On 11-07-2021 02:46, Bruce Kellett wrote:
On Sun, Jul 11, 2021 at 10:21 AM smitra <smi...@zonnet.nl> wrote:
On 11-07-2021 00:59, Bruce Kellett wrote:
On Sun, Jul 11, 2021 at 8:27 AM smitra <smi...@zonnet.nl> wrote:
On 10-07-2021 21:41, 'Brent Meeker' via Everything List wrote:
On 7/10/2021 1:19 AM, Bruno Marchal wrote:
So, in general we can conclude by generalizing this to any
large
number of particles that even with what we consider to be
permanent
records, you don't get rid of the theoretical possibility of
interference between the sectors where those records are
different.
We can if the universe is expanding faster than light beyond the
Hubble
radius.
The expansion of the universe is irrelevant. The information
needed
to see the interference pattern continues to exist outside the
horizon
when it isn't accessible to us anymore. And this is irrelevant
for the
discussions about observations in quantum mechanics. If an
observer
performs a measurement and the claim is that this is a unitary
process with the observer evolving into a superposition, while
the objection
against this claim is that infrared photons are escaping and will
eventually move beyond the Hubble volume, then these photons will
still not have escaped beyond the Hubble horizon by the time the
observer
is aware of the results of the experiment. So, whether or not the
photons will eventually no longer be accessible, cannot be
relevant.
Once the photons escape from the immediate environs of the
experiment,
they are not recoverable. Try shining a torch at night to
illuminate a
tree. Now try to stop the illumination already present. You can
stop
future illumination by covering the torch, or switching it off.
But
once the tree is illuminated it is not reversible. The expansion
of
the universe, and the existence of the Hubble horizon, just makes
the
irreversibility more obvious.
This is only true in practice, not in principle because the escaping
photons be captured and detected in principle.
Not true. You can;t ever catch up with the escaping photons to capture
them.
They can be reflected back using mirrors. One can therefore at least in
principle do experiments that rule out collapse theories for large
systems that are large enough to include observers. And once collapse
theories are ruled out in a few experiments it follows that in a
superposition all the sectors where observers find different results
objectively exist, regardless of whether or not in those particular
situations one could have done the same sorts of experiments that ruled
out collapse theories.
It's also irrelevant,
because the photons that escape continue to exist, the information
contained in the photons continues to exist, even if it were true
that
they could not be recovered. Then if we were to conduct an
interference
experiment with the balls then we wouldn't see an interference
pattern.
But if we write down where each balls lands on the screen then this
information together with information that could be obtained by
performing certain measurements on the escaping photons emitted by
each
ball would still yield an interference pattern.
This is where you go seriously wrong. Simply recording where the
photons land does not quantum erase the information they carried. Once
the photons carry off the which way information, the interference
pattern is restored only if the information carried by the photons is
quantum erased. Simply running the photons into a screen (or the
wall), even if you record where they land, is not quantum erasure.
See, for example, the paper arXiv:1206.6578 on quantum erasure. In
this paper they say "the presence of path information anywhere in the
universe is sufficient to prohibit any possibility of interference. In
other words, the atoms' path states alone are not in a coherent
superposition due to the atom-photon entanglement." This transfers
directly to the buckyball experiment under discussion. Running the
photons into a screen, or the wall, does not destroy the ball-photon
entanglement.
Recording where the photons land on the screen is enough, this is very
easy to see. Let's consider an interference experiment where the
particle gets entangled with another particle that carries away the
which way information. If we work in the position representation then we
have a wavefunction:
psi(x,y) = 1/sqrt(2) [psi_1(x,y) + psi_2(x,y)] (1)
where x is the position of particle 1 just before it hits the screen and
y the position of the other particle at that time, and psi_1(x,y) is the
wavefunction when only slit 1 is open while psi_2(x,y) the wavefunction
with only slit 2 open. We also assume that particle 2 carries the which
way information, this means that psi_1(x,y) and psi_2(x,y) for x kept
fixed and considered as a function of y are eigenfunctions with
different eigenvalues of an observable that corresponds to extracting
the which way information from the second particle. This then implies
that psi_1(x,y) and psi_2(x,y) are orthogonal for every x:
Integral psi_1(x,y)*psi_2(x,y)d^3y = 0 (2)
So, if we compute the probability of observing particle 1 at some
position x, while we don't measure the position of particle 2, we find
by taking the modulus squared of (1) and integrating over y:
Integral |psi(x,y)|^2 d^3y = 1/2 Integral |psi_1(x,y)|^2 d^3y + 1/2
Integral |psi_2(x,y)|^2 d^3y
+ Re Integral psi_1(x,y)*psi_2(x,y) d^3y
And the last term vanishes due to (2), so we don't detect interference.
But now suppose that we perform a simultaneous measurement of the
positions of both particles. The probability of detecting particle 1 at
position x and particle 2 at position y is:
|psi(x,y)|^2 = 1/2 |psi_1(x,y)|^2 y + 1/2 |psi_2(x,y)|^2 + Re
psi_1(x,y)*psi_2(x,y)
So, there now clearly an interference term. If we keep y fixed the
probability distribution of x will show interference., This requires one
to count the dots on the screen for particle 1 for the cases where
particle 2 ends up within some narrow range for some fixed y. If we
don't keep track of y and just add up all the dots at the positions on
the screen for particle 1, then that amounts to integrating the
interference term over all y, which will make it vanish die to
orthogonality.
Saibal
Bruce
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