I How to derive Born's rule for arbitrary observables from Bohmian mechanics?

[Demystifier]

Wigner's analysis indicates to me that this is impossible.

I don't understand that claim, can you explain how Wigner indicates that it is impossible?

 

[A. Neumaier]

I don't understand that claim, can you explain how Wigner indicates that it is impossible?

I had discussed this in post #42.

 

[Demystifier]

I had discussed this in post #42.

It's basically the objection that non-demolition is not a reasonable assumption. But I don't see how is that related to the assumption that detector wave functions are approximately separated in the position space.

 

[A. Neumaier]

It's basically the objection that non-demolition is not a reasonable assumption. But I don't see how is that related to the assumption that detector wave functions are approximately separated in the position space.

No. Wigner's statement (quoted at the end of my post #42) essentially says that nondemolition is a necessary condition to get the wanted decomposition. Of course you assume only that the decomposition is approximate, so the argument by Wigner is not watertight in your case.

But your argument is completely absent - you just write some formulas and then jump without further justification to the desired conclusion [namely to (1) in post #98]!

 

[Demystifier]

But your argument is completely absent - you just write some formulas and then jump without further justification to the desired conclusion [namely to (1) in post #98]!

I totally disagree, but if you think so I don't know what argument to offer without repeating myself.

 

[A. Neumaier]

In (1) [equation label added by me] you assume without justification that the $\Psi_k(\vec{x},\vec{y})$ with different $k$ have approximately disjoint support. This is unwarranted without a convincing analysis.

I totally disagree, but if you think so I don't know what argument to offer without repeating myself.

You could explain in more detail why (1) follows from your argument for the second equality.
The only way this can be concluded seems to me by assume that the $\Psi_k(\vec{x},\vec{y})$ with different $k$ have approximately disjoint support. But you didn't give an argument for the latter, you just said that we must assume it:

Instead of my Eq. (3), more generally we have
$$|k\rangle|A_0\rangle \rightarrow \sum_q a_q |q\rangle |A_{kq}\rangle~~~~ (3')$$
[...]

In the multi-position representation we have
$$\Psi(\vec{x},\vec{y})=\sum_k c_k\Psi_k(\vec{x},\vec{y})$$
where
$$\Psi_k(\vec{x},\vec{y})=\sum_q a_q \psi_q(\vec{y}) A_{kq}(\vec{x})$$
Using the Born rule in the multi-position space we have
$$\rho(\vec{x},\vec{y}) =|\Psi(\vec{x},\vec{y})|^2
\simeq \sum_k|c_k|^2 |\Psi_k(\vec{x},\vec{y})|^2$$
In the second equality we have assumed that $A_{kq}(\vec{x})$ are macro distinct for different $k$, which we must assume if we want to have a system that can be interpreted as a measurement of $k$.

You actually assume that $A_{kq}(\vec{x})$ and $A_{k'q'}(\vec{x})$ have essentially disjoint support whenever $k\ne k'$. How does this follow from the definition of the $A_{kq}$ in (3') above?

 

[Demystifier]

But you didn't give an argument for the latter, you just said that we must assume it:

I said that we must assume that if we want to have a system that can be interpreted as a measurement of $k$. Sure, there are many interactions for which this condition is not satisfied. But such interactions, whatever they may be useful for, are not useful for measurement of $k$. Hence I don't consider such interactions that are not useful for measurement of $k$. I only consider those that are useful.

You actually assume that $A_{kq}(\vec{x})$ and $A_{k'q'}(\vec{x})$ have essentially disjoint support whenever $k\ne k'$. How does this follow from the definition of the $A_{kq}$ in (3') above?

It doesn't follow from that. Instead, it follows from the assumption that an interaction that can be used for measurement of $k$ exists. It is an existence assumption. And it seems to me (correct me if I'm wrong) that you think that such an interaction doesn't even exist.

 

[A. Neumaier]

I said that we must assume that if we want to have a system that can be interpreted as a measurement of $K$. Sure, there are many interactions for which this condition is not satisfied. But such interactions, whatever they may be useful for, are not useful for measurement of $K$. Hence I don't consider such interactions that are not useful for measurement of $K$. I only consider those that are useful.It doesn't follow from that. Instead, it follows from the assumption that an interaction that can be used for measurement of $K$ exists. It is an existence assumption. And it seems to me (correct me if I'm wrong) that you think that such an interaction doesn't even exist.

This amounts to saying that, by definition, a process deserving the label measurement of a selfadjoint operator $K$ with eigenstates $|k\rangle$ is one where
$$|k\rangle|A_0\rangle \rightarrow \Psi_k $$
and the $\Psi_k(\vec{x},\vec{y})$ have essentially disjoint support. With this definition, of course everything is trivial.

This is what I mean that you essentially assume what there is to prove. With this definition, the nontrivial statement would be to show that there is a class of physically meaningful not [missing not added later] nondemolition processes where this property actually holds. What I think Wigner's argument amounts to is that this is impossible.

 

[Demystifier]

With this definition, the nontrivial statement would be to show that there is a class of physically meaningful nondemolition processes where this property actually holds. What I think Wigner's argument amounts to is that this is impossible.

I still don't understand why do you talk about non-demolition. My equation that you called (3') in #106 is not non-demolition. Consequently, my last equation in #106 is also not non-demolition. It would be non-demolition if $a_q$ whre nonzero only for $q=k$, but I do not assume that.

 

[vanhees71]

I think the trouble again is that you don't discuss some specific (idealized toy model of a) measurement of momentum (I guess that's what you mean by $k$). One way to measure the momentum of a charged particle is to use a cloud chamber and a magnetic field and then measuring the curvature of the "trajectory" (i.e., the track of the particle as indicated by the droplets formed). Then all you need is a slightly modified calculation as in the famous Mott paper adding the magnetic field to the Hamiltonian.

 

[A. Neumaier]

I still don't understand why do you talk about non-demolition. My equation that you called (3') in #106 is not non-demolition. Consequently, my last equation in #106 is also not non-demolition. It would be non-demolition if $a_q$ whre nonzero only for $q=k$, but I do not assume that.

Yes, but your demonstration of (1) consists in saying that in order to qualify for a measurement the right hand side of (3') must satisfy the postulate I stated in post #108 (which I abstracted from your treatise by dropping irrelevant calculations). Since this now amounts to a definition of what a measurement process is, the question is which real experiments fall under this definition.

Nondemolition measurements are well-known to satisfy the condition of the definition in post #108. But nondemolition experiments are only a tiny part of the collection of measurements for which the Born rule is claimed. Hence to have a credible derivation of Born's rule you need to show that at least one other physically interesting class of (therefore not nondemolition) measurements is also eligible in this definition. Wigner gives arguments that suggest to me that the latter seems impossible to do. If that were the case your definition would rule out all usual (not nondemolition) experiments from being measurements. Thus there is something nontrivial left to be shown.

 

[A. Neumaier]

I think the trouble again is that you don't discuss some specific (idealized toy model of a) measurement of momentum (I guess that's what you mean by $k$). One way to measure the momentum of a charged particle is to use a cloud chamber and a magnetic field and then measuring the curvature of the "trajectory" (i.e., the track of the particle as indicated by the droplets formed). Then all you need is a slightly modified calculation as in the famous Mott paper adding the magnetic field to the Hamiltonian.

In the paper, $k$ is a label for an eigenstate of the arbitrary Hermitian operator $K$.

A cloud chamber is a nondemolition experiment since the particle continues (after having ionized some atoms leading to the droplets) with essentially the same momentum. For the class of nondemolition measurements the derivation given is ok. But the discussion is about whether Born's rule can also be derived in Bohmian mechanics for measurements not falling into this special class.

 

[Demystifier]

Nondemolition measurements are well-known to satisfy the condition of the definition in post #108.

No, your definition of non-demolition measurements is wrong. Where did you get this definition from?

A nondemolition measurement is a transition of the form
$$|k\rangle |A_0\rangle \rightarrow |k\rangle |A_k\rangle \;\;\; (1)$$
But a transition
$$|k\rangle |A_0\rangle \rightarrow |\Psi_k\rangle \;\;\; (2)$$
is, in general, not a nondemolition meaurement. Instead it is the most general transition possible, where the label $k$ on the right-hand side only means that the final state depends on the initial measured state $|k\rangle$. It is impossible to have a unitary transition that does not have the form (2).

Besides, there is a fine difference between nondemolition measurements and projective measurements. For projective measurements the right-hand side of (1) has the form $|k\rangle |A_k\rangle$ immediately after the measurement at time $t_m$, while for nondemolition measurements it has this form during a long time after the measurement. For projective measurements the right-hand side is really $|\psi_k(t)\rangle |A_k(t)\rangle$ with $|\psi_k(t_m)\rangle=|k\rangle$. Nondemolition measurements are rare, but projective measurements are not. But this distinction is not so important here.

 

[A. Neumaier]

A nondemolition measurement is a transition of the form
$$|k\rangle |A_0\rangle \rightarrow |k\rangle |A_k\rangle \;\;\; (1)$$
But a transition
$$|k\rangle |A_0\rangle \rightarrow |\Psi_k\rangle \;\;\; (2)$$
is, in general, not a nondemolition meaurement. Instead it is the most general transition possible,

I didn't claim anything else than what you just wrote. But the question is whether a physically meaningful class of non-nondemolition measurement in this agreed sense actually satisfies the criterion in post #108 (which I abstracted from your treatise by dropping irrelevant calculations). You just assume that any reasonable measurement satisfies this criterion, while my reading of Wigner suggests that only the very restricted class of nondemolition measurements can satisfy it.

 

[Demystifier]

while my reading of Wigner suggests that only the very restricted class of nondemolition measurements can satisfy it.

Well, my problem is that I still don't have a clue why do you read Wigner that way. It certainly isn't what Wigner explicitly said, is it?

 

[Demystifier]

Anyway, my reading of Wigner is different. He objects that a realistic measurement takes a finite time, so it's not clear to what value of the measured observable the result of measurement refers, unless the measured observable is a conserved quantity. That indeed is a valid objection, but can be easily resolved. Let the duration of the process of measurement be $\tau_m$ and let $\tau_i$ be the characteristic intrinsic time during which the measured observable significantly changes. Then if the process of measurement is sufficiently fast so that $\tau_m \ll \tau_i$, the objection by Wigner is resolved for all practical purposes. What was not known in the Wigner's time is that duration time of measurement $\tau_m$ is indeed typically very short, due to the fast decoherence caused by a large number of the apparatus degrees of freedom. @A. Neumaier I hope it makes sense to you.

 

[A. Neumaier]

Well, my problem is that I still don't have a clue why do you read Wigner that way. It certainly isn't what Wigner explicitly said, is it?

Independent of what Wigner says, the only class of processes where it is known that the postulate I stated in post #108 (which I abstracted from your treatise by dropping irrelevant calculations) is satisfied is the class of nondemolition measurements. You have not shown that there are other experiments that fit this definition but simply assumed that all meaningful measurement settings belong to this class.

On the other hand, Wigner said (p.298 in Wheeler/Zurek) that ''only quantities which commute with all additive conserved quantities are precisely measurable''. Here ''precisely measurable'' means that they satisfy the Born rule to in principle arbitrary accuracy when the detector is constructed appropriately (depending on the requested precision). This precludes the derivation of the Born rule to arbitrary accuracy for quantities that do not commute with all additive conserved quantities. But these are precisely the non-nondemolition measurements.

 

[Demystifier]

This precludes the derivation of the Born rule to arbitrary accuracy for quantities that do not commute with all additive conserved quantities.

Maybe you are right about that, but I don't think that we need a derivation of it to arbitrary accuracy. All we need is a derivation to an accuracy that matches the accuracy in actual experiments. Note that quantum theory is tested with a great accuracy only for some conserved quantities (most notably $g-2$ in QED), while for other quantities it is good but not so great. A typical match between ideal theoretical Born rule and actually measured frequencies looks something like this:

 

[A. Neumaier]

Maybe you are right about that, but I don't think that we need a derivation of it to arbitrary accuracy. All we need is a derivation to an accuracy that matches the accuracy in actual experiments.

Yes. But still it must be shown that this is the case for a nontrivial class of measurements of nonconserved observables. Simply claiming that it must be the case by definition is not enough.

 

[Demystifier]

Yes. But still it must be shown that this is the case for a nontrivial class of measurements of nonconserved observables. Simply claiming that it must be the case by definition is not enough.

Fair enough, but I think in #116 I gave some additional heuristic arguments. If I find a reference where a more serious quantitative analysis is done, I will let you know.

 

[A. Neumaier]

Fair enough, but I think in #116 I gave some additional heuristic arguments.

These only relate to the duration of measurements, not to properties of the resulting wave functions that would be needed to be established.

 

[Demystifier]

These only relate to the duration of measurements, not to properties of the resulting wave functions that would be needed to be established.

Fine, but how about the general argument based on decoherence, essentially saying that wave functions of many-body systems tend to decohere into branches localized in the position space because the interactions (that cause decoherence) are local in the position space? That heuristic argument (that can be supported by some explicit calculations in the literature) helps to explain why macroscopic objects look as being well localized in space. If you can accept that argument (which, admittedly, is still only heuristic), then the readings of measuring apparatuses are just a special case.

For more details with a quantitative analysis see e.g. https://journals.aps.org/prd/abstract/10.1103/PhysRevD.24.1516

 

[A. Neumaier]

Fine, but how about the general argument based on decoherence, essentially saying that wave functions of many-body systems tend to decohere into branches localized in the position space because the interactions (that cause decoherence) are local in the position space?

But in other cases, there is decoherence into coherent states, which are not local in the position space.

That heuristic argument (that can be supported by some explicit calculations in the literature) helps to explain why macroscopic objects look as being well localized in space. If you can accept that argument (which, admittedly, is still only heuristic), then the readings of measuring apparatuses are just a special case.

For more details with a quantitative analysis see e.g. https://journals.aps.org/prd/abstract/10.1103/PhysRevD.24.1516

The abstract of this paper says, ''Thus the environment can be said to perform a nondemolition measurement of an observable diagonal in the pointer basis'', confirming my reading of Wigner's analysis.

Assuming the results of this paper, the question remaining is whether for any selfadjoint operator $K$ of the measured system a suitable measuring apparatus and a suitable environment (suitable = not contrived) can be found such that

1. this operator is diagonal in the resulting pointer basis, and
2. each element of this pointer basis is well localized in space.

Maybe this (in my view nontrivial) question has a positive answer; then I am satisfied.

 

[Demystifier]

But in other cases, there is decoherence into coherent states, which are not local in the position space.

By "local" in position space, I mean small $\sigma_x$, not zero $\sigma_x$. In that sense coherent states can be local too.

Assuming the results of this paper, the question remaining is whether for any selfadjoint operator $K$ of the measured system a suitable measuring apparatus and a suitable environment (suitable = not contrived) can be found such that

1. this operator is diagonal in the resulting pointer basis, and
2. each element of this pointer basis is well localized in space.

Maybe this (in my view nontrivial) question has a positive answer; then I am satisfied.

What do you mean by "can be found"? Found in nature or found mathematically on the paper? If you mean found in nature, then it is certainly not true, which corresponds to the fact that not any self-adjoint operator can be measured in practice. But then I expect that a weaker claim is true, namely that ... for any selfadjoint operator $K$ that can be measured in practice ... operator is approximately diagonal in the resulting pointer basis ...

 

[vanhees71]

I'm a bit lost about the claim that only for "nondemolation measurements" the Born Rule should hold, but then it's trivial, because "nondemolation measurement" basically means that you measure and observable which is determined in the prepared state of the system. Then of course you get with 100% probability the value this observable takes when the system is prepared in this state, and there's no quibble to begin with.

Where a few people still have quibbles is about the question, how it happens that a measurement gives a well defined result when the observable measured is not determined due to the state preparation. My answer is simply that this is just part of the basic postulates of the theory: Given an ideal measurement, measuring an observable accurately, then you get some value in the spectrum of the observable operator with a probability given by Born's rule.

In other words: As far as I understand it, it's impossible to derive Born's rule from the other postulates and that one has to take it as one of the basic postulates of the theory, subject to be tested by experiment as any other part of the theory.

 

[Demystifier]

... basic postulates of the theory: Given an ideal measurement ...

What many people dislike about that is that the set of basic postulates involves a postulate on measurements. A fundamental microscopic theory should only have postulates on fundamental microscopic objects as such, not postulates on macroscopic measurements. Instead of being postulated, properties of macroscopic measurements should be derived from basic microscopic postulates.

Or as put very elegantly and concisely by Sabine Hossenfelder in the item 18.(c) of
https://backreaction.blogspot.com/2...ztZ3XIAz3XFQUBYQCO85sdGmZ0c8Pb1suXN168Ys65AX0"The measurement postulate is inconsistent with reductionism."

 

[A. Neumaier]

What do you mean by "can be found"? Found in nature or found mathematically on the paper? If you mean found in nature, then it is certainly not true, which corresponds to the fact that not any self-adjoint operator can be measured in practice.

My question in post #1 was about arbitrary $K$, and your answer in post #2 was that the proof can be found in many places. Indeed, on p.5 of your paper https://arxiv.org/pdf/1811.11643.pdf (reference 3 in post #2) you didn't have any restriction on the operator K beyond selfadjointness (which is assumed in Born's rule). So you need to qualify your claim there.

But then I expect that a weaker claim is true, namely that ... for any selfadjoint operator $K$ that can be measured in practice ... operator is approximately diagonal in the resulting pointer basis ...

Maybe. But to show this at least for some class of operators measured in practice that is not trivially nondemolition still requires a nontrivial argument.

 

[A. Neumaier]

I'm a bit lost about the claim that only for "nondemolation measurements" the Born Rule should hold, but then it's trivial, because "nondemolation measurement" basically means that you measure and observable which is determined in the prepared state of the system.

You should reread post #42 and Wigner's treatise in Section II.2 of the reprint collection by Wheeler and Zurek , ''Quantum theory of measurement'', which is the background of my discussion with Demystifier.

 

[Demystifier]

But to show this at least for some class of operators measured in practice that is not trivially nondemolition still requires a nontrivial argument.

Agreed!

 

[Demystifier]

My question in post #1 was about arbitrary KK, and your answer in post #2 was that the proof can be found in many places. Indeed, on p.5 of your paper https://arxiv.org/pdf/1811.11643.pdf (reference 3 in post #2) you didn't have any restriction on the operator K beyond selfadjointness (which is assumed in Born's rule). So you need to qualify your claim there.

I added a note in the post #2.

 

[Demystifier]

Lorentz covariance for instrumentalists

Here I would like to sketch the basic idea how BM, with Lorentz non-covariant law for particle trajectories, reproduces Lorentz covariance of measurable predictions. For simplicity I will write the equations for one spatial dimension, but the generalization to 3 dimensions will be obvious.

We start from nonrelativistic QM. In "Bohmian mechanics or instrumentalists" (the link in my signature below) it is explained how BM reproduces the Born rule for any quantum observable $\hat{K}$ in non-relativistic QM (see also post #70 for some generalizations). Here we explore this general result to understand the emergence of Lorentz covariance of the Born rule.

Let $\hat{x}$ and $\hat{p}$ be the position and momentum operator, respectively, and let $\hat{E}\equiv\sqrt{\hat{p}^2+m^2}$, where $m$ is a constant. (Later, in the relativistic context, we shall interpret $m$ as the mass in units $c=1$, but for now $m$ is just a constant without a specific physical interpretation. The square root involved in the definition of $\hat{E}$ is well defined in the basis of momentum eigenstates.) The measurement of time may be described by a clock operator $\hat{t}_{\rm clock}$ which, due to the Pauli theorem, does not obey the canonical commutation relation: $[\hat{H},\hat{t}_{\rm clock}]\neq i\hbar$. Now consider the following 4 observables
$$\hat{t}'_{\rm clock}=\gamma (\hat{t}_{\rm clock} - \beta \hat{x} )$$
$$\hat{x}'=\gamma (\hat{x} - \beta \hat{t}_{\rm clock} )$$
$$\hat{p}'=\gamma (\hat{p} - \beta \hat{E} )$$
$$\hat{E}'=\gamma (\hat{E} - \beta \hat{p} )$$
where $\gamma$ and $\beta$, for now, are just real numbers without a specific physical interpretation. Non-relativistic QM makes well-defined predictions for probabilities of different measurement outcomes of those 4 observables. And whatever those predictions are, the predictions by BM are the same. Note, however, that BM only describes trajectories of the form $X(t)$; it does not describe trajectories of the form $X'(t)$ or $X'(t')$. The position $x'$ makes sense only as a result of measurement of the observable $\hat{x}'$, there is no such thing as the "actual value" $X'$ existing independent of the measurement. Similarly, there is no such thing as the "actual value" $t'$ existing independent of the measurement. The 4 observables above are nothing but 4 examples of the abstract observable $\hat{K}$ studied in "Bohmian mechanics for instrumentalists".

In general, the wave function for a free particle in non-relativistic QM obeys a Schrodinger equation of the form
$$\hat{H}(\hat{p})\psi(x,t)=i\hbar\partial_t\psi(x,t)$$
Usually $\hat{H}(\hat{p})=\hat{p}^2/2m$, but in general $\hat{H}(\hat{p})$ can be arbitrary. So as a special case of non-relativistic QM consider
$$\hat{H}(\hat{p})=\hat{E}$$
where $\hat{E}$ is defined as above and $m$ is interpreted as mass in units $c=1$. One recognizes that this particular Hamiltonian of non-relativistic QM has a hidden Lorentz symmetry, the same symmetry that is typical for relativistic quantum theory. Furthermore, as a special case we consider
$$\beta^2< 1, \;\;\; \gamma=1/\sqrt{1-\beta^2} $$
so one recognizes that the 4 observables above are given by Lorentz transformations corresponding to the hidden Lorentz symmetry recognized above. So for such a theory (with some technicalities which I omit here because they are not so important for the main idea) we see that Bohmian mechanics makes Lorentz covariant measurable predictions, despite the fact that trajectories do not obey Lorentz covariant laws of motion.

 

[A. Neumaier]

Lorentz covariance for instrumentalists

$\hat{E}\equiv\sqrt{\hat{p}^2+m^2}$Usually $\hat{H}(\hat{p})=\hat{p}^2/2m$, but in general $\hat{H}(\hat{p})$ can be arbitrary. So as a special case of non-relativistic QM consider
$$\hat{H}(\hat{p})=\hat{E}$$
where $\hat{E}$ is defined as above and $m$ is interpreted as mass in units $c=1$. One recognizes that this particular Hamiltonian of non-relativistic QM has a hidden Lorentz symmetry, the same symmetry that is typical for relativistic quantum theory.

But the first nontrivial case is that of two particles. Simply substituting the single particle kinetic energies $\frac{p_k^2}{2m}$ by their relativistic versions $c\sqrt{p_k^2+(mc)^2}-mc^2$ does not produce something Lorentz invariant.

 

[Demystifier]

But the first nontrivial case is that of two particles.

This can be done too, it's relatively straightforward. But since I don't want to do all the work alone, here is a deal. You write down the theory within standard quantum theory (please, don't use the thermal interpretation, just the standard theory), and then I will explain how the same works in BM.

 

[A. Neumaier]

This can be done too, it's relatively straightforward. But since I don't want to do all the work alone, here is a deal. You write down the theory within standard quantum theory (please, don't use the thermal interpretation, just the standard theory), and then I will explain how the same works in BM.

Ok. Gven the nonrelativistic multiparticle Hamiltonian
$$H=\sum_k \frac{p_k^2}{2m} +\sum_{j<k} V(|q_j-q_k|)$$
where $V(r)$ is a Lennard-Jones potential, say, what would be the Lorentz covariant relativistic version?

 

[Demystifier]

Ok. Gven the nonrelativistic multiparticle Hamiltonian
$$H=\sum_k \frac{p_k^2}{2m} +\sum_{j<k} V(|q_j-q_k|)$$
where $V(r)$ is a Lennard-Jones potential, say, what would be the Lorentz covariant relativistic version?

That was not the deal. The deal was that you solve everything within standard quantum theory (including the relativistic covariant version of standard quantum theory; it's up to you whether you will use relativistic QM, relativistic QFT, or whatever you want) and make a relativistic covariant measurable prediction (e.g. some probability distribution of measurement outcomes). After you do all this (you can use existing results from the literature), I explain how the same measurable results can be obtained from the point of view of Bohmian mechanics.

If you complain that it's unfair because you must do the hard part while my part is easy, that's exactly my point. Bohmian mechanics is easy, once one understands how standard quantum theory works. Many people think that it is hard or even impossible to reproduce the predictions of standard quantum theory in the relativistic regime by using BM. My point is that it is in fact very easy, provided that one has done the hard part of doing it within standard quantum theory.

 

[A. Neumaier]

That was not the deal. The deal was that you solve everything within standard quantum theory (including the relativistic covariant version of standard quantum theory; it's up to you whether you will use relativistic QM, relativistic QFT, or whatever you want) and make a relativistic covariant measurable prediction (e.g. some probability distribution of measurement outcomes). After you do all this (you can use existing results from the literature), I explain how the same measurable results can be obtained from the point of view of Bohmian mechanics.

Well, then take the textbook description of QED in the book by Peskin and Schroeder, where everything needed to predict the anomalous magnetic moment of the electron is spelled out in Lorentz invariant terms. Your task is to explain how the anomalous magnetic moment of the electron can be obtained from the point of view of Bohmian mechanics.

 

[Demystifier]

Well, then take the textbook description of QED in the book by Peskin and Schroeder, where everything needed to predict the anomalous magnetic moment of the electron is spelled out in Lorentz invariant terms. Your task is to explain how the anomalous magnetic moment of the electron can be obtained from the point of view of Bohmian mechanics.

That's not really an interesting example (in the context of post #131) because $g-2$ is a scalar so it doesn't change under a change of a Lorentz frame. I think you didn't take this example because you think it would help you to understand how BM does the trick. I think you took this example because you don't need to do any work, while my job would be hard so you would set me up. In fact it wouldn't be that hard for me, but since I think you wouldn't learn anything form it (because it was not your intention when you gave me this task), I will not do it here.

But if you really want to understand something (rather than setting me up), you can pick up one segment of the theory (behind the calculation of $g-2$) which is Lorentz-covariant and looks problematic to you from the Bohmian point of view.

 

[A. Neumaier]

you can pick up one segment of the theory (behind the calculation of $g-2$) which is Lorentz-covariant and looks problematic to you from the Bohmian point of view.

Well, all Lorentz covariant QFT looks problematic to me from the Bohmian point of view, because nothing Bohmian survives renormalization.

 

[Elias1960]

Well, all Lorentz covariant QFT looks problematic to me from the Bohmian point of view, because nothing Bohmian survives renormalization.

I would use a lattice regularization and would not try to go to the limit of the lattice distance to zero. Leave it at Planck length. Renormalization between different lattices is unproblematic.

 

[A. Neumaier]

I would use a lattice regularization and would not try to go to the limit of the lattice distance to zero. Leave it at Planck length. Renormalization between different lattices is unproblematic.

Well, how to get the anomalous magnetic moment of the electron from a lattice calculation to the known accuracy? You cannot get even close with present lattice technology!

 

[Elias1960]

Well, how to get the anomalous magnetic moment of the electron from a lattice calculation to the known accuracy? You cannot get even close with present lattice technology!

I do not care about getting high accuracy first. Initially, I care about having a well-defined theory. Once one has a well-defined theory, one can start to improve the approximation methods. So, for theories with low interaction constants like 1/137 or so, it makes sense to look for approximation methods which make use of it, say, using some variant of a power series. Don't forget that this would be quite irrelevant for defining dBB trajectories - it is about methods to compute something well-defined in QT as well as dBB.

 

[Demystifier]

Well, all Lorentz covariant QFT looks problematic to me from the Bohmian point of view, because nothing Bohmian survives renormalization.

It's good to know what really bothers you, so that we don't need to discuss all other technicalities that are not directly related to renormalization.

You cannot get even close with present lattice technology!

Present is the key word. If we had much much stronger computers which can handle lattices with much much bigger number of vertices, then there are no many doubts that $g-2$ could be be computed on the lattice with a great accuracy.

 

[A. Neumaier]

I do not care about getting high accuracy first. Initially, I care about having a well-defined theory.

Standard renormalized QED at 6 loops is a perfectly well-defined covariant quantum field theory that gives excellent predictions. Its only defect is that it (extremely slightly) violates the axioms of Wightman. Since you discard wightman's axioms as well, you have no reasons left to consider QED ad ill-defined. Thus you should care about standard QED.

Once one has a well-defined theory, one can start to improve the approximation methods.

These are already well developed, to the point of giving results with 12cdecimals of relative accuracy!

So, for theories with low interaction constants like 1/137 or so, it makes sense to look for approximation methods which make use of it, say, using some variant of a power series. Don't forget that this would be quite irrelevant for defining dBB trajectories - it is about methods to compute something well-defined in QT as well as dBB.

There are lots of well-defined theories completely unrelated to experiment. They are completely irrelevant. To claim physical content for a theory you need to show that you can reproduce the experimental results!

Thus to make a Bohmian version of QED based on a lattice you need to spell out which precise lattice field theory (at which lattice spacing, with which interaction constants) you want to consider. For lack of computational evidence you would have to prove theoretically (not just say some handwaving words!) - which is probably impossible in the face of QED triviality and the Fermion doubling problem - that you can accurately approximate this lattice theory in some way that reproduces the standard low energy results of QED. Only then you have a substantiated claim.

 

[A. Neumaier]

Present is the key word. If we had much much stronger computers which can handle lattices with much much bigger number of vertices,

Until this is the case (most likely never, since the computers would need more memory than the size of the universe allows) you only have a dream full of wishful thinking.

then there are no many doubts that $g-2$ could be be computed on the lattice with a great accuracy.

According to the studies on triviality, there are even less doubts that $g-2$ would coms out to be zero to whatever great accuracy your imagined supersupercomputer will be able to muster.

 

[Demystifier]

According to the studies on triviality, there are even less doubts that $g-2$ would coms out to be zero to whatever great accuracy your imagined supersupercomputer will be able to muster.

We discussed that in another thread and didn't in fact agreed on this.

 

[A. Neumaier]

We discussed that in another thread and didn't in fact agreed on this.

Well, you didn't demonstrate the truth of your conjecture, it is just a belief. Beliefs don't count in physios, thus there is at present no Bohmian version of QED making contact with experiment, only a hope.

 

[Demystifier]

Well, you didn't demonstrate the truth of your conjecture, it is just a belief. Beliefs don't count in physios, thus there is at present no Bohmian version of QED making contact with experiment, only a hope.

Fine, but the problem is not in Bohmian mechanics itself. Instead, the problem is in the lattice formulation of QED, irrespective of the interpretation (Copenhagen, Bohmian, thermal, or whatever). The standard practice is to work with a non-lattice type of regularization, which gives numbers that agree with experiments, but has its own mathematical problems because such non-lattice regularizations are not mathematically rigorous.

 

[A. Neumaier]

Fine, but the problem is not in Bohmian mechanics itself. Instead, the problem is in the lattice formulation of QED, irrespective of the interpretation (Copenhagen, Bohmian, thermal, or whatever). The standard practice is to work with a non-lattice type of regularization, which gives numbers that agree with experiments, but has its own mathematical problems because such non-lattice regularizations are not mathematically rigorous.

Yes, and the reason is that QED is not defined on the lattice but on the continuum. It is to any fixed loop order Lorentz covariant and mathematically well-defined (in causal perturbation theory, which constructs everything, the S-matrix, the Hilbert space and the field operators). Already loop order 1 gives an excellent match with experiment, though for very high accuracy one needs orders up to six.
The level of rigor is the same as for lattice theories that break the covariance.

If you want to latticise QED for a subsequent Bohmian treatment you need to specify how to do it in a way that preserve its predictive properties at the same level of rigor. Simply speculating that it can be done is not enough.

That was not the deal. The deal was that you solve everything within standard quantum theory (including the relativistic covariant version of standard quantum theory; it's up to you whether you will use relativistic QM, relativistic QFT, or whatever you want) and make a relativistic covariant measurable prediction (e.g. some probability distribution of measurement outcomes). After you do all this (you can use existing results from the literature), I explain how the same measurable results can be obtained from the point of view of Bohmian mechanics.

If you complain that it's unfair because you must do the hard part while my part is easy, that's exactly my point.

All the hard work had already been done in 1948 and was rewarded in 1954 by a Nobel prize. The results of the hard work can be found in any textbook treating QED; many thousands of students learn it every year.
So there is no need for me to do any additional work.

I think you took this example because you don't need to do any work, while my job would be hard

I don't understand how you can call you job hard given that you said before that

Bohmian mechanics is easy, once one understands how standard quantum theory works.

How standard QED works is understood very well. If you don't like the anomalous magnetic moment, pick instead your preferred scattering amplitude.

 

[Elias1960]

Standard renormalized QED at 6 loops is a perfectly well-defined covariant quantum field theory that gives excellent predictions.

Is it a theory at all? It is nothing but an approximation for a particular experiment, namely scattering of particles which start and end with free particles far away.

Its only defect is that it (extremely slightly) violates the axioms of Wightman. Since you discard wightman's axioms as well, you have no reasons left to consider QED ad ill-defined. Thus you should care about standard QED.

No, it is not even a consistent theory. And I do not care about accuracy of an approximation of a not even well-defined theory, I care first about having a well-defined theory.

These are already well developed, to the point of giving results with 12cdecimals of relative accuracy!
There are lots of well-defined theories completely unrelated to experiment. They are completely irrelevant. To claim physical content for a theory you need to show that you can reproduce the experimental results!

Once I have a well-defined theory, which I have if I use a lattice regularization, then I can start about using your renormalized QED at 6 loops to compute approximations. So, no problem. Nobody forbids me to use such not-even-theories as approximations for particular situations like scattering.

Thus to make a Bohmian version of QED based on a lattice you need to spell out which precise lattice field theory (at which lattice spacing, with which interaction constants) you want to consider.

I can consider a particular lattice theory in general, using unspecified constants. Who was it who has referenced that paper where lattice computations have been used to compute the renormalization down to the place where the Landau pole should appear, but it did not appear on the lattice? So, to compute the renormalization is something possible and has been already done, and once in this case all lattice approximations are well-defined theories. All one has to do is to compute with this program the resulting large distance limit of the constants and to compare them with observation.

For lack of computational evidence you would have to prove theoretically (not just say some handwaving words!) - which is probably impossible in the face of QED triviality and the Fermion doubling problem - that you can accurately approximate this lattice theory in some way that reproduces the standard low energy results of QED. Only then you have a substantiated claim.

QED triviality is not a problem of lattice theory, it is a problem which appears only in the limit of the lattice distance going to zero. Which I propose explicitly not to do. To go with the lattice distance below Planck length simply makes no sense at all. Don't forget that a lattice theory remains well-defined if the interaction constant is greater than 1, while you will fail completely with your Feynman diagrams.

Then, fermion doubling is first of all a problem of getting the accuracy. It appears if you approximate the first derivatives in a node n with $\frac{f(n+1)-f(n-1)}{2h}$, but not if you use the less accurate $\frac{f(n+1)-f(n)}{h}$. Just to clarify that it is not unsolvable in principle. But, ok, even if we prefer higher accuracy, we can get rid of unnecessary doublers. In this case, we can use staggered fermions, which reduces the doublers to four. Then, to regularize the theory, we need discretization in space only, not in time. If one uses the original Dirac equation with the $\alpha_i,\beta$, this gives a staggered evolution equation on a 3D lattice. This reduces the doubling problem by another factor two, thus, gives two Dirac fermions. Completely sufficient for the SM, where fermions appear only in electroweak doublets.

For details, with the explicit 3D lattice, see arxiv:0908.0591

 

[A. Neumaier]

Is it a theory at all? It is nothing but an approximation for a particular experiment, namely scattering of particles which start and end with free particles far away.

No, it is not even a consistent theory. And I do not care about accuracy of an approximation of a not even well-defined theory, I care first about having a well-defined theory.

Of course QED at a fixed number of loops is a theory, an established part of theoretical physics. It is mathematically as well-defined and as consistent as lattice field theory, and gives far superior results.
Scharf's book on QED (did you ever try to read it?) defines everything (not only the S-matrix but the Hilbert space and the field operators) in completely rigorous terms.

Its only defect is that we know nothing rigorous about the limit when the number of loops grows indefinitely, but this is no worse than that we know nothing rigorous about lattice QFTs when the lattice spacing goes to zero.

I use a lattice regularization, then I can start about using your renormalized QED at 6 loops to compute approximations. So, no problem. Nobody forbids me to use such not-even-theories as approximations for particular situations like scattering.

The problem is that you need to show that renormalized QED at 6 loops is actually a valid approximation - which is dubious in the light of triviality results!

QED triviality is not a problem of lattice theory, it is a problem which appears only in the limit of the lattice distance going to zero.

QED triviality is a problem of relating the lattice QED to the continuum QED. Lacking this relation means lacking support for the claim that one approximates the other at the physical values of the parameters defining the specific theory.

For details, with the explicit 3D lattice, see arxiv:0908.0591

This says nothing about how well the successful continuum theory for the standard model approximates the proposed lattice theory, hence does not do what you want it to do.