Today I came across a term used by Feynman in his thirteenth lecture: 'probability per unit time' to go from $| 1\rangle$ to $|2\rangle$ while initially being at $|1\rangle$. This is the excerpt from his lecture (emphasis mine):

$$ i\hbar\,\frac{d{C_n(t)}}{dt}=E_0C_n(t)-AC_{n+1}(t)-AC_{n-1}(t).$$ The first coefficient on the right, $E_0$, is, physically, the energy the electron would have if it couldn’t leak away from one of the atoms. [...] The next term represents the amplitude per unit time that the electron is leaking into the $n$th pit from the $(n+1)$st pit; and the last term is the amplitude for leakage from the $(n−1)$st pit. As usual, we’ll assume that $A$ is a constant (independent of $t$).

Then I googled a bit came before Fermi's golden rule which states:

In quantum physics, Fermi's golden rule is a simple formula for the constant transition rate (probability of transition per unit time) from one energy eigenstate of a quantum system into other energy eigenstates in a continuum, effected by a perturbation. $$ \Gamma_{i\to k} = \frac{d}{dt} |a_k(t)|^2 = \frac{2|\langle k | H^\prime | i\rangle |^2}{\hbar^2}\frac{\sin\omega t}{\omega} $$

So, the probability of transition per unit time is given by the derivative $$\frac{d}{dt}|a_k(t)|^2.$$ But isn't it the change in probability of transition with respect to time & not just probability of transition per unit time? After all, I learnt that derivatives represent how fast some quantity change instantaneously with respect to change in some other quantity(ies); applying it here, I couldn't find any reason why it shouldn't be called change in probability of transition with respect to time. I am not understanding why probability of transition per unit time is related to the derivative as derivative does represent the rate of change of some quantity & not the quantity; isn't it ridiculous to call $\dfrac{dv}{dt}$ velocity per unit time? I'm not understanding what probability of transition per unit time actually means. Can anyone please, please explain this to me?

  • $\begingroup$ The phrase you mention doesn't appear in Feynman's text. $\endgroup$ – innisfree Sep 29 '15 at 23:22

I'm not sure what might be confusing you. Assume, as in most cases with the Golden rule, that the transition rate is constant, $\Gamma$. So, for small times, the cumulative transition probability is $W=\Gamma t$.

Think of the transition as leakage from a vessel. At $t=0$, no water has been lost, but with a constant rate of leakage, $\Gamma$, the longer the leakage lasts, the more water is lost to the vessel, so then the leakage rate, the transition rate, is $\frac{\mathrm{d}W}{\mathrm{d}t}=\Gamma$. The picture is the same if you gave $\Gamma$ a time dependence, when $W=\int\Gamma \, \mathrm{d}t$.

As for English usage, what does "elevation gain per unit of time" tell you when going up the mountain in a gondola?

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