On Wed, Mar 16, 2022 at 4:55 PM smitra <smi...@zonnet.nl> wrote: > On 16-03-2022 04:01, Bruce Kellett wrote: > > On Wed, Mar 16, 2022 at 11:55 AM smitra <smi...@zonnet.nl> wrote: > > > >> On 15-03-2022 18:13, Brent Meeker wrote: > >>> > >>> So it's a collapse of the wave function, which Everett was supposed to > >>> banish? Can you give a Schroedinger equation evolution of this > >>> "determined the moment he sees her notes" change? > >>> > >> Alice's notes are in an entangled superposition with the environment. > >> Even though Bob is located in that same environment, and Bob's body will > >> also get entangled, the fact that Bob does not know the content of these > >> notes, means that Bob's mental state described as a bit string > >> containing the information of everything Bob is aware of, can be > >> factored out of this superposition (if this were not true, then Bob > >> could have psychic powers and know what Alice's results are before > >> looking at her notes). From the point of view of a Bob who measured spin > >> up at some polarizer angle beta, the state before he sees the value > >> Alice found, is of the form: > >> > >> |Bob, up, beta> sum_alpha [a(alpha-beta) |Alice, up, alpha> + > >> b(alpha-beta) |Alice, down, alpha>] > > > > I don't think this is correct. You are selecting a particular result > > for Bob, so we can take his polarizer angle for this result as the > > reference, so Alice's polarizer angle is simply an offset from this > > reference. You have taken Alice's position as a superposition over > > different polarizer angles and then summed over this superposition. > > This is not correct. Alice has a definite polarizer angle when she > > makes her measurement. Bob does not know this angle, but he does know > > that Alice is not in a superposition of different angles. > > The physical process that Alice uses to set her polarizer must be > specified.
No, there is no need for this. The final polarizer angle is all that matters. If Alice uses information from the local environment of the > polarizer that was not available to Bob, then the state must be > specified accordingly. The amplitudes depend on the relative angles. The > superposition over the angles is entirely general, you may replace that > by a single term. > There is no superposition over angles -- there is only one relative angle between the polarizers for each entangled pair. Also, given > > the assumptions, the coefficients a(theta) and b(theta) are known. So > > the correct expression is: > > > > |Bob, up>[sin^2(theta/2)|Alice,up> + cos^2(theta/2)|Alice,down>] > > > > You appear to be saying that the angle between the polarizers, theta, > > is not set until Bob looks at Alice's notes. This is, of course, > > wrong. Alice measures a particular result at a particular angle, so > > the relative polarizer orientation is set at the time of her > > measurement. Bob does not know what this is until they meet and > > exchange notes, but that is not relevant to Alice's situation, which > > is given by the superposition of \up> and |down> results as shown, > > with no summation over angles. > > Yes, if they both choose their angles deterministically (no squares in > the amplitudes, b.t.w.). > OK. The squares were a confusion between amplitudes and probabilities. > This is an important distinction, because it leads to important > > interpretational differences. If the relative orientation theta=60deg, > > we have: > > > > |Bob,up> [0.25|Alice,up> + 0.75|Alice,down>]. > > > > If Bob is 'up' in this case, there is a 25% chance that Alice will > > also be found to be 'up', and a 75% chance that Alice's result will be > > seen to be 'down'. There is no problem with this for this single case, > > since we know that all four branches are possible for non-aligned > > polarizers. > > > > Yes (minor point as above: amplities are 1/2 and 1/2 sqrt(3)) > > > The problem arises when Alice's polarizer is aligned with Bob's, so > > theta=0. In that case, sin^2(theta/2)=0, and cos^2(theta/2)=1, and the > > equation reads: > > > > |Bob,up>|Alice,down>. > > > > Again, this appears to be OK since we know that for the spin singlet, > > aligned spin measurements must always give opposite results. The > > question is, when does the |Alice,up> branch vanish? According to > > Saibal's account, everything is local, and the relative orientation > > theta is only obtained locally when Alice and Bob meet. This means > > that the sin^2(theta/2)|Alice,up> component of the superposition can > > only vanish when the observers meet. What makes the |Alice,up> branch > > vanish at that point? There is no appropriate interaction present. > > In the case where both will choose their polarizers that happen to be > aligned, |Alice, up> exists in the branch were Bob found spin down. > Bob's measurement does not change anything for Alice, it only makes > Bob's sector to get located inside Alice's down branch. > "Makes Bob's sector to get located inside Alice's down branch"? What the hell is that supposed to mean? Bob's sector does not get relocated anywhere. If Bob found up, Alice finds down for aligned polarizers. The branch in which Bob found down is the one in which Alice finds up. For aligned polarizers, sin(0) = 0. > The only sensible account is that if the polarizers are aligned, the > > |Alice,up> branch is never formed. Since the formation (or > > non-formation) of this branch happens at the time of Alice's > > measurement, the non-formation of the |Alice,up> branch for aligned > > polarizers must happen at that time. So information about Bob's > > polarizer angle must be available at the time of Alice's measurement. > > Since Alice and Bob are spacelike separated, this information can only > > have been available non-locally. The |Alice,up> branch cannot vanish > > at some later time, because there is no appropriate interaction that > > can make this happen. The only possibility is that the branch was > > never formed. And this is a non-local phenomenon. > > There is also a |Bob, down> branch within which the |Alice, up> branch > exists. One can ask how 4 branches get reduced to 2 branches, a point > that you've invoked frequently. The branches that never form are the > ones where both find spin up or both find spin down. But that's trivial, > as we created the entangled spin pairs this way. In the spin up sector, > the other spin is spin down and all that happens is that Alice gets > entangled with one spin and Bob gets entangled with the other spin, > which are local processes. > Except that when the measurements are local, the relative angle is unknown to both participants. So neither knows locally that the polarizers are aligned. That knowledge can be available only non-locally. And that is how it happens. If the angles are unknown to each other but are still deterministically > fixed to some arbitrary values, then all four branches can form with the > appropriate amplitudes. One can then ask how the values for the > amplitudes get fixed as this looks like a nonlocal process. However, the > state describes a nonlocal situation created by the entangled spin pair. > The nonlocal aspects of this state do not exist in Bob's or Alice's > sectors when they perform their spin measurements. Of course they do. Otherwise the correct correlations cannot be formed.The correlations cannot wait until Alice and Bob meet because in general there is no interaction there that could reset things. Remember, the results could be exchanged by email, with no direct interaction at all. If Alice finds spin > up then that doesn't change anything for Bob because for Bob, the sector > where Alice found spin down will also exist before he measures his spin. > If Alice finds spin up, then the particle that Bob measures is known to have spin down. And this can only be known non-locally. Bob does not know it at the time of his measurement. The same happens if Alice finds spin down. The particle Bob measures is known, non-locally, to be spin up. That is what the entanglement implies. The process is non-local because the entangled, non-separable, wave function is non-local. In contrast, if we don't assume MWI, then Alice finding spin up in case > of parallel polarizers does affect Bob's experiment as there is then no > sector where Alice where she found spin down. > Huh? Of course there is a branch for Alice finding spin down, just as much as there is a branch for Alice finding spin up. For non-aligned polarizers, there are always four branches on an individual trial. Right? It is just that for repeated trials, some of the branches are redundant and need to be eliminated (or never formed). Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLSv5-jjEYcEZt7xdviHEbqCgpAGApdJgW71Cr_zLTcuXw%40mail.gmail.com.
