A derivation in Peskin and Schroeder in chapter 18.

  • #1
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They write on page 618:
If we insert the leading-order expression (18.39) into (18.94), we obtain the familiar result:
$$(18.95)\sigma(e^+ e^- \to \text{hadrons})=\frac{4\pi \alpha^2}{s}\sum_f Q_f^2$$
where for those who don't have the book at hand, I'll write the related equations:

$$(18.94) \ \ \ \sigma(e^+ e^- \to \text{hadrons})=\frac{4\pi \alpha^2}{s} [ Im c^1(q^2)+Im c^{\bar{q}q}(q^2) \langle 0| m\bar{q}q|0\rangle+ $$
$$+Im c^{F^2}(q^2)\langle 0 | (F^a_{\alpha \beta})^2 | 0 \rangle + \ldots ] $$

$$(18.93) \ \ \ c^1(q^2)= - \bigg( 3\sum_f Q_f^2 \bigg) \cdot \frac{\alpha}{3\pi}\log (-q^2)$$

If I insert (18.93) into (18.94) I get:
$$\sigma = -\frac{4\alpha^3}{s}\sum_f Q_f^2 \cdot Im \log(-q^2)$$

If (18.95) is indeed deducible from (18.94) and (18.93) then ##\alpha \cdot Im (\log (-q^2)) =\pi/ \alpha##;
But why is that?

I must confess that I took hiatus from reading PS; so it might be covered before in the book.
 

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  • #3
vanhees71
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As usual, PS is sloppy with the poles. It should read ##\ln(-q^2-\mathrm{i} 0^+)##. Then you correctly get (with the usual meaning of ##\ln## as being on the Riemann sheet, where the logarithm along the positive real axis is defined real and the cut along the negative real axis)
$$-\mathrm{Im} \ln(-q^2-\mathrm{i} 0^+)=+ \pi \Theta(p^2).$$
 
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