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I'm studying second quantisation and I have the following problem concerning the spontaneous emission that corresponds to the decay of an atom from the level $ \lvert2\rangle$ to $ \lvert1\rangle$ and I want to show within the dipole approximation that the energy radiated by the atom per unit time can be written in the form $\frac{E}{t}=\frac{4e^2}{3c^3}\langle1\rvert \frac{d^2\hat{r}}{dt^2}\lvert2\rangle$. Long story short, using the lifetime of the excited state equation I arrived to $\frac{E}{t}=\frac{4e^2}{3c^3}\langle1\rvert \hat{r}\lvert2\rangle$. The solutions suggest that $\langle f\rvert \hat{r}\lvert i\rangle=-\frac{1}{\omega_{fi}^2}\langle f\rvert \frac{d^2\hat{r}}{dt^2}\lvert i\rangle$, where $\omega_{fi}=\frac{E_1-E_2}{\hbar}$ but I fail to see how to get to this last equality. Any hint of how you get it? Thank you.

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    $\begingroup$ In the Heisenberg picture, the time derivative of an operator is related to the commutator of the Hamiltonian with that operator. This leads to the last equality. $\endgroup$ – G. Smith Nov 24 '18 at 5:43
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In the Heisenberg representation, we have that $\mathcal{O}(t)=e^{iHt} \mathcal{O}(0)e^{-iHt}$. For the position operator, $r(t)=e^{iHt} \mathcal{O}(0)e^{-iHt}$, so sandwiching between two energy eigenstates we obtain \begin{align} \langle 1|\hat{r}(t)|2\rangle=\langle 1|e^{iHt} \hat{r} e^{-iHt}|2\rangle=\langle 1|e^{iE_1t} \hat{r}e^{-iE_2t}|2\rangle=\langle 1|\hat{r}|2\rangle e^{i\left(E_1-E_2 \right)t} \end{align} Differentiating both sides with respect to time twice, and inserting the factors of $\hbar$ where appropriate, proves what you're after.

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