# In the adiabatic theorem, how do we know which eigenstate we start on? (STIRAP)

I am aware of the question here, but it doesn't have an answer and also doesn't answer my question. I'm wondering about a specific case in STIRAP, where the 3 eigenstates are $$|\Psi_\pm\rangle = \frac{1}{\sqrt{2(\Omega_P^2+\Omega_S^2)}}\left[\begin{array}{ccc} \Omega_P \\ \pm\sqrt{\Omega_P^2+\Omega_S^2} \\ \Omega_S \end{array}\right], |\Psi_\text{dark}\rangle = \frac{1}{\sqrt{\Omega_P^2+\Omega_S^2}}\left[\begin{array}{ccc} \Omega_S \\ 0 \\ -\Omega_P \end{array}\right],$$ with eigenenergies $$E_\pm = \pm \frac{\hbar}{2} \sqrt{\Omega_P^2+\Omega_S^2},\ E_\text{dark} = 0$$. Then we kick off $$\Omega_S$$ while leaving $$\Omega_P = 0$$ to push the system into $$|\Psi_\text{dark}\rangle$$ and we can apply the adiabatic elimination to slowly nudge the dark state from $$|1\rangle \rightarrow |3\rangle$$.

Here's my question: what if we kick off $$\Omega_P$$ first, and perform transfer through the $$|\Psi_\pm\rangle$$? I assume doing that would put the system into some superposition of the $$|\Psi_\pm\rangle$$? What superposition would it be, $$|\Psi_+\rangle + |\Psi_-\rangle$$ or $$|\Psi_+\rangle - |\Psi_-\rangle$$ or something else?

If the answer is $$|\Psi_+\rangle + |\Psi_-\rangle$$, wouldn't that mean we can also perform coherent transfer through $$|\Psi_\pm\rangle$$? Or would the system fall into a specific eigenstate instead?

With $$\Omega_P = 0$$ (with non-zero $$\Omega_S$$) the dark state is $$| 1 \rangle$$, as you say, and when $$\Omega_S = 0$$ (with non-zero $$\Omega_P$$) then the dark state is $$| 3 \rangle$$. The idea is that the system state should be dark throughout.

If you start with $$\Omega_S \ne 0,\;\Omega_P = 0$$ then the system will first evolve towards $$| 1 \rangle$$ by optical pumping (before the STIRAP sequence commences). If it is not already in $$| 1 \rangle$$ then this involves photon scatting which of course is what we normally wish to avoid in STIRAP.

If you start with $$\Omega_P \ne 0,\;\Omega_S = 0$$ then the system will first evolve towards $$| 3 \rangle$$ by optical pumping (before the STIRAP sequence commences). If it is not already in $$| 3 \rangle$$ then this involves photon scatting which of course is what we normally wish to avoid in STIRAP. The STIRAP sequence (with $$\Omega_P$$ slowly switched off while $$\Omega_S$$ is slowly switched on) will then transfer the system to $$| 1 \rangle$$ while remaining in dark state throughout.

• But if the eigenstates are as given above (which I assume to be correct, from info.phys.unm.edu/~ideutsch/Classes/Phys566F21/ProblemSets/…, question 2, with ordered basis $\{|g_a\rangle, |e\rangle, |g_2\rangle\}$), how can we know the system start and evolve through the dark state, and not $|\Psi_\pm\rangle$? Jan 6, 2023 at 13:41
• You can know that by doing the calculation. It is essentially the same calculation as the one for the system starting in $|1\rangle$. Jan 6, 2023 at 13:45
• Can you give me more details/references of why that is so? Jan 6, 2023 at 13:56

STIRAP was designed to create complete population transfer(CPT) between the first and the third state, as of your notation.

If you apply $$\Omega_p$$ before $$\Omega_s$$ you will populate the intermediate state(since it is a superposition of $$|Ψ_+⟩$$ $$\pm$$ $$|Ψ_−⟩$$ and both have components in $$|Ψ_2⟩$$), which in turn will compromise the goal of CPT, as always some residual population will remain in $$|Ψ_2⟩$$. Not to mention that $$|Ψ_2⟩$$ is usually prone to decay, which might take the population out of the STIRAP subspace.

Another way to visualize the above is by looking at the STIRAP Hamiltonian in the adiabatic basis, which reads:

$$H_{STIRAP}=\left(\begin{matrix} E_+ & 0 & 0 \\ 0 & E_- & 0 \\ 0 & 0 &E_0\end{matrix}\right)$$.

Since $$H_{STIRAP}$$ is diagonal, once you populate a specific state, its population can not leave it, since it does not interact with any other. That is why you want to control the population in such manner that from the initial moment of excitation, all of the population will reside in $$|Ψ_0⟩$$. The way to do that is by having $$|Ψ_0(t=0)⟩=|1⟩$$ and having $$\Omega_p(t=0)=0$$.

From then on, once you have ensured that you reside in $$|Ψ_0(t)⟩$$, you have to control the "angle" $$cos( \theta ) = \frac{\Omega_p}{\sqrt{\Omega_p^2+\Omega_s^2}}$$ such that the component of $$|Ψ_0(t)⟩$$, that is $$cos( \theta )|1⟩$$ starts from 1 and ends at 0.