On 12-07-2021 00:43, Bruce Kellett wrote:
On Mon, Jul 12, 2021 at 3:44 AM smitra <[email protected]> wrote:
On 11-07-2021 02:46, Bruce Kellett wrote:
On Sun, Jul 11, 2021 at 10:21 AM smitra <[email protected]> wrote:
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.
You can't decide to put mirrors in place after the photons have been
emitted.
Of course, you could always decide in advance to perform all
experiments in reflecting boxes, but that is not what is done in
practice. So the general rule is that escaping photons are not
recoverable.
If in one experiment with mirrors they are recoverable and therefore
interference can be detected in principle, then that proves that the
other sectors do objectively exist. Since the objective existence of
these sectors does not depend on whether or not one uses mirrors in the
experiment, one can conclude that they exist in general.
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)
This is the same mistake that you have routinely made in all of these
discussions. You calculate for particles through each slit
independently.
No that's not what I did.
But if that is the case, then no interference can ever
be seen. The possibility of interference relies on coherence between
the amplitudes at the two slits. In your calculations, you have never
allowed for this coherence -- the amplitudes at the two slits are not
independent.
No, I really think you need to refresh you knowledge of quantum
mechanics.
Bruce
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
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