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In Wilson's approach to renormalization we break up a field $\phi_0$ which includes modes up to some cutoff $\Lambda$ into two parts, $\phi_0=\phi+\tilde\phi,$ where $\phi$ only has modes up to some smaller cutoff and $\tilde\phi$ only contains fast modes. Then we integrate out $\tilde\phi$, which modifies the action for the $\phi$ field.

My problem is that when you apply this procedure you generically get terms which are linear in the fast modes, and they seem to lead to problems from a perturbation theory standpoint.

For instance consider the Sine-Gordon model:

$$\mathcal{L}=\frac{1}{2}(\partial\phi_0)^2+\frac{\alpha}{\beta^2}\cos\beta\phi_0$$ $$=\frac{1}{2}(\partial\phi)^2+\partial\phi\partial\tilde\phi+\frac{1}{2}(\partial\tilde\phi)^2+\frac{\alpha}{\beta^2}(\cos\beta\phi\cos\beta\tilde\phi-\sin\beta\phi\sin\beta\tilde\phi).$$

I can argue away the $\partial\phi\partial\tilde\phi$ term because it is integrated over space in the action, and $\phi$ and $\tilde\phi$ are different Fourier components.

Now I'll expand in powers of $\tilde\phi$, $$\mathcal{L}=\mathcal{L}[\phi]+\frac{1}{2}(\partial\tilde\phi)^2-\alpha{\beta}^{-1} \sin\beta\phi\,\tilde\phi-\frac{\alpha}{2}\cos\beta\phi\,\tilde\phi^2 +\frac{\alpha}{3!}\beta\sin\beta\phi\,\tilde\phi^3+\mathcal{O}(\beta^2,\tilde\phi^4)$$

Note the linear term seems to be to the $\beta^{-1}$. If we include it in a perturbation we can get arbitrarily large negative powers of $\beta$ by having it contract with all the legs of another vertex. For instance the $\tilde\phi^3$ vertex is to order $\beta$ but if we contract all three legs we get a connected diagram to the order $\beta^{-2}$.

So this is my question: I don't see why connected diagrams like these shouldn't contribute when we integrate over all $\tilde\phi$ to calculate the effective action for $\phi$. But it doesn't seem to make sense from a perturbation theory point of view. And textbooks and papers always seem to drop these terms without comment.

Note that if I try to get rid of these linear terms by completing the square in the quadratic action, I simply introduce the powers of $\beta^{-1}$ in the higher order vertices. Moreover I don't think completing the square makes sense in this context since a factor like $\sin\beta\phi$ is a slow field while $\tilde\phi$ is supposed to be fast modes only.


ADDENDUM: SIGMA MODELS

In this Sine-Gordon model I can argue that these problematic diagrams don't matter because they come with increasing powers of $\alpha$, which is a dimensionful parameter, so at scales well below the cutoff the terms should not be important (although we are still getting compensating negative powers of $\beta$)

But more problematic is the O(N) sigma model. $$\mathcal {L}= \frac{1}{2g^2}(\partial n_0^i)^2\qquad (n^i)^2=1$$

We can apply a Wilson renormalization scheme following Polyakov, where we take, $$n_0^i=n^i\sqrt{1-(\tilde\phi_a)^2}+\tilde\phi_a e^i_a,$$ where $n^i$ is the renormalized field with slow modes, $\tilde\phi_a$ are a set of fast modes we integrate out, and $e^i_a$ is a slowly varying set of orthonormal vectors orthogonal to $n^i$ (which we have some gauge freedom in choosing). The details are in Polyakov's original paper (Phys. Lett. 59B, 79) or Peskin and Schroeder chapter 13.3.

Here the problematic linear term goes as $\frac{1}{g^2}e^i_a\partial n^i \partial\tilde \phi_a,$ which is $O(g^{-1})$ if we redefine the field to absorb the $g^{-2}$ coefficient of the quadratic part.

I don't have an obvious argument that this leads to irrelevant couplings or that integrating two slow fields with a fast field can be ignored. So why does everyone ignore terms like these?

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