1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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