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Disclaimer: this question may be very stupid. It looks like I am missing some fundamental point.

Let's consider a massive scalar $\pi$

$$ \mathcal{L}_\pi = -\frac{1}{2}(\partial \pi)^2 -\frac{m^2}{2}\pi^2 + g \mathcal{L}_{int}(\pi)\,,\qquad\qquad\qquad Eq.(1) $$ where $g$ is a coupling and $\mathcal{L}_{int}$ includes non-derivative interactions. The field $\pi$ satisfies the usual Klein-Gordon equation $(\square - m^2) \pi=-g\mathcal{L}'_{int}$ where the potential acts as a source.

My question: are we allowed to perform a field redefinition $\pi(x) \rightarrow \pi(x) + f(x)$ where $f(x)$ is an exact harmonic function (i.e. $\square f(x)=0)$?

I am confused because, if we turn off the interactions by setting $g=0$, the field $f(x)$ behaves like an auxiliary field. Indeed the lagrangian becomes

$$ \mathcal{L}_{\pi+f} = -\frac{1}{2}(\partial \pi)^2 -\frac{m^2}{2}(\pi(x)+f(x))^2\,,\qquad\qquad\qquad Eq.(2) $$

and the equations of motion for $f(x)$ imply the constraint

$$ f(x) = -\pi(x). $$

Now, it is clear that something wrong is happening here. This cannot be correct. I started with a propagating free massive scalar and, by doing a (questionable) field redefinition, I shifted the pole of the propagator at $m^2=0$.

EDIT v1

After the first answer, I want to stress the following points.

  • Assumed the field redefinition I proposed makes sense, I should find that the two theories Eq.(1) and Eq.(2) are equivalents. A hint for such equivalence is given, for example, by the position of the poles of the correlation functions computed within the two theories. If the poles are found at different masses, then the theories for sure propagate different degrees of freedom.
  • If one asks you to work with the following partition function $$ Z[J_\pi,J_f] = \int \mathcal{D}\pi \mathcal{D}f e^{iS_{\pi+f} + i \int d^d x J_\pi f(x) + i \int d^d x J_f f(x) } $$ there is no way to establish that $\pi$ and $f$ are dependent fields, unless you compute the equation of motions for both fields. Notice that there is no way to say that $f(x)$ is an harmonic field. The equation of motions are (in the limit $g=0$)

$$ \frac{\partial S_{\pi+f}}{\partial \pi} -\partial_\mu\frac{\partial S_{\pi+f}}{\partial\partial_\mu \pi} = 0 \rightarrow \square\pi = m^2(\pi + f)\,,\qquad \qquad Eq.(3) $$

$$ \frac{\partial S_{\pi+f}}{\partial f} -\partial_\mu\frac{\partial S_{\pi+f}}{\partial\partial_\mu f} = 0 \rightarrow f=-\pi\,,\qquad \qquad Eq.(4) $$ and you see that, combining Eq.(3) and Eq.(4) you get

$$ \square \pi = 0\qquad\rightarrow\qquad \square f = 0 $$

  • The delicate point may be that $f(x)$ is promoted to a constrained field and the integration in the path integral should be performed in a constrained field space. Such information should be putted in the lagrangian by using a Lagrange multiplier $\lambda(x)$. For example, if I add to Eq.(2) a "gauge fixing term" $$ \mathcal{L}_{\pi+f} = -\frac{1}{2}(\partial \pi)^2 -\frac{m^2}{2}(\pi(x)+f(x))^2 +\lambda(x) \square f(x)\,,\qquad\qquad Eq.(5) $$ then, the lagrange multiplier has the form $\square \lambda(x) = m^2(\pi(x) + f(x))$ which, putted back in Eq.(5), removes the dependence on $f(x)$ and we recover the original lagrangian. Is this the solution?
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What you are essentially doing is separating out the part of $\pi (x)$ that has zero $\square$ from the part that has non-zero $\square$. (Or part of the zero $\square$ part, anyway.) You need to be careful to carry the right terms in all the various places. The result isn't what you got but rather

$$\square \phi = m^2 (\phi + f)$$

where $\pi = \phi + f$, $\square f$ = 0, and $\square \phi \neq 0$. There should be nothing wrong with doing that as long as you carry all the terms in the interaction when you get to it. And presuming that if you get solutions you satisfy appropriate boundary conditions or normalizing conditions etc.

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  • $\begingroup$ What do you mean "the result is not what you got"? You are showing the equation of motion for $\phi$. The EOM for $f(x)$ are $f(x) = -\phi(x)$. $\endgroup$
    – apt45
    Jul 5 '18 at 14:49
  • $\begingroup$ No, there is only one equation of motion. $\phi$ and $f$ are not independent. Alternatively you could introduce the constraint $\square f =0$ into the Lagrangian with an undetermined multiplier. But it will come back to the same thing. $\endgroup$
    – user93146
    Jul 5 '18 at 14:55
  • $\begingroup$ Or to put it another way, the equation of motion for $f$ is $\square f = 0$. $\endgroup$
    – user93146
    Jul 5 '18 at 14:57
  • $\begingroup$ The fact that the two fields are not indipendent must follow from the equations of motion. The EOM for $f$ are $f=-\phi$ which implies $\square f = 0$ because of those of $\phi$. $\endgroup$
    – apt45
    Jul 5 '18 at 15:00
  • $\begingroup$ Nope. Try $f=12$ in your system. $\square f$ is zero. But $\phi \neq -12$ because then $\square \phi$ would also be zero and not satisfy the eqn of motion for $\phi$. The eqn of motion for $f$ is not $f=-\phi$ it is $\square f=0$. $\endgroup$
    – user93146
    Jul 5 '18 at 15:06
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Your procedure is inconsistent: You claim that you are doing a field redefinition $\pi(x) \mapsto \tilde{\pi}(x) = \pi(x) + f(x)$ for some harmonic function $f$, but then you suddenly talk about the "equation of motion" for $f$. When you do a field redefinition, the dynamical field after the redefinition is $\tilde{\pi}(x)$, and you should be looking at the dynamical equations for it - you cannot magically double the d.o.f. by a redefinition. In a true redefinition, $f$ is a fixed function, not a parameter of the Lagrangian. Redefinitions cannot change the number of parameters of the Lagrangian.

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