I am slightly confused about the application of Fermi's golden rule. Which during standard derivations indicates a probability of transitioning from the state $|i \rangle$ to the state $|f\rangle$ of: $$P=\frac{2\pi t}{\hbar^2} | \langle f | \hat V | i \rangle|^2 \delta(E_f-E_i-\hbar \omega)$$ This holds for large values of $t$ but for arbitrarily large values of $t$: $$P>1$$ Which (to me anyway) makes no physical sense since there can't be a probability greater then 1 of been in the final state. If my interpretation is right then how is this problem resolved/allowable? and if it is wrong what part of what I have said is wrong?

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    $\begingroup$ The FGR is obtained using TDPT. If you follow the steps carefully, you'll see that the neglected terms are of order $\hat V^2$ and $t^2$, so that the approximation is only valid for small values of $t$. A way better approximation is obtained by neglecting $\hat V^2$ but keeping $t$ arbitrary. The result is an something like $P\sim\exp[w_{if}t]-1$, where $w_{if}\propto |\langle f|V|i\rangle|^2$. If you expand to first order in $t$, you get the FGR back. This is thoroughly discussed in any book on TDPT, and in most books on QM. $\endgroup$ Commented Mar 4, 2016 at 13:29
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    $\begingroup$ @AccidentalFourierTransform But you would only get the form of the probability that I have given in the assumption that $t$ is very large. This is because there is a use of the approximation equivalent to $$\lim_{t\rightarrow \infty} \left( \frac{\sin^2(xt)}{x^2}\right)=\pi t \delta(x)$$ $\endgroup$ Commented Mar 4, 2016 at 13:50
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    $\begingroup$ That's somewhat oversimplified.There are several assuptions on the magnitude of $t$. It should be smaller that some constants and larger than some others. I don't remember the details right now, but the assumptions are along the lines of $\tau\ll t\ll T$ for some constants $\tau,T$.The part $t\gg \tau$ is needed to get the delta, but the restriction $t\ll T$ means you cannot take $t\to\infty$. These assumptions arent truly fundamental but simplify the analysis. If you don't restrict the magnitude of $t$ you get the exponential I was talking about. Again, this is addressed in most books on TDPT $\endgroup$ Commented Mar 4, 2016 at 13:55
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    $\begingroup$ @AccidentalFourierTransform If you put your comments as an answer I will accept it. Further just of note, I think that this answer physics.stackexchange.com/q/91794 of Qmechanic's does answer my question (although possibly not the question it was actually posted in). $\endgroup$ Commented Mar 5, 2016 at 11:40

1 Answer 1


This question has been discussed in the comment section, and OP has already found the answer in this wonderful post by QMechanic. In that post, we can clearly see that the magnitude of $t$ is not arbitrary: it has to be bounded both below and above, $$ \frac{2\pi}{\Delta \omega} \lesssim t \ll \frac{\hbar}{\sup_{f\in F}|V_{fi}|} $$ where the symbols are defined in QMechanic's post. This is the reason we cannot take $t\to \infty$.

It is useless to reproduce QMechanic's post here, or to try to improve it. Instead, I'll try to address my statement that, if we keep the magnitude of $t$ arbitrary, then the transition probability is an exponential.

In the usual proof of the FGR we find, at some point, the integrodifferential equation $$ c_i(t)=1-i\sum_n\int_0^tc_n(\tau)V_{ni}(\tau)\; \mathrm e^{-i\omega_{ni}\tau} $$ and take $c_n(\tau)\approx c_n(0)$ to get the first order correction (from which the FGR follows). This is the step that constrains the magnitude of $t$. If we take $c_n(\tau)\approx c_n(t)$ instead, we find (after some lengthy manipulations) $$ c_i(t)=\mathrm e^{-\Gamma t/2}\times \text{a phase} $$ where $\Gamma\propto V_{fi}$ is the decay width of the state. This formula is valid for arbitrarily large $t$. Expanding to first order ($\Gamma t\ll 1$), we get the usual formulas back. A detailed derivation (and discussion) can be found in the book by Cohen-Tannoudji Quantum Mechanics, vol 2., page 1351, "Another approximate method for solving the Schrödinger equation".


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