurement device". But that doesn't resolve anything, it simply shifts the problem to defining clearly what that means, begging the question where the boundary between measurement apparatus and quantum system begins and why they aren't both on equal footing. A concise summary of your points would improve the answer. $\endgroup$
    – Hans Wurst
    Sep 21 at 14:11
  • $\begingroup$ @HansWurst I had tried to summarize what I claiming/doing in the first four small paragraphs, but I have added some things to try and make this clearer thank you. All my answer was trying to do was to get the reader to realize the cartoon picture of 'projection/collapse' is basically absent/trivial in the authoritative discussion of [1], in that sense I have succeeded, so yes it 'shifts the problem' to something else, which if the reader realizes is a success to me. I recommend carefully checking [1] to see how everything else is justified, my goal here wasn't to also justify all that too. $\endgroup$
    – bolbteppa
    Sep 21 at 16:56
  • $\begingroup$ Yeah your answer is hard to read, possibly because some notation isn’t clear. In any event, I want to point out that L&L is a bit “old” for that sort of stuff and that a lot of work has been done since. Maybe the writings of Asher Peres are more “modern”, in particular on the essentially non-reversible nature of measurements. I will certainly have additional questions later and may try to reach you in chat rather than extend this comment into a discussion. $\endgroup$ Oct 3 at 23:13
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I think the other answers are pretty complicated.

Doing two consecutive measurements of spin always gives the same result. That is a clear reason why we collapse the wave function - so that after the first measurement, the wave function reflects the correct statistics given all macroscopic information. If you want the WF to always give correct statistics for the particle, this is the only way within copenhagen QM.

Spin was just an example.

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The time evolution of a wavefunction during a measurement (the so called wave-function collapse) is discontinuous .

Meanwhile all time evolutions of wavefunction during interactions, according to Schrödinger equation is continuous .

So there are two kind of time evolutions in QM, and measurement is difference to the kind of interaction you've described.

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The measurement problem is that if $|\text{quantum system in some state}\rangle$ evolves into $|\text{you see some reading on an instrument}\rangle$, and $|\text{quantum system in some other state}\rangle$ evolves into $|\text{you see some other reading on an instrument}\rangle$, then

$$Ψ = α|\text{quantum system in some state}\rangle + β|\text{quantum system in some other state}\rangle \qquad (α,β \ne 0)$$

must, by linearity, evolve into

$$Ψ' = α|\text{you see some reading on an instrument}\rangle + β|\text{you see some other reading on an instrument}\rangle$$

which describes a universe containing two copies of you who saw different measurement outcomes. Because it's possible to prepare systems in states of the form $Ψ$, it's possible to create states of the form $Ψ'$. People don't like that conclusion, but it's unavoidable unless you introduce some sort of nonlinearity into the theory. "Wavefunction collapse" is a collective term for all proposals that introduce a nonlinearity with the goal of getting rid of all but one of the predicted copies of you.

I disagree with Andrea's answer, which says

You have (very, very broadly) two solutions. Either you decide that there are special physical systems called "measuring apparata" that obey different rules of evolution than the rest of quantum systems. Or you say "everything can go in a superposition."

because I think that the phrasing of the first option is unfair to collapse proposals. Measurement devices (and human beings) contain a very large number of particles, and it's consistent with all experimental data that that largeness is their only relevant difference from quantum systems that don't collapse. One simple (if not well motivated) form of collapse is just to imagine that individual particles are very occasionally randomly localized in position. That's enough to make measurement devices collapse almost instantly, and the quantum systems that we examine collapse too rarely to be detectable, simply because of the enormous difference in particle counts. Roger Penrose's idea that collapse happens when there is "one graviton's worth" of difference in the matter distribution between alternatives is another example of this.

Decoherence addresses a somewhat different problem, namely, why you only see "some reading" or "some other reading" on the instrument, and never see $α|\text{some reading}\rangle + β|\text{some other reading}\rangle$. Even if there are multiple copies of you, each copy sees only one of the two discrete readings, and it's not immediately obvious why.

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Many of the other answers, especially the highest-voted answer by @Andrea, are excellent and answer the OP's question. However, since this is my kryptonite, I can't help but add my answer. ;-) In order to add something of value to the existing answers, I will try to raise one more point that hasn't been raised by others and elaborate a bit on the point $2$ in @Andrea's answer, namely, I will give a simple proof as to why "an interaction which evolves the wave-function into an eigenstate of the variable being measured" cannot exist in quantum mechanics.


OK, so there are two separate questions:

  • Is an interaction of the type described by the OP, namely, "an interaction which evolves the wave-function into an eigenstate of the variable being measured" is even desirable as a description of what a measurement is?
  • Is an interaction of the type described by the OP, namely, "an interaction which evolves the wave-function into an eigenstate of the variable being measured" consistent with the principles of quantum mechanics?

I will show below why the answer to both these questions is a resounding NO.


Sniff Test Failure

Let's just assume for the moment that the type of interaction that the OP is describing is possible (even if it isn't as I will come to in a moment). So, if we start with a state $\vert\psi\rangle$ of the system and measure an observable $O$ then if the measurement process is just some process that brings the system to an eigenstate of the observable, say $\vert o_i\rangle$ then no matter how many types I prepare a state $\vert\psi\rangle$, this measurement process always ought to give me the result $\vert o_i\rangle$. This must be so because all processes in quantum mechanics are deterministic except for the wave-packet reduction process which OP is refusing to postulate. However, this flies in the face of our experience that even if I prepare the system in an identical state $\vert \psi\rangle$, the results of the measurement of an observable $O$ are different over different trials, i.e., they are not deterministic but rather probabilistic (as long as $\vert\psi\rangle$ wasn't an eigenstate of $O$, to begin with, of course). So, this shows that the kind of process that the OP proposes to be a measurement process doesn't behave like an experimentally observed measurement process at all.


A No-Go Theorem, If You Wish

So, we have established that the kind of interaction that the OP has described can't describe a measurement process because it has no hope of reproducing the experimentally observed behavior of a measurement process. However, we still take it upon ourselves that this kind of interaction simply isn't consistent with two of the most important properties of quantum mechanics, namely, linearity and unitarity.

Let's say we start with a state $\vert\psi\rangle$ and the interaction brings it to the state $\vert o_i\rangle$, an eigenstate of some observable $O$. Similarly, suppose that if we start with a state $\vert \phi\rangle$ and the interaction brings it to the state $\vert o_j\rangle$, an eigenstate of some observable $O$. Thus, for linearity to be preserved, it must be the case that if we start with a state $\frac{1}{\sqrt{2}}(\vert\psi\rangle+\vert\phi\rangle)$ then this interaction process must evolve it to $\frac{1}{\sqrt{2}}(\vert o_i\rangle+\vert o_j\rangle)$. Now, the interaction is always supposed to bring an initial state to an eigenstate of the operator $O$. This can be the case only if $o_i$ and $o_j$ belong to the same eigensubspace of the observable $O$. Or, in general, the proposed interaction brings any initial state to the same eigensubspace of the said observable. However, this violates unitarity unless the said operator is the identity, which, is not a non-trivial observable.

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  • $\begingroup$ Can the downvoter explain? (Apologies if this comment is against the rules and a moderator removed a previous version of this comment, I am not sure if that's the case or my previous attempt at posting this comment failed due to an unstable internet connection.) $\endgroup$
    – Dvij D.C.
    Sep 18 at 23:58
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Collapse is mysterious because the measurement doesn't have to be interactive. As long as the measurement as done ie there is an information of the measurement in the universe, the wave function collapses. This has been shown multiple times in experiments like double slit experiments where the measuring device was placed after the slit. And even then, particle like behaviour was observed as if the electrons knew they were going to be measured in advance and ahead of time.

The most satisfying explanation to this collapse for me is:

  1. The wave function collapses in one universe and doesn't in another universe. That means the universe holding the information of measurement sees particle like behaviour and universe that doesn't have that information, sees the wave like behaviour

  2. The information is available to electrons at/of all the time. They know if there will be measured in future or not. This is sort of obvious given the fact, at c, time basically stops and there is no meaning of past and future

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The wavefunction isn't actually a thing. It's just the probability distribution for where an electron is likely to be. When an observation is made, you now know where the electron is, so you don't need the wavefunction to know where it might be. That's why we say wave function collapse. The wavefunction isn't a physical thing. Just mathematical calculations.

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View from the quantum world
Wave function collapse is not necessary, if one "believes" in quantum mechanics - a measurement can be described as interaction with a macroscopic but quantum object, in terms of evolution of the wave function or the density matrix.

View from the classical world
If we don't know the laws obeyed by microscopic (or otherwise quantum obejcts), we can try to study these laws... but we are forced to use classical equipment and classical mechanics to measure and analyze the results. The collapse then becomes the process in which quantum mechanical phenomenon causes a classical change in our measurement apparatus. This is why it played such an important role in the original formulations of the QM... but, since people nowadays take QM for granted, the concept of the collapse seems redundant.

Classical limit
Finally, to merge the two viewpoints, one can view the collapse as the classical limit - in the same way as we talk about non-relativistic mechanics or thermodynamic limit - neither of these is strictly correct, but in everyday life they work with good enough precision.

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  • $\begingroup$ But I don't get how the wave-function can ever undergo a non-unitary transformation. It's not just about what value our measurement apparatus measures. Even the wave-function of the particle undergoes a non-unitary transformation, which is forbidden by Schrodinger's equation. $\endgroup$
    – Egg Man
    Sep 19 at 3:02
  • $\begingroup$ @EggMan Schrödinger equation is valid in quantum world, not in the classical one. I think misunderstanding here is about what is quantum, what is classical, and how they are connected. $\endgroup$ Sep 20 at 7:06
  • $\begingroup$ You say "The collapse causes a classical change in our measurement apparatus". I take this to mean "The collapse makes the measurement apparatus show a definite value for the measurement". But the measurement also causes a change in the quantum object, by changing its wave-function non-unitarily. This change is forbidden by the Schrodinger's equation. $\endgroup$
    – Egg Man
    Sep 20 at 13:45
  • $\begingroup$ @EggMan On purely logical level, one cannot apply the Schrödinger equation to the interaction with a classical object - this situation obeys different laws/rules, such as collapse. Then, if looked from purely quantum prospective, the non-unitarity may appear only in the classical limit. Like phase transitions appear only in thermodynamic limit - for a finite system the partition function is always analytical. $\endgroup$ Sep 20 at 13:52
  • $\begingroup$ @EggMan I asked the question to clarify specifically this point (physics.stackexchange.com/q/667074/247642)... but I don't hope that somebody will come with any reasonable amount of math. You may try to read something on Caldeira-Legget approach - they have a review in RMP. $\endgroup$ Sep 20 at 14:43
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I am going to try a different sort of answer here since it is not clear to me what the OP's issue is, and this might be it.

Imagine a plane wave function of a uniform photon field striking a piece of old fashioned photographic film for a limited time. The film contains many individual silver halide crystals which, if they interact with a photon, will turn black. Now, the wave function evolution for this system is such that all of the silver halide crystals, after a brief exposure, should be in a superposition state of, let's say, 0.01% being black and 99.99% of remaining white.

However, when I examine the film some time later I see a single black spot in a certain place, and all the rest is white. I definitely do not see all the crystals in a QM superposition. So, why is this? Does my looking at the film somehow cause every crystal to exit its superposition and assume one of its vectors? My eyes and brain aren't really "interacting" with the film - are they?

This is the mystery ...

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    $\begingroup$ Yes, your eyes are interacting with the film via the photons that reflect off the film. And yes, taking QM seriously, you should conclude that you yourself are in superposition after looking at the film. The "mystery" part is only the fact that humans don't like to think of themselves being in superposition, entangled with the dots on a piece of film. I would see the whole wave function collapse thing as a shortcut to make ourselves believe that we can produce pure quantum states. $\endgroup$ Sep 16 at 21:43
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    $\begingroup$ @cmaster-reinstatemonica Your eyes and brain interacted with the film long before you actually looked at it, via thermal emissions and other interactions between the film and the background, and between the background and you. That’s what decoherence is. $\endgroup$
    – Mike Scott
    Sep 17 at 5:30

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