Creating entangled states via Rydberg blockade

Question

I was watching a talk by Prof. Mikhail Lukin and I have a silly question.

In the talk, he discussed that a typical procedure for generating a Bell pair consists of using the Rydberg blockade on two atoms within nearby optical tweezers (assuming that their distance is within the Rydberg radius). Considering the ground state $|g\rangle$ and the Rydberg state $|r\rangle$ as the qubit bases, if the atoms start in the state $|gg\rangle$, one can excite at most one atom. Since we cannot fundamentally determine which atom is excited, we effectively end up with the following state:

$$|\Psi_{\pm}\rangle\propto|g r\rangle \pm |r g\rangle.$$

However, he further mentioned that this is partially evidenced by the oscillations in the probability of detecting 0 and 1 atom in the Rydberg state. Nevertheless, such oscillations are not sufficient to confirm whether or not we have created an entangled state. To demonstrate entanglement, we need to measure the relative phase between the components, not just the overall populations. To achieve this in the experiment, we can introduce a differential phase shift to one atom relative to the other by applying an additional laser field (via the AC Stark effect). If this phase shift is applied when the atoms are in an entangled state, it transforms the system to a state like

$$|\Psi_+\rangle \propto|g r\rangle + e^{i \phi}|r g\rangle.$$

Now, for instance, if we are in the state $|\Psi_-\rangle$, the laser will no longer have the correct phase to de-excite back to $|gg\rangle$. Thus, oscillations in the signal with respect to $|\phi\rangle$ directly probe this phase and allow one to extract the entanglement fidelity.

My question is:

I don't understand why we need to introduce a phase shift, and especially why if we are in the state $|\Psi_-\rangle$, the laser will no longer be in the correct phase to de-excite the system back to $|gg\rangle$. Wouldn't the same apply to $|\Psi_+\rangle$?

Cross-posted on Physics.

Answer 1

The key insight here is to realize $|gg\rangle$ is only coupled with $|rg\rangle+|gr\rangle$ (bright state) under laser driving. The state with minus phase, $|rg\rangle-|gr\rangle$, is called dark state because normally there should not be any population there. This is denoted in Misha's slides from the link you provided, and for more details how this could happen, you can check the reference provided in his slides.

When you are applying some phase in your bright state you will end up with $|rg\rangle+e^{i\phi}|gr\rangle$, which can be decomposed into under the basis of bright state and dark state. When $\phi$ is increased from 0 to $\pi$, you have more portion of dark state, which cannot be transferred back to $|gg\rangle$ as it is not coupled with the dark state. Therefore, the more dark state you have, the less population you will get in $|gg\rangle$ at the end of the day. By ramping the phase $\phi$ and see a oscillating population in $|gg\rangle$, you can verify the entanglement is indeed created, rather than an incoherent mixed state with even distributed population in $|rg\rangle$ and $|gr\rangle$.

Answer 2

Good question! When we apply our laser pulse we get a $|\psi\rangle = |rg\rangle + |gr\rangle$. Apply another pulse properly can induce stimulated emission which causes a Rydberg state to decay back to the ground state, so it'd send $|rg\rangle, |gr\rangle \mapsto |gg\rangle$. Now if we started in $|\Psi_+\rangle \propto |rg\rangle + |gr\rangle$, then after such a process we measure $|\Psi_+\rangle \propto 2|gg\rangle$. That seems reasonable. However when we started in $|\psi_{f,+}\rangle \propto |rg\rangle - |gr\rangle$, then after a stimulated emission process we'd measure $|\psi_{f,-}\rangle \propto |gg\rangle-|gg\rangle = 0$. Oops, if this worked our wavefunction would disappear, and this is exactly why $|\Psi_-\rangle$ is protected from being de-excited.