An argument against Bohmian mechanics?

[Demystifier]

For example, you could say: Let $(P_{t_i})_{t_i}$ be a set of prejectors indexed by real values $t_i$. The evolution law of a state $[\Psi]$ is given by $[\Psi(t)]= [U(t,t_n)P_{t_n}U(t_n,t_{n-1})\cdots\Psi]$. Of course, it would be pointless to do this, but it is in principle possible.

It is not pointless at all. Indeed, it is the basis for the consistent histories interpretation, which some think of as Copenhagen interpretation done right. I don't like consistent histories interpretation for other reasons, but the formal part above is, in my opinion, a great progress over the traditional Copenhagen interpretation.

 

[A. Neumaier]

it is not clear at all what physical processes do and which physical processes don't count as measurements.

Is that any different in Bohmian mechanics?

 

[Demystifier]

Is that any different in Bohmian mechanics?

Yes. Those processes that lead to decoherence, i.e. wave-function splitting into macroscopic approximately non-overlapping branches, can be counted as "measurement". There are some ambiguities concerning the meaning of "macroscopic" and "approximately", but that's not a serious problem because the theory of Bohmian measurements is not a part of the axioms for Bohmian theory. When measurements are part of the axioms, like in some versions of Copenhagen, then measurement ambiguities are much more problematic.

 

[A. Neumaier]

the theory of Bohmian measurements is not a part of the axioms for Bohmian theory.

But it must be part of the Bohmian interpretation since otherwise the connection to quantum mechanics is lost. When making this connection you inherit all the ambiguities of the traditional interpretations! In the ensemble interpretation, one usually also subscribes to your statement that

Those processes that lead to decoherence, i.e. wave-function splitting into macroscopic approximately non-overlapping branches, can be counted as "measurement"

since decoherence is a consequence of the ensemble view.

 

[ShayanJ]

decoherence is a consequence of the ensemble view

That doesn't make sense to me. Can you elaborate?

 

[Strilanc]

By measuring the correlator, one usually means measuring $A$ and $B$ separately and then combining the measurement results. However, if
$$[A,B]\neq 0$$
then measurement of $AB$ is not equivalent to measurement of $A$ and $B$ and subsequent multiplication of the measurement results. In this sense, by measuring $A$ and $B$ separately, one cannot measure the correlator (1).

If you were given a "measure A" black box and a "measure B" black box, and only allowed to use one of them on any prepared state, you could estimate the commutator by searching for input vectors that produced noisier outputs in one observable than the other.

Actually, with just the ability to pick inputs and sample the measurement result, you can do full process tomography of a black-box-measurement. That gives (a good approximation of) the matrix representing A, and separately B, at which point you can in principle just directly compute their commutator with a computer.

(Note that the tomography and the computation could take a very long time. It's not tractable to do this to two unknown 50-qubit observables. Still, you might get some quick and dirty estimate of their commutator by sampling output entropies of random inputs.)

 

[A. Neumaier]

That doesn't make sense to me. Can you elaborate?

I don't understand.

The standard derivation of decoherence properties makes use of standard statistical mechanics arguments, which have always been coached in the ensemble language.

C.A. Fuchs and A. Peres,
Quantum theory needs no interpretation,
Physics Today 53 (2000), 70-71.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.77.8442&rep=rep1&type=pdf

J.P. Paz, S. Habib and W.H. Zurek,
Reduction of the wave packet: Preferred observable and decoherence
time scale,
Physical Review D 47 (1993), 488.

W.G. Unruh and W.H. Zurek,
Reduction of a wave packet in quantum Brownian motion,
Physical Review D 40 (1989), 1071.

 

[ShayanJ]

I don't understand.

Those papers are behind paywalls but from their abstracts I can see that they agree with what I know from decoherence. We can formulate decoherence for an individual system that is interacting with an environment. But ensemble view assumes that QM can only be applied to an ensemble of systems. So clearly decoherence doesn't follow from the ensemble view. Actually its part of the formalism of QM and not any particular interpretation.

 

[A. Neumaier]

We can formulate decoherence for an individual system [...] Actually its part of the formalism of QM and not any particular interpretation.

This is inconsistent since the formalism of QM doesn't tell without an interpretation whether or not it applies to an individual system.

Those papers are behind paywalls

The first paper is freely accessible. Note that Peres holds the ensemble view, neatly presented in his book.

We can formulate decoherence for an individual system

But the interpretation of the formulas assumes a notion of probability, which makes no sense for an individual system - unless you give up the objectivity of physics and allow subjective probabilities.

 

[rubi]

Then let me repeat once again. BM and standard QM are mathematically not equivalent. But it does not mean that they are not observationally equivalent. Moreover, there is a theorem which proves that they are observationally equivalent. In principle it is not impossible that there is a gap in the proof, but then any critique of their observational equivalence should concentrate on finding the gap within that proof itself. Nobody so far have found such a gap within the proof. About 99% of the existing critiques of the equivalence do not refer to that proof at all, which makes such critiques worthless.

Well, apparently, this theorem is not established rigorously, so finding a gap is not necessary to criticize it.

As Godel has shown in his second theorem, not even the standard axioms for arithmetic of integer numbers can prove their own consistency. It is therefore illusory to expect that axioms of any relevant physical theory (including Bohmian mechanics and standard QM) could prove it's own consistency. But even 99% mathematicians are not worried by the fact that they can't prove consistency of their favored axioms. Why should Bohmians be worried then? In practice, for 99% mathematics and for 99.99% physics, it is enough that the axioms look consistent intuitively. And so far, you haven't presented any heuristic argument (let alone proof) that some of the axioms for BM seem inconsistent.

This is something completely different. Consistency proofs in mathematics are of course always relative to some axiom system like ZFC or even some weak subsystem of second order arithmetic (which suffices for separable Hilbert spaces). If a proof of the consistency of BM relative to ZFC cannot be given, it would be a big problem. Copenhagen is of course consistent relative to ZFC. The incompleteness theorem isn't relevant here.

As I said, nobody so far found any evidence (let alone proof) for such inconsistency. Concerning the proof of consistency, can you prove that standard QM is consistent? If you think you can, then you probably contradict the second Godel theorem.

It is of course the task of the BM community to provide a proof for their claims. Mathematical theorems aren't true until they are disproved. It's the other way around. The consistency of Copenhagen (relative to ZFC) can of course be proved easily, since mathematically, it's just the application of the theory of operators in Hilbert spaces, which is of course derived from ZFC. There is no contradiction with Gödel's theorem.

It is not pointless at all. Indeed, it is the basis for the consistent histories interpretation, which some think of as Copenhagen interpretation done right. I don't like consistent histories interpretation for other reasons, but the formal part above is, in my opinion, a great progress over the traditional Copenhagen interpretation.

Yes, I like the consistent histories interpretation much more than Copenhagen and I agree that it's Copenhagen done right. However, mathematical precise language is not the only difference to Copenhagen.

 

[FeynmanFtw]

One additional comment. It is known that quantum field theory has mathematical contradictions due to the infinite number of degrees of freedom in the UV and IR limits. On the other hand, classical Newtonian mechanics does not have such contradictions. Should we then reject quantum field theory and use Newtonian mechanics instead?

I want to convey my own opinion (albeit based on rather limited knowledge of what current research has shown regarding BM or other interpretations), but I'll do that after Christmas :)

However, the above post caught my attention more than most. I assumed that QFT was one of our strongest quantum theories to date, and I cannot find anything about any UV and IR limit problems on the wiki article. What exactly are these contradictions?

 

[rubi]

One additional comment. It is known that quantum field theory has mathematical contradictions due to the infinite number of degrees of freedom in the UV and IR limits. On the other hand, classical Newtonian mechanics does not have such contradictions. Should we then reject quantum field theory and use Newtonian mechanics instead?

I didn't notice this comment earlier. It's not QFT itself, which has problems. Free QFT's are well defined in any dimension and there exist well-defined interacting models in lower dimensions as well. It's not the number of degrees of freedom that causes problems. QFT's have the same number of degrees of freedom as ordinary quantum mechanics, even if the UV and IR regulators are removed (both theories live in separable Hilbert spaces). It's true that known 4-dimensional interacting models are only well-defined if UV and IR regulators are in place, but that's different from being mathematically contradictory. We have a good understanding for why the perturbation expansions still yield physically correct predictions. Of course, the removal of the regulators is still desirable, but that's of course being researched.

 

[A. Neumaier]

I assumed that QFT was one of our strongest quantum theories to date, and I cannot find anything about any UV and IR limit problems on the wiki article. What exactly are these contradictions?

There are no contradiction in the common, renormalized, Poincare invariant setting actually used for predictions. Only the introductory bare treatment is contradictory, but no prediction depends on this purely motivational part. The bare UV divergences are cured by renormalization, while the IR divergences are cured by using appropriate coherent states to describe the scattering.

All this is well understood. There is, however, an unsolved problem - how to phrase QED in a nonperturbative way. The current, perturbative setting only gives asymptotic series for the predictions, though these give excellent approximations when truncated at order up to six.

QFT's have the same number of degrees of freedom as ordinary quantum mechanics

Not in the traditional counting, where each independent harmonic oscillator operator counts as one degree of freedom. With this counting, quantum mechanics is the quantum physics of finitely many d.o.f., and quantum field theory that of infinitely many.

 

[stevendaryl]

Not in the traditional counting, where each independent harmonic oscillator operator counts as one degree of freedom. With this counting, quantum mechnaics is the quantum physics of finitely many d.o.f., and quantum field theory that of infinitely many.

There is a subtle mathematical difference between the quantum mechanics of an unbounded number of degrees of freedom and an infinite number. The Hilbert space used in QFT describes a general state as a pure as a superposition of:

• A state with zero particles
• A state with one particle
• A state with two particles
• etc.

So it describes a situation that involves any finite number of particles. In perturbation theory, the states are described by starting with a vacuum state and applying a finite number of creation operators. But what it doesn't describe is a situation involving an infinite number of particles. Does infinitely many particles make sense? Sure. If the universe is infinite, then it makes sense to have infinitely many particles; there might be finitely many in any finite volume of space, but the volume of space is infinite. As I understand it, trying to describe the quantum mechanics of a truly infinite number of particles leads to a non-separable Hilbert space, and the mathematics is very different.

In practice, it doesn't make any difference, because we can artificially assume that there are finitely many particles at any given time, even if the universe is infinite.

 

[A. Neumaier]

describes a situation that involves any finite number of particles.

But arbitrarily many, which makes the number of degrees of freedom infinite. An $N$-particle system has $3N$ degrees of freedom. Already a free quantum field theory has an indeterminate number of particles. The number of degrees of freedom therefore exceeds $3N$ for any $N$. Thus there are infinitely many degrees of freedom.

Note that we cannot artificially limit the number of particles to be bounded by some large $N$. This destroys the locality of quantum field theory, which is responsible for the cluster decomposition property, without which it would be impossible to do physics. Also, coherent photon states (such as they are produced routinely by lasers) contain an arbitrarily large number of photons with a nonzero probability. These photons are needed on the formal level to give the coherent states the nice quasi-classical properties that are universally exploited when using them.

The main reason for the mathematical difficulties in quantum field theory is precisely that! Many of the techniques from functional analysis valid for quantum mechanics of finitely many degrees of freedom are no longer applicable to quantum field theory since the number of degrees of freedom is infinite.

Does infinitely many particles make sense?

Nothing I said has anything to do with whether the possibility of an infinite number of particles is or is not realized. It isn't, in standard quantum field theory. It might be in quantum gravity, but unlike QED and the standard model, this is largely unexplored territory.

 

[stevendaryl]

But arbitrarily many, which makes the number of degrees of freedom infinite. An $N$-particle system has $3N$ degrees of freedom. Already a free quantum field theory has an indeterminate number of particles. The number of degrees of freedom therefore exceeds $3N$ for any $N$. Thus there are infinitely many degrees of freedom.

Well, I'm uncertain about the terminology, but the distinction that I'm making is between:

1. An unbounded number of particles, and
2. an infinite number of particles.

The distinction is that in the case of an unbounded number of particles, if $|\psi\rangle$ is a state with an unbounded number of particles, then there is at least one nonnegative number $n$ and a state $|\psi_n\rangle$ with exactly $n$ particles such that $\langle \psi|\psi_n \rangle$ is nonzero.

With a truly infinite number of particles, there would be no overlap with any state with just finitely many particles.

 

[A. Neumaier]

Well, I'm uncertain about the terminology, but the distinction that I'm making is between:

1. An unbounded number of particles, and
2. an infinite number of particles.

This distinction is valid, but doesn't affect the fact that - in contrast to the claim of Rubi above - QFT has infinitely many degrees of freedom while QM has only finitely many. This differences is extremely important.

 

[rubi]

This distinction is valid, but doesn't affect the fact that - in contrast to the claim of Rubi above - QFT has infinitely many degrees of freedom while QM has only finitely many. This differences is extremely important.

Well, that's not what I said. I explained in my post, what I meant by d.o.f., i.e. a state in QFT contains as much information as a state in QM. Of course, in the case of QFT, you need to embed a field algebra into the algebra of operators on the separable Hilbert space, instead of a particle algebra. However, this is not what is problematic. You can easily write down a Fock space and field operators on it and you can also write down well-defined interacting Hamiltonians in that Hilbert space. However, you won't be able to write down an interacting Hamiltonian that satisfies the Poincare algebra relations, i.e. produces a Lorentz-invariant theory. Writing down interacting theories that satisfy the Wightman axioms is of course very difficult.

 

[A. Neumaier]

a state in QFT contains as much information as a state in QM.

But this is wrong. In QM, one cannot represent the information that a system is in a superposition of arbitrarily many particles!

 

[rubi]

But this is wrong. In QM, one cannot represent the information that a system is in a superposition of arbitrarily many particles!

That is not relevant to measuring how much information can be encoded into a state. Of course, different theories encode different information in the state, but the amount of information is exactly the same. I can store either Pulp Fiction or Full Metal Jacket on my hard drive, but if I choose to store Full Metal Jacket, the information about what a quarter pounder with cheese is called in France won't be represented.

 

[A. Neumaier]

That is not relevant to measuring how much information can be encoded into a state. Of course, different theories encode different information in the state, but the amount of information is exactly the same. I can store either Pulp Fiction or Full Metal Jacket on my hard drive, but if I choose to store Full Metal Jacket, the information about what a quarter pounder with cheese is called in France won't be represented.

But measuring information in this way is completely irrelevant. Every real number contains in your sense more information than all the physics books in the world. A classical field theory contains in your sense as much information as a quantum field theory.

In any case, it is highly misleading to equate this strange measure of information with degrees of freedom, as you did in your post #47 (without being precise enough there).

 

[rubi]

But measuring information in this way is completely irrelevant. Every real number contains in your sense more information than all the physics books in the world. A classical field theory contains in your sense as much information as a quantum field theory.

In any case, it is highly misleading to equate this strange measure of information with degrees of freedom, as you did in your post #47 (without being precise enough there).

I stated precisely what I meant by it in my post #47, so all of this is just a semantic discussion. Of course, the dimension of the Hilbert space is relevant to what quantum theories can be formulated on it. You cannot formulate QFT in $\mathbb C^2$, but you can (in principle) formulate it in the usual $L^2(\mathbb R^3)$ space of ordinary quantum mechanics. As I explained, the fact that you are working with a field algebra instead of a particle algebra is not a priori problematic.

 

[A. Neumaier]

You cannot formulate QFT in C2C2\mathbb C^2, but you can (in principle) formulate it in the usual $L^2(\mathbb R^3)$ space of ordinary quantum mechanics.

No, you cannot in any sensible way since there is no natural isomorphism between a Fock space for QFT and $L^2(\mathbb R^3)$. The structure imposed on the Hilbert space makes all the difference for its information content! Quantum mechanics does not happen in arbitrary Hilbert spaces but in Hilbert spaces with distinguished group actions!

 

[rubi]

At this point, it seems like you just don't want to admit that you didn't read my post carefully. As a mathematician, you certainly know that I can pull back any structure between isomorphic objects, so of course I can pull back the group action from a Fock space onto $L^2$. I don't need a natural isomorphism for that. (Moreover, it's exactly the other way around. The irreducible representations of the Poincare group that you obtain from Wigner's analysis are naturally defined on $L^2$ spaces. Fock spaces are built from these $L^2$ spaces and only inherit this group action.)

 

[vanhees71]

Well, I'm uncertain about the terminology, but the distinction that I'm making is between:

1. An unbounded number of particles, and
2. an infinite number of particles.

The distinction is that in the case of an unbounded number of particles, if $|\psi\rangle$ is a state with an unbounded number of particles, then there is at least one nonnegative number $n$ and a state $|\psi_n\rangle$ with exactly $n$ particles such that $\langle \psi|\psi_n \rangle$ is nonzero.

With a truly infinite number of particles, there would be no overlap with any state with just finitely many particles.

There's no state with infinitely many particles in the sense that the expectation value of the number of particles should be finite for the state to make sense. However, you are right in saying that there are states which have a non-zero overlap with any Fock state of a definite finite particle number. An example are the coherent states of the electromagnetic field, which are as close to classical em. waves as you can get. The photon-number probabilities are given by the Poisson statistics.

 

[A. Neumaier]

As a mathematician, you certainly know that I can pull back any structure between isomorphic objects

Yes, but $L^2(R^3)$ doesn't have this structure but must be equipped with it in order to have it. Quantum physics is always about Hilbert spaces already equipped with the right group action.

Ignoring that is like saying I can do functional analysis in the set $R_+$ of positive real numbers, since I can equip any set with continuum cardinality with the structure of $L^2(R^3)$. No mathematician in his right mind would talk like this.

 

[A. Neumaier]

There's no state with infinitely many particles in the sense that the expectation value of the number of particles should be finite for the state to make sense.

But a state must only be square integrable. Therefore there are valid states (such as $\sum_{N>0} N^{-1}|N\rangle$) where the expected number of particles is infinite. Nevertheless, upon each Born-style measurement, the number of particles would come out finite.

 

[vanhees71]

Hm, good point. This would be a state without a vacuum contribution. Interesting. I'm a bit in doubt, whether it can be prepared for any type of particles.

 

[A. Neumaier]

I'm a bit in doubt, whether it can be prepared for any type of particles.

It would take an infinite amount of energy to create it, I believe.

This would be a state without a vacuum contribution.

The state $\sum_{N\ge 0} (N+1)^{-1}|N\rangle$ has the same properties but with a vacuum contribution. Thus that's not the source of the difficulties to create such a state. You can add as much vacuum as you like without changing the infinite expected particle number.

 

[rubi]

Yes, but $L^2(R^3)$ doesn't have this structure but must be equipped with it in order to have it. Quantum physics is always about Hilbert spaces already equipped with the right group action.

Ignoring that is like saying I can do functional analysis in the set $R_+$ of positive real numbers, since I can equip any set with continuum cardinality with the structure of $L^2(R^3)$. No mathematician in his right mind would talk like this.

An object must always be equipped with a structure in order to have it. It's just easier to write it down in some cases. I'm of course not suggesting that one should do QFT on $L^2(\mathbb R^3)$, but that one could do it, i.e. a state vector in QM needs as much information to be fully specified as a state vector in QFT. You cannot do QFT in the Hilbert space of a spin-1/2 system. And you cannot do LQG in the Hilbert space of a QFT. The dimension of the Hilbert space is of course a measure for the information content of a quantum state.

Anyway, I have explained what I meant already in post #47 and what I said was correct, so this discussion is a bit pointless.

 

[A. Neumaier]

The dimension of the Hilbert space is of course a measure for the information content of a quantum state.

No. The conventional, and the only reasonable measure of information applicable a quantum state is the entropy. What you talk about does not deserve the name information in the traditional, scientifically established sense.

 

[vanhees71]

You cannot do QFT in the Hilbert space of a spin-1/2 system. And you cannot do LQG in the Hilbert space of a QFT. The dimension of the Hilbert space is of course a measure for the information content of a quantum state.

First I don't understand what you mean by the first quoted sentence, since a "Pauli particle" (spin-1/2 particle in non-relativistic QT) is described by the usual separable Hilbert space and usually reallized as $\mathrm{L}^2(\mathbb{R}^3,\mathbb{C}^2)$ (position-spin representation or "wave mechanics").

For curiosity, what's LQG?

I also don't know what I should make of the third sentence. The information content of a quantum state, or rather our ignorance about it, is given by the (von Neumann) entropy. For a pure state it's 0 (i.e., we have full possible knowledge about the system's preparation).

 

[rubi]

No. The conventional, and the only reasonable measure of information applicable a quantum state is the entropy. What you talk about does not deserve the name information in the traditional, scientifically established sense.

Then please just assign whatever name you like to it, so we don't need to have this discussion about pure semantics. The dimension of the Hilbert space is a classification criterion for quantum systems.

First I don't understand what you mean by the first quoted sentence, since a "Pauli particle" (spin-1/2 particle in non-relativistic QT) is described by the usual separable Hilbert space and usually reallized as $\mathrm{L}^2(\mathbb{R}^3,\mathbb{C}^2)$ (position-spin representation or "wave mechanics").

I said "spin-1/2 system" and not "(Pauli) particle". A spin-1/2 system is conventionally understood as $\mathcal H=\mathbb C^2$ with the standard spin-1/2 representation of $SU(2)$. People study such quantum systems without referencing the coordinates of particles. Especially in quantum information, people rarely use infinite-dimensional Hilbert spaces.

For curiosity, what's LQG?

Loop quantum gravity.

I also don't know what I should make of the third sentence. The information content of a quantum state, or rather our ignorance about it, is given by the (von Neumann) entropy. For a pure state it's 0 (i.e., we have full possible knowledge about the system's preparation).

See above.

 

[A. Neumaier]

The dimension of the Hilbert space is a classification criterion for quantum systems.

A very, very weak one, since the vast majority of quantum systems of experimental interest (except for finite spin systems) have a separable infinite-dimensional Hilbert space, all of which are isomorphic if you don't have additional structure.

 

[rubi]

A very, very weak one, since the vast majority of quantum systems of experimental interest (except for finite spin systems) have a separable infinite-dimensional Hilbert space, all of which are isomorphic if you don't have additional structure.

I agree, but it seemed to me like Demystifier wasn't aware of it, which is why I explained it in my post. At first, it seems unintuitive that a Hilbert space of separable dimension is enough for field variables, because one might think that a field can have values at uncountably many space-time points. A separable Hilbert space suffices, because only smeared fields (smeared with Schwartz functions) are considered. If one wanted to include fields operators defined at sharp space-time points, one would need a Hilbert space of uncountable dimension.