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In a classic Georgi review of EFT, I have read the following quote

The result of eliminating heavy particles is inevitably a nonrenormalizable theory, in which the nontrivial effects of the heavy particles appear in interactions with dimension higher than four. In the full theory, these effects are included in the nonlocal interactions obtained by integrating out the heavy particles. These interactions, because of their nonlocal nature, get cut off for energies large compared to the heavy particle masses. However, in the effective theory, we replace the nonlocal interactions from virtual heavy particle exchange with a set of local interactions, constructed to give the same physics at low energies. In the process, we have modified the high energy behavior of the theory, so that the effective theory is only a valid description of the physics at energies below the masses of the heavy particles. Thus the domain of utility of an effective theory is necessarily bounded from above in energy scale.

Why are the new interactions, obtained in the EFT by integrating out the heavy fields, nonlocal? I can't see this as a direct consequence of the nonrenormalizability of the interaction or of the fact that it depends on the exchanged momentum.

I do understand that interactions mediated by massive particles are shortrange because the corresponding effective potential is damped by a real exponential term -- but don't know if this has anything to do with my doubt and think that perhaps I'm confusing the concepts of 'local' and 'short range'.

Then, in the phrase

However, in the effective theory, we replace the nonlocal interactions from virtual heavy particle exchange with a set of local interactions, constructed to give the same physics at low energies.

I believe that Georgi is talking about the fact that we expand that new effective interaction in powers of $\left( \frac{p^2}{M^2}\right)$.

Why is the expanded perturbative interaction, now, local?

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GaloisFan
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2 Answers2

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Suppose we start with a UV theory \begin{equation} \mathcal{L} = -\frac{1}{2} (\partial \phi)^2 - \frac{1}{2} (\partial \psi)^2 - \frac{1}{2} M^2 \phi^2 + g \phi \psi^2 \end{equation} where $\phi$ is a heavy field with mass $M$, and $\psi$ is a light field (taken to be massless for simplicity).

We integrate out $\phi$ by solving it's equation of motion. In other words, we make the following replacement in the action \begin{equation} \phi = \frac{g \psi^2}{-\square + M^2} \end{equation} The notation $(\square + M^2)^{-1}$ means to use an appropriate Green's function for the operator $\square + M^2$. You can think of it in momentum space, if you like.

Note that the expression for $\phi$ is non-local. This is because integrating $g\psi^2$ against the Green's function involves an integral over space and time. The integral is non-local, since it involves more than one spacetime point.

Plugging this back into the action, we get a non-local Lagrangian for $\psi$ \begin{equation} \mathcal{L} = - \frac{1}{2} (\partial \psi)^2 + \frac{1}{2}\psi^2 \frac{g^2 }{-\square + M^2}\psi^2 \end{equation} At this stage, the Lagrangian is non-local, but the physics is completely equivalent to the original Lagrangian.

Now we Taylor expand \begin{equation} \frac{1}{-\square + M^2} = \frac{1}{M^2} + \frac{\square}{M^4} + \cdots \end{equation} If we truncate the Taylor expansion, we are left with a local expression (local since there are only a finite number of derivatives and no integrations). The resulting effective theory involves only the light field $\psi$, is local, but is only valid while the Taylor expansion can be trusted, that is, for energies much less than $M$ (when the heavy field cannot be directly produced) \begin{equation} \mathcal{L}_{\rm eff} = -\frac{1}{2} (\partial \psi)^2 + \frac{g^2}{2 M^2} \psi^4+ \frac{g^2}{2 M^4} \psi^2 \square \psi^2 \end{equation} We could include additional terms, if we wanted a higher level of precision.

Andrew
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  • Thank you very much! I just don't understand the concept of that interaction being nonlocal because we perform an integration and how expanding the Green function 'kills' it. Could you please clarify these points? – GaloisFan Feb 15 '22 at 05:35
  • @GaloisFan Let's start with the first point. $\phi(x)$, $\nabla\phi(x)$, $\square \phi(x)$ are all local because they involve a field and a finite number of derivatives evaluated at a single point. The function $F(x) = \int d y G(x, y) \phi(y)$ is non-local because to evaluate $F(x)$ requires knowing the field at other points beyond $x$. Does that make sense? – Andrew Feb 15 '22 at 10:26
  • I understood that point! But I don't believe that to be the case. I would say that the original interaction is said nonlocal precisely because we do need an infinite number of derivatives to write the inverse operator, not because it contains a Green integral -- in fact, the integral remains after truncating the series as I said. I believe this to be the case because I have seen authors call functionals of fields 'local' many times. Don't you aggree? – GaloisFan Feb 15 '22 at 15:56
  • @GaloisFan In my example, $F(x)$ is a function, not a functional. A "local functional" is just the integral of a local function. My point is that $F(x)$ is not a local function. I think it is more obvious written as an integral. But if we say $G(x,y)$ is the Green's function for the Klein-Gordon field, you could also write $F = (\square + m^2)^{-1} \phi$ if you prefer. Whichever way you want to look at it, $F$ is a nonlocal function, and if we used $F$ as a Lagrangian, the action would be nonlocal. – Andrew Feb 15 '22 at 16:21
  • I understand that. My point is exactly that these authors call interactions which are functionals of fields (which are integrals) local -- and, again, truncating the series you do not kill the convolution. Sorry if I'm being dense. – GaloisFan Feb 15 '22 at 16:32
  • @GaloisFan $F(x) = (\square + m^2)^{-1} \phi = m^{-1} \sum_{n=0}^\infty (-1)^{n+1}\left(\square/m^2\right)^n \phi$ is a nonlocal function. If you wrote an action that was $\int dx F(x)$, the theory would be nonlocal. On the other hand, for a finite $N$, $F_N(x) = m^{-1} \sum_{n=0}^N (-1)^{n+1}\left(\square/m^2\right)^n \phi$ is a local function. It only involves a finite number of derivatives of $\phi$ at a single point, and isn't a convolution. If you wrote an action $\int dx F_N(x)$, you would have a local theory. – Andrew Feb 15 '22 at 16:36
  • That's exactly my point, no? The difference is the number of derivatives, not the integral. – GaloisFan Feb 15 '22 at 16:38
  • @GaloisFan I genuinely do not understand your point. $(\square + m^2)^{-1} \phi = \int dy G(x, y) \phi(y)$. It is non-local. You can see this in multiple ways. In the first way, you have an infinite number of derivatives. In the second way, you have a convolution. But they are equivalent expressions. On the other hand, $(\square \phi)^n$ for a fixed $n$ is local since it involves a finite number of derivatives, and is not a convolution. You can phrase it either way. To me, the convolution is more obviously non-local, but as I've said, use whatever you find easier to understand. – Andrew Feb 15 '22 at 16:41
  • I can't see how truncating the series kill the integral as you say -- it merely changes the meaning of $G(x,y)$ to the homogeneous solution of a new operator. – GaloisFan Feb 15 '22 at 16:46
  • @GaloisFan One way to interpret these expressions is as inverse Fourier transforms. The inverse Fourier transform of $\tilde{\phi}/(p^2+m^2)$ (which we can write as $(\square + m^2)^{-1} \phi$) is a non-local convolution integral of $\phi$. The inverse Fourier transform of $(p^2/m^2)^n \tilde{\phi}$ is just some derivatives acting on $\phi$, with no convolution integral. There's no magic, you can literally just do the inverse Fourier transforms and check. – Andrew Feb 15 '22 at 16:49
  • @GaloisFan The fact that the IFT of $p^n \tilde{\phi}$ is a local derivative of $\phi$ follows from standard properties of the Fourier transform. To show the IFT of $\tilde{\phi}/(p^2+m^2)$ is a non-local convolution integral more involved -- the only way I know how to do it is as a contour integral -- but it's done in a lot of books. An example is Section 2.7.1 of Tong's QFT notes: https://www.damtp.cam.ac.uk/user/tong/qft/two.pdf (don't forget about the convolution theorem) – Andrew Feb 15 '22 at 16:56
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That's a good question.

The Wilsonian effective action (WEA) $$\begin{align} \exp&\left\{ -\frac{1}{\hbar}W_c[J^H,\phi_L] \right\}\cr ~:=~~~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\} \end{align}$$ is defined by integrating out heavy/high modes $\phi^k_H$ and leaving the light/low modes $\phi^k_L$. Here $S$ denotes the action of the underlying UV theory. It may depend on a UV cut-off $\Lambda_H\gg \Lambda_L\equiv\Lambda$. The UV action $S$ is conventionally assumed to be local at distances $r\gg \Lambda_H^{-1}$, i.e. it is local for all practical purposes, and we will for simplicity view it as local for the remainder of this answer.

Moreover, here $J^H_k$ denote sources for the heavy modes. They are included for generating purposes and are usually put to zero at the end of the calculation.

The WEA is the generator of connected Feynman diagrams made from heavy propagators $G_H$, local interactions in $S$, and background $J^H$, $\phi_L$.

Due to the bi-local propagators, the WEA is non-local by construction; however the non-locality is exponentially suppressed, cf. Ref. [1,2,3,4]. E.g. if the heavy modes have mass $M$, then propagators take the form $$G_H({\bf r})~=~(-\Box+M^2)^{-1}\delta^{d}({\bf r})~\sim~e^{-Mr}\quad{\rm for}\quad r~\equiv~|{\bf r}|~\gg~ M^{-1}.$$

The WEA is only meant to be probed at low momenta $p\sim\partial\ll\Lambda\lesssim M$ below the scale $\Lambda$; equivalently, at IR distances $r\gg M^{-1}$. We will assume it is possible in this region to adequately approximate the WEA by an infinite series of mono-local higher-order derivative terms, perturbatively in $\partial/M\ll 1$.

For tree diagrams in the WEA, we can be more explicit, since there are no loop-momenta. If we impose low momenta in the background $J^H$, $\phi_L$, then the momentum conservation forces low momenta to run through the heavy propagators. In that case, we can expand the propagator perturbatively in $M^{-1}$, $$G_H({\bf r})~=~(-\Box+M^2)^{-1}\delta^{d}({\bf r})~=~\frac{1}{M^2}\sum_{n\in\mathbb{N}_0}\left(\frac{\Box}{M^2} \right)^n\delta^{d}({\bf r}).$$

The upshot is that we can rewrite the WEA perturbatively in $M^{-1}$ to only consist of mono-local terms, i.e. we can define a Wilsonian effective Lagrangian density (WELD). Since the number of spacetime derivatives is unbounded, the WELD is by definition non-local, cf. e.g. this Phys.SE post.

References:

  1. D. Skinner: QFT II, the RG; p. 59.

  2. T.J. Hollowood, 6 Lectures on QFT, RG and SUSY, arXiv:0909.0859; p. 15.

  3. J. McGreevy, QFT Lecture notes, spring 2015; p. 16 + p. 73.

  4. S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; p. 522.

Qmechanic
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