# Field strength renormalization and the energy-momentum tensor

This question is about the connection between the energy-momentum tensor, dilation transformations, and field renormalization.

From a Wilsonian perspective on renormalization we start out with a cutoff scale $$\Lambda$$, a field $$\phi_\Lambda$$, and a Lagrangian, say $$\mathcal{L}=\frac{1}{2}(\partial\phi_{\Lambda})^2+B\phi_\Lambda^2+C\lambda\phi_\Lambda^4+\dots$$ Then we separate $$\phi_\Lambda$$ into a 'slow' piece with only Fourier components up to some smaller cutoff $$b\Lambda$$, which I'll just call $$\phi$$, and a 'fast' piece with the remaining higher components. After integrating out the fast field, our Lagrangian becomes $$\frac{1}{2}(1+\delta Z)(\partial\phi)^2+(B+\delta B)\phi^2+(C+\delta C)\lambda\phi^4+\dots$$ Now we do two things. We rescale the spacetime argument of $$\phi$$ by $$b$$ so that this field is also defined up to the same cutoff $$\Lambda$$. This introduces powers of $$b$$ due to derivatives and the measure in the action. And then we absorb everything in front of the term $$\frac{1}{2}(\partial\phi)^2$$ into the field, which defines the field renormalization.

When we are done we have essentially done a dilation transformation on our original field $$\phi_\Lambda$$. We scale the spacetime coordinate, scale the field itself by the field renormalization, and do a somewhat hidden but crucial coarse-graining step to keep $$\Lambda$$ fixed. This transformation should be generated by the dilation current $$T^{\mu}_\nu x^\nu$$ derived from the energy momentum tensor $$T$$. If after we are done the coefficients $$B,C,\dots$$ are exactly the same, the dilation current is conserved which means the trace of $$T$$ vanishes. If the coefficients change then the trace of $$T$$ will be related to the beta functions.

Now here is my question.

What is stopping me from simply choosing not to absorb the factor involving $$\delta Z$$ into $$\phi$$, or only absorbing part of it? Then we have a different dilation transformation, which should mean a different energy-momentum tensor. What is stopping me from simply considering the factor in front of $$(\partial\phi)^2$$ as an extra parameter in the space of Lagrangians that also renormalizes? How is the correct field renormalization or equivalently (I think) the correct energy-momentum tensor singled out?

• Not an answer to your (very interesting!) question, but one thing to consider is that you're perfectly free not to do field renormalization, but in doing so you make the renormalization problem much more complicated by adding a kinetic-term parameter to the manifold of couplings. You also lose the nice relation between correlation functions at different scales as being related by factors of Z. Presumably if you follow those factors of Z through the usual calculations you'll find what you're looking for, though I haven't thought it through very carefully. – user35736 Sep 10 at 14:29
• @user35736, well I think that relation (the Callan-Symanzik equation) is still there. It is just that the anomalous dimensions have changed. – octonion Sep 10 at 14:33
• I think we're saying the same thing! – user35736 Sep 10 at 14:43
• Wilson renormalization, using a cutoff $\Lambda$, assumes that only field operators with momenta $|p|<\Lambda$ will be used. Fld ops at a point don't satisfy this condition: we must consider smeared fld ops. The fields in the functional integral are just dummy integration variables, so rescaling their magnitude (to maintain a canonical kinetic term, for example) doesn't change anything. The scaling dimensions of field operators in Wilson renorm must come from the effect of scaling on (correlation functions of) smeared fields. This is affected by a rescaling of the spacetime argument. – Chiral Anomaly Sep 11 at 1:42
• @ChiralAnomaly, well you raise a good point that I wanted to avoid by using phi4 as an example. For instance in 2D massless free field theory, $\phi$ has no anomalous dimension, but vertex operators do aquire one, and this can even be seen from Wilson renorm by integrating out the fast fields in the operator (I've tried this before). But in phi4, the scalar fields themselves do pick up an anomalous dimension themselves and it seems this only comes from something like maintaining a canonical kinetic term. – octonion Sep 11 at 12:28