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Consider QM perturbation theory. For the hamiltonian $\hat{H} = \hat{H}_{0} + \hat{V}$, the set of eigenstates $\{|n\rangle\}$ of $\hat{H}_{0}$ and assuming time independence of $\hat{V}$, one has that the differential probability $dP_{if}(t)$ of transition $$ |i\rangle \in \{|n\rangle\} \to |f\rangle \in \{|n\rangle\} $$ in first order of perturbation theory is $$ dP_{if}(t) = |V_{fi}|^{2}\frac{\sin^{2}\left(\frac{\omega_{fi}\Delta t}{2}\right)}{\left( \frac{\omega_{fi}}{2}\right)^{2}} $$ Here $$ \hat{H}_{0} |n\rangle = E_{n}|n\rangle,\quad \omega_{if} = E_{i}-E_{f},\quad \Delta t = t - t_{0}, \quad V_{fi} = \langle f|\hat{V}|i\rangle $$ and $t_{0}$ is a moment of time when the interaction $\hat{V}$ is turned on.

Fermi golden rule is obtained by taking the limit $$ \lim_{\Delta t \to \infty}\frac{dP_{if}(t)}{\Delta t} $$

Weinberg in his QFT Vol. 1, when considering this limit for $|i\rangle$ being one-particle state (independently on perturbation theory), says that one can't take $\Delta t$ larger than the lifetime $\tau$ of the state, and this means that the energy $E$ of the state is defined up to the width $\Delta E = \tau^{-1}$. I don't understand the reason of this.

Could you please clarity it?

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I expect you've got the answer by now, but logically a particle cannot be expected to survive past its lifetime (at which point it will probably decay into an n-particle state). So the lifetime is a sensible limit on the time a single-particle state exists. Then the $\Delta E$ relation is just an application of the time-energy Heisenberg uncertainty principle, i.e. $ \Delta E \Delta t \geq \hbar = 1$ using natural units, since $\Delta t \leq \tau$.

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