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How does Bohmian mechanics explain the working of a polarizer?
Does the description go through what we (human being) observe?vanhees71 said:Then you have to even ask more generally, how does Bohmian mechanics describe photons at all? I've no clue...
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 ...A. Neumaier said:How does Bohmian mechanics explain the working of a polarizer?
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.vanhees71 said:I've strong doubts that there's a Bohmian interpretation for photons.
Yes, that's the basic idea of BM for instrumentalists.microsansfil said:Does the description go through what we (human being) observe?
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.Demystifier said: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 ...
If your point is that it is described by QFT then ...A. Neumaier said:This is a process not described by the Schrödinger equation,
... read "Bohmian mechanics for instrumentalist", as it explains how QFT should be interpreted in Bohmian terms.A. Neumaier said:hence it is not clear to me how it would be described in Bohmian terms.
No, the point is that it is described by dissipative equations that do not seem to have a Bohmian equivalent.Demystifier said: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.
A. Neumaier said:No, the point is that it is described by dissipative equations that do not seem to have a Bohmian equivalent.
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.microsansfil said: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
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 said:No, the point is that it is described by dissipative equations that do not seem to have a Bohmian equivalent.
Where can I read about details? Or is this just a conjecture and not an established fact?Demystifier said: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.
That quantum dissipative equations can be derived from unitary equations is an established fact.A. Neumaier said:Where can I read about details? Or is this just a conjecture and not an established fact?
Well, I know it very well in the context of ordinary QM.Demystifier said: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.
I'd like to see the Bohmian equivalent at a more fundamental level.Demystifier said:the non-unitary dissipative equations emerge from more fundamental unitary evolution, implying that there is a 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 said: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 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.Demystifier said: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.
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 said:So my question amounts to asking for the Bohmian equivalent of Lindblad equations.
Well, there is a simple modification for the latter, which is a classical limit of a suitable Lindblad dynamics. It takes the formDemystifier said:there is no Hamiltonian formulation of a Newton equation with a friction term.
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.A. Neumaier said: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?
I'll check this...Demystifier said: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.
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?Demystifier said: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.
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?A. Neumaier said: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.
Did you read the paper?vanhees71 said: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?
The only question is what the Bohmian trajectories are good for? So why should you calculate them. Everything observable is already given by the solution of the Schrödinger equation.Demystifier said: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.
Which paper?Demystifier said:Did you read the paper?
The one linked in my signature below.vanhees71 said:Which paper?
I didn't refer to QFT, so your interpretation of what I said is unfounded. The process described follows from QED, but is modeled in the analysis of actual quantum optics experiments in a coarse-grained fashion.vanhees71 said: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?
There is no "signature below'' - whether a signature is shown depends on user settings!Demystifier said:The one linked in my signature below.
In "Bohmian mechanics for instrumentalists" I explain that there is no much point in explicit calculation of Bohmian trajectories, yet they are good for having an intuitive conceptual picture of QM. This is somewhat similar to effective field theories, where there is no much point in explicit calculations in the more fundamental theory, yet the idea that there is a more fundamental theory is good for having an intuitive conceptual picture of effective QFT.vanhees71 said:The only question is what the Bohmian trajectories are good for? So why should you calculate them.
I didn't know that. But I think showing signature is the default.A. Neumaier said:There is no "signature below'' - whether a signature is shown depends on user settings!
This is an incorrect view. One often calculates some things from the more fundamental theory (if it is known), to be matched by the coefficients in the effective theory.Demystifier said:effective field theories, where there is no much point in explicit calculations in the more fundamental theory,
Yes, but once you have the coefficients, which what "to have the effective theory" means, then you don't longer need the more fundamental theory.A. Neumaier said:This is an incorrect view. One often calculates some things from the more fundamental theory (if it is known), to be matched by the coefficients in the effective theory.