Stimulated Raman adiabatic passage

Consider the description of three level Ʌ atom having ground states ${\displaystyle |g_{1}\rangle }$ and ${\displaystyle |g_{2}\rangle }$ (for simplicity suppose that the energies of the ground states are the same) and excited state ${\displaystyle |e\rangle }$. Suppose in the beginning the total population is in the ground state ${\displaystyle |g_{1}\rangle }$. Here the logic for transformation of the population from ground state ${\displaystyle |g_{1}\rangle }$ to ${\displaystyle |g_{2}\rangle }$ is that initially the unpopulated states ${\displaystyle |g_{2}\rangle }$ and ${\displaystyle |e\rangle }$ couple, afterward superposition of states ${\displaystyle |g_{2}\rangle }$ and ${\displaystyle |e\rangle }$ couple to the state ${\displaystyle |g_{1}\rangle }$. Thereby a state is formed that permits the transformation of the population into state ${\displaystyle |g_{2}\rangle }$ without populating the excited state ${\displaystyle |e\rangle }$. This process of transforming the population without populating the excited state is called the stimulated Raman adiabatic passage.[5]

Consider states ${\displaystyle |1\rangle }$, ${\displaystyle |2\rangle }$ and ${\displaystyle |3\rangle }$ with the goal of transferring population initially in state ${\displaystyle |1\rangle }$ to state ${\displaystyle |3\rangle }$ without populating state ${\displaystyle |2\rangle }$. Allow the system to interact with two coherent radiation fields, the pump and Stokes fields. Let the pump field couple only states ${\displaystyle |1\rangle }$ and ${\displaystyle |2\rangle }$ and the Stokes field couple only states ${\displaystyle |2\rangle }$ and ${\displaystyle |3\rangle }$, for instance due to far-detuning or selection rules. Denote the Rabi frequencies and detunings of the pump and Stokes couplings by ${\displaystyle \Omega _{P/S}}$ and ${\displaystyle \Delta _{P/S}}$. Setting the energy of state ${\displaystyle |2\rangle }$ to zero, the rotating wave Hamiltonian is given by

${\displaystyle H_{\mathrm {RWA} }=-\hbar \Delta _{P}|1\rangle \langle 1|+\hbar \Delta _{S}|3\rangle \langle 3|+{\frac {\hbar \Omega _{P}}{2}}(|1\rangle \langle 2|+\mathrm {h.c.} )+{\frac {\hbar \Omega _{S}}{2}}(|3\rangle \langle 2|+\mathrm {h.c.} )}$

The energy ordering of the states is not critical, and here it is taken so that ${\displaystyle E_{1}<E_{2}<E_{3}}$ only for concreteness. Ʌ and V configurations can be realized by changing the signs of the detunings. Shifting the energy zero by ${\displaystyle \Delta _{P}}$ allows the Hamiltonian to be written in the more configuration independent form

${\displaystyle H_{\mathrm {RWA} }=\hbar {\begin{pmatrix}0&{\frac {\Omega _{P}}{2}}&0\\{\frac {\Omega _{P}}{2}}&\Delta &{\frac {\Omega _{S}}{2}}\\0&{\frac {\Omega _{S}}{2}}&\delta \end{pmatrix}}}$

Here ${\displaystyle \Delta }$ and ${\displaystyle \delta }$ denote the single and two-photon detunings respectively. STIRAP is achieved on two-photon resonance ${\displaystyle \delta =0}$. Focusing to this case, the energies upon diagonalization of ${\displaystyle H_{\mathrm {RWA} }}$are given by

${\displaystyle E_{0,\pm }=0,{\frac {\Delta \pm {\sqrt {\Delta ^{2}+\Omega ^{2}}}}{2}}}$

where ${\displaystyle \Omega ^{2}=\Omega _{P}^{2}+\Omega _{S}^{2}}$. Solving for the ${\displaystyle E_{0}}$ eigenstate ${\displaystyle (c_{1}\,c_{2}\,c_{3})^{T}}$, it is seen to obey the condition

${\displaystyle c_{2}=0,\;\Omega _{P}c_{1}+\Omega _{S}c_{3}=0}$

The first condition reveals that the critical two-photon resonance condition yields a dark state which is a superposition of only the initial and target state. By defining the mixing angle ${\displaystyle \tan \theta =\Omega _{P}/\Omega _{S}}$ and utilizing the normalization condition ${\displaystyle |c_{1}|^{2}+|c_{3}|^{2}=1}$, the second condition can be used to express this dark state as

${\displaystyle |\mathrm {dark} \rangle =\cos \theta \,|1\rangle -\sin \theta \,|3\rangle }$

From this, the STIRAP counter-intuitive pulse sequence can be deduced. At ${\displaystyle \theta =0}$ which corresponds the presence of only the Stokes field (${\displaystyle \Omega _{S}\gg \Omega _{P}}$), the dark state exactly corresponds to the initial state ${\displaystyle |1\rangle }$. As the mixing angle is rotated from ${\displaystyle 0}$ to ${\displaystyle \pi /2}$, the dark state smoothly interpolates from purely state ${\displaystyle |1\rangle }$ to purely state ${\displaystyle |3\rangle }$. The latter ${\displaystyle \theta =\pi /2}$ case corresponds to the opposing limit of a strong pump field (${\displaystyle \Omega _{P}\gg \Omega _{S}}$). Practically, this corresponds to applying Stokes and pump field pulses to the system with a slight delay between while still maintaining significant temporal overlap between pulses; the delay provides the correct limiting behavior and the overlap ensures adiabatic evolution. A population initially prepared in state ${\displaystyle |1\rangle }$ will adiabatically follow the dark state and end up in state ${\displaystyle |3\rangle }$ without populating state ${\displaystyle |2\rangle }$ as desired. The pulse envelopes can take on fairly arbitrary shape so long as the time rate of change of the mixing angle is slow compared to the energy splitting with respect to the non-dark states. This adiabatic condition takes its simplest form at the single-photon resonance condition ${\displaystyle \Delta =0}$ where it can be expressed as

${\displaystyle \Omega (t)\gg |{\dot {\theta }}(t)|={\frac {|\Omega _{S}(t){\dot {\Omega }}_{P}(t)-\Omega _{P}(t){\dot {\Omega }}_{S}(t)|}{\Omega (t)^{2}}}}$