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I've just completed problem 13.1 in Srednicki in which he tells us to relate the field-strength renormalization $Z_{\phi}$ to the spectral density $\rho(s)$ that appears in the Lehmann representation of the exact propagator. It seems straightforward-- I follow the hint, insert unity using the 0, 1, and multi-particle states of the interacting theory, and make use of canonical equal-time commutation relations to get

\begin{equation}

\frac{1}{Z_{\phi}} = 1+ \int_{4m^2}^{\infty} ds \rho(s)

\end{equation}

My question is this-- How do I reconcile the above result, which implies that $0 < Z < 1$, with the divergent expressions one obtains for $Z_{\phi}$ in standard perturbative calculations?

I'm almost positive I did the problem correctly (in many-body theory there are similar bounds one derives for the residue at the quasi-particle pole of the Green function). Would the bound on Z be restored if I could do a non-perturbative summation of diagrams?

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# Field-strength renormalization problem (13.1) in Srednicki

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