In chapter 7 of Peskin and Schroeder, on page 214 we have the expression of the spectral density function

$$\tag{7.7}\rho(M^2)=\sum_{\lambda} (2\pi)\delta(M^2-m_\lambda^2)|\langle\Omega|\phi(0)|\lambda_0\rangle|^2$$

and we can write alternatively $$\tag{7.8} \rho(M^2)=2\pi\delta(M^2-m^2)\cdot Z+(\text{nothing else until } M^2\geq (2m)^2).$$

I think this means that $Z=\sum_{\lambda}|\langle \Omega|\phi(0)|\lambda_0\rangle|^2$ where the sum is taken over all $\lambda_0$ such that $\boldsymbol{P}\lambda_0=0$ and $H\lambda_0=m\lambda_0$, where $m$ is the physical mass, or the exact energy eigenvalue at rest. However, at the bottom of page 215, Peskin and Schroeder claim $Z=|\langle \lambda_0|\phi(0)|\Omega\rangle|^2$ without telling us if there is a sum.

My question is: should there be a sum? Otherwise, Peskin and Schroeder don't tell us which $\lambda_0$ they mean.


1 Answer 1


P&S assume there is only one 1-particle term in the sum of the Kallen-Lehmann spectral representation (7.7).

  • $\begingroup$ But Peskin and Schroeder write right below equation 7.2 that "where the sum runs over all zero momentum states $|\lambda_0\rangle$, which seems to indicate that there could be more than one $|\lambda_0\rangle$? $\endgroup$ Oct 6, 2022 at 17:08
  • $\begingroup$ They're the multi-particle states. $\endgroup$
    – Qmechanic
    Oct 6, 2022 at 17:24

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