# Branch Cut of Wavepacket - LSZ Reducation Formula Peskin

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.

Here is a quick sketch which might help you. We can identify a branch cut by considering the difference between one-sided limits at each side of the cut. For an analytic function(i.e. when there is no cut), such a difference should vanish. I will define the integral in (7.41) as $$\int \frac{\mathrm{d}^3 k}{(2\pi)^3} \varphi(\mathbf{k}) f(\mathbf{k}) \frac{\mathrm{i}}{(p^0)^2 - \mathbf{k}^2 - m^2 + \mathrm{i}0} =: g((p^0)^2)$$ Now consider $$g(p^2 + \mathrm{i}\epsilon) - g(p^2 - \mathrm{i} \epsilon)$$ as $$\epsilon \to 0$$. This yields \begin{align*} g(p^2 + \mathrm{i}\epsilon) - g(p^2 - \mathrm{i} \epsilon) = \mathrm{i}&\int \frac{\mathrm{d}^3 k}{(2\pi)^3} \varphi(\mathbf{k}) f(\mathbf{k}) \left( \frac{1}{p^2 - \mathbf{k}^2 -m^2 + i \epsilon} - \frac{1}{p^2 - \mathbf{k}^2 -m^2 - i \epsilon} \right) \\ \overset{\epsilon \to 0}{\longrightarrow} &\int \frac{\mathrm{d}^3 k}{(2\pi)^3} \varphi(\mathbf{k}) f(\mathbf{k}) (2\pi) \delta(p^2 - \mathbf{k}^2 - m^2) \end{align*} where we have used the principle value of $$1/x$$ given by $$\frac{1}{x \pm \mathrm{i}0} = P\frac{1}{x} \mp \mathrm{i} \pi \delta(x).$$ From this expression you can readily see that the difference of the one-sided limits actually give a finite value whenever $$\mathbf{k}$$ is in the support of $$\varphi(\mathbf{k})$$, i.e. we have a branch cut whose length is approximately the width of the momentum wavefunction $$\varphi(\mathbf{k})$$, as claimed.