I What makes the interpretations of Quantum Mechanics so important?

  • #61
Some thoughts/questions regarding the physical status of the interpretations of QM.

In GR one can interpret the increasing distances as being due to motion or to expansion of space. As these interpretations are coordinate dependent they are not 'true physics' or not 'real' in a sense.

In contrast do we expect that one of the interpretations of QM is 'true physics' or 'real' and we just don't know which one? If yes is there a slight chance to find the 'real' interpretation e.g. by advanced future technology?

Do you think that a future theory of Quantum Gravity could solve the puzzle?
 
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  • #62
vanhees71 said:
I don't understand the above statement. What do you mean by "cutoff-based perturbation theory". A cutoff is just a way to regularize the theory. After renormalization there's no cutoff anymore. You can also renormalize the theory without introducing any cutoff by just using BPHZ. Another way to regularize is dim. reg. No matter, what you do, you'll have to introduce in the one or the other way an energy-momentum-mass scale (however you name it), namely the renormalization scale, where you define your coupling constants. To change the scale there are the renormalization-group equations. Perturbation theory is of course only a good approximation, where the effective coupling is small, i.e., at low scales for QED and at large for QCD.

I always thought causal PT is at the end yielding the same result as any other perturbation theory. So how can it prevent that the perturbative QED doesn't break down finaly at energy scales defined by the Landau pole?
Causal PT is essentially BPHZ made rigorous.

QCD shows that a Landau pole is irrelevant when you avoid its vicinity. This holds for every renormalized theory. That the Landau pole was considered a fault of QED at high energies was due to the fact that when working with a cutoff, one obtained cutoff dependent summed contributions to the S-matrix that diverged at some large value of the cutoff.

But with cutoff-independend renormalization schemes this argument becomes empty. It gives a valid perturbative formula for the S-matrix ##S(E,\Lambda)## at any energy ##E## and any energy scale ##\Lambda## used in the renormaization condition. If one could resum the series and adjust the coupling constants appropriately, one would get something independent of ##\Lambda##. Different choices of the latter therefore just amount ot chopping in different ways the nonperturbative values into the terms of an infinite asymptotic series.

That a pole appears in these schemes just means that the perturbative expansions resulting for a choice of ##\Lambda## close to the pole gives poor approxmations. But the perturbative expansions at other (higher or lower) energy scales are still well-defined. Extrapolating the results from these to any wanted energy with suitable resumming schemes therefore gives sensible results at all energies.

In particular, the Landau pole is not a sign of that a theory is only effective, only a sign of that the perturb<tion series cannot always be trusted.
 
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  • #63
Well, does this imply that after all non-perturbative QED is shown to have no problems at whatever scale? I thought there's still some unsolved (IR/collinear?) problem left. Has this been resolved in recent years?
 
  • #64
vanhees71 said:
Well, does this imply that after all non-perturbative QED is shown to have no problems at whatever scale? I thought there's still some unsolved (IR/collinear?) problem left. Has this been resolved in recent years?
For standard QED, no problems are expected anywhere. On a nonrigorous level, everything is well understood for QED; even the infrared problem in QED is understood (Kulish-Faddeev) since QED (with photons and electrons only) has no bound states. QED with 2 species of charged particles is already much harder and poorly understood, as even the nonrigorous techniques for handling the probably infinitely many bound states are poorly developed.

What is unsettled for standard QED is the rigorous nonperturbative construction, i.e., the problem of how to give the perturbative asymptotic series definite values. In general, an asymptotic power series is the Taylor expansion of uncountably many different functions. What is missing is finding a recipe that selects the correct one. This is a very hard and unsolved problem in functional analysis.
 
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  • #65
I see. Thanks!

What I don't understand is your statement that QED with just electrons (and then of course necessarily also positrons) has no "bound states". What about positronium? Or is this simply because after all it's unstable due to pair annihilation?
 
  • #66
vanhees71 said:
What I don't understand is your statement that QED with just electrons (and then of course necessarily also positrons) has no "bound states". What about positronium? Or is this simply because after all it's unstable due to pair annihilation?
Yes, it decays already in pure QED. Hence mathematically, positronium is only a resonance, i.e., it corresponds to a pole of the analytically continued Green's function, described by the continuous part of the mass spectrum. In contrast, a bound state is, by definition, a discrete eigenvalue of the mass operator, corresponding to a pole of the physical Green's function.

This is different from the neutron, say, which is a bound state in QCD. It is also unstable, but not as a particle in QCD, only as a particle in the standard model, due to the weak interaction.
 
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  • #67
A. Neumaier said:
This is different from the neutron, say, which is a bound state in QCD. It is also unstable, but not as a particle in QCD, only as a particle in the standard model, due to the weak interaction.
I thought that QCD was part of the standard model, so what exactly do you mean by this statement?
 
  • #68
ftr said:
I thought that QCD was part of the standard model, so what exactly do you mean by this statement?
If QCD alone is considered it is stable, but including the rest of the standard model such as the weak interactions allows it to decay.
 
  • #69
DarMM said:
If QCD alone is considered it is stable, but including the rest of the standard model such as the weak interactions allows it to decay.
Thanks, so can we say that QCD is part of the standard model an approximate statement. How would you like to describe the standard model, say in a wiki.
 
  • #70
ftr said:
Thanks, so can we say that QCD is part of the standard model an approximate statement. How would you like to describe the standard model, say in a wiki.
The QCD langragian is a component of the Standard Model Lagrangian. However conclusions about particles in QCD alone are not strictly accurate for the full Standard Model. This is basic enough QFT however, I'd suggest reading an introductory textbook.
 
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  • #71
timmdeeg said:
Summary: Provided it's correct that the interpretations of Quantum Mechanics can be neither proved nor disproved why then do researchers invest so much time and talent in this field?

Adam Becker wrote an entire book answering that question. Some people believe foundations is worthless for physics, but as Adam points out in his book, the converse is rather true. https://www.physicsforums.com/threads/interview-with-astrophysicist-adam-becker-comments.943015/
 
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  • #72
timmdeeg said:
So it's the pure joy of participating in an intellectual challenge with no quantifiable outcome. :wink:
I think that's sometimes called "curiosity".
 
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  • #73
Jimster41 said:
I think that's sometimes called "curiosity".
You mean the "curisity" to learn which interpretation is the "true" one? Will we ever know?

But wait, how about the "curiosity" how to get rid of assumptions which seem to contradict each other, e.g. the ugly collapse and the bizarre MWI? Would this be a preferred interpretation, perhaps the "true" one? I haven't been following up the Thermal Interpretation of @Neumaier closely. Isn't this interpretation formulated without recourse to the collapse of the wavefunction and to the Many Words? If yes, has it the potential to be the "true" interpretation then?
 
  • #74
timmdeeg said:
... how to get rid of assumptions which seem to contradict each other, e.g. the ugly collapse and the bizarre MWI?
Not sure what "bizarre" means here, but MWI is just QM without the ugly collapse postulate.
 
  • #75
DarMM said:
The QCD langragian is a component of the Standard Model Lagrangian. However conclusions about particles in QCD alone are not strictly accurate for the full Standard Model. This is basic enough QFT however, I'd suggest reading an introductory textbook.
apparently that is true, however, I have read a lot of textbooks for QFT, standard models, QCD( which are mostly about the formalism) but this fact did not seem obvious. I did check these books again and googled but I did not find any clear specific statement in that regard, maybe I should try harder.
 
  • #76
Michael Price said:
Not sure what "bizarre" means here,
Isn't the story told in #1 here a bit bizarre?
 
  • #77
It's not bizarre, it's simply a misleading formulation you read quite often. I've never understood what the intention of the authors making it may be. It's said "a quantum state can be in a superposition". That's just a meaningless intellectual-sounding phrase.

QT is mathematically formulated using a certain kind of vector space, the socalled Hilbert space, i.e., a complex vector space with a scalar product, inducing a norm according to which it is complete (i.e., all Cauchy sequences of vectors converge), i.e., it's a Banach space. Usually one also assumes that the Hilbert space is separable, i.e., that there exist countable complete orthonormal sets. As for any vector space, each vector can be written as a superposition (generalized also to infinite series) with respect to any such complete orthonormal set. That's a mathematical feature and not very bizarre but simply natural for any Hilbert space.

Further a certain special kind of states, the socalled pure states, can be represented by a normalized state vector ##|\psi \rangle##. The true representatives of states are always statistical operators, i.e., self-adjoint positive semidefinite operators with trace one. For a pure state represented by a state vector ##|\psi \rangle## that statistical operator is the projection operator ##\hat{\rho}_{\psi}=|\psi \rangle \langle \psi|##.

The observables are described by self-adjoint operators, defined on a dense subspace of Hilbert space, which define a complete set of eigenvectors, sometimes including generalized states, living in the dual of the domain of the operator. If an observable is represented by such an operator ##\hat{A}## the possible values it can take are the (generalized) eigenvalues (or spectral values) ##a## of this operator. If the system is prepared in a pure state, for which the observable takes a determined value ##a##, it must be represented by an eigenvector of this eigenvalue, i.e., ##\hat{A} |\psi \rangle=a |\psi \rangle##. Any vector, representing a pure state can be written as a superposition wrt. the eigenbasis of ##\hat{A}##. If it's not a eigenstate of ##\hat{A}## the observable ##A## doesn't take a determined value, and then measuring ##A## you only know the probabilities to get each of the possible values ##a##. That's it.

There's nothing bizarre with this, it's simply how physicists over the last 119 years have figured out nature behaves.
 
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  • #78
Thanks for your detailed response.
vanhees71 said:
It's not bizarre, it's simply a misleading formulation you read quite often. I've never understood what the intention of the authors making it may be. It's said "a quantum state can be in a superposition". That's just a meaningless intellectual-sounding phrase.
Perhaps "bizarre" is the wrong wording. What puzzles me is the multiple existence of the observer. Claus Kiefer says in "Der Quantenkosmos" Das mehrfache und gleichzeitige Vorhandensein desselben makroskopischen Beobachters erscheint freilich so ungeheuerlich, daß es nicht verwundert, wenn Alternativen untersucht werden." That's what I intended to express in my recent post.
 
  • #79
Well, esoterics sells, and since the "hippies saved physics" in the 1960ies quantum esoterics is a pretty well selling kind, particularly since people think it would have to do something with science...

I'm a bit surprised that Claus Kiefer nowadays also writes such stuff...
 
  • #80
vanhees71 said:
I'm a bit surprised that Claus Kiefer nowadays also writes such stuff...
So what's the reason that physicists prefer other alternatives? I don't think that Kiefer is wrong here.
 
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  • #81
vanhees71 said:
It's said "a quantum state can be in a superposition". That's just a meaningless intellectual-sounding phrase.
Quantum mechanics says that the state of the particle can be a superposition of both possible measurement outcomes. It’s not that we don’t know whether the spin is up or down; it’s that it’s really in a superposition of both possibilities, at least until we observe it.

To understand you correctly is this "meaningless intellectual-sounding phrase" just a didact tool?
 
  • #82
I don't know what it is. I never could make any sense of it. It's like saying "this vector in Euclidean 3D space is a superposition". Can you tell, what I want to say with this? I don't know, what sense it should make at all!
 
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  • #83
timmdeeg said:
So what's the reason that physicists prefer other alternatives? I don't that Kiefer is wrong here.
Who prefers which alternatives? I know quite many physicists, but I've never met one who takes such esoterical pseudo-science (aka popular-science) writing seriously.
 
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  • #84
timmdeeg said:
Thanks for your detailed response.

Perhaps "bizarre" is the wrong wording. What puzzles me is the multiple existence of the observer. Claus Kiefer says in "Der Quantenkosmos" Das mehrfache und gleichzeitige Vorhandensein desselben makroskopischen Beobachters erscheint freilich so ungeheuerlich, daß es nicht verwundert, wenn Alternativen untersucht werden." That's what I intended to express in my recent post.
tut tut. This forum is meant to be in English. Translation:
"The quantum cosmos. The multiple and simultaneous presence of the same macroscopic observer seems so monstrous, however, that it is not surprising that alternatives are investigated."
The MWI idea may seem "monsterous", but so have many new scientific ideas. If it fits the data, and is not magical, then it should be accepted. The MWI is a return to the classical certainties, where physical things are real.
 
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  • #85
LOL. So just observing this computer screen, i.e., registering the electromagnetic field with the retina of my eyes and processing it in my brain multiplies myself into zillions of copies. Where are all these copies of myself? Can I talk to them or see them or what? It's even a "return to classical certainties" to have all these copies? Last but not least, where in Everett's original writings are all these copies?
 
  • #86
vanhees71 said:
LOL. So just observing this computer screen, i.e., registering the electromagnetic field with the retina of my eyes and processing it in my brain multiplies myself into zillions of copies. Where are all these copies of myself? Can I talk to them or see them or what? It's even a "return to classical certainties" to have all these copies? Last but not least, where in Everett's original writings are all these copies?
The language of "timelines" is more helpful here. Each of the possible outcomes (near copies of you, in this example) are each actualized in their own timeline, which result from the splitting.
The language of splitting is in Everett's writing, even in his Wheeler-censored PhD thesis summary publication (see the footnote he was able to insert).
 
  • #87
Do you have a reference? I only know the 9-page RMP paper (but I've to admit that I've not read it for a long time), which I remember not to have any esoteric statements like the creation of multiple copies of observers.
 
  • #88
vanhees71 said:
Do you have a reference? I only know the 9-page RMP paper (but I've to admit that I've not read it for a long time), which I remember not to have any esoteric statements like the creation of multiple copies of observers.
Since I have the article open in front of me,
"note added in proof"
... all the separate elements of the superposition individually obey the wave equation with complete indifference to the presence or absence ("actuality" or not) of any other elements. This total lack of effect of one branch on another also implies no observer will ever be aware of any "splitting" process...

But it is irrelevant what Everett said. It follows from quantum theory minus the collapse postulate.
 
  • #89
Michael Price said:
But it is irrelevant what Everett said. It follows from quantum theory minus the collapse postulate.

What follows Quantum Theory minus the Collapse Postulate is the 'Bare Theory' not MWI though.
 
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  • #90
Quanundrum said:
What follows Quantum Theory minus the Collapse Postulate is the 'Bare Theory' not MWI though.
The bare theory has no way of making the other elements of the superposition disappear.
 
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