# Wavefunction renormalization in the on-shell scheme

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?

• Who insists? Which page? Commented Oct 12, 2021 at 14:51
• For instance in the book "Quantum Field Theory and the Standard Model" by Schwartz. In page 332 he states the condition on the residue of the two-point function. The same is done in Peskin & Schroeder, pages 324 and 325. In particular equation 10.19 gives the condition on the residue of the two-point function at the mass pole. I mean, while identifying the physical mass with the pole of the two-point function seems justified, this condition on the residue seems quite arbitrary (why set it to $i$ and not some other arbitrary value?).
– Gold
Commented Oct 12, 2021 at 15:01
• @Gold regarding your last sentence in your comment here, without looking into the precise details you quote, I think the reason is that you want a canonically-normalised kinetic term in your quantum effective action. that sign feeds through to that if I remember correctly. you don't need it to be canonically-normalised, but it is convenient when deriving the Feynman rules because it is the standard convention. Commented Oct 13, 2021 at 8:24

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.
• Thanks for your answer. About your first paragraph I think that the way in which I stated that was confusing. You start with a bare field $\phi$ and through (1) in the question you get $Z$. Now you define $\phi^R = Z^{-1/2}\phi$. What I mean is that if you use (1) to relate $\phi^R$ to in/out fields, it will be a relation like $\phi^R(x)\to \sqrt{Z^R}\phi_{\rm in/out}(x)$ with $Z^R = 1$. The point is that $\phi^R(x) = Z^{-1/2}\phi(x)$ and so if $\phi(x)\to Z^{1/2}\phi_{\rm in/out}(x)$ we get $\phi^R(x)\to \phi_{\rm in/out}(x)$. I meant that the field $\phi^R$ has one $Z^R = 1$.