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⟩
and the Rydberg state |r⟩
as the qubit bases, if the atoms start in the state |gg⟩
, one can excite at most one atom. Since we cannot fundamentally determine which atom is excited, we effectively end up with the following state:

|Ψ±⟩∝|gr⟩±|rg⟩.

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

|Ψ+⟩∝|gr⟩+eiϕ|rg⟩.

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

My questions are:

Why can we simply neglect all the other energy levels that are between |g⟩
and |r⟩
and assume that these two form a two-level system?
Why isn't observing the probability of detecting 0 or 1 atom in the Rydberg state enough to certify entanglement? If the state were separable, would one not always obtain either 0 or 1
I don't understand why we need to introduce a phase shift, and especially why if we are in the state |Ψ−⟩
, the laser will no longer be in the correct phase to de-excite the system back to |gg⟩
. Wouldn't the same apply to |Ψ+⟩
?

Answer 1

In quantum physics experiments, two-level systems like qubits simplify real-world scenarios. These systems allow precise control and eliminate negligible energy levels, enabling the creation and confirmation of quantum entanglement by introducing phase shifts to probe phase relationships beyond mere state probabilities.

Step-by-step explanation:

In the realm of quantum physics, experiments often involve the simplification of a system to a two-level model, such as the ground state |g⟩ and a higher-energy Rydberg state |r⟩, to create what is known as a qubit. The selection of only two energy levels, neglecting intermediate ones, is possible due to the resonance conditions achieved by carefully tuned lasers that couple only to these specific transitions, hence other energy levels do not play a significant role under these conditions. In the generation of a Bell pair through the Rydberg blockade, detecting only 0 or 1 atom in the Rydberg state is insufficient to assert entanglement, because these probabilities could arise from non-entangled states as well.

Entanglement is characterized not only by the probability distribution of states but also by the phase relationships between them. This is where the introduction of a differential phase shift comes into play. The phase shift, induced through the AC Stark effect, changes the state to |gr⟩ ± eiφ|rg⟩. In the case of the state |Ψ-⟩, applying a phase shift alters the pathway of deexcitation in a way that is different from |Ψ+⟩, thereby facilitating the confirmation of entanglement through the observation of how the system's oscillations depend on the phase φ. The fact that not all deexcitation pathways are equivalent when a phase shift is introduced to an entangled state is why the system set in |Ψ-⟩ does not return to |gg⟩ under the same conditions that apply to |Ψ+⟩.