Free-Air Transmission and Quantum Effects?

[JDługosz]

Why does the double-slit experiment with a laser work properly in an atmosphere? Likewise, I've seen news reports of entangled photons transmitted in a fiber or in open-air used to generate a shared secret. So, both superposition and entanglement works.

But, the fact that the speed of the light is slower shows that the photon is interacting with the air (or fiber) molecules. (I understand that the photon mixes with quasi-particles called phonons and the resulting mixture, actually a polariton, travels with a speed slower than c.

So, why doesn't the air cause rapid decoherance? You don't need to consider every air atom to perceive the entanglement. Instead, you can seemingly ignore the air (or glass).

 

[JDługosz]

* bump *

 

[genneth]

I'd guess that the interaction isn't strong enough. For things which rely on polarisation or spin, there is essentially no interaction with things like air or dielectrics

 

[JesseM]

I don't understand the details myself, but I was curious about this question myself, so years ago I bookmarked this thread from sci.physics.research which seems to address it...for example, Douglas Natelson writes:

Peter Shor wrote:
> I don't think this really answers the question. The speed of light in
> air is slightly (0.03%) less than the speed of light in a vacuum. This
> means that even when the photons aren't scattering, they're still
> interacting with the air. But this interaction doesn't seem to affect
> the double slit experiment much. Why not? Would the double slit experiment
> work in water or glass, which have much higher indices of refraction but are
> still pretty transparent? Why? Anyone have an answer?

My sense is that this must be because elastic interactions do
not dephase. When people first started thinking about trying
to see quantum interference effects of electrons in metals
(before 1980 or so), there was this intuitive bias that the
elastic mean free path would be the limiting length scale for
coherence -- that is, that every elastic scattering event of
an electron off of a grain boundary or impurity atom would
cause decoherence. We now know that this isn't the case.
Interactions that do not entangle the electron with dynamical
degrees of freedom of the environment do not cause decoherence.
That's why it's possible to see things like weak localization,
universal conductance fluctuations, and the Aharonov-Bohm effect
in solid state systems, even when far from the ballistic limit.
This is all discussed very nicely by people like Yoseph Imry.
It's also been discussed for photons in papers like PRL 81,
5800 (1998); PRL 92, 033903 (2004); etc.

So, my best guess is that the elastic interactions between
the light and the dielectric medium (that give rise to the
dielectric response) are (mostly) reversible, and don't
correspond to changing the internal states of the dielectric.
Now, if you had a material with weird dielectric properties
(e.g. lots of loss, resonances, etc.), then you could run into
decoherence issues.

And Greg Kuperberg writes:

Peter Shor <peterws...@aol.com> wrote:
>I don't think this really answers the question. The speed of light in
>air is slightly (0.03%) less than the speed of light in a vacuum. This
>means that even when the photons aren't scattering, they're still
>interacting with the air. But this interaction doesn't seem to affect
>the double slit experiment much. Why not? Would the double slit experiment
>work in water or glass, which have much higher indices of refraction but are
>still pretty transparent? Why? Anyone have an answer?

I'm not entirely sure that this is a good explanation, but here is what I
think is the point. Because the photon has low energy, it measures very
little about the positions of the molecular dipoles that refract it,
much less their internal state, only with their presence or absence.
And because the photon has a long wavelength, it only measures the
aggregate presence of many molecules in a large volume. Thus the photon
measures almost nothing that has any uncertainty. Here "uncertainty"
means either in the narrow sense of being in a superposition, or in
the ultimately more general sense of being different from one photon
to the next. To say that the photon is measuring something is of
course equivalent to saying that its state is importing entropy.
In the visible wavelength range, you would need an astronomically high
index of refraction or a very unstable medium of propagation to invalidate
these approximations. (E.g. the two-slit experiment might not work well
in an erratic gale wind.) The two-slit experiment does of course take
a different form in the X-ray range.

It may be useful to think of the photon's evolution as a quantum
operation. For the above reasons (I think), the operation is
extremely close to sub-unitary, which is to say a single Kraus term.