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Wigner's friend thought experiment mentions that there is an apparent paradox - 'when exactly did the collapse occur?'. Whether it occurred when the friend made the measurement, or when Wigner asked about it.

Why is it different than if I flipped a coin and observed the result, but hid the result from my friend and so he/she does not know it yet? Isn't it clear that the collapse happened when the measurement was made, and Wigner just does not have the knowledge about it yet.

I am sure that there is a basic property of quantum mechanics that makes this thought experiment qualitatively different from what I am assuming. But my knowledge about quantum mechanics is exceptionally little.

UPDATE: The answers given here supplemented with this explanation of what is deterministic in Schrodinger's Equation and the answers about what unitary time evolution is are clear enough to at least understand that there is an apparent paradox.

For more clarity, I think the only way would be to study the mathematics of quantum theory.

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Wigner's friend is a complicated biochemical system composed of a quite-large collection of atomic nuclei and electrons, and as such s/he is within the remit of quantum mechanics and, particularly, describable by the continuous, deterministic evolution set by the Schrödinger equation.

This means that it is perfectly possible to disagree with your statement:

isn't it clear that the collapse happened when the measurement was made?

No, it isn't clear. There is a description of reality, which is perfectly consistent, in which the friend is in a quantum-mechanical state consisting of an entangled superposition with the original system; this superposition only collapses when observed by Wigner.

Of course, you might disagree that this is a reasonable description, and that's a perfectly tenable position. These are disagreements about the interpretation of quantum mechanics, and not all the paradoxes make equal sense when seen from all of the possible viewpoints. Just keep in mind that just because one of the paradoxes flattens out from your chosen interpretation, it doesn't mean that you are "right".

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    $\begingroup$ When I wrote 'Isn't it clear that the collapse happened...' I meant that I am puzzled as to where the paradox comes from, it was not meant as a declarative statement. $\endgroup$
    – Yash Sharma
    Commented Oct 12, 2020 at 11:29
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    $\begingroup$ However, the question of the OP I believe is, what is it that prevents this: "a quantum-mechanical state consisting of an entangled superposition with the original system" from being considered as simply representing a lack of knowledge about which outcome has occurred? Or where does that interpretation start running into rocks - especially in this particular context? $\endgroup$
    – The_Sympathizer
    Commented Jan 12, 2021 at 0:48
  • $\begingroup$ But it is not a paradox or inconsistency if you assume wave function is reduced at initial measurement and subsequent collapse by the second observer gives results within this reduced wavefunction. It is not superdeterminism, just normal determinism that empirical physics is grounded on. $\endgroup$
    – Duke William
    Commented Jul 1, 2023 at 7:05
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The apparent paradox is that if we replace Wigner's friend by a video recorder then it appears (if you believe in an objective collapse of the wave function) that wave function collapse only happens when Wigner looks at the video. The challenge is then to account for why Wigner's friend makes the wave function collapse but Wigner's video recorder does not.

There are various responses to the scenario:

  1. Wave function collapse is an objective feature of reality but is only triggered by a conscious observer (or, at least, an observer/recorder with a sufficiently high degree of complexity).
  2. Wave function collapse is subjective - it is quite valid and consistent for Wigner and Wigner's friend to think that wave function collapse happened at different times.
  3. There is no wave function collapse - in the many worlds interpretation, each time at which the wave function appears to collapse is simply a node from which different parallel versions of the future branch off.
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  • $\begingroup$ Why we can't just state wave function is reduced by the video recorder? So no paradox whatsoever. $\endgroup$
    – Duke William
    Commented Jul 2, 2023 at 11:39
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This is a good and important question.

Whether there is a paradox or not depends on what measurements can be performed on Wigner's friend. The trick is that there is no difference between a superposition vector

$$|\psi\rangle := \alpha |0\rangle + \beta |1\rangle$$

and an ordinary classical mixture of $|0\rangle$ and $|1\rangle$ with probabilities $|\alpha|^2$ and $|\beta|^2$ if you can only measure in the $\{ |0\rangle, |1\rangle \}$ basis. Thus, we can, if we are only allowed to do that, interpret the superposition as simply meaning that Wigner is ignorant of the state of his friend's mind. It could be something more, but it is equally consistent for it to not be.

What messes things up is if you can do measurements in an incompatible basis. That means measuring Wigner's friend in a basis that, itself, involves superpositions. Whether that is possible or not depends, in theoretical grounds, on two things:

  1. whether the "algebra of observables" of the Wigner's friend system is "tomographically complete" or not, and
  2. whether we can subject Wigner's friend to evolution under an arbitrary Hamiltonian (i.e. a unitary operator).

And I do not know if the answer to both these is known, especially (2). I also believe that while, for a single quantum particle, the algebra of observables generated by $\hat{x}$ and $\hat{p}$ is tomographically complete, when we go to multiple particles, the algebra generated by $\{ \hat{x}_i,\ \hat{p}_i\ |\ i \in I \}$ for some index set $I$, is not necessarily so. A simpler example of this is the spin operators $\hat{S}_x$, $\hat{S}_y$, and $\hat{S}_z$ for a spinning particle. For one particle, this generates a complete set; for two, the algebra generated by both sets of spin operators for each particle is not.

If we can do that, though, then there is a real distinction between Wigner's friend being in a state analogous to $|0\rangle$ above, $|1\rangle$ above, and being in a superposition of the two as emerges from Schrodinger's equation.

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  • $\begingroup$ Regarding measuring the friend in another basis, this indeed seems to be required to get the really interesting "paradoxes". See for example Guérin 2021, "A no-go theorem for the persistent reality of Wigner’s friend’s perception" at doi.org/10.1038/s42005-021-00589-1 $\endgroup$
    – BjornW
    Commented Dec 14, 2022 at 17:45
  • $\begingroup$ @BjornW: Yes. The gist of that paper seems to be if we assume that "behind" the pure state assignment of $\frac{|\text{"I saw a '0'"}\rangle|0\rangle + |\text{"I saw a '1'"}\rangle|1\rangle}{\sqrt{2}}$ that Wigner gives the joint friend-electron system is a "hidden" (to Wigner) perceptual fact that either the friend really "saw a '0'" or "saw a '1'" in the meantime,we can show that if Wigner then does the measurement in the "ok, fail" basis, then despite that it will not change Wigner's state assignment, it will (or at least must stand a chance to) rewrite the perception of the friend. $\endgroup$
    – The_Sympathizer
    Commented Dec 14, 2022 at 19:43
  • $\begingroup$ To me, though, I don't see that as a particularly surprising result. Whatever could do that measurement would have to be a physically hugely invasive process if it is even possible at all, and since Wigner's assignment necessarily obscures from him the information on the friend's perception, its changing on him unawaredly after doing that seems not particularly remarkable. $\endgroup$
    – The_Sympathizer
    Commented Dec 14, 2022 at 19:43
  • $\begingroup$ Yeah I didn't really get how the paper suggested that Wigner should proceed really with the measurement in a, say, Bell basis. Assume the friend outputs his state using photon polarization (a "computational" basis), sure, you can then set up a measurement of these in a Bell state instead but that won't change the friend who will eventually get out to explain his story.. so I assume, like you imply, that you'd have to "rotate" the friend (and everything in that lab) somehow into the other basis..? $\endgroup$
    – BjornW
    Commented Dec 15, 2022 at 1:20
  • $\begingroup$ Isn't this kind of analogous then to what we already are, kind of, OK with in QM, with for example a photon or electron in a dual-slit interference experiment? We can't ask it which slit it went through after we detected it at a bright fringe, and even worse, we can't ask the dark-fringe non-detected electron that just got destructively interfered with its cousin and never happened :) As soon as you start implying that particles are friends of Wigner's and not just stupid particles, you run into these issues.. $\endgroup$
    – BjornW
    Commented Dec 15, 2022 at 1:27

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