Re: Why does EPR need two observables?
- From: "a student" <of_1001_nights@xxxxxxxxxxx>
- Date: Mon, 20 Feb 2006 20:51:44 +0000 (UTC)
Vonny N. wrote:
Now Einstein, as we all know, believed that the world was most
certainly local. What is less frequently noted is that it is clear from
his arguments that he also assumed that *Bohr* would be against
non-locality. Thus he wanted to refute Bohr's interpretation by
demonstrating that it entailed non-locality. And I claim (or, better
perhaps, ask whether) this can be done with a single observable.
I think this is reading rather a lot into the EPR paper. Einstein
later commented that he didn't like Podolski's write-up of the paper.
But he did NOT say that he felt the point was to prove that Bohr's
interpretation was nonlocal. He gave his reading of the EPR scenario
as showing that quantum mechanics gave too many descriptions of the
unmeasured particle -depending on which (complementary) observable was
measured on the first particle (whereas Podolski emphasised
incompleteness). So it might be more reasonable (less shoehorning) to
credit the nonlocality reading as your own, rather than as Einstein's
:) - this doesn't detract from it being of interest.
To do this, suppose you have an entangled state in which the total
momentum, say, is zero, but such that neither of the individual systems
is in a momentum eigenstate. Then a momentum measurement on one system
tells us immediately the result of a subsequent momentum measurement on
the other. As you say, there is nothing unusual about this, billiard
balls do it all the time. However, Bohr claimed that the billiard ball
model is *a priori* incorrect and that a value of momentum is only
meaningful upon measurement. But if Bohr wants to maintain his
'measurement = meaning' interpretation, he appears now to be forced to
concede some type of action-at-a-distance, with or without
consideration of other observables. And that, I think, is all Einstein
was trying to demonstrate.
I probably still don't fully understand your point of view. But it
seems to me that the hypothetical Bohr has no problem at this stage of
the argument - the real Bohr emphasised repeatedly that there is no
phenomenon before it is observed. Hence, for him, the momentum of the
second particle has not been measured/created/made meaningful at this
stage. It has simply been made predictable (should someone actually
measure the momentum of the second particle).
Bohr similarly held that, simply because a system is in an eigenstate
|a> of some observable A (and hence that A is predictable), it does not
mean that A actually "has" a value equal to a. The conditions for
measuring A must first be set up and the measurement carried out. Only
then is A measured, and a measured value created/defined/made
meaningful. Note that the conditions for measuring A will be
physically incompatible with the conditions for measuring some
complementary observable B, and hence only one can be
measured/defined/made meaningful. In particular, if B was measured
instead of of A, there is no sudden "disturbance" of a real value A=a -
there is simply a different phenomenon and a different measurement
result created.
Thus, I don't see Bohr actually conceding any nonlocality, at least on
the basis of your argument as I understand it.
.
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