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?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/210275 $\endgroup$ Commented Sep 10, 2021 at 4:39
  • $\begingroup$ @NiharKarve, related yes but it's a different question. I'm not asking if all the symmetry preserving terms are added. Rather, I'm asking if it is possible to prove that symmetry breaking terms don't get added. $\endgroup$
    – Jase Uknow
    Commented Sep 10, 2021 at 6:47

2 Answers 2


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!

  • 1
    $\begingroup$ Can you provide any more justification for: 1. "internal symmetries do not mix low energy modes with high energy ones."; and 2: The connection between anomalies and measures under symmetry transformations (used in third line of equation)? $\endgroup$
    – Jase Uknow
    Commented Sep 12, 2021 at 2:49
  • $\begingroup$ I've added a longer explanation on why internal symmetries do not mix high and low energy modes. As for the anomalies, that is the definition. A symmetry is not anomalous when it does not change the path integral measure. $\endgroup$
    – Einj
    Commented Sep 12, 2021 at 23:52
  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.


  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 ${\cal D}\phi$.

  • $\begingroup$ 1. Would you agree that $\tilde{\phi}_L$ in (D) is not the mode one gets by restricting $\tilde{\phi}$ from (C) to low momentum modes using the momentum cutoff you have used to define $\phi_L$ ? I think this is where I was going wrong. 2. Basically, (D) is the definition of the transformed low energy mode, $\tilde{\phi}_L$, and it is defined this way because it satisfies the same symmetry as the original field. Right? $\endgroup$
    – Jase Uknow
    Commented Sep 12, 2021 at 13:13
  • $\begingroup$ 1. Essentially no. 2. Essentially yes. $\endgroup$
    – Qmechanic
    Commented Sep 12, 2021 at 15:00
  • $\begingroup$ Notes for later: A mixed quadratic term $\phi_L\phi_H$ is ruled out by momentum conservation. Similarly, $\phi_H$-tadpole terms $\phi_L\ldots\phi_L\phi_H$ are kinematically suppressed. $\endgroup$
    – Qmechanic
    Commented Oct 10, 2021 at 19:36
  • $\begingroup$ Notes for later: $\exp\left\{-\frac{1}{\hbar}W_{\rm int}[J,\phi_L] \right\}=\exp\left\{\frac{1}{\hbar}J b\right\}\exp\left\{-\frac{1}{\hbar}W_{\rm int}[JA^{-1},\widetilde{\phi}_L] \right\}$, cf. physics.stackexchange.com/a/674473/2451 $\endgroup$
    – Qmechanic
    Commented Nov 5, 2021 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.