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

  • #121
Demystifier said:
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.
 
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  • #122
A. Neumaier said:
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
 
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  • #123
Demystifier said:
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.
Demystifier said:
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.
 
  • #124
A. Neumaier said:
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.

A. Neumaier said:
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 ...


 
  • #125
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.
 
  • #126
vanhees71 said:
... 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."
 
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  • #127
Demystifier said:
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.
Demystifier said:
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.
 
  • #128
vanhees71 said:
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.
 
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  • #129
A. Neumaier said:
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!
 
  • #130
A. Neumaier said:
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.
 
  • #131
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.
 
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  • #132
Demystifier said:
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.
 
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  • #133
A. Neumaier said:
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.
 
  • #134
Demystifier said:
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?
 
  • #135
A. Neumaier said:
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.
 
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  • #136
Demystifier said:
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.
 
  • #137
A. Neumaier said:
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.
 
  • #138
Demystifier said:
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.
 
  • #139
A. Neumaier said:
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.
 
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  • #140
Elias1960 said:
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!
 
  • #141
A. Neumaier said:
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.
 
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  • #142
A. Neumaier said:
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.

A. Neumaier said:
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.
 
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  • #143
Elias1960 said:
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.
Elias1960 said:
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!
Elias1960 said:
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.
 
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  • #144
Demystifier said:
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.
Demystifier said:
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.
 
  • #145
A. Neumaier said:
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.
 
  • #146
Demystifier said:
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.
 
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  • #147
A. Neumaier said:
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.
 
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  • #148
Demystifier said:
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.
Demystifier said:
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.
Demystifier said:
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
Demystifier said:
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.
 
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  • #149
A. Neumaier said:
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.
A. Neumaier said:
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.
A. Neumaier said:
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.
A. Neumaier said:
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.
A. Neumaier said:
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
 
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  • #150
Elias1960 said:
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.
Elias1960 said:
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!
Elias1960 said:
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.
Elias1960 said:
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.
 

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