As for your question, usually one assumes that the initial quantum state and the dynamics are known in the Wigner’s friend scenario, so one should assign that quantum state. More generally, if you make a single measurement of a quantum state, the only thing that you know about it is that it was not orthogonal to the projector you measured. This is true for any interpretation.

As for the true universal pure state, there is no hope finding that out, for pretty much the same reasons you’ll never find out the state of the entire universe even in classsical GR.

]]>The paper works within the ontological model framework, where quantum states are epistemic states over an ontic state space.

The usual argument of the Wigner’s friend scenario is that since the friend only sees one outcome, she will assign either $\ket{+}_S\ket{F_+}$ or $\ket{-}_S\ket{F_-}$ to the lab (IF she can assign a quantum state to herself, which is not true for some interpretations).

In such cases the no-go result is obvious.

The paper shows that the no-go result is true even when the friend does not assign a quantum state to herself (but she still has an ontic state), only to the system she measures. This is the standard and common application of quantum theory to the scenario, just like Wigner’s original treatment.

Therefore, the paper does not refute MWI, where the friend assigns the same pure state to the lab as Wigner, as mentioned in your comment.

But I have a question for MWI: which pure state the friend should assign to the lab, if she does not contact to anyone outside the lab? Infinitely many pure states of the lab are consistent with her observations, which differs by a phase between the two terms $\ket{+}_S\ket{F_+}$ and $\ket{-}_S\ket{F_-}$. All such states seem applicable.

Similarly, how do we know which pure state is the true universal pure state in MWI?

More generally, I don’t think reality is in any sense observer dependent. Measurement results are, as defended by Brukner and Bong et al., but reality is not made of measurement results. Reality is made of quantum states, and these are objective and observer independent.

]]>It shows how, if there are ontic states beneath quantum states, such states must be relative to observers.

So this together with the recent Nature paper (https://www.nature.com/articles/s41567-020-0990-x), which is in the same vein as Brukner of your blogpost, says that reality is in some sense observer dependent, whether it is observable or non-observable (ontic states). ]]>

The point is, Gleason’s theorem doesn’t say anything about what probabilities you get from a specific quantum state. It only says that if you want to extract a set of non-negative numbers that sum to one from a set of projectors that sum to identity, and you want your function to be non-contextual, and the Hilbert space dimension is at least 3, then the only choice you have is to take the trace of the projectors with *some* density matrix. It doesn’t say anything about which density matrix you should use.

In other words, it fails egregiously my second criterion. It doesn’t even apply to dimension 2. Even if it did, it can’t tell me anything about a specific quantum state. Even if it did, it can’t tell me anything about relative frequencies.

I didn’t understand your second question. I did expand what I meant by “often” a little below.

]]>as far as I understand, you reject the argument “the law of large numbers is a theorem” by “it asserts that <> sequences have 1/2 as limiting frequencies” and “in the word <> is hidden a probabilistic concept”.

The strong law of large numbers (you seem to quote the weak one) does not assert that the set of “wrong limiting frequencies” has small measure (with respect, by assumption, to the equiprobable Bernoulli infinite product distribution, which can be argued to be the only relevant measure here); it asserts that this set has measure zero.

For example, in quantum mechanics, assume that a system is prepared in a state $\phi$ and that you measure it with an apparatus admitting $\phi$ as a pure state. Then quantum mechanics predicts that the measurement will yield a definite result! Not an “almost definite result”.

Moreover, the mathematical probability theory does not claim to prove anything about the results of throwing coins in real life!

What do you think?

]]>what about Gleason’s theorem? It asserts that any map assigning to any closed subspace of a Hilbert space a number between $0$ and $1$ satisfying some probabilistic postulates must be of the usual form we know from quantum mechanics.

So, one can think of it as stating : “if one wants reasonable probabilities, they should be given according to the Born rule”. How does it not answer your requirements?

Moreover, isn’t your second requirement supposing something contained in the word “often”?

]]>What’s the difference? Neither Monday or Tuesday, nor an order, are mentioned. It does say it takes place over two days, but only that “[the researchers] will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice).” And the only reason for “two days” seems to be a customary sleep cycle. So:

The researchers put you to sleep, then flip two coins, a dime and a quarter. After an appropriate amount of time, they look at the two coins. If either is showing Tails, they perform the following procedure. Then after another appropriate amount of time, they turn the dime over. If either of the coins is now showing Tails, they perform the procedure, possibly for the second time.

Procedure: Wake Sleeping Beauty. Once she is fully awake, ask her for her confidence that the quarter is currently showing heads. After she answers, wipe her memory and put her back to sleep.

Solution: There are four equally-likely states for (Quarter, Dime) at any point in time. They are {(Heads, Heads), (Heads, Tails), (Tails, Heads), (Tails, Tails)}. This is true regardless of whether the dime has been turned over. But the quarter remains in the same state it had when it landed.

But from the fact that she is awake, SB knows that the state (Tails, Tails) is eliminated. The chances for each remaining state is 1/3. That means the chance the quarter is showing Heads is 1/3. Which means the chance the quarter landed Heads is 1/3.

]]>\[v = \frac{at}{\sqrt{1+(at)^2}}. \] Here $v$ and $t$ are the velocity and time from the point of view of the Earth, and $c=1$ as usual. ]]>