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\end{align}</math>
as stated. The next step is to find the Hamiltonian in the [[interaction picture]], <math>H_{1,I}</math>. The required unitary transformation is:
\begin{align} U & = e^{iH_0t/\hbar} \\ & = e^{i \omega_0 t/2 (|\text{e}\rangle \langle\text{e}| - |\text{g}\rangle \langle\text{g}|)} \\
& = \cos\left(\frac{\omega_0 t}{2}\right)
\left(|\text{e}\rangle \langle\text{e}| + |\text{g}\rangle \langle\text{g}|\right) + i \sin\left(\frac{\omega_0 t}{2}\right) \left(|\text{e}\rangle \langle\text{e}| - |\text{g}\rangle \langle\text{g}|\right) \\ & = e^{-i\omega_0 t/2}|\text{g}\rangle \langle\text{g}| + e^{i \omega_0 t/2} |\text{e}\rangle \langle\text{e}| \\
& = e^{-i\omega_0 t/2}\left(|\text{g}\rangle \langle\text{g}| + e^{i \omega_0 t} |\text{e}\rangle \langle\text{e}|\right)
\end{align}
</math> ,where the 3rd step can be proved by using a [[Taylor series]] expansion, and using the orthogonality of the states <math>|\text{g}\rangle</math> and <math>|\text{e}\rangle</math>. Note that a multiplication by an overall phase of <math>e^{i \omega_0 t/2}</math> on a unitary operator does not affect the underlying physics, so in the further usages of <math>U</math> we will neglect it. Applying <math>U</math> gives:
: <math>\begin{align}
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