Peskin & Schroeder, An Introduction to Quantum Field, page 224-225, formula 7.41
\begin{multline}\sum_{\lambda} \int \frac{d^{3} k}{2(2\pi)^3 E_{\mathbf{k}}(\lambda)}\Phi (\mathbf{k}) \frac{i}{p^{0}-E_{\mathbf{k}}(\lambda)+i \epsilon} \sqrt{Z}\left\langle\mathbf{\lambda_{\mathbf{k}}}\left|T\left\{\phi\left(z_{1}\right) \cdots\right\}\right| \Omega\right\rangle \\ \underset{p^0\to+E_{\mathbf{p}}}{\sim} \int \frac{d^{3} k}{(2\pi)^3}\Phi (\mathbf{k}) \frac{i}{p'^2 - m^2 +i \epsilon} \sqrt{Z}\left\langle\mathbf{k}\left|T\left\{\phi\left(z_{1}\right) \cdots\right\}\right| \Omega\right\rangle \tag{7.41} \end{multline}
With $p'= (p^0 , \mathbf{k} )$. $\mathbf{p}$ is the center of the wavepacket $\Phi (\mathbf{k})$. $E_{\mathbf{p}} = \sqrt{ \mathbf{p}^2 + m^2}$. Before we do the limit, $p^0$ is just a number unrelated to $\mathbf{p}$ .
The one-particle singularity is now a branch cut, whose length is the width in momentum space of the wavepacket $\Phi (\mathbf{k})$.
This is the statement I don't understand. If I am not wrong, a branch cut is an interval where the function has a "continuum" of singularities. But I only see a singularity at $\mathbf{k}=\mathbf{p}$. Is it because you can find an arbitrary number of state $\lambda$ with total momentum $\mathbf{k}$ that has energy $E_{\mathbf{p}}$ ? if this is true, isn't $p'= (k^0 , \mathbf{k} )$ with $ k^0 = \sqrt{ \mathbf{k}^2 + m^2}$ a better way of writing it?
Also in that case shouldn't we have a kind of density for the number of states $\lambda$ with momentum $\mathbf{k}$ and energy $E_{\mathbf{p}}$?
Any help will be appreciated.
