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When an observer causes the wave function of a particle to collapse, how can we know that the wave function was not collapsed already before the measurement?

Suppose we measure the z-component of the spin of an electron. After the measurement, it is entirely aligned along the measured direction, e.g. the +z-direction. Before the measurement we need to assume that a probability distribution proportional to $|\Psi|^2$ of the two allowed directions is present.

If we repeat the measurement with many identically prepared electrons, we should see such a distribution finally. For example, we could measure 40% spin-down and 60% spin-up.

However, it seems we could also assume that all of these particles have a defined spin-direction before we measure them.

What is an intuitive (being aware that quantum phenomena as such are rarely intuitive) explanation for why we cannot simply assume that the spin was already aligned completely in that measured direction?


With regards to the suggestion that this two-year old question is a duplicate of the one asked yesterday, I would like to point out that my question isn't limited to entanglement, but asks about a very fundamental principle in quantum mechanics, and as such is not a duplicate.

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Quantum mechanics was developed in order to match experimental data. The seemingly very weird idea that some observables do not have a definite value before their measurement is not something physicists have been actively promoting, it is something that theoretical considerations followed by many actual experiments have forced them to admit.

I don't think there is an intuitive explanation for this. It is closely linked to the notion of superposition. The basic idea is that we do indirectly observe the effects of interference between superposed quantum states, but upon actual measurement we never see superposed states, only classical, definite values. If we suppose these values where there all along, then why would we have any interference? The whole framework of QM would be pointless.

In other words, a quantum state is what it is (whatever that is) precisely because it is in contrast to a classical state: crucially, it only describes a probability distribution for observables values, not actual, permanent values for these observables.

A wavefunction that would always be collapsed would just be a classical state. Now why (and does?) a measurement "collapse" anything at all is an open question, the measurement problem.

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  • $\begingroup$ What would be a simple example of an interference of electrons that contains the spin? $\endgroup$ – ahemmetter Oct 7 '16 at 15:37
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Imagine the following set of experimental data:

Every day, you decide whether to put on your sunglasses before looking at the sky to check the weather. Every day I, a thousand miles away, do the same thing.

After we've made our observations, we call each other on the phone to compare. We discover that on days when we've looked without sunglasses, we always see the same thing (sometimes sunny, sometimes cloudy). On days when one of us wears sunglasses and the other doesn't, we still always see the same thing. But on days when we both wear sunglasses, it is invariably the case that one of us sees a sunny sky and the other sees a cloudy sky.

Now suppose every day, one of four things is true: Either the sky above your house is sunny (and looks sunny with or without sunglasses), or it's cloudy (and looks cloudy with or without sunglasses), or it's in a condition that looks sunny without sunglasses but cloudy with them, or it's in a condition that looks sunny with sunglasses but cloudy without. Likewise for the sky above my house. And suppose each sky is unambiguously in one of these states before we look at it.

Question: What pattern could account for the experimental data? Answer: None. If your sky and my sky are always either both sunny or both cloudy, that accounts for what we see on three out of four days but can't account for what we see when we both wear sunglasses. If there's some much more complicated pattern (e.g. 8% of the time our skies are both sunny, 7% they're both cloudy, 19% yours is sunny while mine is in the state that looks sunny only through sunglasses, etc), you still won't be able to account for that experimental data. It's not hard to prove that no matter what percentages you assign to the sixteen possible pairs of states, the experimental data just don't fit your predictions.

Conclusion: You can't use ordinary probability theory to explain the weather.

Now in real life we don't have this problem with weather, because we never see the kind of experimental data I supposed in the first place. But in quantum mechanics, we do see such data (not exactly as I've supposed here, but close enough so that the same issue arises). Thereefore you can't use ordinary probability theory, in the sense you're trying to use it, to explain the observed facts.

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The precise answer is contained within the Kochen-Spekker theorem and Bell's theorem. (I know it's awkward that one of them has the form "the [name] theorem" and the other has the form "[name]'s theorem". That's a long-standing inconsistency in English math and physics usage.)

The key point is the fact that you can measure in different bases. If you have a fixed state $|\psi\rangle$ (whose time-evolution you neglect), and you agree to always measure in the same fixed orthonormal basis $\{|i\rangle \}$ (e.g. the position basis), then the probability distribution $\left \{ p_i = |\langle i | \psi \rangle|^2 \right \}$ is completely classical, and could absolutely simply reflect that the system was in an unknown but definite state before the measurement.

But it turns out that there's no single classical probability distribution (which could simply reflect uncertainty in the system's definite pre-measurement state) that simultaneously reproduces the Born statistics in every basis.

So if you try to understand what's so weird about quantum mechanics while only considering measurements in a single basis, then you'll fail, because the quantum mechanics of a single state measured in a single basis really is just classical probability theory. To see what's really going on, you need to consider measuring in different bases (or equivalently, allowing yourself to act a non-diagonal unitary operator on the state before measuring it).

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What is an intuitive (being aware that quantum phenomena as such are rarely intuitive) explanation for why we cannot simply assume that the spin was already aligned completely in that measured direction?

For that single particular measurement that is exactly what we assume.

Lets make it simpler : here are measurements of the electrons in the orbitals of the hydrogen atom:

measured hydrogen orbital

Each dot there is an (x,y) measurement from a single hydrogen atom. Orbitals are the probability loci for the electrons about the proton of the hydrogen.When that electron interacted with the detecting system, it was there.

Quantum mechanics is the theory that models and predicts what the accumulation of all the measurements will show ( the probability distribution)

Edit, to address the new title:

How do we know a quantum state isn't just an unknown classical state?

People have been trying from the beginning of the formulation of quantum mechanics, unsuccessfully to find an underlying classical deterministic system from which the probabilities could be calculated classically.

The de Broglie-Bohm theoretical model succeeded to have the wavefunction of non relativistic mechanics as emergent from a deterministic model.

In addition to a wavefunction on the space of all possible configurations, it also postulates an actual configuration that exists even when unobserved. The evolution over time of the configuration (that is, of the positions of all particles or the configuration of all fields) is defined by the wave function via a guiding equation. The evolution of the wave function over time is given by Schrödinger's equation.

The problem is that it cannot be extended to the relativistic regime and equations. In addition the theory violates most physicists intuitions that the simplest mathematically models are preferred over complex ones, Occam's razor:" Among competing hypotheses, the one with the fewest assumptions should be selected" .

There are continuous explorations along the line of seeking a deterministic underpinning for quantum mechanics. G. 't Hooft ( Nobel prize winner)is a member of this site and has written on his efforts towards a deterministic quantum mechanics, see here for example..

The answer is that at present quantum mechanics models successfully all of our experimental observations, whereas there exist no deterministic theory that can show that all the quantum mechanical models emerge from an underlying deterministic frame/level.

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  • $\begingroup$ So it makes no difference to assume an electron has a pre-defined location which prior to the measurement we simply dont know, or to say the electron is described by a wave function, the square of which gives the probability distribution, since both result in the same distribution after many measurements. Is that roughly the idea behind it? $\endgroup$ – ahemmetter Oct 7 '16 at 10:52
  • $\begingroup$ Yes. There have been efforts to try to create the probability distributions from classical deterministic equations, unsuccessfully most of the time. Bohm's pilot theory reproduces non relativistic quantum mechanics in a more complicated way, but fails when relativity has to be included. $\endgroup$ – anna v Oct 7 '16 at 12:26
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Actually there is a simple experiment to distinguish a quantum superposition from a classical one.

Suppose you have two boxes. There are 10000 particles in the state

$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$

In one box; and 5000 and 5000 in the states

$|\phi_0\rangle = |0\rangle$ and $|\phi_1\rangle = |1\rangle$

In the other $|0\rangle$ is the state with negative spin in $ z$ and $|1\rangle$ the state with positive spin in $z.$

This boxes represent exactly what you are asking for, namely, is there any difference between a superposition state and one that is already collapsed according to the probabilities of QM?

If you measure spin in $z$ you'll get the same results in both boxes, as you said. $50\%$ up and $50\%$ down. So it seems that quantum superpositions are indeed just like ensembles of collapsed wavefunctions. But... what happens if we measure in $x\,?$

Since the state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ is the state $|1\rangle_x$ (the state with positive spin in $x$) every measurement in the superposition box will give spin up. However, since both $|0\rangle$ and $|1\rangle$ are linear superpositions of $|0\rangle_x$ and $|1\rangle_x $ with equal probabilities the second box will give $50\%$ up and $50\%$ down. So there is a notorious difference between boxes.

This is what is usually called interference. When the particles are in a superposition of up and down, wavefunctions interfere and modify the probabilities of measuring eigenvalues of other basis.

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Consider 3 such electrons in a state that is a superposition of all 3 spins up in the $z$-direction and all 3 spins down:

$$\left|\psi\right> = \frac{1}{\sqrt{2}} \left[\left |\uparrow\uparrow\uparrow\right> - \left|\downarrow\downarrow\downarrow\right>\right]$$

Consider the 3 observables:

$$\begin{split} A_1 &= \sigma_x^{(1)}\sigma_y^{(2)}\sigma_y^{(3)}\\ A_2 &= \sigma_y^{(1)}\sigma_x^{(2)}\sigma_y^{(3)}\\ A_3 &= \sigma_y^{(1)}\sigma_y^{(2)}\sigma_x^{(3)} \end{split} $$

Here the superscript denotes on which spin the operator acts. So, the observable $A_i$ corresponds to measuring the product of the $x$-component of the $i$th spin and the $y$-components of the two other spins. Using:

$$\begin{split} \sigma_x\left|\uparrow\right> &=&\left|\downarrow\right>\\ \sigma_x\left|\downarrow\right> &=&\left|\uparrow\right>\\ \sigma_y\left|\uparrow\right> &=&i\left|\downarrow\right>\\ \sigma_y\left|\downarrow\right> &=-&i\left|\uparrow\right>\\ \end{split}\tag{1} $$

You then find that:

$$A_i\left|\psi\right> = \left|\psi\right>$$

So, measuring any of the three $A_i$'s will always yield $1$; the product of measuring an $x$ component of one spin and the $y$ component of the two other spins is always equal to 1 despite the individual spin measurements yielding completely random results. If we assume that these results were already determined before they were actually measured, then that means that the counterfactual results of measuring a different spin components are also well defined.

While we cannot tell what the results would have been had we measured different spin components, we do know that all the $A_i$'s yield 1. Whatever the spin measurement results would have been, the product $A_1A_2A_3$ must be equal to 1. But written out in terms of the individual spin measurements, this product equals the product of the result of measuring the three $x$-components and the squares of the three $y$-components. Since these squares are equal to 1, the product of the results of measuring the three $x$-components must yield 1.

Now, it's easy to check using (1) that:

$$\sigma_x^{(1)}\sigma_x^{(2)}\sigma_x^{(3)}\left|\psi\right> =- \left|\psi\right>$$

So, the result of measuring the product of three $x$ components and multiplying the results is always $-1$, and not $1$. This thus proves wrong the assumption that the results of the spin measurements are determined regardless of whether they are actually measured.

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  • $\begingroup$ Just because you pointed me to this answer, I'm trying to understand it, even though this is a more complex example than the one I was asking about. Anyway my doubt here is that you are mixing factual and counterfactual results applying to both of them the same quantum constraints. Counterfactual definiteness means you can include measurements that have not been performed as separate results, but without mixing the observables as if they came from the same measurement. $\endgroup$ – user1892538 Jul 2 '17 at 21:38
  • $\begingroup$ @user1892538 The QM predictions can just be taken to be what an experimentalist who doesn't know QM will find. So, whenever xxx is measured it's always -1 but xyy, yxy and yyx is always 1 and that is enough for the experimentalist to falsify local hidden variables. $\endgroup$ – Count Iblis Jul 2 '17 at 22:16
  • $\begingroup$ Ok thanks (in fact your answer makes perfectly sense here) but I still don't see the equivalent of the above with a two spins singlet by measurements only along the axes x,y and z. If you do, please answer my question. Of course I understand how Bell inequality can work with different angles in that case, as per my own answer. $\endgroup$ – user1892538 Jul 3 '17 at 1:04
  • $\begingroup$ Btw as an experimentalist you also need to make spacelike separated all those 3 times 4 (times n for accuracy) measurements, to close the locality loophole at least. Just out of my curiosity, do you have any reference to this experimental setup (maybe from arxiv)? $\endgroup$ – user1892538 Jul 3 '17 at 1:39
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    $\begingroup$ Yes, you want spacelike separation. The original argument was presented in this paper. $\endgroup$ – Count Iblis Jul 3 '17 at 23:33
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When an observer causes the wave function of a particle to collapse, how can we know that the wave function was not collapsed already before the measurement?

First of all an "observer" is any physical interaction with the system deemed "enough" to produce a measurement. Second, "collapse" is a dangerous term to use. We know that the system isn't only the particle-like component we see because of things like interference patterns and entanglement.

Suppose we measure the z-component of the spin of an electron. After the measurement, it is entirely aligned along the measured direction, e.g. the +z-direction. Before the measurement we need to assume that a probability distribution proportional to |Ψ|2|Ψ|2 of the two allowed directions is present.

If we repeat the measurement with many identically prepared electrons, we should see such a distribution finally. For example, we could measure 40% spin-down and 60% spin-up.

However, it seems we could also assume that all of these particles have a defined spin-direction before we measure them.

You just said they were prepared identically. If they had different intrinsic properties, they wouldn't be identical.

What is an intuitive (being aware that quantum phenomena as such are rarely intuitive) explanation for why we cannot simply assume that the spin was already aligned completely in that measured direction?

One example is Entanglement, where the correlation of measured spins actually depends on the particles "knowing" what axis their twin is measured in.

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As long as you have a single electron to be measured it does not make any sense to ask yourself whether it was in a certain spin direction before being measured or if you forced it to assume one.

If we have many electrons which we assume to be "identical", for example because they are produced in the same way (for example being emitted by an heated metal), and we see different outcomes when we measure their spin, it does make sense ask ourselves if a classical description or a quantum description holds.

For example, they may have different speed. But we could formulate a classical description of this phenomenon: they come out of the metal with a given speed, unknown to us -the experimenter - but having a certain value before the measure. The distribution of these speed may follow a certain classical probability distribution.

But sometime a quantum picture is needed. For example, for the spin. In this case, we must assume that before the measurement the spin has no definite value and that the probability distribution which describes the outcomes are written as |psi|^2.

Usually when a quantum picture holds a naive classical one fails, because probabilities - as computed with wave functions - have different properties than naive classical probabilities. Think for example of the interference pattern in the two slit experiment.

However, since quantum is so different from everyday experience, it is natural to ask whether it holds to a classical description of a certain "quantum" phenomena, but that we are not able of finding it because we miss something. This attempts are called "hidden variable" descriptions. In this descriptions we postulate that some hidden degree of freedom exists, which we ignore (hence the name hidden). It is this degree of freedom, which is completely classical - described by a standard probability distribution - which rule that outcomes which we see and which appear to follow a quantum picture.

It is possible to show, however , that a hidden variable picture always fail to reproduce exactly a quantum picture. So from a theoretical Point of view, quantum theory and classical theory are indeed different. Moreover, experiments have been made (Google Alan EPR), which rule out that a hidden variable description holds for certain phenomena. So it is fair to claim that we have evidence of at least one situation in nature which cannot be described by a classical theory, not matter how many "hidden variables" exist which we ignore.

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What is an intuitive (being aware that quantum phenomena as such are rarely intuitive) explanation for why we cannot simply assume that the spin was already aligned completely in that measured direction?

(emphasis mine) It seems to me that others are explaining why the measurements can't be predicted by classical probabilistic theory but for the question that is actually asked here, there is a simpler answer.

  • If you measure in the $x$ direction, you get the result that the spin is either towards the $x$ direction or opposite.
  • If you measure in the $z$ direction, you get the result that the spin is either towards the $z$ direction or opposite.

Now if the spin already has a well-defined direction before our measurement, it would mean that:

  • If the spin has well-defined direction $+x$ or $-x$, we will measure in $x$ direction.
  • If the spin has well-defined direction $+z$ or $-z$, we will measure in the $z$ direction.

But how could the direction of the spin (that we yet haven't measured!) cause us to measure in that direction?

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