On 12-04-2022 01:17, Bruce Kellett wrote:
On Mon, Apr 11, 2022 at 10:47 PM smitra <smi...@zonnet.nl> wrote:
On 11-04-2022 14:02, Bruce Kellett wrote:
On Mon, Apr 11, 2022 at 5:35 PM smitra <smi...@zonnet.nl> wrote:
The trouble with this is that the entanglement spreads only at the
speed of light or less. If faster than that, then it is npn-local.
Alice and Bob make their separate measurements at spacelike
separations. So if the entanglement resulting from Alice's
measurement
affects Bob's measurement, that that is FTL or non-local.
In the MWI Alice's measurement does not affect Bob's measurement.
That is your contention.
It's not just 'my contention"" it is implied by the rules of QM when we
omit collapse.
But if Alice does not make a measurement, Bob
gets both up and down, 50/50. If Alice does make a measurement, then
the Alices branch in which she got up does not have a Bob who saw
50/50 up/down. So Bob's result is clearly affected by Alice's
measurement.
No it isn't because Bob is not located in one or the other branch before
he observes what Alice has found. Before he makes that observation,
Alice is in a superposition of both possibilities. And we know that, in
general, if you are going to assume that the measurement results exist
prior to measurement that will introduce non-local effects.
Adding the other branch in which Alice got down does not
affect this result, since in that branch, Bob's result is also
affected in that he did not see 5-/50 up/down.
Bob then does end up seeing a 50/50 result for up/down, because the
branch he'll end up with is random.
Since in the local case, Bob's measurement is independent of
Alice's,
he must see |up> and |down> with equal probability (in different
branches). By dropping the entangled part of the wave function
after
Alice's measurement, you rule out the possibility that Bob can get
|up> or |down> with equal probability after Alice's measurement
(as he
certainly can before her measurement). What is it about Alice's
measurement that changes the probabilities for Bob's result? At
spacelike separation, this can only be a non-local influence.
Bob's result is not affected by Alice's measurement because she is
in a
superposition of the two possible results.
The fact that she is in a superposition makes no essential difference.
One can always look at what happens in one branch of the
superposition.
No one can't because for Bob bith results are relevant.
And in each branch, Bob's result is affected by Alice's
measurement. The measurement outcomes are correlated, after all, and
that is what correlation means -- the results are not independent. Not
independent means that if one of the pair does not make a measurement,
then the results for the other are different. (Counterfactuals rear
their ugly head here.)
But because both branches objectively exist, and Bob can end up in
either one, he's identical in both before he observes Alice's result,
his results remains random even after Alice makes her measurement.
The analysis that you give above is the standard analysis for QM
with
collapse, and that is usually regarded as a non-local effect.
No it isn't because in the analysis I gave there is no collapse. So,
as
far as Bob is concerned, Alice has not found one definite result.
Any copy of Alice that Bob meets has only one result.
The Bob she meets is identical to the Bob in another sector where her
copy has found a different result.
The
superposition can always be analysed in terms of the separate
components. You have built a collapse into your analysis without
realizing it -- Bob sees a different wave function according to
whether Alice makes a measurement or not.
No he doesn't, Bob factors out of the superposition.
If she makes a measurement,
the non-compliant part of the original wave function collapses -- it
is not passed on to Bob. If she does not make a measurement, Bob sees
the whole original wave function.
That's not how it works. No noncompliant parts ever arise.
It might be useful here to compare the correlated case with the
picture if the particles that Alice and Bob measure are completely
independent. In the case of independent particles, there is no doubt
that both observers have a 50/50 chance of seeing up or down. When
they meet (or when their forward light cones overlap) the rules of MWI
say that each, being part of the world entangled with the other
party's measurement, splits into two -- one branch for each result
obtained by the other.
That's your Straw Man version of the MWI again. What the MWI says is to
evolve the initial state according to the Schrödinger equation and then
to not invoke any collapse of the wavefunction.
The consequence is that there are four
branches: up-up, up-down, down-up, and down-down.
Yes, that's what yolu will get if you do the math in this case.
When the spins are
correlated, two of these branches do not appear; the up-up and
down-down branches do not conserve angular momentum so they are
absent.
Yes, but the absence of the two states comes out as a result of applying
the rules of the MWI. The real rules, not your straw man MWI rules.
The challenge of any analysis is to give a full and detailed
account of the difference between these two cases. In other words, how
does the fact that the spins are correlated affect the measurements
that the individuals can make?
It has now been explained to you multiple times. If the spins are
correlated you only have two terms of in the superposition to begin
with. Bob gets entangled with his spin and Alice with her spin. There is
really nothing to explain here.
Saibal
Bruce
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