# Energy per unit time of spontaneous emission within second quantisation problem

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.

• 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. – G. Smith Nov 24 '18 at 5:43

## 1 Answer

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.