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I am working with the atom-light interaction, where the atom is considered a two-level system and the total Hamiltonian is given by $$H = H_A + H_{AL} = \hbar\omega_0\hat\sigma_{ee} + \frac{\hbar\Omega}{2}(\hat\sigma_{ge}e^{i\omega t} + \hat\sigma_{eg}e^{-i\omega t})$$ where $\hat\sigma_{ab} = |a\rangle\langle b|$ and $\omega_0,\omega$ correspond to the atomic transition frequency and laser frequency respectively. I would like to transform this into the rotating frame, as seen [here][1], by the unitary transform $U = e^{i\omega t |e\rangle\langle e|}$.

I believe that, for a time-dependant unitary transform, the Hamiltonian transforms as $$\tilde{H} = UHU^\dagger + i\hbar\left(\frac{\partial}{\partial t}U\right)U^\dagger$$

My question is probably rather basic, but I am unsure of how the exponential operator would act on the Hamiltonian. For example, what does $(e^{i\omega t |e\rangle\langle e|})(\hbar\omega_0|e\rangle\langle e|)(e^{-i\omega t |e\rangle\langle e|})$ become? My question is essentially how do the bras and kets in the exponential act on the bras and kets that are not in an exponential?

[1]: https://en.wikipedia.org/wiki/Unitary_transformation_(quantum_mechanics)#:~:text=In%20quantum%20mechanics%2C%20the%20Schr%C3%B6dinger,an%20operator%20called%20the%20Hamiltonian).

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Exponentials of operators are defined by their power series. $$ \exp(\hat{O}) = \sum_{n=0}^{\infty} \frac{\hat{O}^n}{n!}$$ So your question boils down to: what are the exponentials of a projection operator?

We immediately see that $(i\omega t |e\rangle\langle e|)^n = (i\omega t |e\rangle\langle e|)\cdot(i\omega t |e\rangle\langle e|)\dots = (i\omega t)^n|e\rangle\langle e|$ and therefore $e^{i\omega t |e\rangle\langle e|} = e^{i\omega t} |e\rangle\langle e|$.

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