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In Srednicki's Quantum Field Theory chapter 11, the probability of a $2 \to n$ scattering process is calculated to be

$$ P = \frac{|\left<f|i\right>|^2}{\left<f|f\right>\left<i|i\right>} = \frac{(2\pi)^4 \delta\left(k_1+k_2-\sum {k'}_i\right) T}{2E_1 2E_2 V} |\mathcal{T}|^2\prod_{i'} \widetilde{dk}_i'.\tag{11.11+20} $$

He then proceeds to divide this by $T$ to obtain a probability per unit time, before yielding an expression for the differential cross section. However, the factor of $T$ is troubling me: as $T\rightarrow \infty$, $P$ is going to exceed $1$ at some point, isn't it? Does this have to do with dealing with plane waves instead of wave packets? I am aware that this problem also shows up when one uses Fermi's golden rule to calculate decay rates, but I guess I never thought about it too seriously then.

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In chapter 11 of Srednicki, take note of equation 11.14: $$(2 \pi)^4\delta ^4(0) = \int d^4 x e^{i0\cdot x} = VT,$$ where $V$ is total volume of space (which is cancelled by another $V$ that appears in the denominator to $P$), and $T$ is total elapsed time.

Thus $P \propto T$, so you must divide $P$ by $T$ to get a well-defined general expression for probability per unit time that is independent of the interval.

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  • $\begingroup$ Thanks for your reply. I understand how T shows up and how it is divided to get a probability per unit time. It's that this expression for probability is confusing me - it is still a probability and should be bounded by 1, right? And even the dividing by T to get a probability per time sounds a little dubious to me too, if there's a 80% chance of something happening within two days, we can't simply say said event has a 40% chance of happening each day, can we? $\endgroup$
    – Mike
    Feb 18 at 5:11
  • $\begingroup$ Actually you can say that an 80% probability of something happening in two days is equivalent to a 40% probability of something happening in one day or the next because 0.4 + 0.4 = 0.8. This is assuming the events are independent. P(a or b) = P(a) + P(b). Furthermore, the initial expression is the probability that something will happen during interval T, so dividing by T gives the probability of that event happening per unit time. $\endgroup$
    – klippo
    Feb 18 at 6:20
  • $\begingroup$ Then that event happening in three days would have a probability of 120%? $\endgroup$
    – Mike
    Feb 18 at 11:01

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