From: *Bruno Marchal* <[email protected] <mailto:[email protected]>>
On 14 Aug 2018, at 04:12, Bruce Kellett <[email protected]
<mailto:[email protected]>> wrote:
From: *Bruno Marchal* <[email protected] <mailto:[email protected]>>
On 12 Aug 2018, at 14:59, Bruce Kellett <[email protected]
<mailto:[email protected]>> wrote:
No, Price is wrong. He collapses the wave function in a non-local
manner, even though he doesn't seem to realize it. Let me try
again. The state is
|psi>= (|u>|d> - |d>|u>).
Let Alice interact with particle 1 at one end:
|Alice>|psi> = |Alice>|u>|d> - |Alice>|d>|u>
Alice interacts only with particle 1 (locally), so |Alice>|u>|d>
--> |Alice sees u>|u>|d>, and similarly for the other component.
Now Bob interacts with a different state. He does not see |psi> as
above, but rather
|Alice sees u>|u>|d>|Bob> - |Alice sees d>|d>}u>|Bob>
The, if Bob measures along the same axis, he gets down for Alice's
up, or up for Alice's down. If he measures at some different angle,
he gets the appropriate rotated results. But Bob NEVER sees the
original unaltered rotationally symmetric singlet state:
I have never disagreed with this. But Bob’s view is only a part of
the picture. Any particular Bob cannot see the symmetry; because it
is part of the symmetry.
But it is Bob's view here that is relevant to the formation of the
correlations.
It is each Bob’s view. Yes.
Bob measures that state he sees, not some phantom symmetric state.
Which Bob? The whole question is there.
The Bob that receives the partner particle to the one Alice measures.
There is no ambiguity here. It is your insistence that Bob's identity is
not clear that is causing you difficulties.
Alice's measurement destroys the symmetry of the singlet state
In her view, yes. It makes the symmetry no more viable to her. It can
be restored in principle if Alice get amnesic. Nothing is really
destroyed.
That fact that Alice's interaction with the state is unitary and can be
reversed does not mean that the original symmetry still exists in some
sense. If I place a large weight at some point on the circumference of a
bicycle wheel, the rotational symmetry of that wheel is lost. The fact
that I can reverse the process by removing the imposed weight does not
mean that the altered wheel is still rotationally symmetric in some
wider view.
-- it is not preserved magically in any many-worlds picture. There is
no 'outside view' in MWI
There is the universal wave function, at least if you accept the idea
that the cosmology obey QM.
as there is in your classical M/W duplication thought experiments.
There is no view where the symmetry is preserved. Not even Tegmark's
imaginary bird view preserves the symmetry. You are mistaken here.
?
The side view is only a (big) rotation (unitary transformation) on a
(big) Hilbert space or von Neuman algebra.
It seems that you are basing your conviction that all physics is
ultimately local on the idea that all interactions are unitary
transformations of the universal wave function. But that is not
sufficient. You have also to postulate that the wave function itself is
actually local. And we know that that is not true. Because
non-separable, that is, non-local, states do actually exist within the
universal wave function. As Maudlin points out, the basing an argument
for locality on the wave function fails because the wave function itself
is not a local object.
Bruce
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