Why is the Nambu-Goto path-integral ill-defined?

Question

I have found a lot of places saying that the Nambu-Goto action is ill-defined, that the squareroot exponential is a complicated thing to make sense of in a path-integral and so on. Then people go on and introduce the Polyakov action to make sense of things, and this is fine with me.

My question is, can anyone share the details why the Nambu-Goto action is ill-defined? I cannot seem to find any rigorous analysis showing that path-integrals with squareroots are problematic. I am looking for a proof of the ill-defined nature of the Nambu-Goto path-integral, or at least a fairly detailed description.

Generally it's easy to define a path integral for a Gaussian measure (in which case, the Lagrangian is quadratic). Defining it for any other case is a long and harduous process, especially if it's not polynomial. — Slereah, May 05 '20 at 17:51
Possible duplicate: Why are the Nambu-Goto action and Polyakov action equivalent at quantum level? Related: Why does a square root term make the quantisation of action difficult? — Qmechanic, May 05 '20 at 18:07
Thank you, Slereah. I heard this kind of statement before, and what I am really looking for is arguments that it is so. I would appreciate some references. Qmechanic, I read https://physics.stackexchange.com/q/268782/, and indeed it does not give any details why the Nambu-Goto path-integral is ill-defined, it states that it is and what to do about it. Thank you for the related discussion. That is helpful. — FrederikN, May 06 '20 at 08:31

score 1 · Answer 1 · answered May 06 '20 at 10:10

The main problem with the Nambu-Goto path integral is how to obtain a consistent path integral measure (PIM). When people say that it is ill-defined (and not just difficult to work with), they presumably mean that various naive choices of PIMs are inconsistent. The easiest way to obtain a consistent PIM is to go to the Hamiltonian formulation, cf. e.g. this Phys.SE post. One may argue that the Hamiltonian formulation of the Nambu-Goto string is equivalent to the Polyakov string, and hence consistent, cf. e.g. my Phys.SE answer here.

score 0 · Answer 2 · answered Sep 26 '21 at 02:42

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See the complete and unambiguous solution of this problem in : https://inis.iaea.org/search/search.aspx?orig_q=RN:25041294

answered Sep 26 '21 at 02:42

user314900

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Why is the Nambu-Goto path-integral ill-defined?

2 Answers2