= Solution
{c}
Differentiating the specified rotating-frame amplitudes gives the time-independent <Schrödinger equation>
$$
i\frac{d}{dt}\begin{pmatrix}\widetilde c_1\\\widetilde c_2\end{pmatrix}
=\frac12\begin{pmatrix}\delta&\Omega\\\Omega&-\delta\end{pmatrix}
\begin{pmatrix}\widetilde c_1\\\widetilde c_2\end{pmatrix}.
$$
The <characteristic polynomial> is $\lambda^2-(\delta^2+\Omega^2)/4=0$, hence
$$
\lambda_\pm=\pm\frac12\sqrt{\delta^2+\Omega^2}.
$$
For $\Omega\ne0$, corresponding normalized <eigenvectors> are proportional to $(\Omega,2\lambda_\pm-\delta)^T$. Their arbitrary complex coefficients evolve as $e^{-i\lambda_\pm t}$; this solves the amplitude equations for any initial <quantum state>.
Restoring the common <energy> origin $\overline E=(E_1+\hbar\omega+E_2)/2$, the two <dressed state> <energy eigenvalues> are
$$
E_\pm=\overline E\pm\frac\hbar2\sqrt{\delta^2+\Omega^2}.
$$
Thus their positive <energy> separation increases by
$$
\boxed{\Delta(E_+-E_-)=\hbar[\sqrt{\delta^2+\Omega^2}-|\delta|]
=\frac{\hbar\Omega^2}{2|\delta|}+O(\Omega^4/|\delta|^3).}
$$
Equivalently, the branches connected to $E_1'=E_1+\hbar\omega$ and $E_2$ shift by $+\hbar\Omega^2/(4\delta)$ and $-\hbar\Omega^2/(4\delta)$ respectively. Their \i[signed] difference $E_1'-E_2$ changes by $\hbar\Omega^2/(2\delta)$, which is the printed expression with a signed convention. For negative <Rabi detuning>, an actual positive separation still increases; for example at $\delta=-10\Omega$ it cannot have the negative change implied by reading the printed expression as an unsigned gap. These opposite <AC Stark shifts> describe level repulsion.
Back to article page
