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This is something I've dug into for a while, and what I am wondering about is the following argument, which I haven't necessarily seen before.

First, we'll take the Wigner's friend variant, because to me, this one makes the strongest argument for the claim in the titular question. In Wigner's friend, for those not familiar with it, we have a quantum system, then a "friend" who will observe that system, then a "Wigner" who observes the "friend": in particular, the quantum system starts in a superposition state, say it's a spinning electron, and its spin state is

$$|\psi\rangle := \frac{1}{\sqrt{2}} \left(|{\uparrow}\rangle + |{\downarrow}\rangle\right)$$

Now here's the first point: we will not make any ontological claims about this superposition. The trick is the next point: the "friend" now measures the rotation axis of the electron and gets and records at least in their head a result, say "$\uparrow$", and now subsequent measurements will return that same result. No problems ... until we look at how "Wigner" would analyze this.

"Wigner" can, just as well, presumably (under some philosophical assumptions like physicalism of the mind) put the "friend" and electron into a giant Schrodinger equation, with hugely complex Hamiltonian operator covering zillions of atoms and a massively complicated initial state, and run it forward. "Wigner", however, will predict then that the combined "friend"-electron system ends up in something that can roughly be considered as

$$|\psi_\mathrm{friend}\rangle := \frac{1}{\sqrt{2}} \left(|\text{I saw a "$\uparrow$"}\rangle\ |{\uparrow}\rangle + |\text{I saw a "$\downarrow$"}\rangle\ |{\downarrow}\rangle\right)$$

and this is then claimed to be problematic in some way. But why is it? If we put Wigner's and friends' experiences on equal footing, then it seems there is no other conclusion you can have but that this simply means that the superposition here actually should be understood as Wigner's knowledge, and Wigner doesn't know which outcome happened.

The question is: why is that a problematic interpretation, beyond potentially taking issue with the philosophical assumptions just mentioned (physicalism of mind, equal footing of different experiences)? Now I am aware that, of course, if we make certain assumptions as to what a superposition is knowledge about, we can run into problems (as any one of a number of theorems that anyone is familiar with the whole topic of quantum foundations would almost surely know by now), but what I am after is a more minimal thesis:

  1. at bare minimum, quantum states represent knowledge of measurable system parameters,
  2. according to the "real state of the world", the specific parameter "Did I see a $\uparrow$ or a $\downarrow$?" has a classical outcome before Wigner sees it, despite that Wigner assigned a superposed pure state to his friend.

without any presumption of what the "real state of the world" in (2) is/isn't in the case of general physical systems and/or how that the knowledge in (1) does/doesn't relate to it.

So: what is wrong with this thesis?

Now for what seems to be the objection, and my challenge thereto. Pretty much every no-go result on these things (e.g. Bell, PBR, Frauchiger-Renner, etc.) invariably involves at least one measurement that is analogous to measuring the friend in a basis that itself involves superpositions. And while this seems to just be taken for granted in every presentation I've seen, what my question is is how can we justify this measurement?

Keep in mind what the above $|\psi_\mathrm{friend}\rangle$ is: it is actually the state for/of a huge number of atomic particles, and the states going in on the right are actually just representatives drawn from wide subsets of Hilbert space, corresponding to, say, different ways the friend has cocked their head, different jiggles of their atoms in the course of thermal motion, and so forth. More to the point, the "I saw a $\uparrow$" etc. stuff going into the superposition on the right is a simplified representation of what would, if we are consistent in our application of quantum theory be a very complicated proposition based on the atomic particles.

What result tells us that this measurement is even theoretically possible to make? As it would seem that without it, why can't we just write off the difference between the superposition and a classical uncertainty between the two outcomes above as lying in an unphysical domain or, in a sense, simply being a mathematical artifact that has led theorizers astray?


ADD 1: Someone has mentioned that Wigner's friend could send a signal out of the isolated lab. This won't cut the mustard. The EM field will be ascribed a superposition. That's very basic QED. In effect, we just added another "layer" - the EM field - to Wigner and the friend, between them. When Wigner hears the signal, Wigner will either get one or the other frequency, just as if he opens up the lab.

ADD 2: There is a very simple way to see that anything that would "betray" the superposition in the second case as being something nontrivial would also be seriously physically consequential. Presume it is possible, then do a serial measurement as follows with 3 rounds. First, do what Wigner usually does and open up the lab and ask the friend. The friend has a 50% chance to then be found as

$$|\psi_\mathrm{friend}\ 2\rangle := |\text{I saw a "$\uparrow$"}\rangle\ |{\uparrow}\rangle$$

. Suppose that happens. Now Wigner seals up the lab again and does the questionable part, measuring in the basis

$$\mathcal{B} := \left\{ \frac{1}{\sqrt{2}} \left(|\text{I saw a "$\uparrow$"}\rangle\ |{\uparrow}\rangle + |\text{I saw a "$\downarrow$"}\rangle\ |{\downarrow}\rangle\right), \\ \frac{1}{\sqrt{2}} \left(|\text{I saw a "$\uparrow$"}\rangle\ |{\uparrow}\rangle - |\text{I saw a "$\downarrow$"}\rangle\ |{\downarrow}\rangle\right) \right\}$$

Note this either puts friend in the same state as before or else another with identical probabilities.

Thus Wigner goes and opens the lab again. It is possible, with 50% chance, that what Wigner gets this time is the opposite of what he got in his first measurement, as being what friend saw. General 50% overall chance. In particular, friend's mental state was changed in the meantime, and so too was the state of the quantum particle, and indeed everything else in the lab I just omitted for simplicity from my maths above. Something dramatic had to have happened here. This is not a simple thing - my question is whether it makes even physical sense to ascribe $\mathcal{B}$ as carrying information about a physical property of the system, or if measurement in $\mathcal{B}$ and associated dramatic transformation of reality, is just mathematical fiction/spuriousness.

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  • $\begingroup$ Will then not everything remain in a superposition forever? So also Wigner himself upon measuring the state of the friend? You can imagine the system cat-friend-Wigner as a whole system upon which you can make a measurement. $\endgroup$ Apr 27 at 22:13
  • $\begingroup$ @Deschele Schilder : Yep, that's the "paradox". $\endgroup$ Apr 27 at 22:22
  • $\begingroup$ Doesn't this show that knowledge of the state of a system is independent of what is actually going on? That even when no one looks at the superposition of electron spins, the superposition can still collapse (in the right circumstances, say when a certain field passes (independently of the friend). That for the system friend-electron-spins an actual collapse takes place (independently of Wigner). If collapse were dependent on knowledge no sapient (or non-sapient but observing) beings could develop at all. $\endgroup$ Apr 27 at 22:33
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    $\begingroup$ to be fair, almost every statement concerning Schrödinger's cat should be interpreted as ignorance... $\endgroup$ Apr 27 at 23:05
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I presume you are aware that in general a quantum superposition of states is empirically different than ignorance of which state a system is in (e.g. in superpositions there may be interference between the branches). So the "problem" in the Wigner's friend experiment is explaining why the calculations Wigner gets (which yield a superposition) are wrong or if they are indeed wrong. Wigner could just arbitrarily say that his friend observed the situation and therefore the wave function collapsed, but if he treats the friend as a quantum system why should he do this? It's basically a restatement of the measurement problem, which is the root of all of the competing interpretations of quantum mechanics.

As to why we should just ignore the difference... well, in principle Wigner could isolate the laboratory so well that the branches of the superposition containing his friend could interfere with one another... or not, if the friend's observation has already collapsed the wave function. Physicists would love to be able to actually perform this experiment to answer one way or another, but of course it's far too difficult in practice. But much simpler versions are being performed, and so far the question of when or even if the wave function collapses into a definite either/or outcome is open.

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  • $\begingroup$ I do indeed know that point: that's why I mentioned in my answer that it hinges on measuring in unusual basis. If you can measure it in a weird basis, it and, at least, an incoherent mixture, will be empirically different. The question is, can you, in principle do that to Wigner's friend and, if so, what is the theoretical proof, and what would the setup have to involve even if we can't practically build it? So I guess my question could be rephrased as "what would Wigner have to do to cause quantum interference between the two states of his friend?" $\endgroup$ Apr 28 at 1:12
  • $\begingroup$ (Note that simply declaring it by mathematical fiat is not enough - you can do anything you want to maths, and get all manner of results. What you need is to show that it makes physical sense to do something to the friend that would betray that something different was going on in there. Like with the no go arguments I mentioned - what physically would it mean to measure the friend in that superposed basis?) $\endgroup$ Apr 28 at 1:26
  • $\begingroup$ Rather than opening the room the friend could send a signal to Wigner, with a frequency which depends on the result (cat dead or cat alive). If the friend is indeed in a superposition these frequencies would interfere. $\endgroup$
    – Eric Smith
    Apr 28 at 1:51
  • $\begingroup$ How does that work? If the friend sends a signal, the quantum equations give a superposition for that signal - the EM field is described as having a quantum superposition state, not a classical wave state with interfering waves. (Basically, the state of the EM field will be [to be rough] $\frac{1}{\sqrt{2}} \left(|\text{waving with 5 Hz}\rangle + |\text{waving with 10 Hz}\rangle\right)$ and not $|\text{waving with 5 Hz + 10 Hz}\rangle$. Note where things are in/out of the ket symbols.) $\endgroup$ Apr 28 at 2:06
  • $\begingroup$ So when Wigner detects the signal, Wigner will only find a 5 or 10 Hz frequency (I just made those numbers up). Wigner will not find a beat between them. Just as if he opened up the lab. $\endgroup$ Apr 28 at 2:08
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I will answer my own question - but with the caveat this is only a partial answer.

The relevant difficulty centers crucially around whether one can make a measurement of the Wigner friend or Schrodinger's cat in an arbitrary basis. In particular, the quantum-mechanical formalism does not necessarily allow for arbitrary-basis measurement: measurements must be of "observables", or sensible physical parameters, of the system, and which observables are allowed depends on the system in question, just as a classical system may have different number and kind of classical observables. For example, a moving point particle can be considered to have position and momentum but not angular position.

And to distinguish between a superposition ket

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

and a classical statistical uncertainty between $|0\rangle$ and $|1\rangle$, one needs to be able to measure the system in a basis that is incompatible with $\{ |0\rangle, |1\rangle \}$. One such basis is the "Hadamard basis"

$$\mathcal{B}_\mathrm{Had} := \left\{ \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \frac{|0\rangle - |1\rangle}{\sqrt{2}} \right\}$$

In this case, if Schrodinger's cat is in a genuinely different state than either $|0\rangle$ or $|1\rangle$ (i.e. "alive" or "dead", or for Wigner's friend, "I saw a '$\uparrow$'" vs "I saw a '$\downarrow$'", or otherwise), then measurement of the cat in this peculiar basis will betray it.

However, in order for that to make sense, we have to either have one of two things:

  1. the basis in question corresponds to that for a physically meaningful observable in the system, i.e. it is included in its "algebra of observables",

  2. we can subject the system to a suitable Hamiltonian so as cause the Hadamard superpositions to evolve into $|0\rangle$ and $|1\rangle$, then do the measurement there.

In other words, to say there is a difference we need either arbitrary measurement, i.e. every basis corresponds to an observable property of the system, or we need arbitrary Hamiltonian.

And what I do not know is if these criteria are met for the Wigner's friend system - that would require some kind of mathematical and/or physical proof. I may be wrong, and I'm intent on soliciting comments on this, but I believe that the observable algebra for a system of many particles, as we might model Wigner's friend by, is generated by the position/momentum/spin observables of those particles taken individually, or at least something similar to that (i.e. we may have to take into account indistinguishability). The question is whether or not this set of observables, then, is "tomographically complete", i.e. it includes the total set. Valter Moretti, in an answer to a related question I posted here:

Is there a Hermitian operator on $L^2(\mathbb{R})$ which is outside the C*-algebra generated by $\hat{x}$ and $\hat{p}$?

suggests that $\{\hat{x}, \hat{p}\}$ generates the entire observable algebra for a single-particle system, thus includes a tomographically complete set.

But, I am not sure whether $\{\hat{x}_i, \hat{p}_i | i \in I\}$ for many particles indexed with an index set $I$, is tomographically complete. It is easily shown with some simple matrix manipulation that for spin of a spin-1/2 particle like an electron, the algebra generated by $\{\hat{S}_x, \hat{S}_y, \hat{S}_z\}$, given by the Pauli matrices, for a single particle is complete, but that generated by $\{\hat{S}_{x_i}, \hat{S}_{y_i}, \hat{S}_{z_i} | i \in I\}$ for two or more particles is not - there are states we cannot distinguish by measurements of even all the observables within that algebra.

That said, in the latter case, I believe we can "recover" completeness because we can administer effectively arbitrary unitary operators to it - the question is whether this can be done with the moving-particle or even better RQFT quantum field system constituting something as complex as a human in a lab. If this is provably impossible, then a subjective interpretation of Wigner's friend is entirely on the table - otherwise, the relevant paradox is that if Wigner's friend's measurement outcome is decided before Wigner asks, then we have to ask why that Wigner could empirically distinguish between the pre- and post-asking states, via a measurement in the $\mathcal{B}_\mathrm{Had}$ basis, and how this does/doesn't reconcile with a further outside observer ("friend of Wigner")'s account of the whole affair. They would appear to assign empirically different predictions for measurements in certain bases.

Hence why I say it is only a partial answer, because of (1) and (2) above. I don't know what the general consensus on that is.

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