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Required fields*

5
  • $\begingroup$ Essentially for setting up a field theory is about above all which field to choose, a scalar field theory will be always scalar, even if you take the square root of it. And would $\sqrt{L }$ supposed to be? It would not have the right dimensions. Moreover, $L$ has to be Lorentz invariant, which is a very strong constraint. The Lagrangian has to be at least quadratic in its degrees of freedom, otherwise the EQM give absurd results. So square-rooting $\phi D\phi$ would make $L$ linear in $\phi$ with a absurd EQM. $\endgroup$ Commented Mar 4, 2023 at 16:22
  • $\begingroup$ Actually formally one can write the KG-equation as $i\partial_t \phi= \sqrt{-\nabla^2 +m^2}\phi$ and develop the rhs (apart from $\phi$) as Taylor series. However, the resulting equation is non-local and not manifestly Lorentz-covariant. However, locality is an important feature of modern field theory, a principle that cannot easily abandoned. Therefore it is not the way to go. $\endgroup$ Commented Mar 4, 2023 at 21:40
  • $\begingroup$ Thank you very much for your so valuable thoughts. However, your answer, due to imprecision in the formulation of my question, unintentionally tells me things I am already aware of. The true question is, does there exist a square root of Euler-Lagrange equations in the sense it forms a factorization, just like how you proved that Dirac operator squares to Klein-Gordon? $\endgroup$ Commented Mar 4, 2023 at 22:08
  • $\begingroup$ There is no much choice for finding another decomposition. Imagine the search of the roots of a quadratic form. Decomposing the the KG-differential operator in 2 linear factors is quite the same stuff. The pair of roots of a quadratic form is unique, it is weary to search for another pair. Of course one might wonder if the representation (up to equivalence) of the Clifford algebra is really unique. In 4D it is indeed the case which is well documented in the literature. If there were another decomposition, it would be already lively discussed in the physics literature which is not the case. $\endgroup$ Commented Mar 5, 2023 at 0:41
  • $\begingroup$ If we lift our attention to bigger matrices or generally speaking other algebras the coefficients forming a non trivial square root of the KG operator live on some kind of curve (or algebraic variety if you'll indulge). I wonder from a purely mathematical standpoint if a family of differential operators whose coefficients live on a particular variety itself carries interesting mathematical structure. $\endgroup$ Commented Jun 9, 2024 at 18:51