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There is a very clear derivation of Fermi's golden rule (actually Dirac's) here. Everything runs smoothly until, somehow, the equivalence

$$ \Gamma_{a \rightarrow b} = \frac{P_{a \rightarrow b}}{t} $$

is introduced and it is assumed that $\Gamma$ is some kind of rate (e.g. the $\beta$-decay rate of the neutron). However I don't see the equivalence with a rate (i.e. some kind of event counting per unit of time). If a process follows, say, Poissonian statistics, then the mean time until the next event is $1/\lambda$, so $\lambda$ events are happening per unit of time. But here the only thing I see is a probability divided by $t$.

Edit: here it is stated that "the transition probability is also called the decay probability and is related to the mean lifetime $t$ of the state by $P_{a \rightarrow b} = 1/t$"

Could someone help clarifying the concepts?


PS: I have checked previous posts with similar questions, to make sure there is no duplicate. I found this question which, I think, is essentially asking the same as me

Interpretation of "transition rate" in Fermi's golden rule

However, there is no good accepted answer (e.g, it is not very obvious what Qmechanic answer is actually answering).

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    $\begingroup$ A rate, by definition, is how much of a thing happens per time. If you want to know more than that you'll have to be more specific in the question. $\endgroup$ – DanielSank Dec 3 '15 at 23:36
  • $\begingroup$ I don't see how this definition is counting how much of a thing happens per time. If a process follows, say, Poissonian statistics, then the mean time until the next event is $1/\lambda$, so $\lambda$ events are happening per unit of time. But here the only thing I see is a probability divided by $t$. $\endgroup$ – nabla Dec 4 '15 at 8:05
  • $\begingroup$ I think you should explain that in the post. We can't read your mind :) $\endgroup$ – DanielSank Dec 4 '15 at 8:48
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The explanation you are looking for is on the bottom of pg.1 in the notes you linked to:

The central result is that for weak perturbations and long times, the transition probability $P_b(t)$ raises linear in time as $P_b(t) = \Gamma_{a \rightarrow b} \times t$ for $b\neq a$.

Now the transition probability $P_b(t)$ is by definition the average number of transitions into state $b$ during time $t$, in the sense of the average number of systems that end up in state $b$ from among a given initial ensemble of such systems. If all transitions to $b$ come from state $a$ (say the ensemble is initially prepared in state $a$), and if $P_b(t) \sim t$, then the average number of transition events per unit time is genuinely $\Gamma_{a \rightarrow b} = P_b(t)/t$. Hence $\Gamma_{a \rightarrow b}$ is indeed a transition rate in this case.

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  • $\begingroup$ So in a system with $N$ systems, the rate would be $N \frac{\mathrm{d}P_b}{\mathrm{d}t}$, right? $\endgroup$ – nabla Dec 4 '15 at 14:43
  • $\begingroup$ Yes, if by rate you mean the actual number of transitioned systems out of $N$. But it is customary to take the ratio to $N$ to keep things independent of ensemble size, and so the rate is just $dP_b/dt$. $\endgroup$ – udrv Dec 4 '15 at 14:54
  • $\begingroup$ Ok. I had to give a classical interpetation to the probability. Thanks! $\endgroup$ – nabla Dec 4 '15 at 15:01
  • $\begingroup$ This answer still doesn't answer the question at hand, as I see it. The issue is that $P_b (t)$ is not by definition the average number of transitions into state $b$ after a time $t$. Rather, it is the probability of finding the value b, if a single measurement is done at time t, by the usual postulates of QM which one would find in any textbook. If one wishes to reinterpret this quantity, it would cause inconsistencies elsewhere. $\endgroup$ – doublefelix Dec 14 '18 at 15:07

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