Your simulation assumes the quantum mechanical results. In other words,
it assumes non-locality in order to calculate the statistics. Where does
the cos^2(theta/2) come from in your analysis?
Bruce
On 27/04/2016 1:38 pm, Jesse Mazer wrote:
On Tue, Apr 26, 2016 at 9:16 PM, Bruce Kellett
<[email protected] <mailto:[email protected]>> wrote:
On 27/04/2016 1:13 am, Jesse Mazer wrote:
On Tue, Apr 26, 2016 at 6:45 AM, Bruce Kellett
<[email protected] <mailto:[email protected]>>
wrote:
You think that "the state of the other particle" refers to
the quantum state that would be assigned to B given only
knowledge of the state of A (as well as knowledge of how they
were entangled originally). Actually, that is the
interpretation I gave the words, except I teased out what
that actually meant. From the entangled state, given A's
state (result, say |+>), you would assign a state |-> to B.
But this is wrong for spacelike separations -- the state B
actually measures is exactly the same as the state A
measured: |psi> = (|+>|-> - |->|+>)/sqrt(2).
You use the full state if you just want to generate the total
probabilities for various possible *joint* outcomes. But if you
want a conditional probability of various outcomes *just for B*
given knowledge of what measurement A got, this can be done in
QM, in the Schroedinger picture you could project |psi> onto on
eigenstate that corresponds to whatever definite outcome was
measured on A, resulting in a different state vector for the
combined system |psi'> which may lead to different probabilities
of getting various results for B, but which does not assume any
knowledge of what measurement was actually performed on B. I
assume something similar is possible in the Heisenberg picture
which Rubin is using, so I was speculating that he meant
something like this when he talked about a label on one particle
which says something about the state of the other particle.
I am well aware of this, and I also thought that was probably what
Rubin had in mind. The problem is that this simply sneaks
non-locality in the back door -- neither Rubin nor you appear to
realize this. This is often the problem I find with these attempts
to give a local account of EPR -- non-locality is built in
unobtrusively!
That is why I said that, in any strictly local account, if A gets
|+>, B still measures the original |psi>. The measurement by A
cannot /locally/ affect the state that B measures (or vice versa).
OK, let's say experimenter A measures particle 1, and experimenter B
measures particle 2. Any given copy of particle 1 has a "label" that
says something about the state of 2--we can imagine that the copy of
particle 1 carries a little clipboard on which is written down both
its own quantum state, and a quantum state it assigns to particle 2.
When that copy of 1 is measured, it not only adjusts its own state (to
an eigenstate of the measurement operator), it also adjusts the state
it has written down for 2. You seem to be assuming, in effect, that
when a copy of 1 adjusts what it has written down for the state of 2
on its own clipboard, this must mean that copies of 2 also
instantaneously adjust what they have written down about *their* own
state. However, in a copying-with-matching scheme, there's no reason
this need be the case! The state that particle 1 assigns to particle 2
on its clipboard may just be for the purposes of later
matching--deciding which copy of 2 to "partner up with" once it can
meet them (or get some type of causal influence from them). The
fraction of copies of 2 that show a given result when B measure can
still be totally independent of what the various copies of 1 have
written down on their clipboards about the state *they* assign to 2.
For example, say we are using a particular setup where if particle 1
is measured along a spatial vector V (say, one parallel to to the
x-axis and pointing in the +x direction) and gives a result +, that
means if particle 2 is measured at a 120-degree angle from V, it will
have a 75% chance of giving the result + and a 25% chance of giving
the result -. So if a given copy of particle 1 is indeed measured
along V and does give a result +, it can adjust the state it assigns
to particle 2 on its clipboard accordingly, assigning 2 a state (or
reduced density matrix) which has an amplitude-squared of 0.75 for +
at an orientation of 120 degrees from V. It can pass on this clipboard
information (Rubin's 'label') to copies of other systems it interacts
with, like the experimenter, who carry their own clipboards/labels.
Then if that copy of the experimenter later interacts with particle 2
(or with some other particle or system that conveys information about
particle 2), the state assigned to 2 on the experimenter's clipboard
is used to decide which copy of particle 2 it should be matched with.
In this case, this could ensure that if it gets matched to a copy of
particle 2 that was indeed measured at an angle of 120 degrees from V,
there is a 75% chance it'll be matched to a copy of particle 2 that
gave the result + (i.e. a copy of particle 2 that has an
amplitude-squared of 1 for + at an orientation of 120 degrees from V),
and a 25% chance it'll be matched to a copy of particle 2 that gave
the result - (i.e. a copy of particle 2 that has an amplitude-squared
of 0 for + at an orientation of 120 degrees from V).
Is there anything in Rubin's words that clearly rules out the
possibility that the "label" one particle carries about a second
particle's state is only for the purpose of matching in this way, and
that the label has no effect whatsoever on the actual fraction of
copies of the second particle that showed a given measurement result?
Or perhaps this interpretation just didn't occur to you?
It isn't obviously wrong in my interpretation above, and I think
it's wrongheaded to imagine you can be confident about the
interpretation of any verbal statement by a physicist if you
don't have a detailed grasp on the mathematics of the model the
physicist is talking about--if you don't you may miss possible
interpretations, like the ones above that you don't seem to have
considered.
I think your arrogant patronizing here is a bit over the top! I
have perfectly well understood the mathematics of Rubin's paper--
that is why I was confident that the paragraphs I quoted
accurately summarized his detailed arguments and results.
If you understand the mathematics, why not point to the actual
equations that you think are incorrect rather than an ambiguous verbal
summary? I didn't think I was being patronizing, since I said several
times that I myself didn't understand the detailed mathematics of his
paper myself (not having much familiarity with the Heisenberg picture
of quantum mechanics which the paper is using).
Also, do you plan to respond to the rest of my comment? In
particular, do you think you can come up with any simple
numerical examples that show a local-copies-with-matching model
can't correctly reproduce some observed statistics at a given
location if we assume that location has been "shielded" from any
physical influences from Alice or Bob (and assuming 'matching'
between copies of Alice and copies of Bob can only be done in
regions that have received measurable physical signals from
them), as you seemed to claim earlier?
No, I have no intention of replying to any of this because it is
all beside the point. If you can't follow the simple physical and
conceptual arguments that I make, numerical examples are not going
to help much.
Why not try it and see how I respond, instead of immediately assuming
that because I find your English-language arguments ambiguous or
unconvincing, that implies I would find it impossible to follow a
numerical example? (talk about patronizing!) I did get my
undergraduate degree in physics, I'm pretty sure I could follow a
numerical example involving the statistics of entangled particles, and
I think I would find it much *easier* to follow that a purely verbal
argument because there'd be less room for ambiguity in interpreting
what you mean.
If Alice and Bob are truly local, and fully independent, then no
matching scheme can ever have the necessary information to
reproduce the quantum probabilities -- where do you find the
cos^2(theta/2) basis for the probabilities if they are truly
independent?
The numerical example I gave earlier, showing how
copies-with-later-matching could reproduce a certain set of statistics
that violate a Bell inequality, was actually based on an experiment
with entangled particles where the probabilities are derived from the
cos^2(theta/2) relation you mention. In a message I posted on April
19, I said:
'For example, one Bell-inequality-violating quantum experiment would
involve Alice and Bob each choosing from one of three detector angles,
with the result that when they choose the same angle they are
guaranteed to get opposite results with probability 1, whereas when
they measure different angles they only have a 1/4 probability of
getting opposite results (the corresponding Bell inequality says that
in any local realist theory, if they get opposite results with
probability 1 when they use the same setting, the probability of
getting opposite results on different settings must be greater than or
equal to 1/3--if anyone's interested, I did a little derivation of
this in a post at http://physics.stackexchange.com/a/140883/59406 ).
So in the simulation, let's say we have 360 copies of Alice and Bob
each, and 120 copies of Alice used each of the three settings 1,2,3,
likewise with Bob. Of the 120 copies of Alice who used setting 1, 60
got the result + and 60 got the result -. If we just look at the 60
copies of Alice who used setting 1 and got result +, then when the
collection of messages from copies of Bob arrives at the computer
simulating the copies of Alice, it will assign 20 of these copies of
Alice to get the message "Bob used setting 1 and got result -", 15 of
them to get the message "Bob used setting 2 and got result +", 5 of
them to get the message "Bob used setting 2 and got result -", 15 of
them to get the message "Bob used setting 3 and got the result +", and
5 of them to get the message "Bob used setting 3 and got the result
-". So indeed, we find that the Alice-copies who learn that Bob used
the same detector setting as her will always learn that Bob got the
opposite result with probability 1, whereas the Alice-copies who learn
that Bob used a different detector setting will only have a 1/4 chance
of hearing that Bob got the opposite result from their own.'
This particular result can be seen in an experiment where the
stern-gerlach devices that measure polarization can be oriented at
three possible angles at 120-degree intervals--one device could be
oriented in the direction of some vector V, the second orientated
along a vector 120 degrees from V, and the third along a vector 240
degrees from V (all in the same plane). For a pair of spin-entangled
electrons, if both electrons are measured at the same orientation,
they have a probability of cos^2(0) = 1 of giving opposite results (if
one gives result +, the other gives result - with probability 1). On
the other hand if they are measured at different orientations, the
angle between the two detectors must be ±120 or ±240, so they have a
probability of cos^2(±60) = cos^2(±120) = 0.25 of giving opposite
results. And these are exactly the statistics that would be seen by a
randomly-selected copy of Alice or a randomly-selected copy of Bob
using the rules above. These rules first locally determine the number
of copies of Alice and number of copies of Bob that see each result
for their own measurement, without any knowledge of the orientation or
result for the other experimenter (in this case, the rule can be very
simple--however many copies of Alice use a given detector angle,
exactly half of those copies get + and half get -, and likewise for
the Bob-copies). Only later do the rules do any matching to determine
which copy of Alice gets matched to which copy of Bob, and they do it
in a way that gives the statistics above. So, the final probability
that a randomly-selected matched pair will have a given pair of
results will match the probabilities you get if you use the
cos^2(theta/2) rule to predict the real-world probabilities for this
particular experiment (and as I said, the resulting probabilities are
ones which violate a Bell inequality).
Jesse
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