Does Photon Emission Cause Decoherence?

[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.

 

[vanhees71]

No, the usual description is that of an atom moving in an external magnetic field, i.e., a one-particle description. If you assume that there is no gas, then there is no decoherence but unitary time evolution of the single-particle state only. For a complete treatment, see

http://arxiv.org/abs/quant-ph/0409206

 

[naima]

"We know that spin coherence is lost as the state evolves in time and the two spatial parts become orthogonal. Since the measurement process naturally invokes the partial trace over the spatial part of the initially pure global state the remaining spin part becomes a mixture." (Amir Caldeira)
in http://arxiv.org/abs/quant-ph/0608192
Caldeira adds at the end that
"Recovery of coherence should be achieved simply by recombining the two beams"

That is what i said: coherence is fragile when not measured by a macroscopic device (the magnet measures nothing)

 

[vanhees71]

I don't agree with the authors that this is "decoherence" in the usual sense. If you trace out the position information, of course you end up with a mixed state, but that's not decoherence, which is due to interaction of the quantum system with "the environment" (which can be among other sense also the measurement apparatus).

 

[naima]

When we measure a spin with a SG the only thing which matters here is the density matrix in the up down basis. And it is not pure.

 

[vanhees71]

This has nothing to do with decoherence!

 

[naima]

There are several ways to measure entanglement. We can use Neumann's entropy. We can say that there is entanglement if the states are not separable (yes/no answer). There is also linear entropy.
We have an analog thing with decoherence.
Let us us take a two level state. When it is pure its entropy is null and and when it has decohered its entropy may be equal one bit.
In all cases we vave in its density matrix 2 null off diagonal terms.
We usually say then that decoherence is done.
This is a yes/no characteristic.
As the particle was entangled we can consider the global Hilbert space of all the degrees of freedom. Caldeira ended its paper by saying that coherence can be recovered simply by merging the beams of the SG apparatus. Here the spin of the particle was coupled with its z position. The coherence is fragile because we have only to act on one degree to erase the decoherence.
We see that in addition to the nullity of the off diagonal terms, there is another quantity which measures the easiness to recohere. It is the dimension of the global hilbert space (minus 2 here). If it is high we would have to manage numerous parameters to get recoherence and spontaneous recoherence is highly improbable. Robustness of decoherence only appears when macroscopic internal or external numbers of degrees of freedom come into play.
Measurement cannot give outputs if decoherence was done but fragile. That is why you need a screen in front of rhe SG.

 

[naima]

There is another way to say that:
In the SG with no gas the spin decoheres. it is in an improper mixed state. one does not see the outcome. If the particle enters a bubble chamber, its spin density matrix remains unchanged but we see the result. coherence cannot reappear. In this second period the particle is entangled with the gas environment. The mixed state begins to be a statistical (proper) mixture of results and one is observed. Proper and improper mixtures have the same matrix but are physically different. the particle evolves here from an improper to a proper mixture state. There is no collapse. Only a statistical mixture at the end.
Zurek gives a time coefficient Zu(t) which enable to describe the state as
(1 - Zu(t)) (improper) + Zu(t) (proper).
Another related point of view: http://arxiv.org/abs/0901.0795

 

[vanhees71]

Again: If there is no gas and thus the system closed, i.e., a single particle moving in a magnetic field, nothing decoheres. The time evolution is entirely unitary. There cannot be decoherence with a unitary time evolution. That's behind the socalled "measurement problem", discussed vigorously in this forum (forgetting the physics over the philosophy unfortunately ;-)).

 

[naima]

I don't agree with the authors that this is "decoherence" in the usual sense.

You already said that you disagree with the co-author of the Caldeira-Leggett model. Who is mainstream? You or Caldeira?
Caldeira himself says that something differs from the usual sense. It is not decoherence but the notion of environment. Here we have not a particle which decoheres while interacting with other particles but a degree of feedom entangling with other degrees of freedom.
I notice that the other usual forumers are very cautious and silent.