Re: Information transfer without energy transfer ?



On Mar 27, 9:12 am, 5...@xxxxxxx wrote:
If you enjoy this as much as I do : fine.
This is not like the double-slit experiment, there is no
interference of the wavefunction of the photon with itself.
Please check next step :http://powerdown.free.fr/tx/

I still am having trouble with the preparation of a state where a
single photon is constrained to go in the +x or -x direction with
50-50 probability. Do you have an experimental configuration in mind
that will accomplish this?

But in any event, the next step is also problematic. If the detector
at E is *inactivated*, this means that it does not make a measurement.
What this then means is that 50% of the time the photon will be
detected at [delta] and 50% at [delta'], according to your
prescription.


On Mar 27, 2:36 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Mar 27, 8:31 am, 5...@xxxxxxx wrote:

On Mar 27, 2:24 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

Yes, OK, I'm with you so far.

Really, not kidding ?
Okay, now I'm gonna translate my proposal the best I can,
step by step, and we should discuss any divergence
of opinion as we did before, until we found it reasonnable,
then we sould proceed to the next step.
The translation will take some time, tough.

Thanks for your effort.

If I might suggest, your proposal for two detectors is fine, but it is
not really conceptually different than a double-slit experiment. Since
the double-slit experiment is *thoroughly* discussed in both popular
and professional literature, it might be useful to discuss your idea
in that context, so that I can make references to suggested readings.

PD

On Mar 27, 8:06 am, 5...@xxxxxxx wrote:
I already have good books on quantum physics.
My proposal does not concern entangled states
of several particles, there is only one photon involved.

Here is a short description of what I mean :http://powerdown.free..fr/tx/

In this particular case, I *know* that this the
standard interpretation of quantum mechanics.

The point is : no direction in space is determined
at the moment the photon is emitted from the source.
No localization of the photon is available until it is measured.
And the wavefunction is not a surface of energy,
it is a spherical wave of probability
( sum of potentialities written as orthogonal bases )
that will resolve in only one of those potentialities
at the time a measure is made.
At this moment, the wavefunction collapses, and the photon is
localized with 100% certainity where the measure has been made.

This is what I would like both of us to agree.

Yes, OK, I'm with you so far.

On Mar 27, 1:09 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

I think the first thing you need to read up on are:
- what quantum states are, and how they are delineated by certain
(quantum) numbers -- that's why they are called quantum states in the
first place
- what quantum degeneracy means, and how two states can satisfy the
same preparation conditions.
- what superposition means in the case of degenerate quantum states.
- what happens to that degeneracy at the point of observation.

I assume I am ... so, any detector lying at the surface of the
spherical wave
(which can be pretty large) has the same (tiny) chance to catch the
photon.

On 26 mar, 23:34, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Mar 26, 4:48 pm, 5...@xxxxxxx wrote:
The photon is in whatever quantum state it pleased,
providing its wavefunction travels along the straight line.
A photon is emitted at point S : then its wavefunction is a
superposition of the two possible directions : towards E and R .

No, this is not correct.
How is this state prepared?

Do not confuse a scenario with two possibilities as being a
wavefunction.- Hide quoted text -

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.



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