# Is the cluster decomposition equivalent to no phase transitions?

• A
I think the cluster decomposition states that products of space like separated observable decouple when sandwiched with states.

An analogy with statistical mechanics seems to suggest that we are stating there are no phase transitions. For example, in the Ising model all spins are correlated in the phase transition.

Is this true?

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king vitamin
Gold Member
Yes, I would say that a scale-invariant field theory does not satisfy cluster decomposition. In fact, there is a difficulty in even defining the S matrix in such theories because there is no notion of asymptotically non-interacting particles, and since the definition of cluster decomposition I know (from Weinberg) is in terms of the S matrix, I think the property would immediately fail. Note that expanding fields in terms of creation and annihilation operators also generically fails in strongly-interacting scale-invariant field theories.

I'll add that first-order phase transitions are not scale invariant, so I'm only discussing continuous/critical phase transitions here.

jordi
Yes, I would say that a scale-invariant field theory does not satisfy cluster decomposition. In fact, there is a difficulty in even defining the S matrix in such theories because there is no notion of asymptotically non-interacting particles, and since the definition of cluster decomposition I know (from Weinberg) is in terms of the S matrix, I think the property would immediately fail. Note that expanding fields in terms of creation and annihilation operators also generically fails in strongly-interacting scale-invariant field theories.

I'll add that first-order phase transitions are not scale invariant, so I'm only discussing continuous/critical phase transitions here.
Yes, sure, I was talking about continuous phase transitions, not first order ones, thank you for the clarification.

If it is like this, why is the cluster decomposition so important? For sure, under normal circumstances it happens that an experiment performed "here" is uncorrelated from an experiment done "there". But why should this be a general condition?

Continuous phase transitions are ubiquitous in Nature. Why for fundamental fields it should happen that there should never be a continuous phase transition?

king vitamin
Gold Member
It's really just a physical demand from Weinberg's POV. If our universe were such that the couplings were fine-tuned so that the ground state was a quantum critical point, then at low energies measurements everywhere in the universe would all be correlated with each other. But we don't see correlations between ourselves and Alpha Centauri, so apparently we can ignore those field theories when constructing the field theory of our universe.

In some sense, Weinberg is then limiting himself in his construction of QFT in his textbook. But his textbook does take a very unique approach to QFT, motivating it as a "nearly" unique theory of relativistic particle physics in our universe (whereas QFT has many applications outside of that realm). His chapter on cluster decomposition works up to him motivating the expansion of the Hamiltonian in terms of creation/annihilation operators, whereas it is well-known that such an expansion doesn't exist in certain scale-invariant QFTs. So he is not really treating quantum field theories in complete generality in those chapters.

jordi and Haelfix
Yes, I agree.

I see some kind of logical inconsistency in Weinberg's position: he convincingly argues that nobody should consider the Einstein-Hilbert action seriously (irrespective if so far nobody has seen a violation of this Lagrangian). The action should contain all terms allowed by symmetries, so one should not exclude higher powers, with coefficients "suppressed" by the Planck mass.

However, why should the cluster decomposition principle hold? In the same way as the EH action, we have never seen a violation of it. But conceptually, why should it be true? I think this principle is different from relativistic invariance or unitarity, which are reasonable to assume to be true "always".

Why, for example, it could not be the case that in a black hole there is a phase transition? I am not giving any reason why this should be the case, and I am not arguing that this is the case. But why couldn't it be? Is there a rational argument excluding a phase transition in a black hole (or anywhere else) ab initio, apart from the argument "we have never seen it before"?

"Nassim Taleb's Black Swans everywhere ..."

A. Neumaier
2019 Award
It's really just a physical demand from Weinberg's POV. If our universe were such that the couplings were fine-tuned so that the ground state was a quantum critical point, then at low energies measurements everywhere in the universe would all be correlated with each other. But we don't see correlations between ourselves and Alpha Centauri, so apparently we can ignore those field theories when constructing the field theory of our universe.
We cannot check correlations with Alpha Centauri, so we cannot know whether these exist. Ignoring part of the universe certainly results in dissipation, which is observed!

Moreover, quarks do not satisfy cluster decomposition due to confinement, hence Weinberg's argument is seriously flawed. It can apply only to macroscopically measurable fields.

jordi
Maybe a way out on the quarks issue is the following: the cluster decomposition property says that *if* fields have a high spacelike separation, they decouple. Quarks cannot separate far away, due to confinement. So, the cluster decomposition property does not apply to them.

Of course, it is highly unsatisfying that for some fields, such an important property does not apply to them. And we know it does not apply to them only ex-post (when we learn about confinement).

So, it seems Weinberg highlights something important, but he in some way does not nail it. Of course, it is not his fault, he is one of the best physicists ever, just that the subject is hard.

A. Neumaier
2019 Award
the cluster decomposition property says that *if* fields have a high spacelike separation, they decouple.
No, it says ##\langle\phi(x)\psi(y)\rangle-\langle\phi(x)\rangle\langle\psi(y)\rangle\to 0## in every state, when ##x-y## is spacelike and its norm tends to infinity.

jordi
From https://arxiv.org/pdf/1602.00662.pdf , about Haag's theorem:

"On the mathematical physics side, the verdict was accepted and put down to the impossibility to implement relativistic quantum interactions in Fock space."

If I understand correctly, Fock space is the space of creation and annihilation operators, i.e. the space where the "cluster decomposition property lives".

If this claim is true, no interacting quantum field theory can be defined using creation and annihilation operators, i.e. the cluster decomposition is meaningless in QFT.

Is this true?

A. Neumaier
2019 Award
From https://arxiv.org/pdf/1602.00662.pdf , about Haag's theorem:

"On the mathematical physics side, the verdict was accepted and put down to the impossibility to implement relativistic quantum interactions in Fock space."

If I understand correctly, Fock space is the space of creation and annihilation operators, i.e. the space where the "cluster decomposition property lives".

If this claim is true, no interacting quantum field theory can be defined using creation and annihilation operators, i.e. the cluster decomposition is meaningless in QFT.

Is this true?
No. See my post #8 for a formulation that doesn't depend on Fock space.

jordi
OK, it is true your post #8 does not depend on Fock space. But (recalling from memory) Weinberg in his vol I describes the cluster decomposition using creation and annihilation operators.

Maybe it is possible to describe cluster decomposition in another language that does not conflict with Haag's theorem, but at least it seems to me that Weinberg does not do it.

Demystifier
Gold Member
I think the cluster decomposition states that products of space like separated observable decouple when sandwiched with states.

An analogy with statistical mechanics seems to suggest that we are stating there are no phase transitions. For example, in the Ising model all spins are correlated in the phase transition.

Is this true?
Not exactly. The cluster decomposition principle says this:
https://www.physicsforums.com/threa...ion-and-epr-correlations.409861/#post-2773617
This rules out mathematical phase transitions in infinite systems, but it is fully compatible with physical phase transitions in finite systems.

jordi
A. Neumaier
2019 Award
OK, it is true your post #8 does not depend on Fock space. But (recalling from memory) Weinberg in his vol I describes the cluster decomposition using creation and annihilation operators.

Maybe it is possible to describe cluster decomposition in another language that does not conflict with Haag's theorem, but at least it seems to me that Weinberg does not do it.
Well, Weinberg also ignores Haag's theorem when constructing interacting QFTs. At his level of exposition this is excusable, and your arguments are irrelevant.

A. Neumaier
2019 Award
Haag's theorem is relevant only in a nonperturbative setting. Weinberg's is perturbative.

jordi
DarMM
Gold Member
Basically when you do perturbation theory you compute an expansion of the Correlation functions:
$$\mathcal{W}\left(x_1,...,x_n,\lambda\right) \rightarrow \sum_{n}\lambda^{n}\mathcal{G}_{n}\left(x_1,...,x_n\right)$$

It turns out that the ##\mathcal{G}_{n}\left(x_1,...,x_n\right)## functions can be computed from Fock space. So when you are doing perturbation theory you can ignore Haag's theorem.

The problem is that the series doesn't sum back up to the original ##\mathcal{W}\left(x_1,...,x_n\right)## and begins to diverge around ##n \approx \frac{1}{\lambda}##.

There are methods of getting around this such as Borel summation that try to modify the series. A simple case is that you alter the ##\lambda^{n}## to be ##\frac{\lambda^{n}}{n!}##. Let's make this replacement:
$$\mathcal{B(W)}\left(x_1,...,x_n,t\right) = \sum_{n}\frac{t^{n}}{n!}\mathcal{G}_{n}\left(x_1,...,x_n\right)$$
Note I've replaced ##\lambda## with ##t## to avoid confusion later.

Then you can recover the original function with the integral:
$$\mathcal{W}\left(x_1,...,x_n\right) = \int_{0}^{\infty}{e^{-\frac{t}{\lambda}}\mathcal{B(W)}\left(x_1,...,x_n,t\right)dt}$$

This works great in lower dimensions, but unfortunately in 4D in for example Yang-Mills theory the integral has poles for certain values of ##t##.

These poles prevent you from performing the integral. Of course one can shift the contour around the poles (typically there are infinitely many) however this introduces an ambiguity in the evaluation. This ambiguity is roughly ##\mathcal{O}\left(e^{\frac{1}{\lambda}}\right)##, so clearly a nonperturbative effect.

Some poles are instantons and others are renormalons.

Renormalons originate in chains of bubble graphs in perturbation theory. When you renormalize the theory such diagrams have their divergences cancelled by counterterms, but the counterterms also contain finite parts that do not cancel infinities in graphs. These finite counterterm pieces end up contributing a ##n!## term to the perturbative series. It's this rapid growth term that causes the renormalon poles.

So note that the unrenormalized Feynman series lacks such renormalon poles. However due to being unrenormalized they diverge as you remove the cutoff and thus have no continuum limit (obviously since this is why you bother renormalizing at all).

jordi, A. Neumaier and dextercioby
Very interesting, thank you. Is there a reference to understand Borel summation in this setting? (I mean, not a comprehensive one, but one that goes to the basics and does explicit calculations).