A Ballentine: Decoherence doesn't resolve the measurement problem

[thephystudent]

Sorry for not having time to study this long paper in detail. Could you perhaps give a summary of its essential ideas to put your question into a context?

see wikipedia

This is basically the idea that you reproduce the effect of decoherence (Lindblad equation, but also extensions beyond markov exist), by averaging over all possible measurement records.

 

[Demystifier]

Because in BM there is always the uncollapsed wave function, which only undergoes decoherence. So if we trace over that, we will get the same reduced density matrix as collapsing then forgetting.

In BM, there is a concept of conditional wave function. For instance, for two degrees of freedom $x_1$ and $x_2$ described by the full wave function $\Psi(x_1,x_2,t)$, the conditional wave function of the first degree of freedom is
$$\psi_1(x_1,t)=\Psi(x_1,X_2(t),t)$$
where $X_2(t)$ is the Bohmian trajectory. According to BM, $\Psi$ never collapses. What collapses is $\psi_1$. Decoherence, on the other hand, is something that happens with $\Psi$. Actual outcomes, or lack of knowledge of the actual outcomes, is something related to $\psi_1$.

How is it related to density matrices? The reduced density matrix is obtained from $\Psi$ as
$$\rho^{\rm reduced}_1={\rm Tr}_2|\Psi\rangle \langle\Psi|$$
which does not refer to $\psi_1$ at all. The lack-of-knowledge-about-the-outcome density matrix $\rho^{\rm knowledge}_1$, on the other hand, is related to the lack of knowledge about $\psi_1$. So in BM, $\rho^{\rm reduced}_1$ and $\rho^{\rm knowledge}_1$ are conceptually different.

 

[Demystifier]

see wikipedia

This is basically the idea that you reproduce the effect of decoherence (Lindblad equation, but also extensions beyond markov exist), by averaging over all possible measurement records.

I think it's OK as a practical method that, however, does not help to solve the measurement problem. I'm not sure if it answers your question.

 

[zonde]

You said that decoherence is of no use. I say it is of use because it helps to explain why the detector has an outcome at all.

In the paper linked in your signature you postulate an axiom "All perceptibles are beables." So without beables there are no perceptibles.
So which part would you say decoherence helps to explain? Does it help to explain beables or does it help to explain perceptibles (without explaining anything about beables)?

 

[Demystifier]

In the paper linked in your signature you postulate an axiom "All perceptibles are beables." So without beables there are no perceptibles.

Good summary of my work, thanks!

So which part would you say decoherence helps to explain? Does it help to explain beables or does it help to explain perceptibles (without explaining anything about beables)?

That's a good question! It helps to explain the perceptibles. Decoherence is relevant to Sec. 3, where beables are not yet considered. In Sec. 3 I do not talk about decoherence explicitly, but it is implicit in Eqs. (4) and (10).

 

[atyy]

In BM, there is a concept of conditional wave function. For instance, for two degrees of freedom $x_1$ and $x_2$ described by the full wave function $\Psi(x_1,x_2,t)$, the conditional wave function of the first degree of freedom is
$$\psi_1(x_1,t)=\Psi(x_1,X_2(t),t)$$
where $X_2(t)$ is the Bohmian trajectory. According to BM, $\Psi$ never collapses. What collapses is $\psi_1$. Decoherence, on the other hand, is something that happens with $\Psi$. Actual outcomes, or lack of knowledge of the actual outcomes, is something related to $\psi_1$.

How is it related to density matrices? The reduced density matrix is obtained from $\Psi$ as
$$\rho^{\rm reduced}_1={\rm Tr}_2|\Psi\rangle \langle\Psi|$$
which does not refer to $\psi_1$ at all. The lack-of-knowledge-about-the-outcome density matrix $\rho^{\rm knowledge}_1$, on the other hand, is related to the lack of knowledge about $\psi_1$. So in BM, $\rho^{\rm reduced}_1$ and $\rho^{\rm knowledge}_1$ are conceptually different.

Yes, they are conceptually different, but do they have the same form? Non-selective measurement is quite commonly said to be equivalent to decoherence (eg. the references tha @thephystudent gave). Can BM help see the conditions under which it might be correct to be confused?

 

[Demystifier]

Non-selective measurement is quite commonly said to be equivalent to decoherence (eg. the references tha @thephystudent gave). Can BM help see the conditions under which it might be correct to be confused?

I think they are not equivalent, and I guess BM helps to see why.

 

[atyy]

I think they are not equivalent, and I guess BM helps to see why.

But they do have the same form, or they don't have the same form?

In QM, the proper reduced density matrix has the same form as the improper reduced density matrix. Do the BM quantities correspond to each of these?

 

[thephystudent]

I think it's OK as a practical method that, however, does not help to solve the measurement problem. I'm not sure if it answers your question.

I can agree with you that this does also not give a very simple answer to the measurement problem (have we actually properly defined the problem anyway?), but it does seem to reflect some close connection between decoherence and measurement to me.

 

[stevendaryl]

If you are talking about eigenstates of the measured microscopic system, then you are right. But if you are talking about states of the macroscopic pointer (measuring apparatus), they are always well localized in configuration space. In my view, BM is not so much about positions of electrons and photons, as it is about positions of macroscopic pointers (see the paper linked in my signature).

But don't all objects, large or small, have definite locations in BM?

 

[Demystifier]

But they do have the same form, or they don't have the same form?

In QM, the proper reduced density matrix has the same form as the improper reduced density matrix. Do the BM quantities correspond to each of these?

In that sense, they are the same.

 

[Demystifier]

But don't all objects, large or small, have definite locations in BM?

Some obviously don't. For example, a phonon (the quantum of sound).

 

[martinbn]

Some obviously don't. For example, a phonon (the quantum of sound).

But the phonon is not an object. He asked if all objects have locations in BM.

 

[Demystifier]

But the phonon is not an object.

Then what is it? A subject? A verb?
Now seriously, define "object"!
Or more to the point, in what sense is a photon an object and a phonon isn't?

 

[stevendaryl]

Then what is it? A subject? A verb?
Now seriously, define "object"!
Or more to the point, in what sense is a photon an object and a phonon isn't?

Well, I don't actually know about BM applied to anything other than nonrelativistic phenomena. In nonrelativistic QM, the wave function is assumed to be a function of $3N$ dimensional configuration space, where $N$ is the number of particles. So what counts as a particle is sort of baked-in. I assumed that the same is true of Bohmian mechanics.

 

[atyy]

In that sense, they are the same.

So BM justifies the confusion, since in BM both types of reduced density matrices are concurently valid, and by calculating one of them the other is also automatically calculated. In comparison, the confusion is not so easy to justify in Copenhagen, since both types of reduced density matrices are not concurrently valid. Of course as you say, BM also clarifies why they are different (as does loose reasoning around Copenhagen).

 

[Demystifier]

So what counts as a particle is sort of baked-in.

It is often said that photon is a particle while phonon is a quasiparticle. But it is more an expression of theoretical prejudices then an expression of experimental facts. At the level of effective theories, there is no any substantial difference between particles and quasiparticles. When someone says that photon is a particle (rather than a quasiparticle) it is not much more than expression of a naive belief that the current theory of photons is the fundamental theory, and not merely an effective theory that will one day be superseded by a more fundamental one.

 

[stevendaryl]

It is often said that photon is a particle while phonon is a quasiparticle. But it is more an expression of theoretical prejudices then an expression of experimental facts. At the level of effective theories, there is no any substantial difference between particles and quasiparticles. When someone says that photon is a particle (rather than a quasiparticle) it is not much more than expression of a naive belief that the current theory of photons is the fundamental theory, and not merely an effective theory that will one day be superseded by a more fundamental one.

I'm not asking about the philosophical distinction between particle and quasiparticle. In the mathematical formalism of whatever theory, there are some basic entities. In nonrelativistic quantum mechanics, there are point-masses, for example. Clearly, what's modeled as a fundamental particle in one theory might be modeled as a composite particle or a quasiparticle in another theory. But I'm asking about BM.

 

[martinbn]

Then what is it? A subject? A verb?
Now seriously, define "object"!
Or more to the point, in what sense is a photon an object and a phonon isn't?

The definition of a phonon says "A phonon is the quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency." How is that an object?!

 

[andrew s 1905]

The definition of a phonon says "A phonon is the quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency." How is that an object?!

Well I scattered a laser beam off them while doing Raman spectroscopy for my doctorate.
Regards Andrew

 

[Nugatory]

The definition of a phonon says "A phonon is the quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency." How is that an object?!

Is there a reasonably rigorous definition of "photon" that does not invite the same question? I'm asking here, not arguing.

 

[Demystifier]

The definition of a phonon says "A phonon is the quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency." How is that an object?!

You didn't answer my question(s). And by the way, the definition is wrong. A superposition of phonons of different frequencies is a phonon too.

 

[martinbn]

Is there a reasonably rigorous definition of "photon" that does not invite the same question? I'm asking here, not arguing.

It may be that the same applies for the photon. I didn't bring that up, it was Demystifier. I don't know he did bring it up, I thought that Bohmian mechanics cannot deal with photons, after all they are relativistic. But for me there is at least one crucial difference. The photon can exist on its own, it can propagate in vacuum. The phonon on the other hand cannot, if you remove the lattice of atoms/molecules there are no phonons. Also I thought that elementary particles correspond to irreducible representations of so and so group, and there aren't any for phonons.

 

[martinbn]

You didn't answer my question(s).

Giving the definition does answer your questions. You insists that the phonon is an object, you need to justify it.

And by the way, the definition is wrong. A superposition of phonons of different frequencies is a phonon too.

Definitions cannot be wrong. They can be inconstant or useless, but not wrong. In any case in your definition you allow that a superposition of phonons is also a phonon. Doesn't that make it even less of an object?

 

[stevendaryl]

It may be that the same applies for the photon. I didn't bring that up, it was Demystifier. I don't know he did bring it up, I thought that Bohmian mechanics cannot deal with photons, after all they are relativistic. But for me there is at least one crucial difference. The photon can exist on its own, it can propagate in vacuum. The phonon on the other hand cannot, if you remove the lattice of atoms/molecules there are no phonons. Also I thought that elementary particles correspond to irreducible representations of so and so group, and there aren't any for phonons.

Feynman developed a reformulation of electromagnetism that eliminated the electromagnetic field as an independent variable. In his reformulation, the electromagnetic field was completely determined by the motions of charged particles. I don't know whether a Bohmian version of that theory might be possible.

 

[Demystifier]

In any case in your definition you allow that a superposition of phonons is also a phonon. Doesn't that make it even less of an object?

1) Is wave function an object? Is vector an object?
2) Is photon an object?
3) Is Schrodinger cat an object?

 

[stevendaryl]

1) Is wave function an object? Is vector an object?
2) Is photon an object?
3) Is Schrodinger cat an object?

I think that the issue is more well-constrained than that. In Bohmian mechanics (the nonrelativistic version), there is an associated wave function, $\psi(x_1, y_1, z_1, x_2, y_2, z_2, ..., t)$. The triples $(x_j, y_j, z_j)$ are interpreted to be the location of particle number $j$. So it seems that you have to figure out what a "particle" is in order to apply Bohmian mechanics. Quantum mechanics in its Dirac formalism seems a little more general, in that you can stipulate arbitrary states of the system; states don't have to be associated with particle positions.

 

[martinbn]

1) Is wave function an object? Is vector an object?
2) Is photon an object?
3) Is Schrodinger cat an object?

1) No, no. These are not physical objects.
2), 3) Yes, yes.

You keep asking questions, but you give no answers nor explanations.
1) What is an object?
2) Why is the phonon an object?

 

[PeterDonis]

define "object"!

You gave a phonon as an example of an object that doesn't have a definite position in BM. You must have had some definition of "object" in mind in order to make that claim. What definition was it?

 

[Demystifier]

You gave a phonon as an example of an object that doesn't have a definite position in BM. You must have had some definition of "object" in mind in order to make that claim. What definition was it?

For our purposes, an object is anything that, in principle, can cause a click in a detector.