Rabi Oscillation

In a two level quantum system, the energy will oscilate in the two states.

Notes

Two states $|g⟩$ and $|e⟩$. With the interaction $\widehat{H^{\prime }}$， $V=⟨g|\widehat{H^{\prime }}|e⟩$ and detunning$\Delta =⟨e|\widehat{H^{\prime }}|e⟩=-⟨g|\widehat{H^{\prime }}|g⟩$. With Schrodinger Equation $i\hslash \frac{\partial \Phi }{\partial t}=\hat{H}\Phi$ , we can use $\Phi =C_1|g⟩+C_2|e⟩$: And $|g⟩=e^{i\frac{E_g}{\hslash }t}|g_0⟩$:

$$\{\begin{array}{rl}i\hslash \dot{C_1}& =-\Delta C_1+V{C}_2\\ i\hslash \dot{C_2}& =V{C}_1+\Delta C_2\end{array}\text{\hspace{1em}(1)}$$

Here, it could be seen as a matrix equation:

$$i\hslash \dot{\mathbf{C}}=\left(\begin{array}{cc}-\Delta & V\\ V& \Delta \end{array}\right)\mathbf{C}\text{\hspace{1em}(2)}$$

It could be a problem solving and get:

$$C(t)=e^{-i\hat{H}t/\hslash }C(0)\text{\hspace{1em}(2)}$$

Here, $\hat{H}$ could be diagonalizable and $\lambda =\pm \Omega =\pm \sqrt{V^2+{\Delta }^2}$ , and the eigenstate here is $\left(\begin{array}{c}V\\ \Delta +\lambda \end{array}\right)$ . So:

\begin{align} C(t)&=e^{-i\hat{H}t/\hbar}C(0)\\ &=\begin{pmatrix} V & V\\\Delta +\lambda_+ & \Delta + \lambda_-\end{pmatrix} \begin{pmatrix} e^{-i\lambda_+ t/\hbar} & \\ & e^{-i\lambda_- t/\hbar}\end{pmatrix}\begin{pmatrix} V & V\\\Delta +\lambda_+ & \Delta + \lambda_-\end{pmatrix}^{-1}C(0)\\ &=-\frac{1}{2V\Omega}\begin{pmatrix} Ve^{-i\lambda_+ t/\hbar} & V e^{-i\lambda_- t/\hbar}\\(\Delta +\lambda_+)e^{-i\lambda_+ t/\hbar} & (\Delta + \lambda_-) e^{-i\lambda_- t/\hbar}\end{pmatrix} \begin{pmatrix} \Delta + \lambda_- & -V\\-(\Delta +\lambda_+) & V\end{pmatrix}C(0)\\ &=\begin{pmatrix} \cos \frac{\Omega t}{\hbar}+i\frac{\Delta}{\Omega}\sin \frac{\Omega t}{\hbar} & -i\frac{V}{\Omega}\sin \frac{\Omega t}{\hbar} \\ -i\frac{V}{\Omega}\sin \frac{\Omega t}{\hbar} & \cos \frac{\Omega t}{\hbar}-i\frac{\Delta}{\Omega}\sin \frac{\Omega t}{\hbar} \end{pmatrix}C(0)\tag{3} \end{align} $$with $C(0)=\begin{pmatrix}1 \\ 0 \end{pmatrix}$ , we could get:

\begin{align} C(t)=\begin{pmatrix} \cos \frac{\Omega t}{\hbar}+i\frac{\Delta}{\Omega}\sin \frac{\Omega t}{\hbar} \ -i\frac{V}{\Omega}\sin \frac{\Omega t}{\hbar} & \end{pmatrix}\tag{4} \end{align}

Thus, it is an oscillation. between state 1 and state 2. If detunning is 0, it means $\Delta=0$ and $\Omega=V$.

\begin{align} C(t)=\begin{pmatrix} cos\frac{\Omega t}{\hbar}\ -isin\frac{\Omega t}{\hbar} \end{pmatrix} \end{align}

$$Inthiscase,theatomicinversionrateis:$$

W(t)=P_e-P_g=-cos\frac{2\Omega t}{\hbar}

## Conclusions