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I am studying $\phi^4$ theory and so far I understand the mass and coupling constant renormalizations. In these theories, once we expand a diagram in perturbation theory we "cancel" the divergences by demanding the value we get from the diagram equals a measured quantity such as the physical mass or physical coupling.

For field strength renormalization I understand the need for it and how it appears, but because it is not physical like mass or the coupling constant are I am less sure about the corresponding renormalization condition. My guess is that if we define our physical field $\phi$ as $\phi = Z^{-1/2}\phi_0$ where $\phi_0$ is the bare field, then we require that in the UV limit the physical field has coefficient 1. This places a condition on $Z$ so that the physical fields have no dependence on the momentum cutoff.

Is this correct or is there more to it?

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At any loop level, the general form of the dressed propagator of the $\phi^4$ theory is $$\tilde{D}_\text{resummed}(p)=\frac{i}{p^2-m_0^2(\Lambda)-\Sigma(p^2,\Lambda)}\tag{1}\label{1},$$ where $\Sigma$ is the self-energy, i.e. the sum of 1PI diagrams. Although at one-loop level the self energy depends merely on the cutoff scale $\Lambda$ displays no $p$ dependence, it does at higher loop levels. The effect is that not only the pole of the propagator is shifted - giving rise as OP knows to the renormalized mass in the on shell renormalization scheme, but it also changes the behaviour around the pole, i.e. the residue. In fact, if you define $m_R$, the renormalized mass by $$p^2-m_0^2(\Lambda)-\Sigma(p^2,\Lambda)\lvert_{p^2=m_R}=0$$ The expansion of the propagator around the pole yields $$\frac{i}{p^2-m_0^2(\Lambda)-\Sigma(p^2,\Lambda)}\tag{2}\label{2}\approx\frac{iZ}{p^2-m_R}+\text{(finite terms)},$$ where $$Z=1-\frac{d\Sigma}{dp^2}\bigg\lvert_{p^2=m_R}\tag{3}\label{3}$$. As a side note, OP may have recognized that \eqref{2} is the Källén-Lehmann spectral representation of the propagator. Rescaling the fields as OP mentions removes $Z$ from the numerator of \eqref{2}.

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  • $\begingroup$ Thank you for your answer. $\endgroup$
    – CBBAM
    Commented Mar 22 at 19:14

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