Can the double slit experiment distinguish between QM interpretations?

In summary, the paper discusses the use of the double slit experiment to distinguish between different interpretations of quantum mechanics. The authors argue that the correct framework for treating time on the same footing as other observables is relativistic quantum field theory, rather than non-relativistic quantum mechanics. They also point out that none of the existing discussions or comparisons of quantum mechanics interpretations use QFT as their framework. However, the paper also claims that time is a parameter in the Schrodinger equation, not an operator, and therefore fails to consider the solution to this problem in the form of QFT. The author, who has written previous papers on this topic, believes that standard QM makes well defined predictions for the arrival time probability distribution and that the predictions are
  • #1
47,604
23,878
TL;DR Summary
A paper just released on arxiv claims that the double slit experiment can be used to distinguish between QM interpretations. Is its claim correct?
This paper claims that the double slit experiment can be used to distinguish between QM interpretations:

https://arxiv.org/abs/2301.02641

IMO, the paper goes astray right at the start, when it points out that time is a parameter in the Schrodinger equation, not an operator, so that equation gives no way to uniquely derive a probability distribution for measurement results as a function of time--but then fails to note that the solution to that problem is to not use the Schrodinger equation in the first place. The correct framework in which to treat time on the same footing as other observables is relativistic quantum field theory. That is never even mentioned in this paper.

To be fair, I have never seen *any* discussion or comparison of QM interpretations that uses QFT as its framework; they all use non-relativistic QM, even though we know that's just an approximation. But I think it's still an issue even if it's an extremely common one. What do other QM experts here think?
 
  • Like
Likes bhobba, Demystifier, Doc Al and 1 other person
Physics news on Phys.org
  • #2
There is also a Schrödinger equation in QFT and time is just a parameter in that equation. It's just more common to formulate QFT in the Heisenberg picture or using path integrals. However, all of these descriptions are equivalent. So if the paper goes astray (it may very well, I didn't read it in detail), it's not because its using the Schrödinger equation.
 
  • #3
Nullstein said:
There is also a Schrödinger equation in QFT
In non-relativistic QFT, perhaps. But I specified relativistic QFT in the OP.
 
  • #4
PeterDonis said:
In non-relativistic QFT, perhaps. But I specified relativistic QFT in the OP.
There is a Schrödinger equation in any QFT, relativistic or not. Heuristically, it is given by the quantization of the Hamiltonian, which can be obtained by Legendre transformation of the Lagrangian of the theory (see Hamiltonian field theory for an overview). Rigorously, it follows from applying Stone's theorem to the time evolution operator.
 
  • #5
Nullstein said:
There is a Schrödinger equation in any QFT, relativistic or not. Heuristically, it is given by the quantization of the Hamiltonian, which can be obtained by Legendre transformation of the Lagrangian of the theory (see Hamiltonian field theory for an overview). Rigorously, it follows from applying Stone's theorem to the time evolution operator.
All of this either requires a preferred frame, which violates the principle of relativity, or accepting that different frames have different Hamiltonian operators, Schrodinger equations, etc., which, as far as QM interpretation is concerned, undermines the whole point of trying to use those things in the first place. As I said in the OP, the correct relativistic approach for QM interpretation is to treat time and space on the same footing. The approach you describe, however valid it is mathematically, does not do that.
 
  • #6
PeterDonis said:
All of this either requires a preferred frame, which violates the principle of relativity, or accepting that different frames have different Hamiltonian operators, Schrodinger equations, etc., which, as far as QM interpretation is concerned, undermines the whole point of trying to use those things in the first place.
No, none of this applies. The Hamiltonian looks the same in all inertial frames and none of them is preferred. One can pass between them using Lorentz transformations.
PeterDonis said:
As I said in the OP, the correct relativistic approach for QM interpretation is to treat time and space on the same footing. The approach you describe, however valid it is mathematically, does not do that.
Time and space is treated symmetrically in the Hamiltonian QFT as well. The equation is not manifestly covariant, but everything is equivalent to a covariant formulation. It's also commonly taught in QFT courses and explained in standard textbooks such as Peskin & Schröder, so I don't see why it should be avoided. Here's another exposition of the formalism. The first couple of sentences in the paper are certainly correct assertions also about QFT, so if the authors can derive some interesting conclusions from that observation, it should be taken seriously.
 
  • #7
Let me put it differently. Time and space are treated symmetrically in QFT, independent of the concrete formulation. But it's not because time becomes an observable described by an operator. It's because position ceases to be an observable described by an operator. Both time and position are just parameters in QFT. The observables in QFT are given by the fields, which are represented by field operators. So we don't need the Schrödinger equation to conclude that time is just a parameter (in both QM and QFT). Given the introductory sentences, the paper seems to be based on the observation that time is just a parameter and not an observable. In order to challenge that assumption, you would have to provide as a counterexample a quantum theory (particle or field, it doesn't matter), wherein time is in fact an observable described by an operator.
 
  • Like
Likes gentzen and vanhees71
  • #8
This paper is about the arrival time probability distribution problem in QM. Since time is not an operator, it looks as if it's not clear what the standard quantum mechanical prediction for a time distribution is. However, my opinion is that standard QM does make well defined predictions for time distributions. After all, the exponential time distribution of decays, and deviations from the exponential distribution, are well understood in standard QM. In my opinion, when standard QM is analyzed correctly, it also makes well defined predictions on the arrival time problem. And when the calculations are done correctly, the end result is that the predictions of standard QM are the same as those of the Bohmian interpretation.

I (with two collaborators) have written 3 papers on this topic; the first two are published, the third is not published yet, but the final result that the two interpretations give the same results is explicit only in the third paper:
https://arxiv.org/abs/2010.07575
https://arxiv.org/abs/2107.08777
https://arxiv.org/abs/2207.09140
In the first paper we formulated the general framework. In the second paper we further developed the formalism and made some approximate calculations. In the third paper we understood all this from a deeper point of view and made exact analytic calculations.
 
  • Like
Likes bhobba, gentzen, PeroK and 1 other person
  • #9
PeterDonis said:
In non-relativistic QFT, perhaps. But I specified relativistic QFT in the OP.
Relativistic QFT also has a Schrodinger equation
$$H|\psi(t)\rangle = i\hbar\partial_t |\psi(t)\rangle$$
where ##H## is the QFT Hamiltonian.
 
  • #10
PeterDonis said:
The correct framework in which to treat time on the same footing as other observables is relativistic quantum field theory.
Time is not an observable in relativistic QFT.
 
  • #11
PeterDonis said:
To be fair, I have never seen *any* discussion or comparison of QM interpretations that uses QFT as its framework;
Do you mean a comparison of their measurable predictions?
 
  • #12
The interpretational issues like the so-called "measurement problem" or the "wave-particle-dualism problem" are not too much different between relativistic and non-relativistic QT, and indeed time is not an observable in both relativistic and non-relativistic QT. Position can be defined as an observable for all kinds of non-relativistic and all kinds of massive relativistic particles. For massless particles a position operator is definable only for spins 0 and 1/2.

In relativistic QT the localizability of particles is, however, much more restricted than in non-relativistic QT, because of pair production, which is a necessary consequence of microcausality and locality, which is the only known causal description of interacting relativistic quantum systems, i.e., the only way to define a relativistic quantum dynamics, that's compatible with relativistic causality is the use of local quantum fields, which allow for the description of local observables that obey the microcausality condition, which in turn also leads to the Poincare covariance and unitarity of the S-matrix. A particle description is also only possible for (asymptotic) free states.
 
  • Like
Likes LittleSchwinger
  • #13
As an aside re/ QFT and the Schroedinger equation: This previous discussion might be of interest

PeterDonis said:
To be fair, I have never seen *any* discussion or comparison of QM interpretations that uses QFT as its framework; they all use non-relativistic QM, even though we know that's just an approximation.
For posterity:
https://arxiv.org/ftp/arxiv/papers/1805/1805.12246.pdf
https://arxiv.org/pdf/gr-qc/9304006.pdf
Some discussion of decoherent histories as applied to quantum theories with different levels of generality
 
  • #14
Nullstein said:
Time and space are treated symmetrically in QFT, independent of the concrete formulation. But it's not because time becomes an observable described by an operator. It's because position ceases to be an observable described by an operator. Both time and position are just parameters in QFT.
Yes, this is a valid way of treating time and space on the same footing; I would put it that the parameter in QFT is "which spacetime point", and labeling spacetime points requires a 4-dimensional coordinate ##x^\mu##.

However, one can't then pick out just one piece of the parameter ##x^\mu##, "time", and treat it differently; yet that's what the Hamiltonian formulation and the Schrodinger equation do. That is why I object to them as a basis for QM interpretation. Of course I don't object to them as a mathematical framework for doing calculations; but that's not what this thread is about.
 
  • #15
Demystifier said:
Time is not an observable in relativistic QFT.
Yes, I misspoke in that part of the OP. See post #14.
 
  • #16
Demystifier said:
Do you mean a comparison of their measurable predictions?
I mean any discussion of QM interpretations using QFT as a framework.
 
  • #17
PeterDonis said:
I mean any discussion of QM interpretations using QFT as a framework.
But you have seen one of my papers on that, I know because once you correctly summarized the main ideas of it.
 
  • #18
PeterDonis said:
However, one can't then pick out just one piece of the parameter ##x^\mu##, "time", and treat it differently; yet that's what the Hamiltonian formulation and the Schrodinger equation do. That is why I object to them as a basis for QM interpretation.
But what remains then, path integral formulation? How do you define the notion of state in the Hilbert space within the path integral approach?
 
  • #19
PeterDonis said:
Yes, this is a valid way of treating time and space on the same footing; I would put it that the parameter in QFT is "which spacetime point", and labeling spacetime points requires a 4-dimensional coordinate ##x^\mu##.

However, one can't then pick out just one piece of the parameter ##x^\mu##, "time", and treat it differently; yet that's what the Hamiltonian formulation and the Schrodinger equation do. That is why I object to them as a basis for QM interpretation. Of course I don't object to them as a mathematical framework for doing calculations; but that's not what this thread is about.
Well, that's the problem with the operator formulation of relativistic QFT. If you want to use the canonical formalism, you have to use the Hamiltonian description, and this is done in an arbitrary but then fixed inertial frame of reference (I'm talking about standard QFT in SRT, not in curved spacetimes of GRT of course). Then you first get a formulation that is not manifestly Poincare covariant.

A way to stay Poincare covariant at all steps of the calculation is to use the path integral in the Lagrangian formulation, but there you must be very careful, whether you can do this in a naive way. That's only the case if the Hamiltonian is quadratic in the canonical field momenta with field-independent coefficients! For a detailed discussion of the most simple example (free spin-0 field), see Sect. 4.5.1 of

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

For a non-trivial example see the discussion of (scalar) QED in 6.3, when going from the axial-gauge (partially) fixed Hamiltonian (canonical) path-integral formalism to a manifest Lagrangian path-integral formalism using a covariant gauge, showing that both are equivalent.
 
  • Like
Likes LittleSchwinger
  • #20
Demystifier said:
But what remains then, path integral formulation?
That would treat time and space on a equal footing, yes.

Demystifier said:
How do you define the notion of state in the Hilbert space within the path integral approach?
Why do you need to do that?
 
  • #21
Demystifier said:
But you have seen one of my papers on that, I know because once you correctly summarized the main ideas of it.
Was that the paper that treated QFT as emergent from a non-Lorentz invariant underlying theory?
 
  • Like
Likes Demystifier
  • #23
PeterDonis said:
Was that the paper that treated QFT as emergent from a non-Lorentz invariant underlying theory?
Yes.
 
  • #24
PeterDonis said:
Why do you need to do that?
To discuss the (interpretations of) entanglement, for instance. How would you define entanglement without state in the Hilbert space?
 
  • #25
Demystifier said:
How would you define entanglement without state in the Hilbert space?
Nonzero entanglement entropy. I believe that can be computed in the path integral approach.
 
  • #26
PeterDonis said:
Nonzero entanglement entropy. I believe that can be computed in the path integral approach.
Entropy in which state? The path integral approach is developed to compute things (including entanglement entropy) in the vacuum state, but how to define other states with path integrals? Bell state, for instance?
 
Last edited:
  • #28
vanhees71 said:
The path-integral method, together with the Schwinger-Keldysh real-time contour or, for equilibrium, the Matsubara imaginary-time contour, can be applied to any state:

https://itp.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf
Of course it can be applied, as long as one defines states as objects in the Hilbert space. But to define the Hilbert space, one needs canonical quantization.

The question is, can one formulate (not merely apply) QFT without ever mentioning Hilbert space and/or canonical quantization? I don't think that path-integral quantization can do that. Perhaps algebraic quantization goes in that direction, which indeed can be made manifestly Lorentz invariant, but with this formalism it's hard to get concrete measurable results.
 
Last edited:
  • Like
Likes vanhees71
  • #29
I think, it's impossible to avoid Hilbert space, because that's the mathematical framework QT is formulated in. Canonical quantization is a rather heuristic method to guess the possible form of the algebra of observables, Hamiltonians, etc. Here I'd say a more reliable approach is the group-theoretical methods exploiting symmetries.
 
  • Like
Likes Demystifier
  • #30
vanhees71 said:
I think, it's impossible to avoid Hilbert space, because that's the mathematical framework QT is formulated in. Canonical quantization is a rather heuristic method to guess the possible form of the algebra of observables, Hamiltonians, etc. Here I'd say a more reliable approach is the group-theoretical methods exploiting symmetries.
Yes. And for discussing the resulting interpretations, a good starting point is
http://philsci-archive.pitt.edu/8890/
 
  • Informative
Likes vanhees71
  • #31
But to return to the main subject of this thread, quantum theory does not need to be relativistic to study time distributions. For instance, non-relativistic QM has well defined predictions for the time of decay. The question is, does it also has well defined predictions for the time of arrival? I claim that it has, because both can be described by the same theoretical framework (post #8).
 
  • Like
Likes vanhees71
  • #32
Are you talking about the "tunneling time" and related questions? I think, here the solution can only be to look at specific experiments and (try) to describe them with QT. I think the problem with this is that there's not a clear definition of what "tunneling time" means, and this is not restricted to QT but also within classical theory of waves. E.g., some years ago there was a big debate about faster-than-light signals in electromagnetic wave guides. Of course there's nothing faster than light that's not allowed to be faster than light within Maxwell's theory, which is relativistic of course. In this case the question has been answered already by Sommerfeld and Brillouin in 1907-1913 ;-).
 
  • Like
Likes Demystifier
  • #33
vanhees71 said:
Are you talking about the "tunneling time" and related questions?
Yes.
vanhees71 said:
I think, here the solution can only be to look at specific experiments and (try) to describe them with QT.
At first I thought that too. But then I developed a general framework of idealized measurements that can be applied to tunneling time as well, see the first paper in #8.
 
  • Like
Likes vanhees71
  • #34
Demystifier said:
The question is, can one formulate (not merely apply) QFT without ever mentioning Hilbert space and/or canonical quantization? I don't think that path-integral quantization can do that. Perhaps algebraic quantization goes in that direction, which indeed can be made manifestly Lorentz invariant, but with this formalism it's hard to get concrete measurable results.
Algebraic QFT is capable of defining QFTs without the notion of a Hilbert Space. This is increasingly necessary in curved spacetimes, especially for generic spacetimes without the symmetries required to define the notion of particle or where one possibly lacks a global state.

Of course Hilbert spaces are closely tied to this formalism, as they are involved in representations of the abstract observable algebra. Only recently though have researchers found how to compute directly physical quantities using the formalism. See the monograph of Kasia Rejzner.
 
Last edited:
  • Like
Likes Greg Bernhardt and dextercioby
  • #35
The problem with AQFT, however is that for decades nobody was able to describe interacting particles in (1+3) dimensions!
 
  • Like
Likes Demystifier

Similar threads

Back
Top