It is known that for local field redefinitions for which the LSZ formula is valid: $$\langle 0|\phi(x)|p\rangle \neq 0$$ field redefinitions don't change the S-matrix. (See QMechanic's answer to Equivalence Theorem of the S-Matrix)

But is it true for non-local field redefintions? For instance if I take a field redefinition of the form: $$\psi(x)=e^{-l\Box} \phi(x)$$

Will the S-matrix be invariant under this?

From QMechanic's explanation linked above and the answer by AccidentalFourier Transform here it would seem that the answer should be yes.

Edit: My main concerns are

  1. Is such a transformation invertible? Is this a concern?

  2. Any boundary condition on $\psi$ would translate to infinitely many boundary conditions on $\phi$. Is this relevant?


1 Answer 1


Claim 1. If $\psi(x)$ is an arbitrary operator that satisfies \begin{equation} \langle 0|\psi(x)|p\rangle\neq 0\tag1 \end{equation} then it is a valid interpolating field, and as such, it can be used in the LSZ formula. The proof can be found in any introductory text, such as Weinberg.

Claim 2. If we assume that \begin{equation} \langle 0|\phi(x)|p\rangle\neq 0\tag2 \end{equation} and define \begin{equation} \psi(x)\overset{\mathrm{def}}=\mathrm e^{-\ell\partial^2}\phi(x)\tag3 \end{equation} then we have \begin{equation} \langle 0|\psi(x)|p\rangle\neq 0\tag4 \end{equation}

The proof is straightforward. One just need to use $\langle0|\phi(x)|p\rangle=c\mathrm e^{ipx}$ for some non-zero constant $c$, and the trivial identity $\mathrm e^{-\ell\partial^2}\mathrm e^{ipx}=\mathrm e^{\ell p^2}\mathrm e^{ipx}$.

Conclusion: the non-local redefinition $(2)$ is a valid redefinition, and the $S$ matrix is invariant under it.

  • 1
    $\begingroup$ This is what I have as well. However there are a couple of questions: 1. The transformation between the fields is non-invertible. Naively, if I implement the field redefinition as a change of variables in the path integral, a non-invertible transformation would be illegal as the Jacobian is 0. 2. Boundary conditions. $\psi \rightarrow 0$ corresponds to an infinite number of boundary conditions for $\phi$. I'm not quite sure if these are issues or non-issues. $\endgroup$ Jul 17, 2017 at 18:26
  • $\begingroup$ @NirmalyaKajuri note that $\phi\mapsto \mathrm e^{-\ell\partial^2}\phi$ is invertible, with inverse $\psi\mapsto \mathrm e^{+\ell\partial^2}\psi$. $\endgroup$ Jul 19, 2017 at 10:57
  • $\begingroup$ But isn't it true that it will map all solutions of $\partial^{2n}\phi=0$ to zero, and these are all independent functions? $\endgroup$ Jul 19, 2017 at 11:50
  • 1
    $\begingroup$ @NirmalyaKajuri no, it will map all solutions of $\partial^{2n}\phi=0$ to $\phi\mapsto \phi$. $\endgroup$ Jul 19, 2017 at 11:56
  • $\begingroup$ @NirmalyaKajuri I'm glad I could help. Cheers! $\endgroup$ Jul 19, 2017 at 12:13

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