Consider Euclidean Klein Gordon quantum field theory on the toroidal spacetime $X\simeq S^1\times \cdots\times S^1$, with action

$$S(\varphi) = \int_X \varphi(\Delta+m^2)\varphi$$

and scalar field $\varphi\in C^\infty_X$, the space of smooth functions on $X$. Here, $\Delta$ is the Laplace operator corresponding to the standard flat metric on the torus. The claim is that this theory is invariant under the Wilsonian renormalization semigroup.

To fomulate this statement precisely, let $C^\infty_{(a,b)}$ denote the linear span of smooth functions with eigenvalue in $(a,b)$ under application of $\Delta$, and decompose the space of smooth functions according to the eigenvalues of the Laplace operator:

$$C^\infty_{[0,\Lambda)} \simeq C^\infty_{[0, \Lambda')}\oplus C^\infty_{[\Lambda',\Lambda)}.$$

Using the formula for the action of the Wilsonian renormalization semigroup $S[\Lambda]\to S[\Lambda']$ on the space of quantum field theories as given in Renormalization and Effective Field Theory, we have, for $\varphi\in C^\infty_{[0, \Lambda')}$,

$$S[\Lambda'](\varphi)=\frac{\hbar}{i}\log \bigg(\int_{\varphi^\perp \in C^\infty_{[\Lambda',\Lambda)}}d\mu^\perp \exp(iS[\Lambda](\varphi+\varphi^\perp)/\hbar)\bigg),$$

where we have factored the low-energy Feynman measure out from the higher-energy Feynman measure via $d\mu_\Lambda\equiv d\mu_{\Lambda'}\wedge d\mu^\perp$, where the Feynman measure $d\mu_\lambda$ at a general energy scale $\lambda$ is defined by the condition

$$\int_{\varphi\in C^\infty_{[0,\lambda)}}d\mu_\lambda \exp(iS[\lambda](\varphi)/\hbar)\equiv 1. $$

First note that the operator $(\Delta+m^2)$ lies in the algebra generated by the operators $1,\Delta$, and therefore trivially respects the eigenspace decomposition of $\Delta$. Therefore, the modes of the scalar field are uncoupled, and the action factorizes as


and therefore the functional integral also factorizes:

$$S[\Lambda'](\varphi)=S[\Lambda](\varphi)+\frac{\hbar}{i}\log \bigg(\int_{\varphi^\perp \in C^\infty_{[\Lambda',\Lambda)}}d\mu^\perp \exp(iS[\Lambda](\varphi^\perp)/\hbar)\bigg),$$

and therefore the action gains at most a constant offset, which nonetheless leaves the overall theory invariant. Therefore massive Klein Gordon theory, and, as it seems, any other free theory which does not couple low- and high-energy degrees of freedom, must be conformally invariant, that is, invariant under Wilson's renormalization semigroup. This statement, however, seems to contradict the literature, so I guess I am looking for the source of my misunderstanding.


1 Answer 1


The source of your misunderstanding is that you are using the "wrong" RG. I explained this at length in my answer to Wilsonian definition of renormalizability . You are using what I called the nonautonomous version of the RG. whereas to see fixed points etc. it's better to work with the autonomous version which involves integration over fast modes followed by a rescaling that restores the original UV cutoff. This is most intuitive in the lattice block spin approach where you have a random field on $\mathbb{Z}^d$, then you make a new field of block averages where the blocks have linear size say $L$. You need to rescale, i.e., shrink your new lattice ($(L\mathbb{Z})^d$) by a factor of $L$ so the RG becomes an evolution over the fixed space of unit lattice theories. Unless your RG is a "time"-independent dynamical system on a fixed space, it is hard to talk about a strict notion of fixed points.

If you do this change of coordinates to the autonomous setting you will see that the mass term will destroy the fixed point property of the massless Gaussian. Namely, the mass will grow according to $m^2\rightarrow L^2 m^2$ at each RG iteration. Fixed points only correspond to $m^2=0$ (or $m^2=\infty$).


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.