Consider a complex function

$$\tilde{f}(z)=z\int_{M}^{\infty}ds' \frac{\rho(s')}{z-s'} \qquad (1)$$

, where $M>0$ and $$\rho(s')=\frac{1}{s'}\sqrt{1-M/s'}.$$

This function is analytic in the entire complex plane except the branch cut $[M,\infty)$ in the real axis.

I'm trying to calculate $$f(s)=\lim_{\epsilon\to 0+}{\tilde{f}(s+i\epsilon)}$$, for $s>M$, which gives the first order quantum correction of the 2-2 scattering amplitude in the interacting $\phi^4$ scalar quantum field theory.

Note that the imaginary part of $f(s)$ can be calculated easily:

$$Im f(s)=s \int_M^\infty{ds' \rho(s') [-\pi \delta(s-s')]} \\ \quad\quad\quad=-\pi s\rho(s)=-\pi \sqrt{1-M/s} \qquad (2)$$

, due to the (distributional) formula

$$\lim_{\epsilon\to 0+}{ Im \left[\frac{1}{x+i \epsilon}\right] } = -\pi \delta(x). \qquad (3)$$

The answer given by the text (L. S. Brown, Quantum field theory, pp. 172-173) is

$$\tilde{f}(z)=\sqrt{1-\frac{M}{z}} \ln\left(\frac{\sqrt{z-M}+\sqrt{z}}{\sqrt{z-M}-\sqrt{z}}\right)-2 \qquad (4)$$

. The book has not specified the branch cut of the square roots and the log in this expression, but if the branch cut of the log is the negative real axis, it would give

$$ f(s) = \sqrt{1-\frac{M}{s}} \left[\ln\left(\frac{(\sqrt{s-M}+\sqrt{s})^2}{M}\right)-i\pi\right] -2 \qquad (5)$$

, which is consistent with (2).

However, I cannot understand how this answer is obtained. The textbook says that it is due to some "uniqueness property":

In view of the analytic structure of $\tilde{f}$, it is uniquely determined by its discontinuity across the cut, $2i$ $\rm{Im} {\tilde{f}(s)}$, along with its asymptotic behavior and its value at one point.

... Now $$\lim_{|s|\to\infty}{\tilde{f}(s)}=\lim_{|s|\to\infty}{\int_{\rm{const.}}^{\infty}{\left[\frac{1}{s'}-\frac{1}{s'-s}\right]}}=\ln{s}+\rm{const.}$$

, and [$\tilde{f}(0)=0$]. Therefore [(4) holds], since this function has the all the required properties.

Also, I do not understand the calculation of the asymptotic behavior of $\tilde{f}(z)$. I even doubt that (4) satisfies $\tilde{f}(0)=0$, because (4) seems to diverge when $z\to 0$.

QUESTION. Could you elaborate the quoted argument above that (4) indeed holds? Or could you suggest another, more direct, route to solve this problem?


Suppose you are given a function $f$ with some singularities, otherwise holomorphic. Suppose you have a function $g$ that has exactly the same singularities as $f$, including the one at infinity (i.e. the asymptotic behavior). Also suppose that there is a point $z_0$ such that $f(z_0)=g(z_0)$. Then the difference $h=f-g$ is a function such that

  • $h$ is an entire function (holomorphic everywhere)
  • $|h|$ is bounded
  • there is a point $z_0$ such that $h(z_0)=0$

First two conditions ensure that $h=const$ and the third condition ensures that $h=0$, giving $f=g$.

  • $\begingroup$ I know of Liouville theorem, but I wonder how I can apply it for this problem. The equation (4) has branch cuts (whose locations are not prescribed) so I'm not sure how I can extend the difference between the equations (1) and (4) to an entire function. $\endgroup$ – pdfs Jun 1 '14 at 5:25
  • $\begingroup$ You have to specify branch cuts in $(4)$ in a way such that the resulting branch cut of (4) will be the same (including the discontinuity) as of $\tilde f$. Then the difference will be an everywhere continuous function holomorphic everywhere except possibly on the former branch cut and at singularities. Then if it was indeed a branch cut, you could have chosen it to go along a different curve for both $g$ and $f$, and then the same argument shows that the difference is holomorphic on the former branch cut. To the remaining isolated singularities apply Riemann removable singularity theorem. $\endgroup$ – Peter Kravchuk Jun 2 '14 at 3:38

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