In order for observables like scattering amplitudes to be finite, one must implement a renormalization scheme for parameters like the mass $m$, the coupling $\lambda$, and the field $\phi$.

Question: Why isn't the renormalization of the composite operator $\phi^2(x)$ part of the usual renormalization procedure as well? Since its renormalization is required for calculating matrix elements of, say, the energy-momentum tensor, and because of the $\frac{1}{2}m^2\phi^2$ term in the Lagrangian, I don't understand why its not needed.

I imagine it must not be necessary for rendering the S-matrix finite, so I suppose I'm more so confused as to why the $\phi^2$ in the lagrangian isn't a problem, but $\langle \phi^2\rangle$ is. The two-point function is an integral part of the mass renormalization, and it seems like $\langle \phi^2\rangle$ is just just local limit of that. Since this local limit looks like a bubble diagram which doesn't contribute to the S-matrix, maybe that has something to do with it?

I have a follow-up question, though I understand if this post is long enough: If the lagrangian is $$\mathcal{L}=\frac{1}{2}\partial_\alpha\phi\partial^\alpha\phi-\frac{1}{2}m^2\phi^2+V(\phi)$$ Could you renormalize just the field $\phi$ for the kinetic term $\partial_\alpha\phi\partial^\alpha\phi$, renormalize $\phi^2$ for the mass term, and then whatever composite operator the interaction lagrangian is described by?


1 Answer 1


The renormalization applies to the parameters of the theory. Usually that means the masses, the charges (or coupling constants) and so on. That is, the parameters that you would measure to determine the theory. The usual way that happens is loop corrections.

Mass gets measured. It correspoonds to a line in a Feynman diagram, and a propagator in the mathematical formalism. You add an extra line coming out of the original line then getting re-absorbed. That makes a loop, and loops diverge because you have to integrate the momentum around the loop. So you renorm the mass in the theory so as to have the remaining mass after the loop is calculated be the observed mass.

Charge (or a coupling constant) gets measured. It corresponds to a vertex in the theory. In QED it's a photon coming out of a charged line, a "three point" vertex. In other theories the vertex would be different, gluons and quarks or whatever. Then you add an additional line, say a photon emitted by the charge before the vertex, and re-absorbed after the vertex. Another loop, another divergence, removed by renorm of the charge. The charge in the theory is set so the effective value is what we measured.

If the theory is renormalizable, then this process allows the N-loop infinities to be pushed to the N+1-loop level. In QED this adds another factor of $\alpha$, and we hope that this converges. If it is not renormalizable then this fails somehow. In GR, for example, each order of loop expansion produces more infinities, requiring more parameters, destroying predictions.

But $\phi$ is not a paramter of the theory. So we don't have a measured value to target to be able to renorm it. The $\phi$ is something calculated by using the theory for a given situation rather than something adjusted as a parameter of the theory. This is very different from m or e, which are measured and then used to make other predictions.

  • 3
    $\begingroup$ To be clear though, composite operators do get renormalized. This is because $\lim_{x\rightarrow0}\langle\phi(x)\phi(0)\rangle$ is generically divergent and hence not well-defined. To define the operator $\phi(x)^2$, one would need to define the operator as not simply two copies of $\phi(x)$ on top of each other. $\endgroup$ Nov 9, 2021 at 2:45
  • $\begingroup$ @RichardMyers Maybe I didn't articulate my primary confusion well enough. I understand that composite operators get renormalized, but for some reason it isn't necessary when calculating S-matrix elements in a scalar field theory even though there are composite operators in the lagrangian (both interaction and non). My real question is why can we get away with not considering the renormalization of composite operators when they're in the lagrangian itself? $\endgroup$
    – Adots005
    Nov 9, 2021 at 18:30
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    $\begingroup$ @user143854 the Lagrangian is not operator-valued. Indeed, nothing inside the path integral is. As such, there is no reason to replace the terms in the Lagrangian with something else. If you wanted to replace them anyway, of course you would be allowed to do so, but you may change the theory you're studying depending on what you change. $\endgroup$ Nov 10, 2021 at 0:39
  • $\begingroup$ @RichardMyers "the Lagrangian is not operator-valued" I need to think about this more but I think that's the key point I was missing. Thank you:) $\endgroup$
    – Adots005
    Nov 10, 2021 at 15:57

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