I Would studying MWI be a waste of time?

[Denis]

Just like the universe can simultaneously exist and not exist.

Sorry, but this would be the ultimate end of science.

 

[vanhees71]

Yes, but it's also important to write carefully. In post #150 you wrote:
"Only, because the value of an observable is indetermined it doesn't mean that it doesn't exist."
I assumed that "it" refers to "value of an observable". Obviously, by "it" you only meant "observable".

So, do you say that the value (in a non-eigen-state) doesn't exist before measurement? Or are you just agnostic about that? How about the value in an eigen-state before measurement?

Scripsi scripsi! I say the value of an observable for a system not prepared in a state of the form
$$\hat{\rho}=\sum_{\beta} p_{\beta} |a,\beta \rangle \langle a,\beta|,$$
where $|a,\beta \rangle$ are a set of eigenvectors to the eigenvalue $a$ of the self-adjoint operator $\hat{A}$, that represents the observable $A$, is indetermined. For me the phrase "The value does not exist" doesn't make any sense.

Of course, if the system is prepared in a state of the form given above, then the observable has the determined value $a$ before observation.

I'm agnostic about what in general happens when measuring the observable on a system not prepared in a state as given above. It depends on the construction of the measurement apparatus. Only for "filter measurements" you end up, by definition, in a state of the above form.

 

[Demystifier]

Of course, if the system is prepared in a state of the form given above, then the observable has the determined value a before observation.

I think now it's the right time to ask you how do you interpret the EPR case. Let Alice and Bob be two spatially separated observers. Let Alice has measurement apparatus which projects a state to either $|A1\rangle$ or $|A2\rangle$. Likewise, let Bob has measurement apparatus which projects a state to either $|B1\rangle$ or $|B2\rangle$.

At time $t_0$ let the state before the measurement be prepared in
$$|A1\rangle|B1\rangle + |A2\rangle|B2\rangle$$
At that moment $t_0$, both $A$ and $B$ are in an indetermined state, do you agree?

At the next moment $t_1$, assume that Alice performs her measurement and founds that the system is in the state $|A1\rangle$. But Bob does not yet perform his measurement. At that time $t_1$, after the measurement by Alice but before the measurement by Bob, can we say that the Bob's system is in the determined state $|B1\rangle$?

If your answer is no, then how is it consistent with your statement quoted above?

If your answer is yes, then what has caused this change (from indeterminate at $t_0$ to determinate at $t_1$) of the $B$-state? If it was caused by the measurement by Alice, then how can it be compatible with locality?

 

[vanhees71]

This I've also answered very often. Here you have the usual entanglement issue, where you have a composed system, for which the values of observables concerning the parts are indetermined. The correlations described by the entangled state are, however, there from the very beginning. So when Alice finds the value $A1$ she knows that Bob must find $B1$ and vice versa. For Bob $B1$ is still undetermined, and he finds the values according to the probabilities when measuring an ensemble. There's no action at a distance (at least not within local microcausal relativistic QFT), i.e., A cannot affect instantaneously the values of the observables of B's part of the system. The correlations are revealed only when A and B compare there measurement protocols.

 

[stevendaryl]

This I've also answered very often.

I would say that you've responded many times, rather than answered many times. Your response is an answer to a different question, which is whether Alice can affect Bob's measurement results. But you never answered the original question, which is:

If Alice has measured her particle to have spin-up (along the z-axis, say), does that mean that Bob's particle's state has a definite value for its z-component of spin (even if he hasn't measured it yet)?

 

[PeterDonis]

Thread closed for moderation.