Re: Information transfer without energy transfer ?

On Mar 26, 6:17 pm, 5...@xxxxxxx wrote:
All right. I'll try to get it step by step.
Imagine a single photon is being emitted (say, from an electron)
For now let's not talk about its polarisation or frequency or whatever
- please.
While travelling, this photon has no spatial localisation, it should
be described
by its wavefunction, a spherical wave (probability amplitude)
travelling at speed c,
until it is "catched" (say, by another electron) and then at this
moment,
its wavefuntion collapses immediately and the location of our photon
is defined.
Am I right to this point ?

No. You don't have what wavefunction collapse really means, and you're
confusing a classical model with a wavefunction.
A photon wavefunction is not represented by a classical, spherically
radiating energy shell traveling at c.

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.

.

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