Interaction picture and Rabi's problem

\( \def\ket#1{{ \left| #1 \right> }} \) \( \def\bracket#1{{ \left( #1 \right) }} \)

Usually in a quantum mechanics course, Schrodinger picture is introduced along with the Schrodinger equation 
\begin{equation}
\hat{H} \ket{\psi(t)} = i \hbar \frac{\partial}{\partial t}\ket{\psi (t)},
\end{equation}
where the quantum state $\ket{\psi(t)}$ evolves in time but quantum operators, like Hamiltonian $\hat{H}$, remains unchanged. Heisenberg picture approaches in an opposite way: operators evolve in time while states remain constant. The time evolution of an operator $\hat{A}$ satisfies the following identity
\begin{equation}
\hat{A}(t) = e^{i \hat{H} t/\hbar} \hat{A}(0) e^{-i \hat{H} t/\hbar}.
\end{equation}
Interaction picture (sometimes it is called as Dirac picture) is a hybrid one where both states and operators incorporate time dependence. The interaction picture provides a useful tool when it comes to examine quantum systems placed in external field. 

To elaborate interaction picture, here we consider a simple example of $1/2$-spin particle subjected to an oscillating magnetic field (Rabi's problem). The external field is composed by two parts: one part is parallel to $Oz$ with constant magnitude of ${B_z}$ and the other rotates clockwise on $xy$ plane with angular frequency of $\omega $ and has magnitude of $B$.  The Hamiltonian of system, which is also called Rabi Hamiltonian, could be written as
\begin{equation}
\hat{H} =  - {B_z} \hat{\sigma}_z - B\cos \left( {\omega t} \right)\hat{\sigma} _x + B\sin \left( {\omega t} \right) \hat{\sigma }_y,
\label{Hamiltonian}
\end{equation}
where $\hat{\sigma}_j\left( {j = 1,2,3} \right)$ are conventional Pauli matrices. In interaction picture formalism, transformations of Hamiltonian and state are can be written as 
\begin{equation}
\ket{ \psi(t) }_I = \hat{U}_0^\dagger(t) \ket{ \psi(t) },
\label{intket}
\end{equation}
and
\begin{equation}
\hat{H}_I = \hat{U}_0^\dagger (t) \hat{H} \hat{U}_0 - \hat{U}_0^\dagger \bracket{i\hbar {\partial _t}} \hat{U}_0(t),
\label{intHam}
\end{equation}
where 
\begin{equation}
\hat{U}_0 (t) = \exp \left( {{{i{B_z} \hat{\sigma}_z t}/ \hbar }} \right)
\label{Uop}
\end{equation}
 is time-evolution operator corresponding to time-dependent part of Hamiltonian and the subcript $I$ stands for \textit{interaction}. Notice that ket state and Hamiltonian in Eqs. (\ref{intket}) and (\ref{intHam}) satisfy following Schrodinger-like equation
\begin{equation}
\hat{H}_I \ket{\psi (t)}_I = i\hbar \frac{\partial}{\partial t} \ket{\psi (t)}_I.
\label{Schrolike}
\end{equation}
Let us briefly discuss the form of operator $U_0$. Here the system's Hamiltonian is well separated into time-dependent and time-independent terms, hence it is recommended to choose such $\hat{U}_0$ operator like Eq. (\ref{Uop}). If someone insists $\hat{U}_0$ to correspond to time-dependent terms or even mixed terms, the Schrodinger-like equation still holds. However, that is not the spirit of interaction picture. The choice of time-evolution operator $\hat{U}_0(t)$ should cancel the effect of time-independent Hamiltonian part shown in Eq. (\ref{Hamiltonian}) and leave the time-dependent part to govern time-evolution of ket state, by which the problem is simplified.

Let us ignore the aforementioned recommendation and seek another unitary transformation $\hat{U}_0$ which its physical meaning can be deduced. From basic notion of relativity, the external oscillating magnetic field looks stationary in basis rotating with same direction and same frequency, and thus Hamiltonian is time independent in such basis. We therefore choose $\hat{U}_0$ as rotation operator about $Oz$ in clockwise direction 
\begin{equation}
\hat{R}_z (-\omega t) \equiv e^{i \omega t \hat{\sigma}_z t /2} = 
\begin{pmatrix}
e^{i \omega t/2} & 0\\
0 & e^{i \omega t/2}
\end{pmatrix}.
\end{equation}
Straightforward calculation of interaction-picture Hamiltonian in the rotating frame, yields
\begin{equation}
\begin{aligned}
\hat{H}_R &= \hat{R}_z^\dagger (t) \hat{H} \hat{R}_z - \hat{R}_z^\dagger \bracket{i\hbar {\partial _t}} \hat{R}_z(t) \\
&= \begin{pmatrix}
-Bz + \hbar \omega/2 & -B\\
-B & Bz - \hbar \omega/2
\end{pmatrix}.
\end{aligned}
\end{equation}
The obtained Hamiltonian $\hat{H}_R$ is time-independent as we expect. 

We now solve the eigenvalue problem. The ket state in terms of rotating frame can be expanded as
\begin{equation}
\ket{\psi(t)}_R = d_0(t) \ket{0} + d_1(t) \ket{1},
\end{equation}
where $\left| 0 \right\rangle $ and $\left| 1 \right\rangle $ denote two eigenstates of Pauli-$z$ matrix and coefficients satisfy the following differential matrix equation derived from Schrodinger-like equation 

\begin{equation}
i\hbar \frac{\partial }{{\partial t}}\left( {\begin{array}{*{20}{c}}
{{d_0}}\\
{{d_1}}
\end{array}} \right) = \begin{pmatrix}
-Bz + \hbar \omega/2 & -B\\
-B & Bz - \hbar \omega/2
\end{pmatrix} \left( {\begin{array}{*{20}{c}}
{{d_0}}\\
{{d_1}}
\end{array}} \right).
\end{equation}

Assume that the particle's state is initially prepared as $\ket{0}$, we obtain
\begin{equation}
\begin{aligned}
{\left| {{d_1}\left( t \right)} \right|^2} & = \frac{{4{B^2}}}{{4{B^2} + {\alpha ^2}}}\,{\kern 1pt} {\sin ^2}\left( {t\frac{{\sqrt {4{B^2} + {\alpha ^2}} }}{{2\hbar }}} \right), \label{d0}\\
{\left| {{d_0}\left( t \right)} \right|^2} & = 1 - {\left| {{d_1}\left( t \right)} \right|^2}, 
\end{aligned}
\end{equation}
where $\alpha  = \hbar \omega  - 2{B_z}$. 
The probability to find system in state $\left| 1 \right\rangle $ achieves maximum when 
\begin{equation}
\omega  = 2B_z / \hbar,
\label{resonance}
\end{equation}
which means the angular frequency characteristic of system equals to angular frequency of oscillating magnetic field, implying condition of \textit{Rabi resonance}. In resonance regime, the interaction-picture Hamiltonian is proportional to Pauli-$x$ matrix which indicates that particle state evolves back and forth between $\ket{0}$ and $\ket{1}$.

Acknowledgment: This note was initially created in 2018 when I entered Korea University. I acknowledge my lab senior Park Seung, who helped me a lot with quantum mechanics at that time.