Problems with Paper on QM Foundations

[vanhees71]

There are Hilbert spaces of holomorphic functions, but they have nothing to do with field theory.

You can take holomorphic functions as scalar fields (or their real and imaginary parts) in 2D Euclidean space. These are the "harmonic functions" with the properties described. It was only an example. Of course this mathematical phenomenon is more generally just Stokes's theorem for differential forms.

And that's wrong. You can start from a vector field $A^\mu$ too.

There will be some problems with the implementation of the Lorenz gauge, but similar problems appear in condensed matter theories too if you have a continuity equation: If, say, the fundamental theory has exact particle conservation, and you want to have a field theory based on the density $\rho$ together with the continuity equation, it is not easy to reach that $\hat{\rho} \ge 0$ and that the conservation law is not even violated even by vacuum oscillations. Quantum condensed matter theory nonetheless works nicely.

I'm talking about massless spin-1 fields in relativistic field theories. They are necessarily gauge fields, as can be derived from the representation theory of the Poincare group.

Please quote me in a meaningful way. I do not claim "a violation of gauge invariance" but usually write complete sentences. Like the following: Theories with vector fields $A^\mu$ with some approximate gauge invariance are possible. They may be non-renormalizable, but this does not make them invalid as effective field theories. But these theories will certainly not use the Gupta-Bleuler resp. BRST approach, but start with a definite Hilbert space.

You said repeatedly that local gauge symmetries can be approximate, but that's not true, because then they loose their physical meaning. This has nothing to do with (Dyson-) renormalizability or non-renormalizable effective theories.

And that's simply wrong. You can make physical sense of vector fields if you start with a definite Hilbert space. Of course, you need exact gauge invariance to be able to factorize, and without factorization you cannot make physical sense of the whole construction build on the indefinite Hilbert space. But you are not at all obliged to start with some indefinite Hilbert space.

No, massless vector fields must be necessarily quantized as gauge fields. For massive vector fields you are right.

The original approach to QED did not use an indefinite Hilbert space. The references to the original approach:
Dirac P.A.M. (1927). The Quantum Theory of the Emission and Absorption of Radiation. Proc Roy Soc A114, 243-265
Fermi, E. (1932). Quantum Theory of Radiation. Rev Mod Phys 4(1), 87-132
But I would nonetheless recommend instead
Akhiezer, A.I., Berestetskii , V.B. (1965). Quantum Electrodynamics.
which give also the formulas for the original approach.

Of course, in the early days they completely fixed the gauge before quantizing. That's another equivalent way to quantize the em. fieeld, which is even preferrable if you learn the subject for the first time. It's only disadvantage is that it is not manifestly covariant, which makes calculations of higher-order perturbative corrections a nightmare.

Nevertheless also there gauge invariance is needed to make physical sense of the theory. There's no way out: The math of the Poincare group tells you that a massless vector field must be a gauge field.

 

[Sunil]

I'm talking about massless spin-1 fields in relativistic field theories. They are necessarily gauge fields, as can be derived from the representation theory of the Poincare group.

Given that I do not claim that the result of the limit $m\to 0$ of massive vector fields will be something different in its observable predictions, I see no reason to care about such results. If that result tells us that both methods give, in the result, the same theory, fine. If you name them gauge fields or not does not matter.

No, massless vector fields must be necessarily quantized as gauge fields. For massive vector fields you are right.
...
You said repeatedly that local gauge symmetries can be approximate, but that's not true, because then they loose their physical meaning.

And here I simply disagree. AFAIU this is a misunderstanding based on the memories of time when people thought that non-renormalizable theories make no physical sense.

Take a naive discretization of a chiral lattice theory. It will not have exact gauge invariance on the lattice, but will be a well-defined theory. A finite number of degrees of freedom on a finite lattice (say on a large cube) and no infinities. Quantized in a straightforward canonical way. What will be the large distance limit?

Of course, in the early days they completely fixed the gauge before quantizing. That's another equivalent way to quantize the em. fieeld, which is even preferrable if you learn the subject for the first time. It's only disadvantage is that it is not manifestly covariant, which makes calculations of higher-order perturbative corrections a nightmare.

Ok, that's already much better.

I think that I have already said that I don't care about the mathematical tricks for approximate computations. You use "dimensional regularization" with physically completely meaningsless "dimensions" $4-\varepsilon$ in your renormalization? Fine, as long as you don't sell this as being something physically meaningful. You use an indefinite Hilbert space for computing your scattering coefficients? Fine, as long as you don't sell this as being something physically meaningful.

If this particular trick needs gauge invariance, it follows that you cannot apply this trick if there is no such exact gauge invariance, as for a naive chiral gauge field on the lattice. But this is a problem for computations, not for the meaningful definition of the theory itself.

There's no way out: The math of the Poincare group tells you that a massless vector field must be a gauge field.

And what makes the difference for me if I have a well defined theory with massive gauge fields with observable potentials and take the limit $m\to 0$? The result will be a gauge theory? Fine. What's the problem?
(BTW, the math of the Poincare group may be also irrelevant on the fundamental lattice level.)

 

[vanhees71]

Again. This has nothing to do with Dyson renormalizability or effective field theories with an infinite number of parameters ("low-energy coupling constants"). A gauge theory, for which the local gauge symmetry is broken has no physical interpretation, because it leads to acausality and a non-unitary S-matrix.

The limit $m \rightarrow 0$ for massive vector fields is non-trivial. You can describe the massive vector field as a Proca field. Then the limit is not well defined (as you can already see looking at the free propagator of this field, containing a piece $\propto p_{\mu} p_{\nu}/m^2$. Another possibility is to describe it as an Abelian gauge field, which is the Stueckelberg approach. Then the limit $m \rightarrow 0$ can be taken and leads to the usual gauge theory for a massless vector field. Again, whenever you deal with massless vector fields you end up with a gauge theory!

 

[Sunil]

You have not got the point that I'm not afraid of ending up with a gauge theory? My point is that this does not require that the $A^\mu$ should be handled as completely unphysical fields living in an indefinite Hilbert space of the Gupta-Bleuler resp. BRST approach.

One possibility would be the one you have accepted with your "in the early days they completely fixed the gauge before quantizing". Another one, canonical quantization on a lattice of a chiral gauge field which does not have exact lattice gauge symmetry on the lattice. It is nonetheless well-defined, unitary, and has some large distance continuous limit.

I'm quite happy if these examples all lead to the same gauge theory. In this case, the gauge potentials would be normal physical fields living in physical definite Hilbert space. Even if the operators measuring their values would not be observables because of gauge symmetry in the continuous limit, they would be handled with the same mathematics as usual observables. And there would be the BRST approach which would be preferable for computations because some manifestly covariant integrals are easier to compute.

 

[vanhees71]

You have not got the point that I'm not afraid of ending up with a gauge theory? My point is that this does not require that the $A^\mu$ should be handled as completely unphysical fields living in an indefinite Hilbert space of the Gupta-Bleuler resp. BRST approach.

I try one last time: There is no indefinite Hilbert space. The entire point of the covariant operator quantization of gauge theories (BRST) is that there is no such thing!

One possibility would be the one you have accepted with your "in the early days they completely fixed the gauge before quantizing". Another one, canonical quantization on a lattice of a chiral gauge field which does not have exact lattice gauge symmetry on the lattice. It is nonetheless well-defined, unitary, and has some large distance continuous limit.

If you quantize on a lattice it's by construction clear that there's nothing observable which is not gauge invariant.

I'm quite happy if these examples all lead to the same gauge theory. In this case, the gauge potentials would be normal physical fields living in physical definite Hilbert space. Even if the operators measuring their values would not be observables because of gauge symmetry in the continuous limit, they would be handled with the same mathematics as usual observables. And there would be the BRST approach which would be preferable for computations because some manifestly covariant integrals are easier to compute.

The values of the gauge fields cannot be observables, because the operators do not obey the microcausality condition.

 

[Sunil]

I try one last time: There is no indefinite Hilbert space. The entire point of the covariant operator quantization of gauge theories (BRST) is that there is no such thing!

Ok, then I give up.

"The scalar photons are treated by using an indefinite metric"
Gupta, S.N. (1950). Theory of longitudinal photons in quantum electrodynamics, Proc Phys Soc A 63(7), 681-691

"Gupta has introduced an alternative method of quantization for the Maxwell field which differs from the usual one in that the scalar part of the field is quantized by means of the indefinite metric of Dirac. It is shown that this method can be extended into a general and consistent theory, including the case of interaction with electrons."
Bleuler, K. (1950). Eine neue Methode zur Behandlung der longitudinalen und skalaren Photonen. Helvetica Physica Acta 23(V), 567-586

Of course, as I have written already many many times, at the end of the construction some definite Hilbert space is constructed. But this does not make the Hilbert space where the potential operators $A^\mu$ live definite.

If you quantize on a lattice it's by construction clear that there's nothing observable which is not gauge invariant.

No. There are gauge-invariant lattice theories, namely Wilson lattice gauge fields. But it works only for vector gauge fields. If you use a naive lattice approximation for chiral gauge fields then there will be no gauge invariance in the lattice theory. But this does not forbid you to define the lattice theory in a reasonable canonical way (without indefinite Hilbert spaces).

The values of the gauge fields cannot be observables, because the operators do not obey the microcausality condition.

In the Lorenz gauge the equations are Lorentz-covariant and classically Einstein-causality holds. So there is no base to assume that the corresponding quantum operators will not fulfill the microcausality condition.

Then, there is no base for assuming that a lattice regularization fulfills relativistic symmetry requirements. A lattice regularization is the most straightforward way to regularize quantum gravity at Planck scale and to obtain a regularized theory with a finite number of degrees of freedom which one can quantize canonically. To expect from such a lattice regularization relativistic symmetry would be strange.

 

[A. Neumaier]

In the Lorenz gauge the equations are Lorentz-covariant and classically Einstein-causality holds.

No. Without gauge fixing, the coupled Maxwell-Klein Gordon equations (the nearest classically to QED) are not hyperbolic, hence do not satisfy Einstein-causality.

 

[Sunil]

No. Without gauge fixing, the coupled Maxwell-Klein Gordon equations (the nearest classically to QED) are not hyperbolic, hence do not satisfy Einstein-causality.

$\square A^\mu = j^\mu$ not hyperbolic?

Then, what has your "without gauge fixing" to do with my "in the Lorenz gauge"? Are you about the remaining gauge freedom after the Lorenz gauge?

 

[A. Neumaier]

$\square A^\mu = j^\mu$ not hyperbolic?

The coupled equations must be hyperbolic. Without coupling you only have a free field.

 

[Sunil]

The coupled equations must be hyperbolic. Without coupling you only have a free field.

The point being? I have included a source term. I would ask you to show me which part of the coupled system makes it non-hyperbolic. Is the equation for the matter fields not hyperbolic?

 

[A. Neumaier]

The point being? I have included a source term. I would ask you to show me which part of the coupled system makes it non-hyperbolic. Is the equation for the matter fields not hyperbolic?

I think it is not the coupling to the scalar field (as I first thought) but the constraint $\partial_\mu A^\mu =0$ that spoils hyperbolicity. I don't remember the details and don't have the time now to check; thus maybe I am wrong.

 

[Sunil]

I think it is not the coupling to the scalar field (as I first thought) but the constraint $\partial_\mu A^\mu =0$ that spoils hyperbolicity.

Ok, this has at least some plausibility. To introduce conservation laws into a field theory is known to be problematic.

I know (memory, without good sources, sorry) that this creates some problems in condensed matter theory too. Take a particle theory with exact conservation law. Then you can use, in a large distance approximation, a classical field theory with fields $\rho(x,t)$ and the corresponding momentum $\pi^i(x,t) = \rho(x,t) v^i(x,t)$. Try to define a quantum field theory for these four fields such that it is the large distance limit of the fundamental theory. How to reach that $\hat{\rho}(x,t) \ge 0$ given that particle number in the fundamental theory is always positive? How to reach that $\partial_t \hat{\rho}(x,t) + \partial_i \hat{\pi}^i(x,t) = 0$ exactly, without even vacuum oscillations around zero, given that particle number is exactly conserved in the fundamental theory? AFAIR there are no good answers for this.

So it seems quite plausible that there are similarly no good answers how to preserve Einstein-covariance of a classical theory with such an exact conservation law during quantization.

 

[Demystifier]

I know (memory, without good sources, sorry)

Now @PeterDonis will come at you.

 

[zonde]

I was on you-tube and saw a video from Oxford on QM foundations. I didn't agree with it, but that is not an issue - I disagree with a lot of interpretational stuff. The video mentioned a paper they thought essential reading:
https://www.mathematik.uni-muenchen.de/~bohmmech/BohmHome/files/three_measurement_problems.pdf

Really nice idea in this paper. If you want to understand what are the problems with QM you have to investigate the areas where QM is weak and that is individual events.

Is it me, or has the author not shown the appropriate care? In particular, they claim theories that violate 1a are hidden variable theories. I thought - what - how does that follow. It may simply mean nature is fundamentally probabilistic. Or, in other words, the three assumptions are not inconsistent.

Specifically, I do not think the following is logically justified:
'And since we are interested in individual cats and detectors and electrons since it is a plain physical fact that some individual cats are alive and some dead, some individual detectors point to "UP" and some to "DOWN", a complete physics, which is able at least to describe and represent these physical facts, must have more to it than ensemble wave-functions.'

My response is - that might be your idea of what compete physics is, but it might be best not to assume that is everyone's idea of complete physics. Einstein, of course, thought QM incomplete, but I am not sure that is necessarily why.

Hmm, to me it seems rather clear just by reading 1A statement: "1.A The wave-function of a system is complete, i.e. the wave-function specifies (directly or indirectly) all of the physical properties of a system."
Complete/incomplete means whether wave-function is or is not best possible description of individual system. And if it's not then additional variables (HV) can improve description of individual system.
It's just a definition of a term. And it is related to history of QM so it does not appear out of the blue.

Anyways, I thought of slightly modified three statements after reading Maudlin's paper.
Given single preparation of quantum state and measurement of the state that gives two easily distinguishable groups of events, we can say that considering each event and taking into account that there are two possible outcomes at least one of the options is true:
1. Measurement event revealed something physical.
2. Measurement event created something physical.
3. Measurement event is not physical.

 

[stevendaryl]

The ensemble interpretation, analogous to the frequentist interpretation of probability (which has the issue that the law of large numbers is only valid for infinite 'experiments'), is a 'practical' interpretation that can't answer fundamental matters. However, it is used all the time in applications.

I don’t understand how the ensemble interpretation of QM is an interpretation.

In classical statistical mechanics, the meaning of an ensemble of systems is a collection of systems that are macroscopically identical but microscopically different.

But in quantum mechanics, an electron that is in a superposition of spin-up in the z-direction and spin-down in the z- direction cannot be interpreted in that way. If you have an ensemble of electrons in that state, it cannot be interpreted as some fraction of them being spin-up and some fraction being spin-down.

Maybe after a measurement of the spin in the z-direction, you can interpret it that way. But that seems to me equivalent to a “collapse” interpretation of measurement.

 

[A. Neumaier]

But in quantum mechanics, an electron that is in a superposition of spin-up in the z-direction and spin-down in the z- direction cannot be interpreted in that way. If you have an ensemble of electrons in that state, it cannot be interpreted as some fraction of them being spin-up and some fraction being spin-down.

In the ensemble interpretation, a superposition is supposed to describe an ensemble of electrons in the superposition, not an ensemble of electrons in one of two particular states.

 

[PeterDonis]

In classical statistical mechanics, the meaning of an ensemble of systems is a collection of systems that are macroscopically identical but microscopically different.

But in quantum mechanics, an electron that is in a superposition of spin-up in the z-direction and spin-down in the z- direction cannot be interpreted in that way. If you have an ensemble of electrons in that state, it cannot be interpreted as some fraction of them being spin-up and some fraction being spin-down.

In the QM ensemble interpretation as compared with classical statistical mechanics, "macroscopically the same" corresponds to "all prepared by the same process", and "microscopically different" corresponds to "will not all give the same result when measured in the same way". Whether or not this is a good enough correspondence to justify the term "ensemble interpretation" is a matter of choice of words; the interpretation itself is clear about what it's saying, and does not claim to be in exact correspondence with the classical statistical mechanics definition of an ensemble.

 

[zonde]

Anyways, I thought of slightly modified three statements after reading Maudlin's paper.
Given single preparation of quantum state and measurement of the state that gives two easily distinguishable groups of events, we can say that considering each event and taking into account that there are two possible outcomes at least one of the options is true:
1. Measurement event revealed something physical.
2. Measurement event created something physical.
3. Measurement event is not physical.

Speaking about option 3. which basically is about MWI. In Maudlin's paper objections to MWI are related to Born's rule. I would say that there is more serious philosophical problem.
MWI denies that my perception of certain consistent observations as objective physical facts is justified.

I will explain. I make an observation, say I look at a track in Wilson cloud chamber. Now I ask: am I seeing things or is this objective physical fact that there is a track in that cloud chamber?
So I ask somebody else to look at this track and ask him if he is seeing the same. But even if he agrees with me he could be influenced by unconscious bias say because he tends to consider more seriously the option that I am right. Or he just lies because it benefits him somehow.
So I ask him to do the experiment according to given instructions and to write down what he is seeing. Then to avoid possibility that my memory is cheating me I make a record myself and after he has made the experiment we compare our records.
If we do that and the records are consistent with each other I can say with great confidence that I am not seeing things and I should consider the track in cloud chamber as a objective physical fact.

So, what does MWI say about all that? It says I can't relay on all these things as a method of getting objective physical facts. Measurement records are not facts at all, they are just illusions of facts because some QM aspect of me is aligned with some QM aspect of measurement record. And communication with other person gives me consistent experience because some QM aspect of me is aligned with certain QM aspect of other person including this QM aspect of all the communication media between us.

So what I can do instead? Well, I should pretend that measurement records are objective facts and repeat to myself the magic word "decoherence" while remembering at the back of my mind that deep down it's a fake certainty.

How this approach is different from superdeterminism?

 

[zonde]

Speaking about option 1. it is actually quite simple - it contradicts observations.
It is obvious with quite simple three polarizers experiment:

 

[Morbert]

Speaking about option 1. it is actually quite simple - it contradicts observations.
It is obvious with quite simple three polarizers experiment:

For posterity:

Consistent Histories as described by Robert Griffiths would assert option 1, and would do so without recourse to hidden variables or similar extension to QM.

When I get a chance I can lay out how CH would approach this experiment.

 

[Morbert]

Call the first, middle, and last filters E G and F respectively.

A conventional description of the photon as is passes through the three filters would present these 4 possibilities*:

i) A photon is absorbed by filter E
ii) A photon passes through filter E and is forced into a vertical polarisation, and is absorbed by filter F
iii) A photon passes through the filter E and is forced into a vertical polarisation, passes through the filter F and is forced into a 45 degree polarisation, and is absorbed by the filter G
iv) A photon passes through the filter E and is forced into a vertical polarisation, passes through the filter F and is forced into a 45 degree polarisation, and passes through the filter G and is forced into a horizontal polarisation

We can see that, in these descriptions, the interaction with a filter is what creates the corresponding polarisation property.

QM let's us write down each of these possibilities as a string of time-ordered projectors. If the time of interaction for E G and F are t1, t2, and t3 then the possibilities can be written down as

i) $\left[E',F_0,G_0\right]_{t_1+\delta t}\odot I_{t_2 + \delta t}\odot I_{t_3+\delta t}$
ii) $\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[E,F',G_0\right]_{t_2 + \delta t}\odot I_{t_3 + \delta t}$
iii) $\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow,E,F,G_0\right]_{t_2 + \delta t}\odot\left[E,F,G'\right]_{t_3+\delta t}$
iv)$\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow,E,F,G_0\right]_{t_2 + \delta t}\odot\left[\rightarrow,E,F,G\right]_{t_3+\delta t}$

QM will return probabilities for these possibilities no problem. But a consistent historian** would say you can move the polarisation projectors to before the filter interactions like so:
i) $\left[E',F_0,G_0\right]_{t_1+\delta t}\odot I_{t_2 + \delta t}\odot I_{t_3+\delta t}$
ii) $\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[E,F',G_0\right]_{t_2 + \delta t}\odot I_{t_3 + \delta t}$
iii) $\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow\right]_{t_2-\delta t}\odot\left[E,F,G_0\right]_{t_2 + \delta t}\odot\left[E,F,G'\right]_{t_3+\delta t}$
iv)$\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow\right]_{t_2-\delta t}\odot\left[E,F,G_0\right]_{t_2 + \delta t}\odot\left[\rightarrow\right]_{t_3-\delta t}\odot\left[ E,F,G\right]_{t_3+\delta t}$

These, according to CH, would correspond to the possibilities:

i) A photon is absorbed by filter E
ii) A photon passes through filter E already having a vertical polarisation and is absorbed by F.
iii) A photon passes through the filter E already having a vertical polarisation, passes through F already having a 45 degree polarisation, and is absorbed by the filter G
iv) A photon passes through the filter E already having a vertical polarisation, passes through F already having a 45 degree polarisation, and passes through G already having a horizontal polarisation.

and QM will still return consistent probabilities for these possibilities.

So what happens when you remove the filter F? You lose the ability to infer anything about 45 degree polarisation between filters E and G because the possibilities will no longer decohere, and QM will refuse to return consistent probabilities, analogous to the way that, in the double-slit experiment, paths that specify one slit or the other will not decohere when a detector is not present. It's not that removing the filter retroactively changes the polarisation of photons. It's that you lose the ability to address polarisations at certain times without certain filters in place.

* For expediency I've omitted possibilities that are consistent but return probability 0

** I should make it clear that Griffiths's CH account of measurements revealing pre-existing properties is not standard among all CH proponents.