A Effective Dynamics of Open Quantum Systems: Stochastic vs Unitary Models

[stevendaryl]

Yes, that's the point. There is no collapse, and the wave function evolves unitarily. However, Bohmian mechanics reproduces the predictions of quantum mechanics without hidden variables and with collapse. So that is a way of deriving collapse from unitarity.

Somehow, I'm not being clear. I understand that that's the claim, but I'm questioning whether it's true. Or rather, I'm question how it's true--how Bohmian mechanics makes predictions equivalent to collapse.

I understand the basic Bohmian idea:

1. Assume that particle positions are initially distributed according to |\psi(x)|^2
2. Use the wave function \psi(x) to compute the "quantum potential" that influences particle motion.
3. Prove that particle motion, together with the quantum potential, insures that probability distribution remains |\psi(x)|^2

So without collapse, Bohmian mechanics has the same probabilistic predictions as the standard interpretation. But now, if we introduce collapse for the standard interpretation, but NOT for the Bohmian interpretation, then the two interpretations will be using different functions \psi(x). The Bohmian analysis will be using \psi_{uncollapsed} and the standard analysis will be using \psi_{collapsed}. So it's not immediately clear that the two interpretations give the same result: Bohmian analysis will be using a probability distribution |\psi_{uncollapsed}|^2, while the standard interpretation will be using a probability distribution |\psi_{collapsed}|^2. So they predict different probabilities for future position measurements (or at least, seem to). It would seem to me that for Bohmian mechanics to be equivalent to the standard interpretation, they would have to use \psi_{collapsed}, rather than \psi_{uncollapsed} in computing the quantum potential.

Now, I think that the answer might be something like entanglement. When you measure a particle, the particle becomes entangled with the measuring device. So the actual "quantum potential" that should be used afterward is not derived from the wave function of the particle alone, but from the wave function of the composite particle + measuring device. This more sophisticated analysis may reproduce the same predictions as if they used \psi_{collapsed}, but it certainly isn't at all obvious, and the equivalence (if they are equivalent) is not particular easy to see.

 

[A. Neumaier]

This leads to a derivation of the Born rule for position. Consequently, if we use a final readout that is position

Everything is obviously built in for position. But I was asking for the Born rule for operators with a discrete spectrum.

In particular, I was asking for the projection operators that project the wave function of the $i$th electron is a macroscopic array of independent, distinguishable electrons to its bound part. There is no macroscopic pointer for measuring this projection operator. Only one for measuring an unknown one of them. But the Born rule for each of it is needed in the conventional arguments used to predict the correct multiphoto counting statistics. It is also needed for applications to quantum computing - where one actually measures only once at the end, but invokes the generalized Born rule for designing the computing equipment.

 

[atyy]

Yeah, I'll have to hope Demystifier answers that.

 

[stevendaryl]

I am tired of explaining how Bohmian mechanics cannot violate the Born rule for this or that specific type of measurement.

I'm not sure what this quote is in response to, but the issue that I have with Bohmian mechanics is not really about Bohmian mechanics (except indirectly), but is about the equivalence of assuming and not assuming the collapse hypothesis.

We can ask the question purely in terms of the standard interpretation of QM, rather than the Bohmian interpretation. Ostensibly, the collapse hypothesis seems to have empirical content:

Initially, the wave function is \psi_{uncollapsed}(x). After a measurement of an observable O is found to give value \lambda, the collapse hypothesis claims that the wave function is now given by \psi_{collapsed}(x), which is obtained from \psi_{collapsed} by projecting onto the subspace of eigenstates of O with eigenvalue \lambda. In subsequent experiments, you use \psi_{collapsed}.

Now, surely there is a difference between using \psi_{collapsed} for prediction of the probabilities of results of future experiments and using \psi_{uncollapsed}. So the collapse hypothesis seems to have testable consequences. So shouldn't we be able to decide, once and for all, whether collapse happens, or not?

Well, I think the answer is not so simple, and the reason is that once you've performed a measurement, from that point on, the particle being measured has become entangled with the system doing the measurement. So a noncollapse interpretation (whether Bohmian, or Many-Worlds, or minimalist interpretation) has to consider the total system of particle + detector, rather than the particle alone. So the key theorem making collapse and non-collapse interpretations equivalent is that the effect of using the uncollapsed wave function for the total system is equivalent to using the collapsed wave function for the particle alone. My point is not that such a theorem is not particularly helped by the Bohmian interpretation, since it's a theorem about wave functions, not about particles.

 

[Demystifier]

But I was asking for the Born rule for operators with a discrete spectrum.

If you ask about operators such as spin, take my analysis in the paper linked in post #30, and in all equations replace $^*$ with $^\dagger$.

 

[Demystifier]

Now, surely there is a difference between using ψcollapsed\psi_{collapsed} for prediction of the probabilities of results of future experiments and using ψuncollapsed\psi_{uncollapsed}. So the collapse hypothesis seems to have testable consequences. So shouldn't we be able to decide, once and for all, whether collapse happens, or not?

Interpretations without collapse have some other assumptions which make them experimentally indistinguishable from interpretations with collapse.

 

[A. Neumaier]

Please, read the general theorem!

The theorem is often stated but I want to read its proof! Please point to a public source where the general theorem is really proved in full generality (rather than outlined only), and I'll read it.

One place where the general theorem is presented is my own paper
http://arxiv.org/abs/1112.2034
Secs. 2.1-2.2. The word "theorem" is not explicitly used, nevertheless the proof ends with the proof of the Born rule in Eq. (13).

You already start with assuming that the operator to be measured has a nondegenerate spectrum. This does not cover measuring a spin component of a spinning particle.

Moreover, after (11), you assume without argument that the state of the universe (the only existing state in Bohmian Mechanics) is effectively a tensor product of the state of system+apparatur and the remainder of the universe, which is an assumed collapse!

Thus your proof only works assuming the collapse and then only for observables with nondegenerate spectrum. But your claim is far more general!

I had read the article http://arxiv.org/pdf/1206.1084v2.pdf pointed to by atyy, which had claimed in its introduction

This chapter provides a comprehensive overview of the Bohmian formulation of quantum mechanics.

but failed to explain why this general theorem or even to give a formal reference to a proof. iIt is the introductory article to a book ''Applied Bohmian Mechanics'', so that I'd be allowed to take it as authoritative. But apparently not...

 

[stevendaryl]

Interpretations without collapse have some other assumptions which make them experimentally indistinguishable from interpretations with collapse.

Yes, I assumed that. But my point, which was addressed to atyy, originally, is that the argument that Bohmian mechanics is equivalent to standard quantum mechanics (with Von Neumann collapse) is not as simple for multiple consecutive measurements as it is for a single measurement. I'm not saying that it's not equivalent, but that the argument that it is equivalent is a lot more complicated than the argument for a single measurement.

 

[Demystifier]

You already start with assuming that the operator to be measured has a nondegenerate spectrum. This does not cover measuring a spin component of a spinning particle.

Yes, but it's trivial to make an appropriate generalization. Instead of eigenstates $|k\rangle$ with eigenvalues $k$ one can introduce eigenstates $|k,l\rangle$ with eigenvalues $k$, and in subsequent equations replace label $k$ with label $k,l$ when appropriate.

Moreover, after (11), you assume without argument that the state of the universe (the only existing state in Bohmian Mechanics) is effectively a tensor product of the state of system+apparatur and the remainder of the universe, which is an assumed collapse!

This is not an assumed collapse. I still have the superposition in (6), so I don't have collapse. The remainder of the universe that appears as a product is that part of the universe that did not play any role in the process of measurement. When you measure spin on Earth, the state of Andromeda galaxy does not have any (significant) influence on it, so the state of Andromeda appears as a product. This, of course, is an approximation, but a very good one.

 

[Demystifier]

Please point to a public source where the general theorem is really proved in full generality (rather than outlined only), and I'll read it.

Even rigorous mathematicians practice outline of certain generalizations of proofs when the generalization is trivial. (For instance, even Godel only outlined his second incompleteness theorem.) But of course, to see that something is trivial requires understanding. That's why textbooks have exercises, to test your understanding. So I challenge you to do the generalizations and nitpicking details by yourself.

 

[A. Neumaier]

Yes, but it's trivial to make an appropriate generalization. Instead of eigenstates |k⟩|k\rangle with eigenvalues kk one can introduce eigenstates |k,l⟩|k,l\rangle with eigenvalues kk, and in subsequent equations replace label kk with label k,lk,l when appropriate.

No, because the decomposition is no longer unique. Thus your subsequent argument depends on which decomposition you choose.

This is not an assumed collapse. I still have the superposition in (6), so I don't have collapse

No. Without collapse you have a similar decomposition as (6) but with the state $\Phi_k$ replaced by a superposition of detector and environment states. Ignoring the environment is collapse - indeed, it changes the dynamics since there is an associated decoherence effect!

When you measure spin on Earth, the state of Andromeda galaxy does not have any (significant) influence on it, so the state of Andromeda appears as a product.

I am not worried about Andromeda but about the immediate neighborhood of the detector - the air and the photons that interact with it. They are not far enough a way for your argument to count. (And if you'd prepare Bell states that extend to Andromeda even Andromeda would have to count as being close...)

Your alleged proof (as written) is therefore not more than handwaving.

the general theorem that measurement of any observable in non-relativistic Bohmian mechanics leads to probabilities given by the Born rule. The theorem is given in most books and reviews on Bohmian mechanics

Given this multiplicity and the importance of the result, someone should have stated and proved such a fundamental general theorem with impeccable arguments. Please point me to such a treatment.

 

[A. Neumaier]

even Godel only outlined his second incompleteness theorem

But 20 years after Goedel there were plenty of impeccable proofs.

Much more than 20 years have passed since Bell popularized Bohm's theory.

 

[Demystifier]

No, because the decomposition is no longer unique. Thus your subsequent argument depends on which decomposition you choose.

The final result does not depend on it. It is easy to show it, so I leave it as an exercise for you.

No. Without collapse you have a similar decomposition as (6) but with the state $\Phi_k$ replaced by a superposition of detector and environment states. Ignoring the environment is collapse - indeed, it changes the dynamics since there is an associated decoherence effect!

I am not worried about Andromeda but about the immediate neighborhood of the detector - the air and the photons that interact with it. They are not far enough a way for your argument to count. (And if you'd prepare Bell states that extend to Andromeda even Andromeda would have to count as being close...)

All the decoherence and neighborhood of the detector can be included into a redefinition of the concept of "apparatus". So its all there, just use the expression "apparatus+neigborhood" instead of "apparatus", if that will make you happy.

Your alleged proof (as written) is therefore not more than handwaving.

Your objections are trivial nitpicking.

Someone should have stated and proved such a fundamental general theorem with impeccable arguments. Please point me to such a treatment.

Physics is not mathematics. A mathematically more rigorous proof is presented in
http://arxiv.org/abs/quant-ph/0308039
but even there such trivial irrelevant omisions can be found by someone who cannot see the forest from the trees.

 

[Demystifier]

But 20 years after Goedel there were plenty of impeccable proofs.

Much more than 20 years have passed since Bell populaized Bohm's theory.

Fully rigorous proofs on Bohmian mechanics are missing only because there is no many mathematical physicists dealing with Bohmian mechanics. As with other rigorous proofs in physics, theoretical physicists (who are not mathematical physicists) consider it irrelevant nitpicking.

 

[Demystifier]

Your alleged proof (as written) is therefore not more than handwaving.

In mathematics, that would be a sufficient reason to dismiss the proof. But in physics, the point is whether you understand that handwaving? If you don't, I can't help you. If you do, as a person who likes rigorous proofs you can turn the handwaving into a rigorous proof by yourself.

 

[A. Neumaier]

just use the expression "apparatus+neigborhood" instead of "apparatus", if that will make you happy.

Now the problem is shifted to the missing interaction with the neighborhood of "apparatus+neigborhood". You are entering a Wigner's friend argument, hence you have he same problem as ordinary QM has with the cut in the Copenhagen interpretation.

 

[Demystifier]

Now the problem is shifted to the missing interaction with the neighborhood of "apparatus+neigborhood". You are entering a Wigner's friend argument, hence you have he same problem as ordinary QM has with the cut in the Copenhagen interpretation.

Again, you are missing the point. If you wish, you may call the whole universe "the apparatus" and say that $n$ is the number of particles in the whole universe. Nothing important in my analysis will be changed and no Wigner's friend problem will remain.

I have also said something more about the Wigner's problem in (the last paragraph of)
http://arxiv.org/abs/1406.3221

 

[atyy]

This more sophisticated analysis may reproduce the same predictions as if they used \psi_{collapsed}, but it certainly isn't at all obvious, and the equivalence (if they are equivalent) is not particular easy to see.

Yeah, it's not obvious to me either. I have simply worked through enough special cases to believe at the non-rigourous level that BM solves the measurement problem, the same way I believe in the Wilsonian effective field theory picture.

 

[A. Neumaier]

A mathematically more rigorous proof is presented in
http://arxiv.org/abs/quant-ph/0308039

I found there neither the general theorem (general enough to apply for measuring spin) nor the proof.

But I found unphysical arguments that affect whatever is done: In going from (5.12) to (5.14) it is claimed that if the support of the initial state is a union of two disjoint regions, this remains so in the future ''for a substantial amount of time''. But for laboratory distances, these times are typically extremely short, of the order of the time one of the light particles involved needs to cross the lab. Thus the effective wave functions (which they later simply call wave functions - see bottom of p.29) are not at all guaranteed to exist only for a substantial amount of time, although the authors claim on p.29 that ''the qualifications under which we have established (5.22) are so mild that in practice they exclude almost nothing''.

 

[A. Neumaier]

I have simply worked through enough special cases to believe at the non-rigourous level

Since Demystifier copped out, could you please point to a paper or bookchapter you worked through, where the case of measuring spin (certainly the most important special case) is treated? Or if this example wasn't needed to make you ''believe at the non-rigourous level that BM solves the measurement problem'', how did you convince yourself of the latter?

 

[A. Neumaier]

it's trivial to make an appropriate generalization. Instead of eigenstates |k⟩ with eigenvalues k one can introduce eigenstates |k,l⟩ with eigenvalues k, and in subsequent equations replace label k with label k,l when appropriate.

No, because the decomposition is no longer unique. Thus your subsequent argument depends on which decomposition you choose.

The final result does not depend on it. It is easy to show it, so I leave it as an exercise for you.

In the case of a spin, $l$ is the position of the measured particle, hence a continuous index. Therefore (3) cannot be valid after your suggested change. Otherwise it would be valid in the limit where the position parts of $k_1$ and $k_2$ tend to each other, which contradicts (4).

Thus the proposed exercise is ill-conceived.

 

[atyy]

Since Demystifier copped out, could you please point to a paper or bookchapter you worked through, where the case of measuring spin (certainly the most important special case) is treated? Or if this example wasn't needed to make you ''believe at the non-rigourous level that BM solves the measurement problem'', how did you convince yourself of the latter?

At the non-rigourous level, I was happy with a non-degenerate case, because I think I can always add a term that breaks the degeneracy by an arbitarily small amount so that it is mathematically non-degenerate, but the difference is physically undetecable.

 

[A. Neumaier]

At the non-rigourous level, I was happy with a non-degenerate case, because I think I can always add a term that breaks the degeneracy by an arbitarily small amount so that it is mathematically non-degenerate, but the difference is physically undetecable.

But Born's rule in its standard discrete form is robust only under deformations that preserve the discreteness of the spectrum. This cannot resolve the infinite degeneracy of a spin component operator in a single spinning particle.

How would you perturb $A=\sigma_3$ by a tiny amount to get an operator with a nondegenerate spectrum? I don't know of any reasonable perturbation that achieves this. For example, $A+\epsilon x_3$ is still doubly degenerate at each generalized eigenvalue.

Even if you find one, the perturbed version would have to be treated with Born's rule in its continuous form, which doesn't collapse the wave function to a nonexistent normalizable eigenstate.

Thus your perturbation argument is far from convincing!

 

[atyy]

But Born's rule in its standard discrete form is robust only under deformations that preserve the discreteness of the spectrum. This cannot resolve the infinite degeneracy of a spin component operator in a single spinning particle.

How would you perturb $A=\sigma_3$ by a tiny amount to get an operator with a nondegenerate spectrum? I don't know of any reasonable perturbation that achieves this. Even if you find one, the perturbed version would have to be treated with Born's rule in its continuous form, which doesn't collapse the wave function to a nonexistent normalizable eigenstate. Thus your perturbation argument is far from convincing!

You can use the Born rule in continuous form, and collapse it to a normalizable state.

I admit I cannot readily construct a suitable perturbation in all cases.

But just to make sure I understand you - your concern is that eg. with spin, the simplest Coulomb potential hydrogen atom treatment has degeneracy, in the sense that spin up and spin down wave functions can have the same energy?

 

[A. Neumaier]

You can use the Born rule in continuous form, and collapse it to a normalizable state.

How do you do that? The typical reasoning is that one cannot measure the continuous spectrum exactly but only approximately. Thus one splits the continuum into a discrete union of intervals, each representing an uncertain measurement, and applies the corresponding projectors to get the collapse. But with the finite resolution, each of these projectors has again an infinite degeneracy! Thus one cannot maintain both reduction to a normalizable state and nondegeneracy.

But just to make sure I understand you - your concern is that eg. with spin, the simplest Coulomb potential hydrogen atom treatment has degeneracy, in the sense that spin up and spin down wave functions can have the same energy?

Of course. Most discrete energy eigenstates of the nonrelativistic hydrogen electron are highly degenerate. This is the reason why one gets a fine splitting in the relativistic treatment, and (since some degeneracy still persists) a hyperfine splitting (= Lamb shift) in the QED treatment. The continuous spectrum of the hydrogen electron remains degenerate even in the QED version.

For a multiparticle system that (unlike the hydrogen atom) can dissociate into more than two pieces, part of the continuous spectrum is even infinitely degenerate!

 

[stevendaryl]

How do you do that? The typical reasoning is that one doesn't measure the continuous spectrum exactly but only approximately. Thus one splits the continuum into a discrete union of intervals, each representing an uncertain measurement, and apply the correponding projectors to get the collapse. But with the finite resolution, each of these projectors has again an infinite degeneracy! Thus you cannot maintain both reduction to a normalizable state and nondegeneracy.

That's an issue that is interesting in its own right, apart from discussions about various interpretations of quantum mechanics.

Let's take position as the most familiar example of a continuous observable. Any scheme for measuring position has a limitation in accuracy. So suppose you are using a procedure that only determines position to an accuracy of \pm \delta x. Then in a sense, you're not measuring position, but some related observable \hat{X} that returns a discrete set of possible results: n \delta x where n is an integer. What then, is the complete set of eigenstates of this operator? (Realistically, there is a distinction between returning a fuzzy result with accuracy \pm \delta x and returning the precise result that is n \delta x. But I'm going to use the latter, because it's easier to analyze mathematically.)

Well, the answer is that a complete set of eigenstates would be of the form \psi_{n,m} where:

• \psi_{n,m}(x) = 0 if x < n \delta x
• \psi_{n,m}(x) = sin(\frac{m \pi x}{\delta x}) if n \delta x < x < (n+1) \delta x
• \psi_{n,m}(x) = 0 if x > (n+1) \delta x

So the index m in \psi_{n,m} is this infinite degeneracy that you're talking about. However, when \delta x is very small, then the expectation value for energy increases rapidly with increasing m so for practical purposes, can't we assume that only the m=1 eigenstate is relevant?

 

[A. Neumaier]

Let's take position as the most familiar example of a continuous observable. Any scheme for measuring position has a limitation in accuracy. So suppose you are using a procedure that only determines position to an accuracy of \pm \delta x. Then [...] a complete set of eigenstates would be of the form \psi_{n,m} where:

• \psi_{n,m}(x) = 0 if x < n \delta x
• \psi_{n,m}(x) = sin(\frac{m \pi x}{\delta x}) if n \delta x < x < (n+1) \delta x
• \psi_{n,m}(x) = 0 if x > (n+1) \delta x

So the index m in \psi_{n,m} is this infinite degeneracy that you're talking about. However, when \delta x is very small, then the expectation value for energy increases rapidly with increasing m so for practical purposes, can't we assume that only the m=1 eigenstate is relevant?

No, because even the first eigenstate in each interval will already have a very high energy, so should be ignorable by the same argument. But one cannot ignore all basis states! This shows that the energy argument is faulty.

What counts is the energy of the superposition, not of a basis state itself. This energy should be small; an example is the state with $\psi(x)=x$ for $x\in[0,1]$ and zero otherwise. But its projection to any measurable interval inside $[0,1]$ has lots of non-negligible Fourier components with high energy. The very slow convergence of the Fourier series plays havoc with any calculations done subsequent to the approximation.

 

[atyy]

How do you do that? The typical reasoning is that one cannot measure the continuous spectrum exactly but only approximately. Thus one splits the continuum into a discrete union of intervals, each representing an uncertain measurement, and applies the corresponding projectors to get the collapse. But with the finite resolution, each of these projectors has again an infinite degeneracy! Thus one cannot maintain both reduction to a normalizable state and nondegeneracy.

One allows a continuous spectrum to be measured exactly. Then a collapse rule suitable for a continuous variable is given in Eq (3) and (4) of http://arxiv.org/abs/0706.3526.

Of course. Most discrete energy eigenstates of the nonrelativistic hydrogen electron are highly degenerate. This is the reason why one gets a fine splitting in the relativistic treatment, and (since some degeneracy still persists) a hyperfine splitting (= Lamb shift) in the QED treatment. The continuous spectrum of the hydrogen electron remains degenerate even in the QED version.

For a multiparticle system that (unlike the hydrogen atom) can dissociate into more than two pieces, part of the continuous spectrum is even infinitely degenerate!

I have never worked through this, but googling came up with an attempt in the spirit of my thinking: https://www.ma.utexas.edu/mp_arc/c/12/12-59.pdf

For an attempt to rigourously show that BM reproduces non-relativistic QM (I believe they treat degenerate eigenvalues also), try:
http://arxiv.org/abs/quant-ph/0308039
http://arxiv.org/abs/quant-ph/0308038

 

[Demystifier]

In the case of a spin, l is the position of the measured particle, hence a continuous index.

Someone much more clever than me had a perfect response to that:
"No, no, you're not thinking; you're just being logical."
Niels Bohr

Let me give you a hint (but again not all the details). I did not say that $|k,l\rangle$ are a complete basis. They are just states that in a given experimental setup can be distinguished.

Applying Born rule is like cooking. Either you understand the general principles or you ask for a precise recipe for each possible case.

 

[A. Neumaier]

They are just states that in a given experimental setup can be distinguished.

If it is only finitely many (as in any real experimental setup) it doesn't resolve the infinite degeneracy.