1
$\begingroup$

In the Optical Theorem section, in the middle paragraph of p235, it is said

"What is left over in expression (7.52) is just the factor $\lambda^2$, the square of the leading-order scattering amplitude, and the symmetry factor (1/2)".

However in (7.52) (which is the invariant amplitude of the 1-loop order diagram in $\phi^4$ theory) there is no $|\mathcal{M}(k)|^2$. Therefore, where does $|\mathcal{M}(k)|^2$ come from in the last equation, p235 which is

$$\operatorname{Disc}\mathcal{M}=2i\operatorname{Im}\mathcal{M}=\frac{i}{2}\int\frac{d^3p_1}{(2\pi)^3}\frac{1}{2E_1}\frac{d^3p_2}{(2\pi)^3}\frac{1}{2E_2}\vert\mathcal{M}(k)\vert^2(2\pi)^4\delta^{(4)}(p_1+p_2-k)?$$

$$i\delta\mathcal{M}=\frac{\lambda^2}{2}\int\frac{d^4q}{(2\pi)^4}\frac{1}{(k/2-q)^2-m^2+i\epsilon}\frac{1}{(k/2+q)^2-m^2+i\epsilon}\tag{7.52}$$

$\endgroup$
  • $\begingroup$ P&S are showing two ways to derive the imaginary part of the 1 loop diagram in phi^4 scattering. The first way is the systematic replacement of propagators with a delta function c.f Cutkosky cutting rules. The second way is via the optical theorem. P&S show of course that both ways are equivalent - the last equation on p.235 is a statement of the optical theorem. $\endgroup$ – CAF Aug 18 '18 at 11:06
  • $\begingroup$ Why are they saying that what is left over in (7.52) is the square of the leading-order scattering amplitude? I cannot see it in (7.52). $\endgroup$ – ketherok Aug 18 '18 at 12:45
  • $\begingroup$ ‘The square of the leading order scattering amplitude’ in the middle paragraph of p.235 is simply a parentheses. That is to say, $|\mathcal M(k)|^2 = \lambda^2$. $\endgroup$ – CAF Aug 18 '18 at 13:55
  • $\begingroup$ Does this address your concern? $\endgroup$ – CAF Aug 20 '18 at 7:26
1
$\begingroup$

P&S are showing two ways to derive the imaginary part of the $1$ loop diagram in $\phi^4$ scattering. The first way is the systematic replacement of propagators with an on shell constraining delta function c.f Cutkosky cutting rules. This originates in $$\text{Im} \frac{1}{k^2-m^2+ \mathrm{i}\epsilon} = -\pi \delta(k^2-m^2).$$

The second way is via the optical theorem. P&S show of course that both ways are equivalent - the last equation on p.235 is a statement of the optical theorem.

The issue in question may actually just be a confusion in the english - ‘The square of the leading order scattering amplitude’ in the cited paragraph above is simply a parentheses to $\lambda^2$, that is to say, $$|\mathcal M(k) |^2 = \lambda^2,$$ no $k$ dependence at tree level on the rhs because we are in the scalar field theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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