# Could QM Arise From Wilson's Ideas

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Mentor
In adding some detail to a question about mass, I gave a link to an article by Sean Carrol:
https://www.preposterousuniverse.com/blog/2013/06/20/how-quantum-field-theory-becomes-effective/
'Nowadays we know you can start with just about anything, and at low energies, the effective theory will look renormalizable. This is useful, if you want to calculate processes in low-energy physics; disappointing if you’d like to use low-energy data to learn what is happening at higher energies. Chances are, if you go to energies that are high enough, spacetime itself becomes ill-defined, and you don’t have a quantum field theory at all. But on labs here on Earth, we have no better way to describe how the world works.'

If QFT is the low energy approximation of just about anything, and QM is a limiting case of QFT, it struck me that could possibly be the 'why' of QM?

Thanks
Bill

dRic2, vanhees71, sandy stone and 2 others

Staff Emeritus
In adding some detail to a question about mass, I gave a link to an article by Sean Carrol:
https://www.preposterousuniverse.com/blog/2013/06/20/how-quantum-field-theory-becomes-effective/
'Nowadays we know you can start with just about anything, and at low energies, the effective theory will look renormalizable. This is useful, if you want to calculate processes in low-energy physics; disappointing if you’d like to use low-energy data to learn what is happening at higher energies. Chances are, if you go to energies that are high enough, spacetime itself becomes ill-defined, and you don’t have a quantum field theory at all. But on labs here on Earth, we have no better way to describe how the world works.'

If QFT is the low energy approximation of just about anything, and QM is a limiting case of QFT, it struck me that could possibly be the 'why' of QM?

Thanks
Bill
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.

Mentor
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.
Quantum Gravity is not renormalisable. But using the Wilsonian view that is not an issue. You know we can find a low energy approximation that calculations can be done with (an effective field theory):
https://blogs.umass.edu/donoghue/research/quantum-gravity-and-effective-field-theory/

We now think that all our QFT theories are effective field theories - even the standard model. It could break down earlier (indeed QED becomes the electroweak theory) but none are generally trusted when we get to the Plank Scale.

The interesting thing and the details are beyond my current knowledge, is you can start with just about anything to get that effective field theory. Lubos, who I normally do not like giving links to, wrote a deeper article about it:
https://motls.blogspot.com/2013/06/kenneth-wilson-rip.html

As I said it is above my level.

Thanks
Bill

Mentor
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.
Yes, but at low enough energies, the quantum aspects of gravity are negligible; you just have a fixed classical spacetime geometry. I think that is the context in which to understand Carroll's remark.

bhobba
If QFT is the low energy approximation of just about anything, and QM is a limiting case of QFT, it struck me that could possibly be the 'why' of QM?
No, the Wilsonian view of renormalization is just about actions. How one uses the actions (eg. as part of a quantum theory) has to be put in by hand.

For example, QFT may arise from string theory, so string theory may provide an explanation for QFT. However, it does not provide an explanation for the "quantum" part of QFT, since string theory is itself a quantum theory.

dextercioby, Demystifier and bhobba
Mentor
Yes, but at low enough energies, the quantum aspects of gravity are negligible; you just have a fixed classical spacetime geometry. I think that is the context in which to understand Carroll's remark.

Way back in the day when I posted on sci. Physics. Relativity I asked Steve Carlip the same question. Certainly, we have an EFT theory of gravity thought valid to about the Plank scale. His views are interesting:

The problem is it does not tell us much. It predicts a few interesting things, e.g., Hawking Radiation can be derived from it, but it does not get us terribly much. The link I gave to John Donoghue's work attempts to see exactly what such a theory may tell us.

Thanks
Bill

Gold Member
If QFT is the low energy approximation of just about anything, and QM is a limiting case of QFT, it struck me that could possibly be the 'why' of QM?
As @atyy said, the Wilson effective theory framework requires that the fundamental microscopic theory should be quantum, but apart from that, it can in principle be any quantum theory. In the paper linked in my signature, I propose that the fundamental theory is a nonrelativistic quantum theory, which resolves many conceptual problems in standard and Bohmian quantum theory.

atyy and bhobba
As @atyy said, the Wilson effective theory framework requires that the fundamental microscopic theory should be quantum, but apart from that, it can in principle be any quantum theory. In the paper linked in my signature, I propose that the fundamental theory is a nonrelativistic quantum theory, which resolves many conceptual problems in standard and Bohmian quantum theory.
In a way we can also link that to Wilson by his pioneering work on lattice gauge theory, which in some philosophies may provide a non-relativistic non-perturbative definition of quantum field theories (of course, we don't yet have a consensus lattice standard model - it'll be interesting to see if the muon g-2 is really helped by the lattice!)

As I have said before, lattice research is secretly Bohmian

bhobba and Demystifier
Gold Member
bhobba and atyy
Staff Emeritus
Yes, but at low enough energies, the quantum aspects of gravity are negligible; you just have a fixed classical spacetime geometry. I think that is the context in which to understand Carroll's remark.

My problem is not with the claim that we can have a sensible low-energy theory of quantum gravity, but with the claim that at low energies it "looks renormalizable". What does it mean to look renormalizable?

My problem is not with the claim that we can have a sensible low-energy theory of quantum gravity, but with the claim that at low energies it "looks renormalizable". What does it mean to look renormalizable?
Carroll's explanation:
"But we don’t care about high energies! We are trying to construct an effective theory at low energies, so we care about the terms for which N≤4 — those are the ones that dominate at low energies. In fact, we have lingo to encapsulate this importance.
...
And that’s it! Those pieces give you the important low-energy description of any theory of a single scalar field, no matter what new particles and crazy nonsense might be going on at higher energies. Of course we don’t work at strictly zero energy, so the “irrelevant” parts might also be interesting and useful, but Wilsonian effective field theory gives you a systematic way of dealing with them and estimating their importance.
...
The old-school idea that a theory is “renormalizable” maps onto the new-fangled idea that all the operators are either relevant or marginal — every single operator is dimension 4 or less."

bhobba
Mentor
My problem is not with the claim that we can have a sensible low-energy theory of quantum gravity
I didn't say we did; I said that at low energies the quantum aspects of gravity are negligible, so the whole idea of a "low-energy theory of quantum gravity" is unnecessary. All we need is the classical theory of gravity we already have.

but with the claim that at low energies it "looks renormalizable".
For gravity, "looks renormalizable" at low energies really means "looks classical". See above.

The old-school idea that a theory is “renormalizable” maps onto the new-fangled idea that all the operators are either relevant or marginal — every single operator is dimension 4 or less."
And for the "naive" quantum theory of gravity, i.e., the massless spin-2 field theory whose field equation turns out to be the Einstein Field Equation, there are no such operators--that's why the theory is said to be non-renormalizable. The coupling constant for this theory has units of mass to the power -2, which makes it "irrelevant" in Wilson's terminology. There aren't any relevant or marginal terms in the Lagrangian at all. So when you look at this theory at low energies, it says there's no quantum stuff going on at all; as I said above, all you have left is a fixed classical spacetime geometry.

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Staff Emeritus
So when you look at this theory at low energies, it says there's no quantum stuff going on at all; as I said above, all you have left is a fixed classical spacetime geometry.
A theory with a fixed geometry can’t describe the gravitational interaction between two objects.

So the meaning of Wilson’s statement is that at low energies, the only relevant interactions are renormalizable ones. (I’m not using “relevant” in the technical sense, which I don’t understand…) Nonrenormalizable interactions can only come into play as fixed background fields.

Mentor
A theory with a fixed geometry can’t describe the gravitational interaction between two objects.
Sure it can. The interaction is encoded in the spacetime geometry.

So the meaning of Wilson’s statement is that at low energies, the only relevant interactions are renormalizable ones. (I’m not using “relevant” in the technical sense, which I don’t understand…) Nonrenormalizable interactions can only come into play as fixed background fields.
That seems correct to me, yes.

bhobba
Staff Emeritus
Sure it can. The interaction is encoded in the spacetime geometry.
No, it can’t be. You can’t have a fixed background geometry to represent an electron, for example, if the electron is treated as a quantum particle.

So the particles that are described by QM don’t interact with each other gravitationally.

gentzen
Mentor
You can’t have a fixed background geometry to represent an electron, for example, if the electron is treated as a quantum particle.

So the particles that are described by QM don’t interact with each other gravitationally.
I don't think this viewpoint is correct.

If it were really true that you can't have a fixed background geometry with a quantum particle, that would apply just as much to flat Minkowski spacetime ("no gravitational interaction") as any other spacetime. So you would be unable to have a model that included both a spacetime geometry and quantum particles at all. All of quantum field theory would go out the window.

What physicists actually do in these types of situations is to use an effective stress-energy tensor for the quantum matter (usually the expectation value of some appropriate operator) as the source in the Einstein Field Equation, and obtain a self-consistent solution in which the gravitational interactions encoded by the spacetime geometry are the "right" ones for the matter that is present. This approach assumes that whatever quantum stuff is going on does not produce any effects that would invalidate the effective stress-energy tensor being used at the classical level. Basically, it means you don't have any superpositions of macroscopically different distributions of matter.

bhobba
Staff Emeritus
I don't think this viewpoint is correct.

If it were really true that you can't have a fixed background geometry with a quantum particle,

It’s not that you can’t have a background geometry, but that geometry cannot take into account quantum particles.

You can have electrons moving in a background geometry but by definition that background doesn’t include the effect of those electrons. The background geometry would (contrary to the spirit of Newton’s third law) act on the electrons but would not be acted on by them.

My problem is not with the claim that we can have a sensible low-energy theory of quantum gravity, but with the claim that at low energies it "looks renormalizable". What does it mean to look renormalizable?
Carroll is probably not claiming that gravity is renormalizable in the sense that he's using, since he qualifies his statement with "just about" ("Nowadays we know you can start with just about anything"). He also has other qualifications in his explanation like "(Strictly speaking, even “irrelevant” operators can be important. In the Fermi theory of the weak interactions, the lowest-order operator you can construct that gives rise to any interaction at all is dimension 6. So you have to keep that interaction to have anything interesting happen — but we say that the resulting theory is “non-renormalizable.”) (And while we’re speaking strictly, this dimensional analysis gives the leading behavior, but not the whole story. In QCD, for example, the coupling is marginal, but it doesn’t remain exactly constant with energy, but rather changes slowly [logarithmically]. If all of your couplings are exactly constant, you have a conformal field theory.)"

So for his "just about" he probably was thinking about the interactions of the standard model (without gravity) which were historically obtained by considerations of "renormalizability". He's saying why although it doesn't really make sense to insist on "renormalizability" ("Pre-Wilson, it was all about finding theories that are renormalizable, which are very few in number"), it was historically a very successful approach.

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Demystifier and bhobba
Mentor
It’s not that you can’t have a background geometry, but that geometry cannot take into account quantum particles.

You can have electrons moving in a background geometry but by definition that background doesn’t include the effect of those electrons. The background geometry would (contrary to the spirit of Newton’s third law) act on the electrons but would not be acted on by them.
I'm sorry, but you are not responding to the part of my post that specifically addressed exactly this issue: yes, the background geometry can take into account the presence of quantum particles, through their effective stress-energy tensor. If you are claiming that it is impossible to have a self-consistent solution in which the source of gravity in a curved spacetime is the effective stress-energy tensor of quantum particles, so that the spacetime geometry includes gravitational interactions between different lumps of stress-energy made of quantum particles, you are simply wrong. There is plenty of literature on this approach; I first learned of it in Wald's 1993 monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics.

Staff Emeritus
I'm sorry, but you are not responding to the part of my post that specifically addressed exactly this issue: yes, the background geometry can take into account the presence of quantum particles, through their effective stress-energy tensor.

No, it doesn't address the issue. The approach you're talking about doesn't describe the gravitational effect of one electron on another. You can't calculate the gravitational scattering of one particle by another.

gentzen
Staff Emeritus
No, it doesn't address the issue. The approach you're talking about doesn't describe the gravitational effect of one electron on another. You can't calculate the gravitational scattering of one particle by another.

Let me illustrate with an analogous approach for electrodynamics. Before QED was developed, one could attempt to compute the electromagnetic interaction of two electrons this way:
1. Start with an ansatz ##J^\mu## for the current due to the two electrons.
2. Use Maxwell's equations to calculate a corresponding electromagnetic potential ##A^\mu##.
3. Calculate the wave functions for the two electrons using ##A^\mu## as a background vector potential.
4. Using the results of step 3, adjust ##J^\mu##.
5. Repeat 1 through 4 until you have a self-consistent solution.
(Actually, I think that an approach like this may have actually been used to calculate energy levels for many-electron atoms.)

This approach might give good results for some circumstances. But it's not at all clear to me that it is some kind of low-energy limiting case of QED.

Mentor
The approach you're talking about doesn't describe the gravitational effect of one electron on another.
It's not just the gravitational effect that you are talking about, it's the quantum aspect of such a gravitational effect. (The classical gravitational effect is described perfectly well by GR.) And, since we are talking about the low energy regime, any such quantum aspects are negligible, as I have already said. (For electrons, we can't even detect the classical gravitational interaction between them, let alone any quantum aspects of it.)

Furthermore, the quantum aspect of gravitational interaction is not the same as background geometry taking into account the presence of quantum particles, which is what you were saying was not possible; the latter is much broader. The Earth is made of quantum particles, but we can describe its gravitational effects just fine using a background spacetime geometry. I understand how that that is not the case you were trying to focus on, but that wasn't clear to me before.

This approach might give good results for some circumstances. But it's not at all clear to me that it is some kind of low-energy limiting case of QED.
The approach you describe here is not the same as the approach I was describing for quantum fields in curved spacetime. An EM analogy to the approach I was describing would be computing the wave functions for electron orbitals in the hydrogen atom by treating the proton purely as a source of an external potential in the Hamiltonian and ignoring all of its quantum properties. (IIRC, in the Feynman Lectures on Gravitation, Feynman tried to make a similar computation for a hypothetical electrically neutral proton and electron, interacting only through gravity, and came up with a Bohr radius for the lowest orbital of the electron larger than the observable universe and a binding energy for that orbital of something like ##10^{-70}## Rydbergs.)

Again, I understand now that this "semiclassical" approach is not what you want to focus on; but in the low energy regime, there isn't anything else there, as I've already said. An interaction like quantum gravitational scattering of one electron off of another simply doesn't appear at all in the low energy effective theory.

WernerQH
If QFT is the low energy approximation of just about anything, and QM is a limiting case of QFT, it struck me that could possibly be the 'why' of QM?
Thank you for bringing this up, and hopefully we can return to the more immediate problem of understanding Q(F)T in flat spacetime before drifting into quantum gravity. I´ve been thinking about this for a long time, and I´ve become convinced that the measurement problem will dissolve once we´ve learnt to view QFT from the right angle. For many physicists QFT is fundamental, but it is probably better seen as a kind of phenomenological theory: a neat way to describe correlations. Also plain QM is just a machinery for calculating correlation functions (e.g. in Bell-type experiments). Of course it must be explained what it is that is "correlated", for example phenomena in liquid helium or magnetic materials. It´s not sufficient to say it´s "measurements" or "quantum fields". The correlations are features of the real world. Otherwise it would be inexplicable why people at CERN haven´t turned to measuring numbers that are less costly to obtain. :-)

For a homogeneous, tenuous medium it is fairly straightforward to derive a Kubo formula for the photon absorption coefficient: $$\kappa = \frac {\mu_0 c} {2 \hbar \omega} \sum_{\mu\nu} {\bf e}_\mu^* {\bf e}_\nu \int dt \int d^3x \ e^{-i (kx - \omega t)} \langle j_\mu(x,t) j_\nu(0,0) - j_\nu(0,0) j_\mu(x,t) \rangle$$ It is suggestive to think of the current commutator as absorption minus stimulated emission, and ## j_\mu(x,t_<) j_\nu(0,0_>) ## as signifying absorption, ##
j_\nu(0,0_<) j_\mu(x,t_>) ## as emission events. Currents are real, and photons can be counted (at least when their frequencies are well above 60 Hz). One should drop the classical picture of "current (density)" as a continuous field spread out smoothly over spacetime. Quantum theory takes into account the actual graininess of matter, and it is more natural (!) to consider events localized in space and time. Fields retain their meaning only in a statistical sense, as expectation values expressing correlations between events. Since there is only one average, currents cannot flow into two different directions at the same time, but this is a very weak constraint on the microscopic events contributing to the average.

Schrödinger´s equation describes the continuous, deterministic evolution of a "wave function" and clearly doesn´t square with the abruptness and randomness of processes occurring in the real world. Von Neumann patched this up with intermittent "measurements". But this has created the "measurement problem". The problem is the lopsided view of the wave function and Schrödinger´s equation as the essence of quantum theory. There is another Schrödinger equation that has ## i ## relaced by ## -i ## and decribes the (backward) evolution of bras. It is usually considered a redundant mirror image of the normal Schrödinger equation. But it is not. Events on the forward evolving sheet of spacetime are just tightly correlated with the events on the other sheet evolving backwards. At the classical level (perhaps down close to the Planck scale) they coincide, but in quantum theory we have to consider bras and kets separately. They are combined using Born´s rule, and only the three together give the complete picture of quantum theory. The Schwinger-Keldysh formalism encapsulates all that in a coherent way and doesn´t need the concept of measurement.

bhobba
Mentor
The Schwinger-Keldysh formalism
Can you give a reference?

WernerQH
Can you give a reference?
For example Landau / Lifshitz, vol X, ch X.
Or Stefanucci / van Leeuwen, "Nonequilibrium Many-Body Theory of Quantum Systems" (CUP, 2013).
An early paper of Schwinger's is J. Math. Phys. 2, 407 (1961).

Mentor
I´ve been thinking about this for a long time, and I´ve become convinced that the measurement problem will dissolve once we´ve learnt to view QFT from the right angle.

So do I. But it has proven difficult to get a consensus on that 'right' angle. I think of QM as a generalised probability theory and the measurement problem as just a carryover of issues at the foundations of probability theory. They are still being 'debated', even today.

Thanks
Nils

WernerQH
I think of QM as a generalised probability theory and the measurement problem as just a carryover of issues at the foundations of probability theory.
This generalised probability theory has arisen because people have been looking only at one half of the picture (the "wave function"). When kets and bras are taken together, normal probability works just fine and makes perfect sense. (At least to me. :-)

bhobba
Gold Member
I´ve been thinking about this for a long time, and I´ve become convinced that the measurement problem will dissolve once we´ve learnt to view QFT from the right angle.
For a long time I've been thinking exactly the opposite. But then I learned about the condensed matter view of QFT, where the field is often an effective long distance non-fundamental thing. Then Wilson's ideas started to make more sense to me and the idea that relativity is not fundamental started to look like a very natural idea. With relativity being non-fundamental, Bell nonlocality suddenly becomes non-problematic. And when non-locality ceases to be a problem, additional variable approaches to the measurement problem like Bohmian mechanics become more natural.

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bhobba
WernerQH
With relativity being non-fundamental, Bell nonlocality suddenly becomes non-problematic.
With propagators reaching backwards in time, Bell nonlocality poses no problem for QFT either. And I see the measurement problem as a non-problem.

bhobba
Staff Emeritus
With propagators reaching backwards in time, Bell nonlocality poses no problem for QFT either. And I see the measurement problem as a non-problem.

The measurement problem has a pragmatic answer, but it doesn't (yet, as far as I know) have a mathematically consistent answer that doesn't go beyond accepted quantum mechanics.

Suppose you have an electron that has a spin state ##\alpha |U\rangle + \beta |D\rangle## where ##|\alpha|^2 + |\beta|^2 = 1## and ##|U\rangle## and ##|D\rangle## are eigenstates of the z-component of spin with spin +1/2 and -1/2, respectively. Does this mean that the electron has a probability of ##|\alpha||^2## of being spin-up in the z-direction, and a probability of ##|\beta|^2## of being spin-down?

I think most people would say "no". Until it's measured, the particle doesn't have a spin in the z-direction. It's not that we just don't know what its spin is. As a matter of fact, you could say that there is nothing at all uncertain about the electron's spin: It has a definite spin of +1/2 along the axis ##\vec{S}## satisfying

##S_z = \frac{1}{2} (|\alpha|^2 - |\beta|^2)##
##S_x = \frac{1}{2} (\alpha \beta^* + \alpha^* \beta)##
##S_z = \frac{i}{2} (\alpha \beta^* - \alpha^* \beta)##

So now put that electron through a Stern-Gerlach device, so that electrons that are spin-up in the z-direction are deflected to the left, to make a dot on the left side of a photographic plate, and electrons that are spin-down in the z-direction make a dot on the right side.

Most people would say that now probabilities come into play. There will be a probability of ##|\alpha|^2## of a dot on the left, and a probability of ##|\beta|^2## of a dot on the right.

But why? Presumably, Stern-Gerlach devices and photographic plates are made up of electrons and protons and photons and neutrons. Each of these constituents is like the original electron, in having a quantum state. So why isn't it the case for this huge system that there are no probabilities until someone measures ITS state? (And whatever you use to measure its state, you can ask why isn't it described by a quantum state, requiring yet another measurement to give probabilities.)

The measurement problem is basically why probabilities apply to measurements (which presumably are just quantum mechanical interactions involving a huge number of particles) but not to interactions involving a small number of particles (one or two or three electrons, for example)?

eloheim and physika
Staff Emeritus