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 questions are:

  1. Why can we simply neglect all the other energy levels that are between $|g\rangle$ and $|r\rangle$ and assume that these two form a two-level system?
  2. 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
  3. 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 Quantum Computing.


1 Answer 1


This question should probably be closed because it asks too many questions at once, so I'll give one complete answer and then partial answers:

  1. In complete generality we cannot. However, in this physical scenario, the laser exciting the Rydberg atoms has a frequency that is close to the transition frequency (the energy difference between the $|r\rangle$ and $|g\rangle$ state). The laser also has a bandwidth that is sufficiently small such that there is not much probability of any light having a frequency close to any other transition frequency for any other level. Atomic and optical physics then say that only the transitions close to resonance (with atomic frequency close to the optical frequency) are capable of being excited. For example, read about the Rabi cycle and look at the effect of detuning on the maximum possible probability of exciting the atom in eg the plot on wiki.
  2. There are other possible separable/entangled states. One could have the overall separable states $$|g\rangle\otimes|g\rangle;\quad |g\rangle\otimes|r\rangle; \quad\frac{|g\rangle+|r\rangle}{\sqrt{2}}\otimes \frac{|g\rangle+|r\rangle}{\sqrt{2}};\quad |r\rangle\otimes |r\rangle, \quad\mathrm{etc.}$$ In some cases the first state is always $|g\rangle$, in some cases it is never $|g\rangle$, in some cases it is $|g\rangle$ 50 percent of the time... these measurements alone do not distinguish. There could be one total excited state, two total, zero total... all in cases without entanglement! And one can list many types of entangled states that all have different properties.
  3. The goal is to distinguish between $|gr\rangle+|rg\rangle$ and the mixed state that is a 50 percent combination of each of $|gr\rangle$ and $|rg\rangle$: $$\rho=\frac{|gr\rangle\langle gr|+|rg\rangle\langle rg|}{2}.$$ Phases are a good way to distinguish.
  • $\begingroup$ Thank you very much for the nice answer. However, I still don't see how one can experimentally use the phase to distinguish between pure and mixed states. Could you recommend any reading on this topic? I got the impression from the talk that there was an attempt to distinguish between $|\psi_{+}\rangle$ and $|\psi_{-}\rangle$. $\endgroup$
    – Cicero
    Commented Nov 7, 2023 at 19:28
  • 1
    $\begingroup$ If you want to distinguish between a mixture of $|0\rangle$ and $|1\rangle$ vs a superposition $(|0\rangle+|1\rangle)/\sqrt{2}$, you look at the overlap-squared of your state with some $(|0\rangle+e^{i\phi}|1\rangle)/\sqrt{2}$ for a variety of values of $\phi$. The mixed state's overlap will always be the same, while the pure state's overlap will be $|1+e^{i\phi}|^2/4=(1+\cos\phi)/2$. If you measure this overlap for a few values of $\phi$ and see that it changes, you learn about the purity of your state $\endgroup$ Commented Nov 8, 2023 at 9:27

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