# B Decoherence by Emission?

1. Mar 2, 2016

### 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?

Last edited: Mar 2, 2016
2. Mar 2, 2016

### 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.

3. Mar 2, 2016

### Feeble Wonk

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

4. Mar 2, 2016

### Staff: Mentor

Last edited by a moderator: May 7, 2017
5. Mar 2, 2016

6. Mar 2, 2016

### Feeble Wonk

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

Last edited by a moderator: May 7, 2017
7. Mar 3, 2016

### Demystifier

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.

8. Mar 3, 2016

### 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)?

9. Mar 3, 2016

### Demystifier

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

10. Mar 3, 2016

### A. Neumaier

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.

11. Mar 3, 2016

### Feeble Wonk

Could you please expand on this?

12. Mar 3, 2016

### Demystifier

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.

13. Mar 3, 2016

### Feeble Wonk

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

14. Mar 3, 2016

### naima

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?

15. Mar 3, 2016

### Demystifier

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.

16. Mar 3, 2016

### naima

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.

17. Mar 3, 2016

### vanhees71

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

18. Mar 3, 2016

### A. Neumaier

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).

19. Mar 3, 2016

### 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.

Last edited: Mar 3, 2016
20. Mar 4, 2016

### 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!

21. Mar 4, 2016

### 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?
You wrote "of course you need a detector..."
This "of course" is the core of the problem.

22. Mar 4, 2016

### 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.

23. Mar 4, 2016

### 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.

24. Mar 4, 2016

### Demystifier

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.

25. Mar 4, 2016

### vanhees71

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.