QFT renormalization and Haag's theorem

In summary, the Wilsonian effective field theory viewpoint says that there is a cut-off below which we write an action involving the low energy degrees of freedom and every conceivable term consistent with the symmetries we know. Then hopefully, if we only look coarsely at low energies, we will only need a few terms for good accuracy, and from limited low energy data we can fit those few terms and be able to make good predictions for other low energy data that we haven't seen.
  • #1
eoghan
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Hi all!
I'm a beginner in QFT. I've read a lot of posts here about Haag's theorem, but I haven't found one which can answer simply and briefly to my question (if such an answer exists):

Do UV divergencies appear because of the Haag's theorem?

Thank you
 
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  • #2
Haag's theorem is a theorem of rigourous QFTs.

Most QFTs are not rigourous (as far as we know), and need a cut-off. Once there is a cut-off, there are no UV divergences. The cut-off means there are new degrees of freedom that become important above the cut-off, but we don't need to know about them if we only do coarse grained experiments at low energy, so our QFTs are low energy effective theories.

Conceptually, the Wilsonian effective field theory viewpoint says there is a cut-off. Below the cut-off we write an action involving the low energy degrees of freedom and every conceivable term consistent with the symmetries we know. Then hopefully, if we only look coarsely at low energies, we will only need a few terms for good accuracy, and from limited low energy data we can fit those few terms and be able to make good predictions for other low energy data that we haven't seen.

In practice, people don't write down "every conceivable term" and see what happens at low energies. Instead, there is the traditional mysterious practice of writing down counterterms to cancel divergences. In fact, this can be shown to produce the same results as the Wilsonian viewpoint of starting with "every conceivable term". So one can use the counterterm technique while believing conceptually in the Wilsonian effective field theory point of view.

The Wilsonian viewpoint is discussed by Srednicki in http://web.physics.ucsb.edu/~mark/qft.html (chapter 29).

In special cases, we can run the "renormalization flow" upwards to higher energies, and find that the flow converges to a fixed point, without introducing new degrees of freedom. This is called Asymptotic Safety. A special case of Asymptotic Safety is Asymptotic Freedom, which just means the theory becomes weakly coupled at the high energy fixed point. If physicists find a QFT that is Asymptotically Safe, then they believe it has a chance of being rigourous. Because Yang-Mills is Asymptotically Safe, making Yang-Mills rigourous has been set as a Clay Millennium Problem, as described in http://www.claymath.org/sites/default/files/yangmills.pdf. That article also gives references to rigourous QFTs in fewer spacetime dimensions. But it is important to understand that the Wilsonian viewpoint provides a framework for doing QFT that makes physical common sesne, without needing our QFTs to be one of these special Asymptotically Safe cases.
 
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  • #3
eoghan said:
Do UV divergencies appear because of the Haag's theorem?

Just to add to what Atty said the following explains what's going on with renormalisation pretty well:
http://arxiv.org/pdf/hep-th/0212049.pdf

Thanks
Bill
 
  • #4
eoghan said:
I'm a beginner in QFT.
Are you familiar with "secular perturbation theory" in good ol' classical mechanics? E.g., Jose & Saletan's treatment of the quartic anharmonic oscillator by perturbation around the ordinary harmonic oscillator?

If you're not already familiar with it, my advice is to go read up on it, since it shows (in a much simpler scenario than QFT) how perturbation theory can be a delicate business. Naive perturbation for the case I mentioned quickly leads to solutions that run away to infinity, even though the potential is obviously confining. "Secular" perturbation theory is a technique for controlling that behaviour by Taylor-expanding the "constants" of the theory, and adjusting things at each order.

Do UV divergencies appear because of the Haag's theorem?
I wouldn't have said "because" of Haag's thm. G. Scharf (in "Finite QED") offers the opinion that the problems arise because one multiplies operator-valued distributions by a discontinuous step-function when doing the Feynman time-ordering stuff. Such an operation is ill-defined. Scharf spends much of the book explaining how to correct for this order-by-order. (But I certainly wouldn't call that a quick answer.)
 
  • #5
strangerep said:
I wouldn't have said "because" of Haag's thm. G. Scharf (in "Finite QED") offers the opinion that the problems arise because one multiplies operator-valued distributions by a discontinuous step-function when doing the Feynman time-ordering stuff. Such an operation is ill-defined. Scharf spends much of the book explaining how to correct for this order-by-order. (But I certainly wouldn't call that a quick answer.)

If I understand some comments from DarMM in a previous thread, Scharf's method (Epstein-Glaser) doesn't remove the need for a cut-off, so it doesn't cure all the problems associated with the UV divergences. Rather, Scharf's method says that if there is a theory that makes sense, there is a way to do perturbation series correctly. However, if there are UV divergences and the theory doesn't make sense, then Scharf's series are formal and lack physical meaning. If that is correct, then the UV divergences indicate that either a cut-off with new degrees of freedom in the UV or Asymptotic Safety is needed.
 
  • #6
strangerep said:
since it shows (in a much simpler scenario than QFT) how perturbation theory can be a delicate business.

Perturbation theory is a delicate business in any area area of applied math that uses it eg numerical analysis.

Intuitively perturbation theory looks pretty benign but, as mathematicians learned long ago, intuition often leads us astray.

Thanks
Bill
 
  • #7
eoghan said:
Hi all!
I'm a beginner in QFT. I've read a lot of posts here about Haag's theorem, but I haven't found one which can answer simply and briefly to my question (if such an answer exists):

Do UV divergencies appear because of the Haag's theorem?

Thank you
See
https://www.physicsforums.com/showthread.php?t=728744&page=1
especially post #30.
 
  • #8
eoghan said:
Hi all!
I'm a beginner in QFT. I've read a lot of posts here about Haag's theorem, but I haven't found one which can answer simply and briefly to my question (if such an answer exists):

Do UV divergencies appear because of the Haag's theorem?
The direct answer is no.

UV divergences appear because powers of the field in the Hamiltonian are not well defined. Take for instance the Hamiltonian of ##\phi^4## theory, with the interaction only occurring in a box, in the interacting picture:

##H = H_{0} + \frac{\lambda}{4!}\int_{\Lambda}{\phi_{0}^4 dx}##

##H_{0}## being the free Hamiltonian and ##\Lambda## the box. ##\phi_{0}## refers to the free field, in the interaction picture you use it instead of the actual field ##\phi## in the Hamiltonian. When I say the actual field, I mean the interacting field that obeys the equation of motion:

##\partial^2 \phi - m^2 \phi = \chi_{\Lambda}\frac{\lambda}{3!}\phi^3##

##\chi## is a function equal to ##1## inside ##\Lambda## and zero outside. From this you can see the field evolves like a free field outside of the box.

Now the problem is that ##\phi_{0}^{4}## is not a well-defined mathematical object. If you attempt to use it, it produces ultraviolet divergences.

Now in two-dimensions, we can use Wick ordering to create a new object denoted:

##:\phi_{0}^4:##

This is the Wick-ordered fourth power and now the Hamiltonian is well-defined.

In three-dimensions, we can use the same trick and Wick-order the power and make the Hamiltonian well-defined, but it will not be self-adjoint (which it must be for sensible time-evolution and real eigenvalues). This will cause the unitary time evolution operator to develop ultraviolet infinities/divergences. It turns out it is impossible to define a sensible version of ##\phi_{0}^{4}## in three dimensions which leaves the Hamiltonian well-defined and self-adjoint.

This means the interaction picture is not sensible, i.e. one cannot use ##\phi_{0}## in place of ##\phi##. However since the fields are just related by a unitary operator ##V \phi V^{-1} = \phi_{0}##, this implies this operator does not exist and hence the fields are not unitarly related.
Eventually after hard analysis (there is a paper by Glimm on the subject in 1968 where he does this analysis) You can prove that the interacting field ##\phi## lives in a separate Hilbert Space, so there was no chance it and ##\phi_{0}## could be unitarly connected.

Now let us go back to two-dimensions. I still have the box in place. Let us remove it, by sending ##\Lambda \rightarrow \infty##. It turns out that once again the Hamiltonian is not well-defined and the real interacting field lives in another Hilbert Space.

So even though we were able to sensibly define a self-adjoint Hamiltonian in two dimensions in a box using the free field, we cannot do so in infinite volume. We must use the interacting field living in a separate Hilbert space.

Haag's theorem is the statement that for all interacting QFTS, the real field will never live in the same Hilbert space as the free field and hence you can not use the free field instead of it (i.e. interaction picture does not exist) unless the interaction is restricted to a box ##\Lambda##.

However even when restricted to a box, ultraviolet divergences can mean the free field is still unusable, such as in three dimensions discussed above, but Haag's theorem does not say anything about that.

By the way ultraviolet divergences can from not just a power of the free field being undefined, but from products of different free fields being undefined. For example in Yukawa theory in two dimensions we have:

##\bar{\psi_{0}}\psi_{0}\phi_{0}##

in the Hamiltonian, with ##\psi_{0}## a spinor free field and ##\phi_{0}## a scalar free field. Even with Wick ordering this cannot be made well-defined. However it turns out that there is a way of altering it, it's a little more complicated, that does make it well defined. Of course the theory must be in a box, otherwise Haag's theorem will prevent the Hamiltonian using free-fields from being well-defined and self-adjoint regardless.

Ultraviolet divergences beyond ones correctable by Wick ordering, or logarithmic divergences in the mass, like Yukawa theory above are too severe to allow the Hamiltonian with the free field to be well-defined and the interacting field would have to be used.

In perturbation theory, these issues can be ignored, all you need to be concerned about are removing the infinities and you can always use the free-field. However if you were to look at finite time evolution non-perturbatively you would see problems and notice that you would need to switch Hilbert space.
 
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  • #9
Hi DarMM,

which article do you mean exactly ?

Glimm, J. (1967a). The Yukawa coupling of quantum fields in two dimensions. I,
Commun. Math. Phys. 5, 343-386.
Glimm, J. (1967b). The Yukawa coupling of quantum fields in two dimensions. II,
Commun. Math. Phys. 6, 61-76.
Glimm, J. (1968a). Boson fields with non-linear self-interaction in two dimensions,
Commun. Math. Phys. 8, 12-25.
Glimm, J. (1968b). Boson fields with the ifJ4 interaction in three dimensions, Commull.
Math. Phys. 10, 1-47.
Glimm, J. and Jaffe, A. (1968a). A l(ifJ4)2 quantum'field theory without cutoffs. I,
Phys. Rev. 176, 1945-1951.
Glimm, J. and Jaffe, A. (1968b). A Yukawa interaction in infinite volume, Commun.
Math. Phys. 11,9-18.
 
  • #10
It's the one you have labelled as 1968b: Boson Fields with the ##\phi^4## Interaction in Three Dimensions.

It's not a light read though!

After this paper it still wasn't proven that the Hamiltonian was bounded below on the new Hilbert space, this was proven in:
Positivity of the ##\phi^{4}_{3}## Hamiltonian, Fortschritte der Physik, 21 (1973), 327–376

This is probably the most complicated paper in the whole area of rigorous field theory.
 
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  • #11
Thank you all for your replies. It's a little bit involved for a beginner, but I'll try anyway to read the papers you linked and to understand all the replies!
 
  • #12
eoghan said:
Thank you all for your replies. It's a little bit involved for a beginner, but I'll try anyway to read the papers you linked and to understand all the replies!
Just to tell you there isn't much to be gained from the paper I referenced at the beginner level, it would probably be impossible to understand, unless you know QFT and functional analysis to a high level. Just try to understand the replies.
 
  • #13
atyy said:
If I understand some comments from DarMM in a previous thread, Scharf's method (Epstein-Glaser) doesn't remove the need for a cut-off, so it doesn't cure all the problems associated with the UV divergences.

The Epstein-Glaser method uses only an IV cutoff, no UV-regularization. A readable account is in Scharf's book on QED (the second edition is stronger than the first). See also the discussion in http://www.physicsoverflow.org/20325/
 
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  • #14
A. Neumaier said:
The Epstein-Glaser method uses only an IV cutoff, no UV-regularization. A readable account is in Scharf's book on QED (the second edition is satronger than the first). See also the discussion in http://www.physicsoverflow.org/20325/

But there you also write that the series is only asymptotic, and you reference an argument of Dyson. Doesn't this mean that Epstein-Glaser alone is not enough to define a UV complete theory?
 
  • #15
atyy said:
But there you also write that the series is only asymptotic, and you reference an argument of Dyson. Doesn't this mean that Epstein-Glaser alone is not enough to define a UV complete theory?
UV and IR problems are concepts associated with the perturbative approach. Thus although Epstein-Glaser has no UV-problems it is a (rigorous) perturbative theory only, and hence says nothing about convergence of the loop expansion.

The nonperturbative construction of a nontrivial 4-dimensional Poincare invariant quantum field theory is still an unsolved problem.
 
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  • #16
A. Neumaier said:
UV and IR problems are concepts associated with the perturbative approach. Thus although Epstein-Glaser has no UV-problems it is a (rigorous) perturbative theory only, and hence says nothing about convergence of the loop expansion.

The nonperturbative construction of a nontrivial 4-dimensional Poincare invariant quantum field theory is still an unsolved problem.

As I understand it, the summability problem is a UV problem also. There are two UV problems indicated by traditional perturbation series. The first is term-by-term UV divergence, and the second is UV completeness which means the series cannot be summed, and is related to the non-perturbative definition of the theory. Epstein-Glaser solves the first problem, but leaves the second, if I have correctly understood DarMM, and I think it is pretty much what I think you are saying also, with different terminology.

The other idea indicating that UV completeness is linked to a non-perturbative definition of a theory is that asymptotic safety or freedom, which is a UV property, is heuristically taken to indicate that the non-perturbative definition is possible.
 
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  • #17
atyy said:
As I understand it, the summability problem is a UV problem also. There are two UV problems indicated by traditional perturbation series. The first is term-by-term UV divergence, and the second is UV completeness which means the series cannot be summed, and is related to the non-perturbative definition of the theory.

UV means caused by short distance behavior, and summability is not of this sort - one has nonconvergence even for anharmonic oscillators, where UV problems are completely absent.

UV completion has nothing to do with the summability of the loop expansion (series in powers of ##\hbar##). It just means that you add enough fields (or strings) to give a theory a (RG improved perturbative) fixed point. But everything there is still perturbative, though sometimes at advanced levels (such as the ERGE approach, which is called nonperturbative, but perturbs around a coherent state).

There is not the slightest positive advance since Dyson in the question of summability of the series, though the difficulties are somewhat better understood (renormalons).
 
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  • #18
atyy said:
As I understand it, the summability problem is a UV problem also. There are two UV problems indicated by traditional perturbation series. The first is term-by-term UV divergence, and the second is UV completeness which means the series cannot be summed, and is related to the non-perturbative definition of the theory. Epstein-Glaser solves the first problem, but leaves the second, if I have correctly understood DarMM, and I think it is pretty much what I think you are saying also, with different terminology.

The other idea indicating that UV completeness is linked to a non-perturbative definition of a theory is that asymptotic safety or freedom, which is a UV property, is heuristically taken to indicate that the non-perturbative definition is possible.

As used in the literature UV completion in itself is more a buzzword than a precise concept, it certainly doesn't mean summability.
(DarMM also says in
https://www.physicsforums.com/threads/haags-theorem-in-qft.728744/page-3#post-4611358
that UV completion and summability are separate issues.) In
https://www.physicsforums.com/threads/haags-theorem-in-qft.728744/page-3#post-4611431
he points out that proving that a theory is its own UV completion implies existence of the theory. But the point here is that proving this of course proves existence as a byproduct. Still, the loop expansion is likely to diverge, as Dyson's argument doesn't depend on the existence of the theory.

But if you take UV completion to be synonymous with nonperturbative existence, the ''UV'' loses its meaning.
 
  • #19
atyy said:
As I understand it, the summability problem is a UV problem also.
No, these are unrelated problems. For example, in string theory there is no UV problem (each string Feynman diagram is UV finite), but the sum of all diagrams is infinite (the expansion is only asymptotic).
 
  • #20
A. Neumaier said:
UV means caused by short distance behavior, and summability is not of this sort - one has nonconvergence even for anharmonic oscillators, where UV problems are completely absent.

UV completion has nothing to do with the summability of the loop expansion (series in powers of $\hbar$). It just means that you add enough fields (or strings) to give a theory a (RG improved perturbative) fixed point. But everything there is still perturbative, though sometimes at advanced levels (such as the ERGE approach, which is called nonperturbative, but perturbs around a coherent state).

There is not the slightest positive advance since Dyson in the question of summability of the series, though the difficulties are somewhat better understood (renormalons).

A. Neumaier said:
As used in the literature UV completion in itself is more a buzzword than a precise concept, it certainly doesn't mean summability.
(DarMM also says in
https://www.physicsforums.com/threads/haags-theorem-in-qft.728744/page-3#post-4611358
that UV completion and summability are separate issues.) In
https://www.physicsforums.com/threads/haags-theorem-in-qft.728744/page-3#post-4611431
he points out that proving that a theory is its own UV completion implies existence of the theory. But the point here is that proving this of course proves existence as a byproduct. Still, the loop expansion is likely to diverge, as Dyson's argument doesn't depend on the existence of the theory.

But if you take UV completion to be synonymous with nonperturbative existence, the ''UV'' loses its meaning.

Demystifier said:
No, these are unrelated problems. For example, in string theory there is no UV problem (each string Feynman diagram is UV finite), but the sum of all diagrams is infinite (the expansion is only asymptotic).

I agree UV completion is not a precise term. and there is no proof that asymptotic freedom rigourously implies non-perturbative existence (otherwise the Clay Millenium problem would be solved). However, the heuristics are that the summability is a UV problem in the sense that there are examples of the presence of infrared and UV cutoffs making a renormalized series summable. Also, there are cases where asymptotic freedom renders the renormalized series Borel summable. Asymptotic freedom is also a heuristic indication that a non-perturbative construction is possible. Perhaps I understood wrongly, but I think this is how one of the "UV problems", and its relationship to summability (renormalons), asymptotic freedom and non-perturbative construction is discussed in Rivasseau's "From perturbative to constructive Renormalization" (p7-9) http://www.rivasseau.com/3.html.
 
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1. What is QFT renormalization and why is it necessary?

Quantum Field Theory (QFT) renormalization is a mathematical process used to remove infinities that arise in the calculations of certain physical quantities in quantum field theory. These infinities result from the fact that the theory is based on continuous space and time, which can lead to infinitely large or small values. Renormalization is necessary in order to make meaningful predictions and ensure that the theory is consistent and well-defined.

2. How does QFT renormalization work?

QFT renormalization involves redefining certain parameters in the theory in order to cancel out the infinities that arise in calculations. This is typically done through a process called "subtracting infinities" where the infinite terms are subtracted from the calculation and replaced with new, finite parameters. These new parameters can then be adjusted to match experimental data, ensuring that the theory agrees with observations.

3. What is Haag's theorem and how does it relate to QFT renormalization?

Haag's theorem is a mathematical proof that shows that it is impossible to have a consistent, interacting quantum field theory without renormalization. This theorem highlights the importance of renormalization in QFT and explains why it is necessary to remove the infinities that arise in calculations.

4. What are the consequences of not using QFT renormalization?

If QFT renormalization is not used, the theory would produce nonsensical and meaningless results. This would make it impossible to make accurate predictions and the theory would not be able to accurately describe the physical world. Additionally, without renormalization, the theory would not be mathematically consistent and would break down at certain energy scales.

5. Are there any challenges or limitations to QFT renormalization?

While QFT renormalization is a powerful tool for removing infinities and ensuring the consistency of the theory, it is not without its challenges. One limitation is that it can be a computationally intensive process, requiring complex calculations and often leading to approximate solutions. Additionally, there are certain situations where renormalization may not be applicable, such as in theories with non-local interactions. However, overall, QFT renormalization remains an essential tool for understanding and predicting the behavior of quantum systems.

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