Does Photon Emission Cause Decoherence?

[Feeble Wonk]

In the never ending quest to understand a complicated idea for which I completely lack the academic requirements to do so, I'd like to ask a few basic questions that I hope will allow simple yes/no responses (to ease the frustration of the physicist/mathematician contributors). Sadly, the answer of "Your question makes no sense" is also a very real possibility, so we might abbreviate that response to "Ugh" for simplicity sake.

First question: In terms of decoherence, does the emission of a photon from a particle constitute a physical interaction with the environment/system in a similar manner to the absorption of a photon?

 

[naima]

When excited atoms passing thru the Young slits, emit a photon it changes the density matrix . It gives some which-path information the visibility of the franges decreases.

 

[Feeble Wonk]

When excited atoms passing thru the Young slits, emit a photon it changes the density matrix . It gives some which-path information the visibility of the franges decreases.

Does this provide information with regard to the location of the photon emitting particle?

 

[DrClaude]

Does this provide information with regard to the location of the photon emitting particle?

Yes.

There is a very nice experiment showing this using molecules:
Lucia Hackermüller, Klaus Hornberger, Björn Brezger, Anton Zeilinger & Markus Arndt, Decoherence of matter waves by thermal emission of radiation
http://www.nature.com/nature/journal/v427/n6976/full/nature02276.html
http://arxiv.org/abs/quant-ph/0402146

 

[naima]

Information is about the path. And so about the position.

 

[Feeble Wonk]

Yes.

There is a very nice experiment showing this using molecules:
Lucia Hackermüller, Klaus Hornberger, Björn Brezger, Anton Zeilinger & Markus Arndt, Decoherence of matter waves by thermal emission of radiation
http://www.nature.com/nature/journal/v427/n6976/full/nature02276.html
http://arxiv.org/abs/quant-ph/0402146

Information is about the path. And so about the position.

Thank you both. Let me work through the sources you've provided. I might offer a follow up question later, if that's OK.

 

[Demystifier]

First question: In terms of decoherence, does the emission of a photon from a particle constitute a physical interaction with the environment/system in a similar manner to the absorption of a photon?

To have a decoherence-like interaction, you need interaction with a large number of degrees of freedom. If the particle (assumed to be an elementary particle or a particle with a very simple structure) and the emitted photon do not interact with the environment, nothing like decoherence will happen. The process of emission will be continuous and unitary, nothing like a quantum jump will occur. The quantum jump requires environment with many degrees of freedom. The environment may be external (e.g. surrounding gas or measuring apparatus) or internal (a complex emitting particle).

The same is true when emission is replaced by absorption.

 

[naima]

This point is very interesting.
It is often said that the visibility interference pattern decreases because of a possible detector of photons. Do you say that the origin is the actual presence of another existing environment (the gas or something else)?

 

[Demystifier]

It is often said that the visibility interference pattern decreases because of a possible detector of photons. Do you say that the origin is the actual presence of another existing environment (the gas or something else)?

Yes. To decrease visibility of interference pattern you need some environment, which can be a detector or something else.

 

[A. Neumaier]

Information is about the path. And so about the position.

Knowing which path gives information about the momentum in the direction of the path and about the position in two orthogonal directions orthogonal to it. Thus it gives only partial position information.

 

[Feeble Wonk]

The environment may be... internal (a complex emitting particle).
The same is true when emission is replaced by absorption.

Could you please expand on this?

 

[Demystifier]

Could you please expand on this?

For internal decoherence (in a somewhat different context) see
http://lanl.arxiv.org/abs/1406.3221 [Eur. J. Phys. 36 (2015) 045003]
and references therein.

 

[Feeble Wonk]

For internal decoherence (in a somewhat different context) see
http://lanl.arxiv.org/abs/1406.3221 [Eur. J. Phys. 36 (2015) 045003]
and references therein.

Thank you. Most of the math is over my head, but the discussion and conclusion summary is excellent.

 

[naima]

To have a decoherence-like interaction, you need interaction with a large number of degrees of freedom. If the particle (assumed to be an elementary particle or a particle with a very simple structure) and the emitted photon do not interact with the environment, nothing like decoherence will happen. The process of emission will be continuous and unitary, nothing like a quantum jump will occur. The quantum jump requires environment with many degrees of freedom. The environment may be external (e.g. surrounding gas or measuring apparatus) or internal (a complex emitting particle).

The same is true when emission is replaced by absorption.

In the case of a Stern Gerlach device, I see several steps in the decoherence;
1) the degrees of freedom of the electron (spin, position) and those of the magnetic field interact and decoherence occurs. What does this "occurs" mean?
We can have at any time t many copies of the electron prepared in the same state (mixed here). We can measure say their Wigner function. It needs that they interact with macroscopic apparatus to get an image. Varying t, we see how the Wigner function and decoherence evolve.
This leads to the second point.
But before measuring the state with a macroscopic device we cannot say that decoherence completely occured. It has nothing to do to the remaining presence of little off diagonal terms. It is still possible to erase what was done. the two output beams of the SG can be the inputs of a beam merger whose output is the initial electron's state.
Things are reversible.
2) registration of the decoherence.
It is what happened when we measured the Wigner function. It is different from a screen in front of the SG. We get here a macroscopic spot which gives the value of spin measurement. In one case we measure the density matrix and in the other the spin. In both cases there is irreversibility.
This scheme can easily understood in the case of an emitted witness photon. Is it the same when the atom absorbs the photon? What is then the environment?

 

[Demystifier]

But before measuring the state with a macroscopic device we cannot say that decoherence completely occured.
...
Is it the same when the atom absorbs the photon? What is then the environment?

How do you know that absorption of the photon by the atom has occured? Before detecting the absorption by a macroscopic device, you cannot know that. So this is where environment is in the case of absorption.

 

[naima]

2) registration of the decoherence.
It is what happened when we measured the Wigner function. It is different from a screen in front of the SG.

As i read again these lines, i am no more sure that they are different. When one gets a tomogram of the density matrix he has to measure quadratures and for each of them he gets random outputs just like with a screen.
Bhobba wrote that measurement (with its random output) occurs when decoherence is done. I see it from the other side: Decoherence is done when measurements occured.

 

[vanhees71]

To have a decoherence-like interaction, you need interaction with a large number of degrees of freedom. If the particle (assumed to be an elementary particle or a particle with a very simple structure) and the emitted photon do not interact with the environment, nothing like decoherence will happen. The process of emission will be continuous and unitary, nothing like a quantum jump will occur. The quantum jump requires environment with many degrees of freedom. The environment may be external (e.g. surrounding gas or measuring apparatus) or internal (a complex emitting particle).

The same is true when emission is replaced by absorption.

Well, decoherence is very efficient (to the dismay of quantum-computer afficionados). It was demonstrated by Zeilinger at all doing double-slit/grating experiments with Bucky balls that the emission of a few black-body photons is sufficient to make the interference go away

http://arxiv.org/abs/quant-ph/0402146

 

[A. Neumaier]

If [...] the emitted photons do not interact with the environment

This is nearly impossible, unless the experiment is done in outer space and the photon escapes into the vacuum, to be absorbed only by a very distant star (which can be ignored due to the finite speed of light).

 

[naima]

You see that a few photons caused the (first step of) decoherence when you look at the interferences.
You need then a macroscopic device.
Demystifier answered to my question about absorbed photons, that to know if the photons were absorbed you had to measure the atom with a macroscopic device. without that decoherence is reversible.

 

[vanhees71]

Where do you need a macroscopic device? Of course, to measure the buckyballs you need a detector. The decoherence through the random emission of thermal photons is not a macroscopic process. The interesting thing about the buckyballs is that they are "mesoscopic", i.e., with 60 or 70 atoms in a molecule they have on the one hand a lot of degrees of freedom and a quasi-continuous energy spectrum, being in this sense on the edge of becoming "macroscopic" objects. On the other hand they are not that large that you can't control them. So you can cool them down so much that they show quantum-mechanical coherence effects in their passage through gratings. Heating them up in a controlled way these interference effects cease more and more the more thermal photons are emitted. That's the demonstration of decoherence and gradual loss of quantum behavior in a pretty well managable system!

 

[naima]

I am afraid that you anwer about another topic. What i said is about the irreversibility of decoherence. Everybody agree that decoherence is a measurable time evolving process.
Is there some thing in your book about possible reversibility of decoherence?
Read my post 16.
You wrote "of course you need a detector..."
This "of course" is the core of the problem.

 

[vanhees71]

It doesn't matter whether you detect the radiated photons or not. The decoherence is due to the electromagnetic interaction of the charges making up the buckyball and the electromagnetic (quantum) field, leading to the emission of photons. Thus, the buckyball is coupled to the "environment", i.e., the electromagnetic field, and this leads to decoherence. Only the groundstate of the buckyball is strictly stable, i.e., no photons can be emitted due to energy conservation. Decoherence has nothing to do with the observation of the photons. Particularly you don't need a "conscious being" to take note of the results of measurements to cause decoherence.

 

[naima]

One more time you did not read post 16.
I talked about the electron in the SG. (the equivalent of you buckyball).
The electron decoheres in the magnetic field. WE AGREE.
You refuse to talk about what occurs after.
If the electron is not detected by a macroscopic screen or a detector, there is a possibility for recoherence.
When you have a decaying atom things only get irreversible decoherence if you measure the atom OR the decay product with a macroscopic device. Here you only talk about one of the two things.

 

[Demystifier]

Well, decoherence is very efficient (to the dismay of quantum-computer afficionados). It was demonstrated by Zeilinger at all doing double-slit/grating experiments with Bucky balls that the emission of a few black-body photons is sufficient to make the interference go away

http://arxiv.org/abs/quant-ph/0402146

Note that Zeilinger et al talk about complex molecules, while my statement contains an explicit note that it refers to particles with a very simple structure.

 

[vanhees71]

One more time you did not read post 16.
I talked about the electron in the SG. (the equivalent of you buckyball).
The electron decoheres in the magnetic field. WE AGREE.
You refuse to talk about what occurs after.
If the electron is not detected by a macroscopic screen or a detector, there is a possibility for recoherence.
When you have a decaying atom things only get irreversible decoherence if you measure the atom OR the decay product with a macroscopic device. Here you only talk about one of the two things.

Why should in a SG apparatus something decohere as long as there's no interaction of the electron with "the environment" (let's ignore that a SG experiment with a charged particle is very difficult if not impossible)? Just running through an external field doesn't decohere anything. There the time evolution is described by a unitary time evolution of the single electron. The SG apparatus doesn't lead to decoherence but to an entanglement between position and spin-projection.

 

[Demystifier]

Why should in a SG apparatus something decohere as long as there's no interaction of the electron with "the environment"?

Because the SG apparatus itself counts as environment. The SG apparatus is a macroscopic device, so most fine degrees of freedom of it cannot be controlled or monitored. Hence decoherence!

The SG apparatus doesn't lead to decoherence but to an entanglement between position and spin-projection.

The SG apparatus leads to both. Let me also remind you that decoherence always involves entanglement.

 

[vanhees71]

A single particle moving in a (inhomogeneous) magnetic field is described by a unitary time evolution (in the non-relativistic approximation). So no decoherence occurs in the usual setup of (idealized) SG experiments before the particle hits the screen, where it is registered.

 

[Demystifier]

A single particle moving in a (inhomogeneous) magnetic field is described by a unitary time evolution (in the non-relativistic approximation). So no decoherence occurs in the usual setup of (idealized) SG experiments before the particle hits the screen, where it is registered.

Yes, but I consider the screen to be a part of the SG apparatus. Otherwise, you are fully right.

 

[naima]

If you believe that there is no decoherence in the SG (with no screen) read
A study of decoherence effects in the Stern-Gerlach experiment using matrix Wigner functions
by Alejandro Perez.
Decoherence occured, but as i wrote it this decoherence is fragile it only become robust when the particle is measured and give a maroscopic spot on the screen.
@vanhees71
You did not say if there is something in your book about recoherence.

 

[vanhees71]

In which book? I've nothing about decoherence in my manuscripts. Concerning decoherence in the SG experiment in the above quoted preprint they write already in the abstract

"We include the interaction with the environment, as described by the Caldeira-Leggett model."

Then of course you have decoherence. No surprise!

 

[naima]

So no decoherence occurs in the usual setup of (idealized) SG experiments before the particle hits the screen.

Then of course you have decoherence. No surprise!

It is hard to follow you. Are you saying that here we have not an idealized SG device?

 

[vanhees71]

Of course not. That's taking into account the rest gas in the vacuum tube. Then of course you have decoherence.

 

[naima]

Thanks for this explanation.
We have 4 types of freedom degrees:
spin and position of the particle.
those of the air and of the magnetic field.
Suppose there is no gaz.
is there still entanglement of the 3 remaining degres and is there decoherence when we trace out the magnetic field?

 

[vanhees71]

I don't know what you mean to "trace out the magnetic field". The magnetic field is at the heart of the whole experiment, no matter whether you take into account the interaction with the rest gas (air) in the vacuum tube or not.

 

[naima]

I trace out the magnetic field when i take the partial trace on its degrees of liberty. We then get the density matrix of the particle in the spin and position basis.
It is the usual thing. And i suppose there is no gas.