# Relevance of Haag's theorem

1. Oct 8, 2014

### Arsenic&Lace

[Mentor's not: Split out from another thread because the question does indeed warrant its ownn thread]

I'm not sure if the following question warrants its own topic, but how relevant do people think Haag's theorem actually is to solving physics problems involving QFT?

Last edited by a moderator: Oct 8, 2014
2. Oct 8, 2014

### atyy

For many things one doesn't need rigourous QFT, and you can put the system on a torus and put a UV cutoff. This is the Wilsonian viewpoint of post #2.

However, it is interesting to ask whether there are rigorous special relativistic QFTs in infinite volume and with no UV cutoff. As DarMM and Demystifier have pointed out in posts #6 and #7, Haag's theorem means that for the "interaction picture" to exist, one has to put the system in finite volume. The interaction picture is mainly used for deriving Dyson's formula etc in non-rigourous QFT. The formula also holds in rigourous field theory, and Haag's theorem just means the usual derivation is sloppy. Rigourous field theory has non-sloppy methods of deriving the same formula.

3. Oct 9, 2014

### Arsenic&Lace

Why is it interesting to ask whether there are rigorous relativistic field theories in infinite volume? I'm not an expert on QFT by even the remotest stretch of the imagination.

What is the use of deriving it more rigorously? Is there some doubt about the Dyson' formula?

4. Oct 9, 2014

### atyy

The interest is intellectual.

5. Oct 10, 2014

### Demystifier

First, because the Universe might be infinite.
Second, because analytical calculations are easier in an infinite volume than in the finite one.
Third, because in a finite volume you must choose some definite value of the volume, and it is not obvious what value should you choose. (The physically measurable result may not depend on this choice, but that's also not obvious.)

6. Oct 10, 2014

### atyy

Another way to ask the question is: Haag's theorem also applies to condensed matter in the thermodynamic limit. However, all condesned matter experiments involve finite volume (also all conceivable accelerator experiments). Do we really care if the formulas we use are wrong (ie. suppose the correct derivation in thermodynamic limit gives a different formula, and that using the ineraction picture for finite N, there should be an N dependence), but all experiments don't have enough resolution to distinguish between 10^23 and 1+10^23 particles?

Actually, how does one consider Haag's theorem and perturbation series for finite N? Usually I naively imagine that there are series in 1/N where N is the number of particles, and I also naively imagine that the first term is the infinite N limit, and higher order terms are corrections for finite N. Since we are dealing with finite N, the interaction picture strictly speaking does exist. Does this mean that there are no perturbations series in 1/N, or that it is wrong to think of the first term as the large N limit? (I don't have a quantum calculation in mind, but I'm naively imagining there is a quantum interaction picture counterpart to finite size scaling, which is a method for extracting critical exponents from numerical simulations with finite numbers of particles.)

Last edited: Oct 10, 2014
7. Oct 10, 2014

### atyy

Here's DarMM's description of Haag's theorem:

Here's Wikipedia's
http://en.wikipedia.org/wiki/Haag's_theorem

It looks like translation invariance is needed. So could we get out by assuming a lattice model - say lattice QED and lattice quantum gravity? Maybe even lattice QCD? These would not be relativistic, but we just need the lattice to be fine enough.

I think the only problem with lattice models is chiral interactions. But I don't know if that is an insoluble problem, or just one where we don't know the full solution yet.

Last edited: Oct 10, 2014
8. Oct 13, 2014

### Demystifier

9. Oct 16, 2014

### DarMM

Not really that relevant. It's important conceptually for realising what an interacting field theory actually is nonperturbatively, but outside of that it's not too important. It doesn't really invalidate anything in the "everyday" QFT toolkit, it just means that the standard derivations of those tools (LSZ theorem, Gell-man Low Theorem) is logically incomplete, such as the derivation you'd see in Peskin and Schroeder and other intro books. However they can be derived anyway using Haag-Ruelle theory for example, so they're not false or anything like that.

Also Haag's theorem doesn't really say anything about perturbation theory (this is in relation to the $\frac{1}{N}$ expansion question above).

10. Oct 16, 2014

### atyy

The idea behind the question was that in a non-rigourous derivation of a perturbation series, one may make use of the interaction picture. However, Haag's theorem says this doesn't exist under certain circumstances. If I have a lattice model, could I say that Haag's theorem doesn't prevent the interaction picture from existing even in infinite volume, because the lattice model breaks translational symmetry?

11. Oct 16, 2014

### DarMM

Yes, Haag's theorem would not prevent the interaction picture from existing in an infinite volume lattice model. Although I can still think of examples where the interaction picture doesn't exist in such models, although it wouldn't be due to Haag's theorem.

12. Oct 16, 2014

### atyy

What are some other reasons the interaction picture may not exist in such models? Would those apply to say lattice QED or lattice QCD?

13. Oct 16, 2014

### DarMM

The reason the interaction picture does not exist is always the same as such. That is, when the Hilbert space of the free theory and that of the interacting theory are not unitarily related. In a physicists approach you could say it occurs when the interacting vacuum requires an energy renormalization (you will not see this in perturbation theory, which is why Haag's theorem does not show up there). So let's say that $\Omega_{0}$ is the ground state of the free theory and $\Omega_{g, \Lambda}$ is the ground state of the interacting theory in a finite volume $\Lambda$. If the energy is normalised so that $H \Omega_{0} = 0$, then the interacting vacuum will have some energy $E(\Lambda,g)$. This is the interacting vacuum's energy "relative to the free theory" let's say (Of course in its own theory it is the ground state and so would have energy zero). If you have $\lim_{\Lambda \rightarrow \infty} E(\Lambda,g) = \infty$, then there will be no interacting picture.

Yes, indeed they would.

14. Oct 16, 2014

### atyy

So basically the whole usual "condensed matter" derivation of second quantization as an exact alternative way of writing the Schroedinger equation for many identical particles will typically fail in the infinite volume limit (say even for a lattice model of interacting spins stuck to each lattice point)?