A Could QM Arise From Wilson's Ideas

[kith]

Another answer that I have heard is that the basic probabilities are for macroscopic collective properties. In the limit as the number of particles goes to infinity, collective properties such as total momentum and center of mass commute, even though single-particle momentum and position do not.

Do you have some pointers to where I can read more about this?

It might also be related to the thermal interpretation of @A. Neumaier which I don't really understand yet.

 

[EPR]

That just raises the question: knowledge about what?

Knowledge about observables that the respective quantum field will produce in accordance with the calculated probabilities.

 

[stevendaryl]

Knowledge about observables that the respective quantum field will produce in accordance with the calculated probabilities.

The difficulty with that answer is that the concept of “knowledge about the value of this observable” assumes that the observable has a value. You can’t know information if that information doesn’t exist.

So that answer doesn’t actually answer anything. Does the observable have a value before it is measured? If yes, that’s a hidden variable theory, which is hard to make consistent. If no, then you have the question of how measurement beings these values into existence.

 

[EPR]

The difficulty with that answer is that the concept of “knowledge about the value of this observable” assumes that the observable has a value. You can’t know information if that information doesn’t exist.

So that answer doesn’t actually answer anything. Does the observable have a value before it is measured? If yes, that’s a hidden variable theory, which is hard to make consistent. If no, then you have the question of how measurement beings these values into existence.

We may have to look for clues outside quantum physics for these answers.

It's a relative reality - everything is a worldline moving through spacetime. Observables thus would appear to have values before measurement and they seem to be hidden(hidden variables) before measurement.
You can counter this by asking how conscious experience could arise in a blockworld Universe and I would say that it probably wouldn't. And it probably didn't.

 

[PeterDonis]

everything is a worldline moving through spacetime

This is true in classical relativity, but it's not necessarily true in QM; it depends on which interpretation you adopt. Interpretation-dependent discussion belongs in the QM interpretation forum, not this one.

 

[Demystifier]

The problem with QM is its ontology, the talk of "quantum objects" with or without well defined properties, or only when "measured". These objects can emerge as special patterns of events in spacetime, just as pixels on your screen may form the letter "Q" without Q-ness being a fundamental property of the screen.

Are you saying that position is a preferred observable?

 

[WernerQH]

Are you saying that position is a preferred observable?

Yes. I don't believe that position and momentum can have exactly the same status.

 

[WernerQH]

the features that make probabilities problematic in quantum mechanics

Do you also object to probabilities derived using Fermi's Golden Rule, like decay rates and cross sections? Those should be unproblematic. They always refer to a finite interval of time.

I suspect that you are looking for something that quantum theory does not provide: an "instantaneous" probability characterizing the present "state" of a system, which could replace the Born rule. Schrödinger's equation only gives the appearance of Markovian evolution. It is not the whole truth.

 

[stevendaryl]

I suspect that you are looking for something that quantum theory does not provide: an "instantaneous" probability characterizing the present "state" of a system

No, I specifically said that I'm looking for a clear statement of what events quantum probabilities are probabilities for. I gave several candidate answers, but there are issues with all of them.

The standard quantum recipe says that they are probabilities for measurement outcomes.

 

[Demystifier]

Schrödinger's equation only gives the appearance of Markovian evolution. It is not the whole truth.

What's the rest of the truth?

 

[A. Neumaier]

(2) The Keldysh closed time-path formalism. It eliminates the need for "measurements".

How does it achieve this claimed fact??

The CTP formalism only calculates q-expectations (n-point functions) but does not relate these to measurements. Thus it does not even touch the questions associated with measurements, let alone eliminate the need for them.

 

[WernerQH]

The CTP formalism only calculates q-expectations (n-point functions) but does not relate these to measurements. Thus it does not even touch the questions associated with measurements, let alone eliminate the need for them.

Indeed, it doesn´t mention measurements. Why should it?
Through the fleeting presence of a virtual W-boson, the collision between two protons in the interior of the sun occasionally produces a deuteron. We have a good theory for that. What´s the place of "measurement" in a theory of the interior of the sun?

I think John Bell has made a strong case "Against Measurement". Von Neumann may have hoped to make a vague term like "measurement" precise by embedding it in a rigid set of axioms. But the continuing debates on the measurement problem indicate that this hasn´t happened. Mathematicians delight in the formal structure of a theory, but physicists are more interested in what it is that is being measured.

 

[WernerQH]

What's the rest of the truth?

A stochastic element is obviously missing. If evolution is perfectly continuous, you would have to conclude that the click of a Geiger counter is only a trick played on us by our senses.

 

[EPR]

Indeed, it doesn´t mention measurements. Why should it?
Through the fleeting presence of a virtual W-boson, the collision between two protons in the interior of the sun occasionally produces a deuteron. We have a good theory for that. What´s the place of "measurement" in a theory of the interior of the sun?

You already assume the existence of a classical Sun made of classical particles that bumb into each other. The only issue with this is that these classical particles do not exist
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[Demystifier]

A stochastic element is obviously missing. If evolution is perfectly continuous, you would have to conclude that the click of a Geiger counter is only a trick played on us by our senses.

So you mean something like Nelson interpretation? (Particles have stochastic trajectories $x(t)$ which are continuous but non-smooth.)

 

[stevendaryl]

Indeed, it doesn´t mention measurements. Why should it?
Through the fleeting presence of a virtual W-boson, the collision between two protons in the interior of the sun occasionally produces a deuteron. We have a good theory for that. What´s the place of "measurement" in a theory of the interior of the sun?

I agree that quantum mechanics SHOULDN'T be about measurements. But the clearest statement about how probabilities come into play in QM does involve measurements. When you measure an observable, you get an eigenvalue of the corresponding operator, with probabilities given by the Born rule.

Decoherence seems like a replacement for measurement in the formalization of QM. You take the full density matrix and trace out the environmental degrees of freedom and what's left looks approximately diagonal. As if the system had "collapsed" to one configuration with probabilities given by the Born rule. But that's a little unsatisfying to me because it seems very subjective to choose which degrees of freedom are the environment.

 

[stevendaryl]

I agree that quantum mechanics SHOULDN'T be about measurements. But the clearest statement about how probabilities come into play in QM does involve measurements. When you measure an observable, you get an eigenvalue of the corresponding operator, with probabilities given by the Born rule.

That's one way to state the "measurement problem": formulate QM in a way that doesn't mention measurements at all, but which gives the same probabilities as the standard recipe. Bohmian mechanics does that.

 

[A. Neumaier]

it doesn´t mention measurements. Why should it?

Because the problems are at the point where an experiment records permanent data - where the talk about probabilities ends. So not in the sun, but when the light emitted from the sun enters our instruments and leaves permanent traces.

The CTP formalism does not help in the least to understand how this can happen.

 

[WernerQH]

Because the problems are at the point where an experiment records permanent data - where the talk about probabilities ends. So not in the sun, but when the light emitted from the sun enters our instruments and leaves permanent traces.

Can nuclear reactions be understood without quantum theory? And don't you think of the accumulation of helium in the sun as a "permanent trace"?

 

[WernerQH]

So you mean something like Nelson interpretation? (Particles have stochastic trajectories $x(t)$ which are continuous but non-smooth.)

No. There must be discontinuities. The number of photons, for example, cannot increase by 0.77.

 

[Sunil]

I don’t understand the claim that “we know you can start with just about anything, and at low energies, the effective theory will look renormalizable”. I thought that the whole reason that quantum gravity is so hard is because the most naive way to quantize GR leads to something that is non-renormalizable.

But for low energy, gravity is extremely weak, so weak that it safely can be ignored. What remains observable in all those particle colliders are only renormalizable theories.

So, the point remains correct. While the whole theory of physics, gravity + SM, is not renormalizable because of GR, it looks renormalizable in particle colliders.

It is only because gravity has no negative charges, that all masses add up, which makes gravity visible in comparison with the SM fields.

 

[A. Neumaier]

Can nuclear reactions be understood without quantum theory? And don't you think of the accumulation of helium in the sun as a "permanent trace"?

How do we know there is accumulated helium in the sun? Even to measure the amount of helium accumulated in the sun, measurements are required. The understanding of the latter in terms of the microscopic quantum description of the equipment is missing.

 

[PeterDonis]

The number of photons, for example, cannot increase by 0.77.

Sure it can. There are plenty of states which are not eigenstates of photon number, and it's easy to find two of them whose expectation values of photon number differ by 0.77.

 

[WernerQH]

Sure it can. There are plenty of states which are not eigenstates of photon number, and it's easy to find two of them whose expectation values of photon number differ by 0.77.

That's interpretation-dependent. You're assuming that the wave function represents an individual system.

 

[PeterDonis]

That's interpretation-dependent. You're assuming that the wave function represents an individual system.

No, I'm just pointing out that your "number of photons" is not the simple thing you appear to think it is. Nothing I said was interpretation dependent: states and expectation values are part of the basic math of QM.

 

[Sunil]

Sure it can. There are plenty of states which are not eigenstates of photon number, and it's easy to find two of them whose expectation values of photon number differ by 0.77.

Yes. Given the beautiful mathematics necessary for doing this, it is useful to learn it. The tool is the holomorphic representation of the canonical commutation relations. (Formulas out of bad memory, thus, modulo signs, p replaced with q, and factors $2, \sqrt{2}, \sqrt{2\pi}$.) On the complex plane $z=p+iq$ states are holomorph functions $f(z)$ with the scalar product
$$ \langle f,g \rangle \sim \int \bar{f}(z) g(z) e^{-z\bar{z}}$$
The probability density $\rho(z)\sim\bar{f}(z) f(z) e^{-z\bar{z}}$ has a quite simple physical interpretation. Make an approximate common measurement of $\hat{p}$ and $\hat{q}$ by measuring instead the communting operators $\hat{p}+\hat{p}_1$ and $\hat{q}-\hat{q}_1$, where $\hat{p}_1$ and $\hat{q}_1$ describe a second test particle prepared in its harmonic oscillator ground state. So, if you want to measure energy, you can measure that p, q and compute H(p,q) as defined by classical physics.
Then, for every point $z_0$ of the plane you have a corresponding state localized around it, named coherent states, $f(z)\sim e^{z-z_0}$, which gives $\rho(z)\sim e^{-(z-z_0)(\bar{z}-\bar{z}_0)}$. Remarkably, in the harmonic oscillator with $H=\frac12 z\bar{z}$ these coherent states follow exactly the classical trajectory. And the energy eigenstates are simply $f_n(z)\sim z^n$.

 

[Demystifier]

No. There must be discontinuities. The number of photons, for example, cannot increase by 0.77.

Discontinuity in the number of photons is one thing, discontinuity in the trajectory $x(t)$ is another.

 

[Interested_observer]

Einstein was right. QM is useful, but it is not complete.

 

[WernerQH]

Discontinuity in the number of photons is one thing, discontinuity in the trajectory $x(t)$ is another.

Of course. I dislike Nelson's theory because I think the notion of a "trajectory" is basically flawed.

 

[Demystifier]

Of course. I dislike Nelson's theory because I think the notion of a "trajectory" is basically flawed.

Why is it flawed?