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Given that the probability of an atom being in an excited state when it is applied by a resonant field is:

$$P=\sin^2(\frac{\Omega_0t}{2})$$

Find an analogous formula in the case of non-resonant excitation when there is detuning $\delta$ given that $\Omega=\sqrt{\Omega_0^2+\delta^2}$.

Is there a quick way to do this without solving the time-dependent Schrodinger equations? I know that the new probability should look like$$P=A\sin^2(\frac{\Omega{t}}{2})$$ so I was trying to do a normalisation condition where integrated the probability over the period should give me one. But this does not give me the correct answer which looks like: $$P=\frac{\Omega_0^2}{\Omega^2}\sin^2(\frac{\Omega{t}}{2})$$

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I would try to approach this problem from a time-independent perspective. For example, the resonant oscillation at frequency $\Omega_0/2$ can be seen from the Hamiltonian of a resonantly-coupled two-level system (in the basis $|0\rangle$, $|1\rangle$): $$ H = \frac{1}{2}\begin{pmatrix} 0 & \Omega_0 \\ \Omega_0 & 0 \end{pmatrix} $$ Diagonalizing this Hamiltonian gives that the two eigenstates are $|\pm\rangle = \frac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle)$, with energy splitting $\Omega_0$. Therefore when the system begins in $|0\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle)$, phase accumulates at a rate $\Omega_0$. This phase accumulation is equivalent to oscillations between $|0\rangle$ and $|1\rangle$.

Now, in the detuned case the Hamiltonian can be written as $$ H = \frac{1}{2}\begin{pmatrix} 0 & \Omega_0 \\ \Omega_0 & \delta \end{pmatrix} $$ In this case, the eigenvectors are similarly linear combinations of $|0\rangle$ and $|1\rangle$ with weights that depend on $\delta$ and $\Omega_0$. The two eigenvectors will have an energy splitting of $\frac{1}{2} \sqrt{\delta^2 + 4 \Omega_0^2}$. Therefore when the system is initialized in $|0\rangle$ (a superposition of the two eigenvectors), phase will accumulate at the rate of the eigenvector energy splitting, which corresponds to oscillations between $|0\rangle$ and $|1\rangle$ at a faster frequency but lower amplitude.

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