A. Neumaier
Science Advisor
- 6,741
- 2,686
How does Bohmian mechanics explain the working of a polarizer?
Does the description go through what we (human being) observe?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 ...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.I've strong doubts that there's a Bohmian interpretation for photons.
Yes, that's the basic idea of BM for instrumentalists.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.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 ...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.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.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.
How about considering the polarizing beam splitter which has a unitary description?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.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 S^{2} 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 S^{2}.
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.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?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.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.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.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.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.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.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 formthere 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.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....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?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?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.