# QFT renormalization and Haag's theorem

1. Sep 17, 2014

### eoghan

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

2. Sep 17, 2014

### atyy

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.

Last edited: Sep 17, 2014
3. Sep 17, 2014

### bhobba

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. Sep 17, 2014

### strangerep

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.

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. Sep 17, 2014

### atyy

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. Sep 17, 2014

### bhobba

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. Sep 18, 2014

### Demystifier

See
especially post #30.

8. Sep 19, 2014

### DarMM

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.

Last edited: Sep 20, 2014
9. Sep 19, 2014

### dextercioby

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. Sep 20, 2014

### DarMM

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.

11. Sep 24, 2014

### eoghan

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. Sep 24, 2014

### DarMM

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. Sep 30, 2014

### A. Neumaier

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/

Last edited: Nov 19, 2015
14. Sep 30, 2014

### atyy

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. Oct 1, 2014

### A. Neumaier

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.

16. Oct 1, 2014

### atyy

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.

Last edited: Oct 1, 2014
17. Oct 2, 2014

### A. Neumaier

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).

Last edited: Nov 19, 2015
18. Oct 2, 2014

### A. Neumaier

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
that UV completion and summability are separate issues.) In
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. Oct 2, 2014

### Demystifier

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. Oct 2, 2014

### atyy

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.

Last edited: Oct 2, 2014