5
$\begingroup$

I just stumbled across a presentation by Tachikawa about "What is Quantum Field Theory". He has an interesting perspective that we should think of (at least a subset of) quantum field theories as being organized by

  • 1-ality $\implies $ A theory with a unique Lagrangian
  • 0-ality $\implies$ A non-Lagrangian QFT
  • 2-ality $\implies$ Two seemingly distinct Lagrangians that generate the same list of correlation functions
  • N-ality $\implies$ There are $N$ distinct Lagrangians that give rise to the same correlators.

I can think of specific examples of $N=0,1,2$, but I'm not familiar with examples of n-ality (defined in the above sense) with $N\neq 0,1,2$. I suppose my question is:

  • Are there concrete examples for any other $N$?
  • For arbitrary $N$?
  • Can there be a countably infinite number?
  • A continuum?

I know that certain string compactifications on $T^d$ have, for example, a large T-duality group such as $O(d,d;\mathbb{Z})$, but I'm not sure how that should thought of in the above (proposed) classification scheme. Does it even fit in this way of organizing things? Basically, I just want some help understanding if "N-ality" partitions the space of QFTs in a useful way, or if it was just a cute trick to describe non-Lagrangian QFTs.

I have in mind that we are, e.g., defining a Lagrangian as a top-form so issues of total derivatives are excluded. I'm also not including integrating out a field as being genuinely different because the partition functions are the same in an obvious way. There might be some other "trivial" equivalences I'm not thinking of at the moment, but my current understanding is that there's something genuinely different about N-ality from these kinds of examples I just mentioned. Finally, I'm not as interested in duality in the sense of IR-duality, as I think I already understand that case via universality classes. Maybe this last paragraph just shows I'm being naive, but I would be interested to learn more if that's the case.

$\endgroup$
5
  • $\begingroup$ +1. If this does not take much space, I suggest mentioning of referencing the examples of 0 through 2-ality (the question is great without it, but I'm saying this because I'd like to see them and I don't think I'll be the only one hahahaha). Alternatively, this could of course be delegated to an answer, eventually $\endgroup$ Commented May 20 at 23:17
  • 3
    $\begingroup$ I'm not sure how well-defined this concept is... Given a Lagrangian QFT, there are an infinite number of field redefinitions one can do that do not change the physical theory but do change the Lagrangian. Arguably, dualities are an example of (very non-obvious) field redefinitions and maybe redefinitions of coupling constants. At least, I would want an answer to give a clear criterion that distinguishes a duality from a field redefinition. $\endgroup$
    – Andrew
    Commented May 20 at 23:20
  • $\begingroup$ Reading through the slides, the main point I take away is that the existence of "0-ality" (non-Lagrangian) theories is important, and I agree with that. I think the "n-ality" idea is more of an oversimplified story and my personal take is that I wouldn't spend too long making it overly precise, because (a) to end up with something non-trivial you'll have to spend a while coming up with some definition that separates transformations of the path integral you don't want to ones you do want, and (b) for the cases that do exist, we probably aren't very confident about $n$. $\endgroup$
    – Andrew
    Commented May 20 at 23:26
  • $\begingroup$ I agree that field redefinitions make this concept ambiguous, which was something I was trying to get around haha. One way to try and maybe get around that is to fix a partition function's source J, and say that we only include the equivalence class of Lagrangians in which the source term remains linear in the action with the canonical normalization. For example, I believe AdS/CFT has this property if you identify $J_{bdry}$ (source) with $\phi_{blk}$ (bulk field profile at boundary) despite having radically different Lagrangians. I wonder if that works, or maybe I'm still missing something. $\endgroup$
    – 11zaq
    Commented May 20 at 23:26
  • $\begingroup$ But I take your point that it might not be a very well-defined concept. In some sense, that's an answer to my question in itself! $\endgroup$
    – 11zaq
    Commented May 20 at 23:28

0