Graduate Non-perturbative description of QFTs

[dx]

TL;DR

gauge theories and non-perturbative description of QFTs

The partition function of a free QFT is of the form

$$ \left( \frac{1}{\mathrm{det \ D}} \right)^{\frac{1}{2}}$$

As far as I know, gauge theories said that the free theory contains in some sense also the interacting theory. Even general relativity which is not generally considered to be a gauge theory also has this property, that you can get the motion of a particle in an arbitrary gravitational field by considering the free motion in local inertial frames. In string theory also, the 'free lagrangian' which describes a world-sheet also contains the interactions, since the 'vertices' in string diagrams are not any different from a smooth world-sheet which looks the same at any point. Mainly any theory therefore seems to be a gauge theory in some sense. So my question is, is the expression above in some sense a non-perturbative description of any kind of QFT?

 

[Demystifier]

I think Higgs field is not a gauge field in any sense.

 

[dx]

I think Higgs field is not a gauge field in any sense.

Perhaps. The Higgs lagrangian has a discrete symmetry Z₂. Maybe we don't know how to regard it as a gauge theory because gauging discrete symmetries is more unfamiliar than gauging continuous symmetries. But I have seen people talk about gauging discrete symmetries and people say6 that it can be done.

Either way, I still would like to know an answer to my question in the case of other usual gauge theories. Can the expression above be regarded in some sense as a non-perturbative description of an interacting gauge theory like QED?

 

[vanhees71]

If you just get a functional determinant, it's the result of doing a Gaussian path integral, i.e., the action is a bilinear functional of the fields and their derivatives, but this describes non-interacting particles. If you want interactions you need at least terms with three fields in the Lagrangian like the interaction term in QED with the Term $\mathcal{L}=-q A^{\mu} \bar{\psi} \gamma^{\mu} \psi$.

 

[Demystifier]

If you want interactions you need at least terms with three fields in the Lagrangian like the interaction term in QED with the Term $\mathcal{L}=-q A^{\mu} \bar{\psi} \gamma^{\mu} \psi$.

Yes, but that can be viewed as included in the definition of the covariant derivative D.

 

[vanhees71]

You can of course formally integrate over $\bar{\psi}$ and $\psi$, but then of course you cannot evaluate the functional determinant exactly anymore as in the free case. For a systematic treatment using the "heat-kernel method", see (there's also a newer edition of this great book):

J. F. Donoghue, E. Golowich and B. R. Holstein, Dynamics of
the Standard Model, Cambridge University press (1992).

 

[dx]

Also, if you have an action with an interaction term jx with the current j, then you can complete the square and write x²/2 + jx as 1/2 (x - j)² + j²/2