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This is a question I am wondering about because the answer to it seems to have some interesting - but perhaps already long considered and dismissed because it's been settled - implications for the measurement problem and question of what the "best" interpretation is of the formalism of quantum theory. Yet it may seem oddly elementary to ask.

Of course, I know that observables in quantum theory are Hermitian operators on a Hilbert space, but my question is: given a quantum system, what set of Hermitian operators correspond to its observables? The reason I ask this is that there are arguments, such as those in this video:

https://www.youtube.com/watch?v=OmaSAG4J6nw

that the Hilbert space can be considered in some sense a term that contains redundancy similar to a gauge theory, and that what "really matters" is the observables and associated probabilities around measuring them, which are taken as forming a structure called a C*-algebra.

Now, if we take that view, and the "observables come first", it suggests that the set of observables is something we define in advance, and then the Hilbert space comes later as a sort of "model" thereof (in a sense close to the one used in mathematics, where we talk of a "model of a set of axioms"). However, I also have seen in other places people saying something different: that the set of observables is all Hermitian operators on a Hilbert space. That is, you make a Hilbert space, then boom, you get the set of observables.

And the thing is: these two things are not equivalent. C*-algebras are not in one-to-one correspondence with Hilbert spaces: instead, they may be isomorphic to proper subsets of the sets of all Hermitian operators on a given Hilbert space.

And to top it all off, textbooks often seem to gloss over this point: many often only say that observables are Hermitian operators, then go on to introduce some (e.g. position, momentum, etc.), but "are Hermitian operators" does not imply its converse, any more than "cars are machines" implies "machines are cars".

Is there a consensus on which view is the correct one, and if so, what is it, and why is the other one considered wrong?


You may ask as to what this has to do with the measurement problem. I've asked here earlier regarding Schrodinger's cat and Wigner's friend, and this is why. If it is the case that the set of observables of a system is not necessarily comprehensive of that on some Hilbert space, then there is a very natural interpretation of quantum states as being knowledge states, that goes like this.

Consider a Wigner's friend set up, and think about the experience of both Wigner and his friend. The friend is shut in a Schrodinger cat box with a quantum experiment. He runs the experiment and sees the result. Wigner then opens the box and asks the friend and the friend replies. But if Wigner models the friend using quantum mechanics and the Schrodinger equation, he gets a superposition state. Under the seemingly sensible philosophical assumption that Wigner's friend and Wigner himself both are having experiences throughout this and Wigner's friend experienced a single outcome, then it is rational to say that Wigner's superposition assignment reflects his ignorance of what outcome transpired, leading to the notion the quantum state is proximately a state of knowledge, belief or otherwise of information on an agent's part, as opposed to a property of the system - through the two may be isomorphic in cases.

But this falls apart if Wigner can do a measurement on his friend on a basis incompatible with. Then it seems hard to avoid that the superposition has ontological character, and we have a real conundrum if we accept those philosophical assumptions about the experience.

Yet all that goes away if, in fact, the incompatible measurement is not an observable of his friend, which is possible if the observables of a system are primary to the Hilbert space instead of the other way around. Indeed, pretty much every paradox and no-go result in this vein - not just this one, but results like the Frauchiger-Renner paradox, too - seem to depend on agents being able to do incompatible measurements on their experiences.

I can't help but think this must have been considered before, because it seems just too easy. But I have had a hard time finding it and, moreover, given it is not even commonly referenced in these discussions, that there must be something trivially wrong with it. What is that?

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  • $\begingroup$ Every Hermitian operator is certainly an observable. They might not all be interesting observables, but in principle, their eigenvalues are things that can be physically measured. For instance, we can define an observable which assigns to the $n$-th energy level of a harmonic oscillator the $2n+1$-th Fibonacci number, and we can measure this number. Other observables might be even more absurd or random, but they could still be measured. $\endgroup$ Aug 12, 2021 at 8:38
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    $\begingroup$ The operators in a $C^*$ algebra are bounded. Most interesting operators (the energy or momentum, say) are not bounded. $\endgroup$
    – mike stone
    Aug 12, 2021 at 14:32
  • $\begingroup$ @Chiral Anomaly. I agree with what you say in principle -- the eigenmodes and associted resolution of the identity are the key thing, but surely the non-locality of operators like $e^{-tH}$ is more than an inconvenience when writing dowe interaction terms? $\endgroup$
    – mike stone
    Aug 12, 2021 at 15:27
  • $\begingroup$ @ChiralAnomaly: field theories are outside of my comfort zone, so please take this as a serious question, not a rhetorical one: doesn't microcausality simply dictate how all the possible observables can be combined into fields, rather than restricting what the possible observables are in the first place? Meaning that any two Hermitian operators $A,B$ can be observables, but unless they commute, $\pi(u_1)=A$ and $\pi(u_2)=B$ requires the separation of $u_1$ and $u_2$ to be time-like, if $\pi$ is a valid field? $\endgroup$ Aug 12, 2021 at 16:23
  • $\begingroup$ @Vercassivelaunos In QFT, observables are expressed in terms of field ops, but the field ops themselves are not necessarily observables. Fermion field ops don't commute with each other at spacelike separations, so microcausality doesn't allow treating them as observables. If we trim the Hilbert space down so that all remaining ops qualify as observables, then it can't include states with different total electric charge (using QED as an example), which we normally prefer to do. $\endgroup$ Aug 12, 2021 at 17:36

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This is a sort of mostly-answer and elaboration to the other.

As @Chiral Anomaly mentions, the set of observables must be specified ahead of time as part of modeling the system in question. Defining its set of observables is like defining what properties it has, as a physical object - just like not all objects have, say, the property of "having an engine" or "intelligence", likewise not all of them will have some particular weird quantum operator. Then, once a set of "basis" observables is put down, additional ones are then generated by algebraic combinations.

Thus, regarding Wigner's friend, the answer does indeed depend on how we model the friend:

  1. If we attribute it only the property "what did it see?", then we have no contradiction as to when Wigner ascribes it the entangled-superposition state, that in an information-based interpretation of the quantum state, this could mean the friend did indeed see an outcome but the Schrodinger equation that Wigner used to obtain it simply isn't good enough to predict the actual outcome.

  2. But if we now consider a friend which does have a complementary (incompatible) property to the above, then we run into difficulties. If the "what-see" property was decided by this point, then Wigner presuambly could determine this by doing an ensemble measurement upon the incompatible-with-"what-see" property, and would falsify the Schrodinger equation prediction empirically. Even if Wigner's quantum state, to be consistent, should still be taken proximally as knowledge, in this situation it seems to hold more "objective weight" because it tells him something is tangibly different in this case.

Yet the Schrodinger equation has never been seen to be falsified empirically. This suggests either there is something conceptually wrong with quantum theory and/or how we are applying it in the situation described (e.g. maybe there is an extra nonlinear governing law, e.g. GRW theory, or there are some sort of exotic hidden variables, or ...), or that real friends don't have properties incompatible to their "experience" property. Discussion (and production of no-go theorems and results) mostly centers around the first outlet - interestingly, I have not seen much mention of the second outlet, even to argue for its falsity. And that was what I was hoping to gain with this question - has that prong both been addressed and is so widely-known as so that it never needs to be, or has something been overlooked here, perhaps because of how quantum theory has been taught or propagated? I still have no answer for that.

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  • $\begingroup$ Based on the comments and on the self-answer, it seems that I misunderstood the question. This answer says: "I have not seen much mention of the second outlet... And that was what I was hoping to gain with this question..." Can you clarify which option you mean by "second outlet"? (Is the first "and/or" meant to separate two different outlets? Or are those two different parts of the first outlet?) $\endgroup$ Aug 15, 2021 at 13:38
  • $\begingroup$ @Chiral Anomaly: You are correct in your assessment that part of this question is about what the observable set constitutes. However, the other part of the question is that this seems to have bearing on interpretational issues for the mathematical formalism, and I want to know if there is something there, or it's long been known to be irrelevant for some reason and, if so, what that reason is and what I am just missing as a result of ignorance and naivete. $\endgroup$ Aug 16, 2021 at 5:25
  • $\begingroup$ And the second part actually inspires the first, because part of what I'm wondering if I'm getting wrong is that there is in fact good reason to assume that the whole Hermitian set is theoretically observable, and thus we have a genuine problem when it comes to things like Wigner's friend. Because if it is not, then it seems that if only the quantum system in the setup has noncommuting parameters but the friend's "experience parameter" has no other incompatible parameter, there is a straightforward solution, but if the friend does have such a parameter, then things indeed become problematic. $\endgroup$ Aug 16, 2021 at 5:27
  • $\begingroup$ And what I'm asking about is if this has even been observed or remarked upon somewhere and if it has, I'd be curious to read about that. If we are indeed deciding observables by modelling, then it seems intuitive at least to me that if we're going to posit Wigner's friend, we'd "naturally" only include for the sake of such an observable corresponding to "I saw a '0'/'1'". And if we do that, Wigner can just interpret his quantum state assignment as a knowledge state and the superposition as merely reflective of his ignorance. $\endgroup$ Aug 16, 2021 at 5:36
  • $\begingroup$ However, if we adduce an incompatible observable (parameter) into the "friend", then this situation changes and there seems to be more non-trivial meaning to the superposition and tension with the idea the "friend" saw an outcome. (Which of course, then, suggests another idea: maybe a friend with incompatible parameters to its "experience" and a friend without them are different kinds of beasts and we aren't philosophically justified in imputing to the former the same kind of "experience" as the latter.) $\endgroup$ Aug 16, 2021 at 5:38

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