Is Bohmian mechanics a convenient ontological overcommitment?

[Demystifier]

What do you need an "ontology"?

It helps to develop an intuitive mental image. It's a very useful thinking tool.

 

[gentzen]

What do you need an "ontology"?

For me, the original Bohmian mechanics is an existence proof of a mathematical model. In a sense this is the minimal requirement for an "ontology", but some people expect more.
But not all mathematical models are equally nice as an "ontology". Often mathematical models contain quotient structures. If too much work is hidden in those implicit equivalence relations, then the model risks to become trivial and meaningless. But some of those quotient structures are there to stay, especially those caused by topological constraints.

Understanding such details can give hints why certain things are difficult. For example, A. Neumaier thinks it is possible that non-separable Hilbert spaces might be required for QFT. But that non-separability could also be a hint that some topological information is missing (for example the surfaces in solid state physics), and that different possibilities for that missing information could lead to different results.

 

[WernerQH]

At the end you can only observe what you can observe. It's at least not subject of the natural sciences to speculate what further may be "behind" that observable facts.

Also experiments need interpretation! It is not clear what exactly the observable facts are in experiments that claim to simulate "Wigner's friend" or reveal "photon trajectories".

 

[Sunil]

That many people after 100 years of modern QT still have a problem to accept that Nature is behaving fundamentally random is not a problem of physics but rather of psychology and philosophy of these individual people.

Essentially nobody has a problem with Nature behaving random. I have never seen anybody with conceptual problems in thermodynamics because of randomness.

You seem to have a problem not understanding that your "fundamentally" is your personal metaphysical preference, nothing more. This can be even proven: There exists a deterministic interpretation, thus, fundamental randomness is not interpretation-independent, thus, a metaphysical preference.

The natural sciences are not to please your epistemological prejudices but to figure out how Nature behaves as objectively and quantitatively observed.

Looks more like you think they have to please, instead, your own epistemological prejudices.

 

[WernerQH]

What's your favored interpretation?

It's a blend of the statistical and the transactional interpretation. But without the "offer" and "confirmation" waves, and also without particles. In my view, it's only the interactions between photons and electrons (events) that are real. As John Bell once wrote:

A piece of matter then is a galaxy of such events.

("Are there quantum jumps?")

 

[WernerQH]

I guess that "randomness" can still be a bit better defined than in the currently dominant frequentist interpretations and subjective Bayesian interpretations.

Yes, there are still debates about the meaning of probability. I consider probability theory an indispensable ingredient of almost all physical theories, much like geometry. My understanding of probability was profoundly changed after my encounter with Jaynes's book "Probability Theory. The Logic of Science". His position is that, given the same facts, all users of probability, if they apply it correctly, must draw the same conclusions.

For a convincing ontology, accepting true indistinguishabiliy and developping appropriate mental images for both bosons and fermions seems crucial to me.

Absolutely. I think of photons and electrons as useful fiction: lines that we draw between short-lived localized currents (events) that are the real stuff of which the world around us is made. Photons and electrons exist only as correlation functions (propagators) helping us to describe the patterns of events that we observe in space-time. The S-matrix is a tool to predict the likeliness of certain patterns of events, and I think it's significant that all the different diagrams that can be drawn connecting a given set of vertices contribute. There is no fact of the matter whether "this" or "that" electron interacted with "that" photon.

 

[vanhees71]

Also experiments need interpretation! It is not clear what exactly the observable facts are in experiments that claim to simulate "Wigner's friend" or reveal "photon trajectories".

Well, you have a setup and the data. One should be aware that particularly for photons "trajectories" are not defined, because as you say you need also a theory for the interpretation of these data, and the definition of the photon today is QED and not the old quantum theory a la Einstein, where photons are misinterpreted as localizable particles, but the modern definition of a photon implies that there is no position operator for photons to begin with. Even for massive particles, as defined by relativistic QFT, there are limits of their localizability due to the creation and annihilation processes inevitably occurring when trying to localize them, although they admit the definition of a position observable.

 

[Demystifier]

Even for massive particles, as defined by relativistic QFT, there are limits of their localizability due to the creation and annihilation processes inevitably occurring when trying to localize them, although they admit the definition of a position observable.

Is this definition Lorentz covariant?

 

[vanhees71]

Yes, that's the important point. Have a look at Arnold Neumaier's physics FAQ and the references quoted therein:

https://arnold-neumaier.at/physfaq/topics/position.html
https://arnold-neumaier.at/physfaq/topics/localization

 

[Demystifier]

Yes, that's the important point. Have a look at Arnold Neumaier's physics FAQ and the references quoted therein:

https://arnold-neumaier.at/physfaq/topics/position.html
https://arnold-neumaier.at/physfaq/topics/localization

As far as I can see, this is not Lorentz covariant. Indeed, it explicitly says: "Note that the position operator is always observer-dependent, in the sense that one must choose a timelike unit vector to distinguish space and time coordinates in the momentum operator. This is due to the fact that the above construction is not invariant under Lorentz boosts"

 

[vanhees71]

In this sense any split of space and time is of course not manifestly Poincare invariant. This holds for classical relativistic physics too.

 

[Demystifier]

In this sense any split of space and time is of course not manifestly Poincare invariant. This holds for classical relativistic physics too.

But classical position can be defined without a split, as a position in 4-dimensional spacetime. Can a position without split be defined in quantum physics? For instance, if it is defined through $|x\rangle \equiv \hat{\phi}(x)|0\rangle$, then why exactly such a construction does not work in the massless case?

 

[vanhees71]

It doesn't work on the operator level: You cannot construct a position operator, fulfilling the commutation relations of the operator algebra, from the generators of the massless irreps. of the proper orthochronous Poincare group for spin $\geq 1$.

 

[Demystifier]

You cannot construct a position operator, fulfilling the commutation relations of the operator algebra

What algebra are we talking about here? Heisenberg algebra $[x_i,p_j]=i\hbar\delta_{ij}$?

from the generators of the massless irreps. of the proper orthochronous Poincare group for spin $\geq 1$.

What about massless spin=0; is there a position operator for relativistic massless scalar particle?

 

[vanhees71]

There are position operators for massless scalar and spin-1/2 particles. See Arnold Neumaier's physics FAQ and the papers quoted therein:

https://arnold-neumaier.at/physfaq/topics/position.html
https://arnold-neumaier.at/physfaq/topics/localization

 

[Demystifier]

There are position operators for massless scalar and spin-1/2 particles.

Is there a physically intuitive hand-waving argument why position operator doesn't exist for massless higher spins?

 

[vanhees71]

I don't know of any intuitive argument. It may have to do with the fact that the irreps of higher spin are necessarily "gauge fields". The reason is that the little group for the massless representations is ISO(2), i.e., the symmetry space of 2D Euclidean affine space. The little group is the subgroup of the Poincare group which leaves one arbitrarily choosen four-momentum, which for the massless representations is light-like, invariant.

Now you need the unitary transformations of the little group to build the irrep. of the full orthochronous proper Poincare group a la Frobenius/Wigner. The ISO(2) is generated by 2 translations and 1 rotation. The corresponding generators for the translations have of course continuous a continuous spectrum. Realizing this would mean you'd describe massless particles with continuous polarization-like degrees of freedom, which is physically not observed. Thus you consider only such representations, for which the translations of this ISO(2) are represented trivially. This leaves only the rotations of the ISO(2) as non-trivial (the trivial one leads to scalar massless fields though). From the point of view of the entire Poincare group these are the rotations with the rotation axis around the three-momentum defined by the arbitrarily chosen light-like four-momentum, i.e., the generator of these rotations is the helicity, and together with the triviality of the translations of the said little group this implies that you'd have all possible helicities $h \in \mathbb{R}$, but at the end you use the Frobenius-Wigner method to build the representation of the entire Poincare group including the usual rotations as a subgroup, and this means that the helicities can only be $0$ (scalar fields), $\pm 1/2$, $\pm 1$, etc. leading to the representations corresponding to spin $s=0$, $s=1/2$, etc. for the rotation group as a subgroup of the Poincare group. So for any non-zero spin of your massless field you get only 2 polarization-degrees of freedom instead of $2s+1$. For spin 1/2 you get both $\pm 1/2$ for each spin component. For $s \geq 1$ you get more spin-like degrees of freedom than the two physical polarization referring to the helicities $\pm s$ for each energy-momentum eigenmode, i.e., you have to deal with unphysical degrees of freedom, making your theory a "gauge theory" when expressed in terms of local fields.

That's why for the electromagnetic field for each energy-momentum eigenmode you have only the two transverse polarization states corresponding two left- and right-circular polarization (i.e., $h=\pm 1$ helicities) and not 3. Of course you want to work with local fields (in both classical and quantum electrodynamics). That's why you introduce a four-vector $A^{\mu}$, but all fields connected by a "gauge transformation" $A^{\prime \mu}=A^{\mu} +\partial^{\mu} \chi$ with an arbitrary scalar field $\chi$ are considered to describe the very same physical situation. Only such a description with redundant, unphysical field degrees of freedom allow for a local description, and that's why we have to deal with the problems connected with gauge invariance when quantizing the theory. One way is to fix the gauge completely (e.g., by choosing the Coulomb gauge) and loosing manifest covariance (leading nevertheless to a covariant field theory as far as the physical fields and the observables built from them are concerned) of to use the tricky Faddeev-Popov quantization by introducing more unphysical ghost fields which precisely cancel the contributions from the also unphysical parts of the vector potentials. This modern path-integral method is equivalent to the manifestly covariant operator method. For the Abelian gauge-theory case of QED that's equivalent to Gupta-Bleuler quantization, but for the non-Abelian case you need the Faddeev-Popov ghosts for a gauge-fixed manifestly covariant formulation and the associate Becchi-Rouet-Stora-Tyutin (BRST) symmetry of the gauge-fixed action. Here, I'd say, the original path-integral formulation aka Fadeev-Popov quantization is more straight-forward.

For massive particles the little group is simply SO(3), and there you have all $(2s+1)$ spin eigenvalues as physical states.

Maybe it's this difference between the massless and massive irreps. which allow for the development of some intuitive argument for the non-existence of proper position observables in the massless case for spin $s \geq 1$.

 

[Killtech]

Photons cannot have trajectories. They don't even have a position observable. I don't think that we have a convincing Bohmian reinterpretation of local relativistic QFT yet. For non-relativistic QT in the first-quantization formulation BM is a consistent theory without much additional merit compared to standard statistically interpreted QT. At least I've not seen any experiment that depicts the Bohmian trajectories of slow massive particles, e.g., in a double-slit experiment or something equivalent.

Is there a physically intuitive hand-waving argument why position operator doesn't exist for massless higher spins?

Is that so? I mean i know it's not formally defined, but it doesn't seem hard to derive a working definition from the framework i would think.

Looking at how Maxwell is quantized and taking a look at what creation/annihilation operators do in the example of a Fock space. For the EM-field these are mostly a solutions of Maxwell. It is convenient that Maxwell knew Born's rule long before Born did (not really) just if you look at how energy density of a EM-field ist calculated. That allows it to make the 6-dim vector composed of the $E$ and $B$ fields work very analogue to a wave function and normalizing it to the energy of a single photon makes sense. Like in QT this means either putting it into a finite box (or localize it by deforming it into a wave package with a gaussian profile?)

Anyhow, classic EM-energy density shows just the interference patterns we need when superposing solutions... so call me naive but i would think that looking for a photon where its energy is located is a canonical idea. Thus you can just assume the position probability distribution to be directly proportional to the energy density. That said, it implies how to define the position observable. Given how impulse observable is related and knowing what impulse we expect it seems clear how it needs to be set up, too.

 

[Demystifier]

so call me naive but i would think that looking for a photon where its energy is located is a canonical idea.

It makes perfect sense in laboratory, but I wouldn't call it canonical. Canonical has something to do with canonical commutation relation between position and momentum.

 

[Killtech]

It makes perfect sense in laboratory, but I wouldn't call it canonical. Canonical has something to do with canonical commutation relation between position and momentum.

Canonical is a word relative to the perspective i guess.

Anyhow, The energy density of a EM-wave and it's continuity equations are well established in every frame, not just the laboratory. Since this approach just derives the observables directly from there, this statement applies to them as well. It's merely that their transformation behavior is somewhat untypical because the normalization needed for probabilities changes and since those observables do that translation, they have to also hold that scaling factor.

But as Hamilton, position and momentum operators go, their commutators work as expected. On a curios note, when i wrote down Maxwell equations in terms of a Hamilton operator for this E-B-field-wave function, i just noticed they look almost like a massless Dirac equation. Or rather the mass term is replaced by a coupling to the charge field.