Proof that Wilsonian renormalization only generates terms consistent with the symmetry of the action

Question

In the Wilsonian approach to renormalization it's easy to see that integrating out high momentum dofs in the path integral generates an infinite number of terms in the renormalized action. It's often said (in textbooks) that all terms consistent with the symmetries are generated.

How does one see that these new terms must be consistent with the symmetries? Is there a proof that symmetry breaking terms cannot be generated by this procedure?

Answer 1

I think it's quite easy to prove with functional integrals. In Wilson renormalization one splits the field in low energy modes and high energy ones, say $\phi = \varphi + \Phi$, where $\varphi$ only has support on small momenta and $\Phi$ on large ones. The action will be $S[\varphi,\Phi] = S_1[\varphi] + S_2[\Phi] + S_\text{int}[\varphi,\Phi]$.

Imagine now that the full theory (high energy) has a symmetry under $\varphi \to \tilde\varphi[\varphi]$ and $\Phi \to \tilde\Phi[\Phi]$. This means that the action is invariant: $S[\tilde\varphi,\tilde\Phi]=S[\varphi,\Phi]$. Note also that I've used the fact that internal symmetries do not mix low energy modes with high energy ones. EDIT: To understand this one can work, for example, in momentum space. The low and high energy modes are defined such that

$$ \varphi(k) = \begin{cases} \phi(k) \quad \text{ for } k \leq \Lambda \\ 0 \quad \quad \,\text{ otherwise}\end{cases}\,, \qquad \Phi(k) = \begin{cases} 0 \quad\quad\,\, \text{ for } k \leq \Lambda \\ \phi(k) \quad \,\text{ otherwise}\end{cases}\,. $$ Since internal symmetries do not act on the momenta, but only on the indices of the fields, the transformed of, for example, $\varphi(k)$ will still have only support on $k\leq \Lambda$, i.e. it will still be a low energy mode. END OF EDIT

Now, the effective action obtained integrating over high energy modes, is defined as

$$ e^{iS_\text{eff}[\varphi]} = e^{iS_1[\varphi]}\int \mathcal{D}\Phi \, \exp \left( iS_2[\Phi]+iS_\text{int}[\varphi,\Phi] \right)\,. $$

Now, perform a symmetry transformation on $\varphi$,

\begin{align} e^{i S_\text{eff}[\tilde\varphi]} &= e^{iS_1[\tilde \varphi]} \int \mathcal{D}\Phi \exp\left( iS_2[\Phi] + i S_\text{int}[\tilde\varphi,\Phi]\right) \\ &= e^{iS_1[\tilde \varphi]} \int \mathcal{D}\tilde\Phi \exp\left( iS_2[\tilde\Phi] + i S_\text{int}[\tilde\varphi,\tilde\Phi]\right) \\ &= e^{iS_1[\varphi]} \int \mathcal{D}\Phi \exp\left( iS_2[\Phi] + i S_\text{int}[\varphi,\Phi]\right)= e^{iS_\text{eff}[\varphi]}\,, \end{align}

where in the second equality I've just performed a change of integration variable $\Phi\to\tilde\Phi$, and in the third I've used the fact that the action is invariant and the symmetry is not anomalous (i.e. $\mathcal{D}\Phi = \mathcal{D}\tilde\Phi$).

Therefore, if the original theory is invariant, the effective low energy one is invariant as well, i.e. no symmetry breaking terms can be generated in the renormalization procedure.

I hope this answers your question!

Answer 2

1. The Wilsonian effective action
   $$\begin{align} \exp&\left\{ -\frac{1}{\hbar}W_c[J^H,\phi_L] \right\}\cr ~:=~~~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\} \end{align}\tag{A}$$ is defined by integrating out heavy/high modes $\phi^k_H$ and leaving the light/low modes $\phi^k_L$. Here $J^H_k$ denotes sources for the heavy modes. The (possibly non-local) Wilsonian effective action $W_c[J^H,\phi_L]$ is the generating functional of connected $\phi_H$ Feynman diagrams in a background $J^H,\phi_L$.

2. Assume the action $$S[\widetilde{\phi}]~=~S[\phi]\tag{B}$$ is invariant under an invertible affine transformation$^1$ $$ \widetilde{\phi} ~=~A\phi+b. \tag{C}$$

3. It is natural to associate the inhomogeneous translation $b$ in eq. (C) with the light modes. This leads to the partial transformation laws $$ \begin{align} \widetilde{\phi}_L ~=~&A\phi_L+b, \cr \widetilde{\phi}_H ~=~&A\phi_H, \cr \widetilde{J}^H ~=~&J^HA^{-1}. \cr \end{align}\tag{D}$$

4. Assume that the path integral measure $${\cal D}\widetilde{\phi}_H~=~{\cal D}\phi_H\tag{E}$$ is also invariant.

5. Then the Wilsonian effective action $$\begin{align} \exp&\left\{ -\frac{1}{\hbar}W_c[\widetilde{J}^H,\widetilde{\phi}_L] \right\}\cr ~\stackrel{(A)}{=}~~~&\int \! {\cal D}\frac{\widetilde{\phi}_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\widetilde{\phi}_L+\widetilde{\phi}_H]+\widetilde{J}^H_k \widetilde{\phi}_H^k\right)\right\} \cr ~\stackrel{\text{linearity}}{=}~&\int \! {\cal D}\frac{\widetilde{\phi}_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\widetilde{\phi_L+\phi_H}]+\widetilde{J}^H_k \widetilde{\phi}_H^k\right)\right\}\cr ~\stackrel{(B)+(E)}{=}~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\}\cr ~\stackrel{(A)}{=}~~~&\exp\left\{ -\frac{1}{\hbar}W_c[J^H,\phi_L] \right\} \end{align}\tag{F}$$ is invariant as well.

References:

1. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4 p. 77 + Section 17.2 p. 84.

--

$^1$ The possible extension of eq. (F) to non-affine symmetries relies on an effective separation of light and heavy modes, cf. Einj's answer. See also a related discussion in Ref. 1, where it is shown that the effective/proper action $\Gamma[\phi_{\rm cl}]$ inherits affine symmetries of the action $S[\phi]$ and the path integral measure ${\cal D}\phi$.