2

In the book "Quantum Field Theory of Point Particles and Strings" by Hatfield, in pages 114-118 the author discusses the proper relation between interacting fields and in/out fields and the spectral representation of the two-point function. In particular \begin{eqnarray} \lim_{t\to -\infty}\varphi(x)&=&\sqrt{Z}\varphi_{\rm IN}(x)\\ \lim_{t\to \infty}\varphi(x)&=&\sqrt{Z}\varphi_{\rm OUT}(x)\tag{7.10} \end{eqnarray} and then studying the spectral representation one finds that $iZ$ is the residue of the propagator at the pole corresponding to the mass of the particle. Moreover the author shows that $0 < Z \leq 1$ and that $Z=1$ if and only if $\varphi(x)=\varphi_{\rm IN}(x)$. In page 118 he further comments that $Z=1$ implies $\varphi=\varphi_{\rm IN}$ and we end up with a free field theory.

Now consider the renormalized Lagrangian for the renormalized field $\varphi^R(x)$. In the on-shell scheme one sets the residue of the propagator at the mass pole to be $i$. In particular, comparing to the above discussion, this means that for the renormalized field $\varphi^R(x)$ we have one $Z^R=1$.

But now this is rather confusing because Hatfield says that a field for which the residue of the propagator at the mass pole is $i$ coincides with in/out fields and is a free field. But the renormalized field certainly can't be free (we are in the interacting theory after all!).

So what is reallly going on here? How can the renormalized field $\phi^R(x)$ have $Z^R=1$ and this still be an interacting theory? How to reconcile this renormalization condition with the remarks by Hatfield that a field whose corresponding $Z=1$ is free?

Gold
  • 38,087
  • 19
  • 112
  • 289

1 Answers1

2

You have pointed out that, in an interacting theory, it is standard to define the renormalized field as $Z^{-1/2}$ times a bare field. However, if you demand that the residue of the two-point function is $i$, you will certainly not get $Z = 1$. In all but very special finite theories, $Z$ has $\frac{1}{\epsilon}$ poles in dimensional regularization.

It is also important to remember that there is nothing special about wanting the residue at the particle mass to be $i$. This is called the on-shell scheme which is merely the first of many schemes that textbooks introduce. I discussed in a previous answer why it's also fine to use completely different conditions for making things finite. In short, QFTs make predictions about scheme independent quantities. Cross section as a function of energy, for example, is scheme independent. But cross section section as a function of coupling and coupling as a function of energy on their own are not.

Connor Behan
  • 10,659