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.


1 Answer 1


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.

  • $\begingroup$ Thank you very much for your answer. I understand now. $\endgroup$ Jan 18, 2022 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.