Question about why QFT in 4D is undefined

[HomogenousCow]

I'm a beginner in QFT and I'm always hearing people say that QFT is undefined in 3+1 dimensions or that the Hamiltonian cannot be rigorously defined, what do these statements mean exactly? In particular I'd like to know what makes the 3+1D case that much worse than non-relativistic QFTs like the ones in condensed matter theory, or even just regular QM where perturbation series can also be divergent (and almost always are). Thanks.

 

[DarMM]

Very simply, because it gets quite technical, it's two connected issues:

1. The fields in quantum field are distributions so there are issues with defining their products.
2. You can't write down different Hamiltonians on the same Hilbert space. So in non-relativistic QM any Hamiltonian of $N$ particles is defined on the same Hilbert space. In QFT every Hamiltonian requires a different Hilbert space.

They're connected because basically the products can only be defined correctly to give a well-defined version of the Hamiltonian once they are defined as operators on the correct Hilbert space.

Nobody has managed to do this for 3+1D QFTs. In lower dimensions people have managed to "find" the correct Hilbert space through limiting procedures from cutoff theories and prove everything is well-defined on them.

 

[HomogenousCow]

Why are these specific to 3+1D though? Aren't the field operators in NR QFT also distributions? And why does changing the time evolution change the entire hilbert space?

 

[A. Neumaier]

Why are these specific to 3+1D though? Aren't the field operators in NR QFT also distributions? And why does changing the time evolution change the entire hilbert space?

So in non-relativistic QM any Hamiltonian of $N$ particles is defined on the same Hilbert space.

Not always - each phase requires a different Hilbert space. For example, superconductivity is based on a non-implementable Bogoliubov transformation that changes the Hilbert space even in the nonrelativistic case.

 

[DarMM]

Why are these specific to 3+1D though? Aren't the field operators in NR QFT also distributions?

These issues are not specific to 3+1D, it's just that the singularities in the distributions are more severe in 3+1D so nobody has managed to show they give well-defined Hamiltonian on an appropriate Hilbert space.

And why does changing the time evolution change the entire hilbert space?

Basically via the Stone theorem you can prove that many typical uses of non-relativistic QM have the same Hilbert space, in the sense that different Hilbert spaces are unitarily related.
In QFT it is possible to have the smeared field $\phi(f)$ has representations as an operator on different Hilbert spaces not unitarily related to each other. For a given Hamiltonian only one choice is correct.

 

[HomogenousCow]

These issues are not specific to 3+1D, it's just that the singularities in the distributions are more severe in 3+1D so nobody has managed to show they give well-defined Hamiltonian on an appropriate Hilbert space.

I see, so there remains hope that one could find a good Hilbert space to have the Standard Model (Or a more fundamental theory) live in?

 

[DarMM]

I see, so there remains hope that one could find a good Hilbert space to have the Standard Model (Or a more fundamental theory) live in?

Yes certainly. They're not proven to be divergent or ill defined, the question has simply yet to be settled.

For example most people think that this is true for QCD. In the Pure Yang-Mills case we essentially have results strong enough that nobody seriously doubts it (there is a finite volume continuum limit for the action).

 

[weirdoguy]

Are there any textbooks that deal with those issues? Or is it mainly spread out in papers?

 

[HomogenousCow]

Yeah I'm really interested in reading something comprehensible for non-experts, like at the level of Peskin&Schroeder or something like that.

 

[DarMM]

The closest thing to a textbook on this stuff in Glimm and Jaffe's monograph "Quantum Physics: A Functional Integral Point of View" or Rivasseau's "From perturbative to constructive Renormalization".

I don't think they're incredibly hard to read if you can already handle Peskin and Schroeder, but that might be a bias from me being overly familiar with them. I think Rivasseau's is the easier of the two.

 

[DarMM]

An interesting simple example of this stuff is $\phi^{4}_{2}$, i.e. scalar with quartic interaction in 2D.

So you have the free path integral measure:
$$d\mu_{C}$$
The $C$ is the covariance, in standard physicists language the two point function:
$$C(x - y) = \int{\phi(x)\phi(y)d\mu_{C}}$$
Though note to be rigorous you must use smeared values as the values of a field at a point are ill-defined:
$$\int{C(x - y)f(x)g(y)d^{2}x d^{2}y} = \int{\phi(f)\phi(g)d\mu_{C}}\\
\phi(f) = \int{\phi(x)f(x) d^{2}x}$$
You denote the measure with this because once you know the covariance you know the free path integral in full detail, i.e. different free theories will only differ by their covariance.

In general we can obtain the values of observables by integrating:
$$\langle \mathcal{O}(\phi) \rangle = \int{\mathcal{O}(\phi) d\mu_{C}}$$

So then we have the quartic interaction. If we bound it in space:
$$\int_{\Lambda}{\phi^{4}(x)d^{2}x}$$

Then the path integral measure for a quartic interaction theory is:
$$e^{-\int_{\Lambda}{\phi^{4}(x)d^{2}x}}d\mu_{C}$$

So we can compute the normalization of this measure by:
$$\langle e^{-\int_{\Lambda}{\phi^{4}(x)d^{2}x}} \rangle = \int{e^{-\int_{\Lambda}{\phi^{4}(x)d^{2}x}}d\mu_{C}}$$

However because $\int_{\Lambda}{\phi^{4}(x)d^{2}x} = +\infty$ for almost all fields since they are distributions then for almost all fields:
$$e^{-\int_{\Lambda}{\phi^{4}(x)d^{2}x}} = 0$$
And so the path integral vanishes.

(More rigorously you can show it vanishes by using Holder's inequality with integration by parts)

However if we renormalize, which is simply Wick ordering in $d = 2$, the quartic term becomes:
$$\int_{\Lambda}{:\phi^{4}:(x)d^{2}x}$$

By Jensen's inequality for a convex function $F$ we have:
$$F(\langle A \rangle) \leq \langle F(A) \rangle$$

So
$$e^{-\langle \int_{\Lambda}{:\phi^{4}:(x)d^{2}x}\rangle} \leq \langle e^{-\int_{\Lambda}{:\phi^{4}:(x)d^{2}x}} \rangle$$

The expectation term on the left is easy to evaluate:
$$\bigg \langle \int_{\Lambda}{:\phi^{4}:(x)d^{2}x} \bigg\rangle = 0$$
It's just a simple Feynman diagram problem in the free theory.

So we have:
$$1 \leq \langle e^{-\int_{\Lambda}{:\phi^{4}:(x)d^{2}x}} \rangle$$

Thus the interacting path integral no longer vanishes. However we have to show the Wick ordering doesn't cause it to diverge.

Note how this is different from the perturbative picture. Nonperturbatively unrenormalized theories vanish. The renormalization process ensures they don't vanish, but you have to check it doesn't cause them to diverge.

 

[A. Neumaier]

The closest thing to a textbook on this stuff

You might refer to the threads here on PF on your 'ladder' ...