In the beginning of chapter 3 on scattering theory in Weinberg's QFT book there is a use of the Cauchy residual theorem that I just cannot get.

First some notation, we are looking at states that are effectively non-interacting, and are considered to be a direct product of one-particle states described by their momenta $p_i$ and a bunch of (possibly discrete) indexes $n_i$. To simplify notation the sum over all indexes and integral over all momenta is written:

$$\int ~\mathrm d\alpha\, \ldots \ \equiv \sum_{n_1\sigma_1n_2\sigma_2\cdots} \int ~\mathrm d^3p_1~ \mathrm d^3p_2\, \ldots\tag{3.1.4} $$

The energy of such a state $\alpha$ is denoted $E_\alpha$ and is the sum of 1 particle energies corresponding to the momenta: $$E_\alpha = p_1^0 +p_2^0+\ldots\tag{3.1.7}$$ Now in the book we are looking at some integrals that look like this:

$$\int ~\mathrm d\alpha~ \frac{e^{-i E_\alpha t} g(\alpha) T_{\beta \alpha}}{E_\alpha - E_\beta \pm i \epsilon}\tag{3.1.21b} $$

$g(\alpha)$ is a smooth function that is non-zero on a finite range $\Delta E$ of energies. $T_{\beta \alpha}$ can probably also be assumed to be smooth.

Now the author extends the integral to a semi-circle in the upper half plane of the energies, uses the Cauchy residual theorem and takes $t \to - \infty$ to get the result 0. My problems are:

  1. The integral is not actually over the energies, but over the momenta. The energy is a function of the momenta, so I'm sure we can do some kind of a substitution to get an integral over energy, but this integral will not be over $\mathbb{R}$ since the energy of each particle is positive. So we cannot close a semi-circle.
  2. To use the Cauchy theorem $g(\alpha)T_{\beta \alpha}$ must be analytic after doing all integrals except for the energy integral, but if at the same time $g(\alpha)$ is supposed to be zero outside of some finite range of energies, this is not possible.

1 Answer 1


Heuristically, one approach to justifying Weinberg's application of Cauchy's formula is to treat the non-analytic integrand as the boundary behavior of a meromorphic function (kind of like Fourier series):

  1. perform all internal integrations until you are left with an integral over energy,
  2. solve the Cauchy-Riemann equations within the upper half plane (possibly with a singular set removed), with the integrand as a boundary condition on the real line,
  3. approximate the original integral using the solution just obtained (e.g. on a nearby contour within the domain of analyticity),
  4. estimate the approximation using the residue formula.

Of course it would be a miracle if all of these steps could be carried out without errors. Fortunately, it's easy to make sure that these errors decay in the large $t$ limit: in the integral $$ I=\int ~\mathrm dE~\frac{e^{-iEt}f(E)}{E-E_\beta\pm i\epsilon}, $$ the worst behavior is near $E=E_\beta$. We can eliminate this simply by separating $I$ into two parts: $$ I^\prime_\pm=f(E_\beta)\int~\mathrm dE~\frac{e^{-iEt}}{E-E_\beta\pm i\epsilon},\quad I_\pm^\textrm{reg} \equiv I-I^\prime_\pm. $$ Cauchy's integral formula can be applied to show $I_\pm '$ vanishes when $t\rightarrow \mp\infty$, while $I^\textrm{reg}_\pm$ decays to zero as $|t|\rightarrow\infty$ by the usual arguments of the Riemann-Lebesgue lemma when $f(E)$ is sufficiently 'nice'. Since $|I|\leq |I^\textrm{reg}|+|I^\prime|$, it follows that $I_\pm$ vanishes (with the appropriate choice of $\pm\epsilon$.)

  • $\begingroup$ Do I understand correctly, that the idea at the start is to approximate the smooth, but not analytic function in the integrand by a meromorphic function, do the usual complex analysis routine and then take a limit? Also I think my first problem is not actually a problem and just a result of me misunderstanding the text, I imagine the semi-circle is not supposed to have "boundary $\mathbb{R}$" (in the limit), but boundary "wherever the energies start till infinity". $\endgroup$
    – s.harp
    Aug 27, 2015 at 11:31
  • $\begingroup$ Actually I guess the functions that approximate the smooth one need only be piece-wise analytic on the real axis, as in this case you can replace the integral over an analytic function on $\mathbb{C}$ with one semi-circle to an integral with one semi-circle and many integrals over the borders of "half annuli", as all parts of the integral upper half plane will vanish as $t \to \infty$. $\endgroup$
    – s.harp
    Aug 27, 2015 at 11:57

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