I Polarization in Bohmian mechanics

A. Neumaier

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How does Bohmian mechanics explain the working of a polarizer?
 

vanhees71

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Do you mean for photons? Then you have to even ask more generally, how does Bohmian mechanics describe photons at all? I've no clue...
 
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Then you have to even ask more generally, how does Bohmian mechanics describe photons at all? I've no clue...
Does the description go through what we (human being) observe?

/Patrick
 

vanhees71

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I don't know. What we observe is completely described by standard Q(F)T without any Bohmian mechanics. It's a somewhat curious possibility to interpret non-relativistic quantum theory in terms of a non-local deterministic theory, but the additional pieces, like the trajectories, do not tell as anything else than what standard QFT tells us.

I've strong doubts that there's a Bohmian interpretation for photons. Photons are the least particle-like quanta directly observable to us. A position observable makes only a much reduced sense. All we know are detection probabilities given the state of the em. field, where the position does not directly refer to a photon but only to the location of the detector used to register the photon having interacted with it at its position.
 

Demystifier

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How does Bohmian mechanics explain the working of a polarizer?
Why do you think it's a problem? If you wonder how Bohmian mechanics explains a measurement of polarization, which is all what matters from an instrumental point of view, then "Bohmian mechanics for instrumentalists" gives a straightforward explanation. Just let the operator in Eq. (2) be the polarization operator and proceed with the analysis there ...
 

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I've strong doubts that there's a Bohmian interpretation for photons.
In "Bohmian mechanics for instrumentalists" there is no need for that. All what one needs is a Bohmian interpretation of the detection of photons, which is quite straightforward.
 

A. Neumaier

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Why do you think it's a problem? If you wonder how Bohmian mechanics explains a measurement of polarization, which is all what matters from an instrumental point of view, then "Bohmian mechanics for instrumentalists" gives a straightforward explanation. Just let the operator in Eq. (2) be the polarization operator and proceed with the analysis there ...
Well, a polarizer doesn't quite measure polarization; instead it absorbs a fraction of the photons and forces the remaining into a particular polarization. This is a process not described by the Schrödinger equation, hence it is not clear to me how it would be described in Bohmian terms.
 
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Demystifier

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This is a process not described by the Schrödinger equation,
If your point is that it is described by QFT then ...
hence it is not clear to me how it would be described in Bohmian terms.
... read "Bohmian mechanics for instrumentalist", as it explains how QFT should be interpreted in Bohmian terms.
 

A. Neumaier

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If your point is that it is described by QFT then ...

... read "Bohmian mechanics for instrumentalist", as it explains how QFT should be interpreted in Bohmian terms.
No, the point is that it is described by dissipative equations that do not seem to have a Bohmian equivalent.
 

atyy

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hi;

This paper describes the evolution of the polarization state during its propagation inside a polarizer.

The Stokes vector s determines a point located on the Poincaré sphere S2 of radius s. The direction of the vector s, characterizes the polarization. Thus, the polarization state of the light wave corresponds to a unique point on the Poincaré sphere S2.

The study of the trajectory does not require the use of dissipative equations!!

/Patrick
 

A. Neumaier

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hi;

This paper describes the evolution of the polarization state during its propagation inside a polarizer.

The Stokes vector s determines a point located on the Poincaré sphere S2 of radius s. The direction of the vector s, characterizes the polarization. Thus, the polarization state of the light wave corresponds to a unique point on the Poincaré sphere S2.

The study of the trajectory does not require the use of dissipative equations!!

/Patrick
The dissipation is in the length, not the direction of the Stokes vector. This length represents the intensity and shrinks unless the input has the same polarization as the output. You can see it in Figure 3 of your reference. This shrinking is not unitary.
 

Demystifier

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No, the point is that it is described by dissipative equations that do not seem to have a Bohmian equivalent.
This is like saying that dissipation in classical mechanics does not have an equivalent in Hamiltonian mechanics. Dissipation is always an effective description emerging when one does not keep track of all degrees of freedom, while the underlying more fundamental dynamics is non-dissipative. So in your case the non-unitary dissipative equations emerge from more fundamental unitary evolution, implying that there is a Bohmian equivalent at a more fundamental level.
 

A. Neumaier

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This is like saying that dissipation in classical mechanics does not have an equivalent in Hamiltonian mechanics. Dissipation is always an effective description emerging when one does not keep track of all degrees of freedom, while the underlying more fundamental dynamics is non-dissipative. So in your case the non-unitary dissipative equations emerge from more fundamental unitary evolution, implying that there is a Bohmian equivalent at a more fundamental level.
Where can I read about details? Or is this just a conjecture and not an established fact?
 

Demystifier

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Where can I read about details? Or is this just a conjecture and not an established fact?
That quantum dissipative equations can be derived from unitary equations is an established fact.

For a basic idea see e.g. Schlosshauer, Decoherence and the Quantum to Classical Transition, Chapter 4. In particular, compare Eqs. (4.1) and (4.2). I am not aware that someone studied it in detail in the Bohmian context (but check Sec. 8.5.2 in the book above), but once one knows the unitary description, adding Bohmian trajectories is in principle straightforward.

For a more detailed analysis see the book by Breuer and Petruccione (I think you cite it in your papers on thermal interpretation), Sec. 3.1.3. and Sec. 3.3.
 

A. Neumaier

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That quantum dissipative equations can be derived from unitary equations is an established fact.

For a basic idea see e.g. Schlosshauer, Decoherence and the Quantum to Classical Transition, Chapter 4. In particular, compare Eqs. (4.1) and (4.2). I am not aware that someone studied it in detail in the Bohmian context (but check Sec. 8.5.2 in the book above), but once one knows the unitary description, adding Bohmian trajectories is in principle straightforward.

For a more detailed analysis see the book by Breuer and Petruccione (I think you cite it in your papers on thermal interpretation), Sec. 3.1.3. and Sec. 3.3.
Well, I know it very well in the context of ordinary QM.

But not how it is treated in the Bohmian context. It seems that there one must first throw away the particle positions to get the standard setting and then proceed from there. Thus no insight can be gained from having assumed a dynamics for definite positions.
the non-unitary dissipative equations emerge from more fundamental unitary evolution, implying that there is a Bohmian equivalent at a more fundamental level.
I'd like to see the Bohmian equivalent at a more fundamental level.
 

Demystifier

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Well, I know it very well in the context of ordinary QM.

But not how it is treated in the Bohmian context. It seems that there one must first throw away the particle positions to get the standard setting and then proceed from there. Thus no insight can be gained from having assumed a dynamics for definite positions.

I'd like to see the Bohmian equivalent at a more fundamental level.
In the ordinary QM one first has to solve the Schrodinger equation in some representation. Assuming that one has done that (in practice that's very hard because of many degrees of freedom), the rest is easy. All what one has to do is to represent the wave function in the position basis and then compute the Bohmian trajectories by the straightforward formula. I don't know what exactly seems problematic to you, but the only hard part is solving the Schrodinger equation with standard QM, the intrinsically Bohmian part is easy.
 

A. Neumaier

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In the ordinary QM one first has to solve the Schrodinger equation in some representation. Assuming that one has done that (in practice that's very hard because of many degrees of freedom), the rest is easy. All what one has to do is to represent the wave function in the position basis and then compute the Bohmian trajectories by the straightforward formula. I don't know what exactly seems problematic to you, but the only hard part is solving the Schrodinger equation with standard QM, the intrinsically Bohmian part is easy.
In quantum mechanics, dissipative systems are usually modeled by Lindblad equations for a density operator. With few degrees of freedom these are easily solved numerically. In particular, this handles passing polarization filters. One doesn't need to go to a unitary many-particle description to use and solve Lindblad equations, and never does in the applications.

So my question amounts to asking for the Bohmian equivalent of Lindblad equations.
 

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So my question amounts to asking for the Bohmian equivalent of Lindblad equations.
There is no such thing. It's somewhat related to the fact that there is no Hamiltonian formulation of a Newton equation with a friction term.
 

A. Neumaier

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there is no Hamiltonian formulation of a Newton equation with a friction term.
Well, there is a simple modification for the latter, which is a classical limit of a suitable Lindblad dynamics. It takes the form
$$\dot q=\partial H(q,p)/\partial p, ~~~~\dot p=-\partial H(q,p)/\partial q -C(q)\dot q.$$
For positive definite ##C(q)##, the energy decreases with time.

How would one have to modify Bohmian dynamics in the dissipative case?
 
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Demystifier

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Well, there is a simple modification for the latter, which is a classical limit of a suitable Lindblad dynamics. It takes the form
$$\dot q=\partial H(q,p)/\partial p, ~~~~\dot p=-\partial H(q,p)/\partial q -C(q)\dot q.$$
For positive definite ##C(q)##, the energy decreases with time.

How would one have to modify Bohmian dynamics in the dissipative case?
The above is not Hamiltonian mechanics, but a generalization of it. Perhaps Bohmian mechanics can be generalized in a similar sense. But if the point of Bohmian mechanics is to give fundamental microscopic ontology, then there is no point in making such a generalization. On the other hand, there are also practical applications of Bohmian mechanics, and in this sense generalization of Bohmian mechanics to dissipative systems might be useful.

Here is a sketch how this could be done. All one needs is a probability density ##\rho(x,t)## where ##x## is a set of particle positions. This ##\rho(x,t)## can be determined by the Lindblad equation. Once one has ##\rho##, one can proceed as in http://de.arxiv.org/abs/quant-ph/0302152 Eqs. (43)-(58). One does not have Eq. (42), but that's not a problem because one has the Lindblad equation instead.
 

A. Neumaier

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there are also practical applications of Bohmian mechanics, and in this sense generalization of Bohmian mechanics to dissipative systems might be useful.

Here is a sketch how this could be done. All one needs is a probability density ##\rho(x,t)## where ##x## is a set of particle positions. This ##\rho(x,t)## can be determined by the Lindblad equation. Once one has ##\rho##, one can proceed as in http://de.arxiv.org/abs/quant-ph/0302152 Eqs. (43)-(58). One does not have Eq. (42), but that's not a problem because one has the Lindblad equation instead.
I'll check this....
 

vanhees71

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In "Bohmian mechanics for instrumentalists" there is no need for that. All what one needs is a Bohmian interpretation of the detection of photons, which is quite straightforward.
Obviously I'm too stupid to see, how this is straightforward. How can you describe the detection of photons without describing the photons to begin with?
 

vanhees71

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Well, a polarizer doesn't quite measure polarization; instead it absorbs a fraction of the photons and forces the remaining into a particular polarization. This is a process not described by the Schrödinger equation, hence it is not clear to me how it would be described in Bohmian terms.
Are you saying the interaction of photons (i.e., the electromagnetic field) with matter is not described by QED (of course not the Schrödinger equation since this is a non-relativistic approximation, which cannot describe photons of course)? How do you come to this conclusion?
 

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