Section 7.3 ("The Optical Theorem") in Peskin and Schroeder's QFT text contains a leading order verification of the optical theorem in $\phi^4$ theory by calculating the (discontinuity across the branch cut in) the imaginary part of the Feynman amplitude $\mathcal{M}$ of the following diagram

one-loop 2 -> 2 s-channel diagram in \phi^4 theory

$$ i\mathcal{M} = \frac{\lambda^2}{2} \int \frac{d^4q}{(2\pi)^4} \frac{1}{(k/2-q)^2 - m^2 + i\epsilon} \frac{1}{(k/2 + q)^2 - m^2 + i\epsilon} $$

Working in the center-of-mass frame, where $k = (k^0, \vec 0)$, there are two poles below the real $q^0$ line and two above

location of poles of amplitude (in complex q^0 plane)

The textbook here mentions that, if the contour is closed downward,

only the pole at $q^0 = -(1/2)k^0 + E_{\mathbf{q}}$ will contribute to the discontinuity.

I cannot come up with an argument as to why the other pole's residue does not contribute to the discontinuity in $\mathcal{M}$. In fact, from the symmetry between $q$ and $-q$ in the integral above, the two propagators should contribute equally. By factorizing the denominators and calculating residues one can see that indeed the two residues are equal.

What gives? Is P&S wrong here? Or am I making an elementary slip-up somewhere?

  • $\begingroup$ I looked it up and I'm puzzled too. In particular because then they continue to compute the integral and indeed also the second propagator is replaced by a $\delta$, so it looks like both of the poles contributed after all... $\endgroup$
    – MannyC
    Feb 12, 2019 at 16:43

1 Answer 1


In Peskin, it only says that:

only the pole at $q^0 = -(1/2)k^0+E_{\vec{q}}$ will contribute to the discontinuity.

So when you do the integration, you either close the integration loop upward or downward. In the book, the latter is chosen and you only have two poles for the evaluation: $q^0 = -1/2 k^0+E_q$ and $q^0 = 1/2 k^0+E_q$.

The pole $q^0 = 1/2 k^0+E_q$ is neglected because it doesn't contribute to the singularity. This can be shown once you evaluate the integration using this pole. In the end, you will obtain a similar result as equation (7.54), with $$\frac{1}{k^0(k^0-2E_q)}$$ changed into $$\frac{1}{k^0(k^0+2E_q)}$$ Here no matter $k^0$ is larger or smaller than $2m$, as long as $k^0$ and $E_q$ are larger than $0$, there is no singularity for the integrand. And of course $k^0$ and $E_q$ should be larger than $0$ since they represent the energy of particles.


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