Rydberg Atoms/Excitation of Rydberg States

Let us now turn to the question of producing Rydberg atoms in an experiment. The most commonly used way is to excite ground state atoms optically using laser light. This can be done using one-photon processes, where the ground state is directly coupled to the Rydberg state, or via multi-photon schemes using one or two intermediate (non-Rydberg) states. We will first focus on the case of direct excitations. In the case of rubidium, the ground state has an ionization energy corresponding to a wavelength of $$\lambda = 297\,\text{nm}$$, i.e., direct excitation requires UV lasers to access the Rydberg states. These lasers have a frequency resolution that allows to select a single Rydberg state for excitation. Then, we can write the dynamics of the excitation process as a two-level system, involving only the electronic ground state $$|g\rangle = |5s\rangle$$ and a single Rydberg state $$|r\rangle = |np\rangle$$. Within the electric dipole approximation, we can treat the laser as an oscillating electric field with peak strength $$E_0$$ and couple it to the dipole matrix element $$d_{gr}$$ of the transition between the ground state and the Rydberg state. Then, we can write the pertubation by the laser field using the Hamiltonian
 * $$H = \Delta|r\rangle\langle r| + [\Omega \cos(\omega t)|g\rangle\langle r| + \mathrm{h.c.}],$$

where we have introduced the energy difference $$\Delta$$ between the ground state and the Rydberg state and the Rabi frequency $$\Omega = d_{gr} E_0$$. We now go into the rotating frame of the laser field, i.e., we make the transformation $$|r\rangle \to |r\rangle \exp(i\omega t)$$. Inserting this into the Schrödinger equation shifts the $$|r\rangle$$ level by the frequency $$\omega$$ and leads to the detuning $$\delta = \Delta-\omega$$. In the rotating-wave approximation, we neglect fast rotating terms on the order of $$2\omega$$ and obtain the effective Hamiltonian
 * $$H = \delta |r\rangle \langle r| + \left[\frac{\Omega}{2}|g\rangle\langle r| + \mathrm{h.c.}\right].$$

When the excitation laser is resonant with the Rydberg transition (i.e., $$\delta = 0$$, we can diagonalize the Hamiltonian using the two eigenstates
 * $$|+\rangle = \frac{1}{\sqrt{2}}(|g\rangle + |r\rangle)$$
 * $$|-\rangle = \frac{1}{\sqrt{2}}(|g\rangle - |r\rangle).$$

In this new basis, the Hamiltonian becomes
 * $$H = -\frac{\Omega}{2} |-\rangle \langle -| + \frac{\Omega}{2} |+\rangle\langle +|.$$

The system undergoes Rabi oscillation between the ground state and the Rydberg state, i.e., the probability to find the system in the Rydberg state is given by
 * $$P_r(t) = \cos^2(\Omega t).$$

The value of the Rabi frequency depends on the strength of the laser as the peak field strength $$E_0$$ is related to the laser power $$P$$, $$E_0 \sim \sqrt{P}$$. Equally important, however, is the asymptotic scaling of the Rabi frequency with the principal quantum number. To obtain the relation, we assume that the ground state wave function is essentially a pointlike object compared to the extension of the Rydberg state. Then, we can approximate the ground state wave function by a Dirac delta function at the origin. The corresponding dipole moment is then proportional to the value of the Rydberg wave function at the origin,
 * $$d_{gr} = \langle 5s |d|np\rangle = \psi_{np}(\mathbf{r} = 0).$$

From the normalization factor of the wave function we obtain $$d_{gr} \sim {n^*}^{-3/2}$$. This means that the higher the Rydberg state we want to excite, the lower the Rabi frequency will be, i.e., it takes longer to reach the Rydberg state.

As Rydberg states are excited atomic states, they always have a finite lifetime $$\tau$$, since the coupling to the vacuum of the electromagnetic field provides a natural decay channel. In principle, this decay can happen via two different processes. The first possibility is a sequence of decays through other Rydberg states, by lowering the principal quantum number at most by one during each individual decay event. The second possibility is a direct decay into the ground state (or the lowest electronic state allowed by selection rules) by the emission of a single photon. We will now calculate the scaling behavior with $$n^*$$ to see which of the two processes is more important.

In both cases, we will use Fermi's golden rule to calculate the decay rate. This is well justified, as the frequency of the emitted photons is always much larger than the corresponding decay rate. According to Fermi's golden rule, the decay rate is given by
 * $$\gamma = 2\pi |\langle f|V|i\rangle|^2 \rho(\omega),$$

where $$|i\rangle$$ and $$|f\rangle$$ are the initial and final state of the decay process, respectively, $$V$$ is the operator describing the interaction leading to the decay, and $$\rho(\omega)$$ is the density of states of the environment at its final energy $$\omega$$. The radiation field has a density of states of $$\rho(\omega) = 8\pi\omega^2$$, and for electric dipole transitions we have $$\langle f|V|i\rangle = \langle f| d E|i\rangle \sim \sqrt{\omega}$$. Consequently, we have an overall $$\omega^3$$ dependence of the decay rate. In the case of decay to adjacent Rydberg levels, the squared dipole matrix element scales as $${n^*}^4$$, but this factor is suppressed by a factor of $$n^{-9}$$ from the $$\omega$$-dependence. Therefore, the total scaling of the decay rate is $$\gamma \sim {n^*}^{-5}$$.

In the case a direct decay to the ground state, the frequency of the emitted photon is essentially independent of the principal quantum number $$n^*$$. The decay rate is therefore only determined by the square of the dipole matrix element of the transition, yielding a scaling according to $$\gamma \sim {n^*}^{-3}$$ for the decay rate. Consequently, the direct decay to the ground state is the most relevant process, resulting in a scaling of $$\tau \sim n^3$$.

However, there are two exceptions to this scaling behavior. If the Rydberg atoms is brought into a circular Rydberg state with $$l=n-1$$, then the only dipole allowed transition is to the adjacent Rydberg state. In this case, the lifetime is only determined by the first process and therefore scales as $$\tau \sim {n^*}^5$$. The other exception arises from finite temperature effects. Instead of being a vacuum, the electromagnetic field is populated with blackbody photons, which can drive transitions between Rydberg states. Then, the lifetime is given by
 * $$\tau = \frac{3{n^*}^2}{4\alpha^3 T},$$

where $$\alpha$$ is the fine structure constant and $$T$$ is the temperature of the environment. Note that while it is possible also to ionize a Rydberg atom by blackbody ration, this process scales only as $$\gamma \sim n^{-7/3}$$ and can therefore be neglected compared to blackbody-induced transitions to other Rydberg states.

However, it is not always possible to use a single laser to perform Rydberg excitations because of the challenges arising from the relatively short laser wavelength. Therefore, it can be more convenient to user two or three lasers to couple the ground state to the Rydberg state. Because of selection rules, the final Rydberg state will be either a $$s$$ state or a $$d$$ state. In rubidium, the laser wavelengths associated with a two-step excitation process are $$\lambda_1 = 780\,\text{nm}$$ and $$\lambda_2 = 480\,\text{nm}$$, respectively. This involves a near-resonant coupling to the intermediate state $$|5p\rangle$$, which is the first electronically excited state. Since this is not a Rydberg state, the intermediate state will decay very fast with a rate of $$\gamma_p = 2\pi\times 6\,\text{MHz}$$, corresponding to a lifetime of only $$\tau_p = 26\,\text{ns}$$. We will therefore consider a large detuning $$\Delta$$ from this intermediate state $$|p\rangle$$, while the energy difference between the ground state $$|g\rangle$$ and the Rydberg state $$|r\rangle$$ differing the sum of the two laser frequencies by a two-photon detuning $$\delta$$. After performing the rotating wave approximation, the Hamiltonian for the three-level system is of the form
 * $$H = \Delta |p\rangle\langle p| + \delta|r\rangle\langle r| + \frac{\Omega_p}{2} (|g\rangle \langle p| + \text{H.c}) + \frac{\Omega_c}{2} (|p\rangle \langle r| + \text{H.c}) = \left(\begin{array}{ccc} 0 & 0 & \frac{\Omega_p}{2}\\ 0 & \delta & \frac{\Omega_c}{2}\\ \frac{\Omega_p}{2} & \frac{\Omega_c}{2} & \Delta\end{array}\right).$$

As the next step, we will perform an adiabatic elimination of the intermediate state. For this, we expand the wave function $$|\psi(t)\rangle$$ according to $$|\psi(t)\rangle = \psi_g(t) |g\rangle + \psi_r(t) |r\rangle + \psi_p(t)|p\rangle$$. The corresponding Schrödinger equation then reads
 * $$i\dot{\psi}_g = \frac{\Omega_p}{2}\psi_p$$
 * $$i\dot{\psi}_r = \delta\psi_r + \frac{\Omega_c}{2}\psi_p$$
 * $$i\dot{\psi}_p = \frac{\Omega_p}{2}\psi_g + \frac{\Omega_c}{2}\psi_r +\Delta\psi_p.$$

We can then eliminate the intermediate state by setting its time derivative to zero, i.e., $$\dot{\psi}_p = 0$$. From the solution of the equation of motion for $$\psi_p$$ we obtain
 * $$\psi_p = -\frac{\Omega_p}{2\Delta} \psi_g - \frac{\Omega_c}{2\Delta}\psi_r.$$

This expression can only be correct in the limit $$\Omega_{p,c} \ll |\Delta|$$, i.e., in the limit of large detuning and therefore small population of the intermediate state. We can then insert the expression for $$\psi_p$$ into the equations of motion for $$\psi_g$$ and $$\psi_r$$, yielding
 * $$i\dot{\psi}_g = -\frac{\Omega_p^2}{4\Delta} \psi_g - \frac{\Omega_p\Omega_c}{4\Delta}\psi_r$$
 * $$i\dot{\psi}_r = \left(\delta-\frac{\Omega_c^2}{4\Delta}\right) \psi_r - \frac{\Omega_p\Omega_c}{4\Delta}\psi_g.$$

These equations of motion are exactly equivalent to those generated by an effective Hamiltonian $$H_\text{eff}$$, which is given in the $$\{|g\rangle,|r\rangle\}$$ basis as
 * $$H_\text{eff} = -\left(\begin{array}{cc}\frac{\Omega_p^2}{4\Delta} & \frac{\Omega_p\Omega_c}{4\Delta}\\ \frac{\Omega_p\Omega_c}{4\Delta} & \frac{\Omega_c^2}{4\Delta}-\delta\end{array}\right).$$

Note that the coupling via the intermediate state shifts the resonance condition for the two-photon transition between the ground state and the Rydberg state. It occurs when the two-photon detuning $$\delta$$ cancels the differential AC Stark shift, i.e., $$\delta = (\Omega_c^2-\Omega_p^2)/4\Delta$$. On resonance, the system undergoes Rabi oscillations with the effective Rabi frequency $$\Omega_\text{eff} = \Omega_p\Omega_c/2\Delta$$.

Finally, we will discuss the optimal choice of the detuning $$\Delta$$. On the one hand, we want to make $$\Delta$$ very large, as then the unwanted radiative decay from the intermediate state is strongly suppressed. On the other hand, making $$\Delta$$ large also reduces the effective Rabi frequency, eventually to the point where it becomes smaller than the decay of the Rydberg state. Therefore, we have to find an optimal balance between these two effects. In the following, we assume that the Rabi frequencies for the two transitions are identical, i.e., $$\Omega_p=\Omega_c\equiv \Omega$$. Then, we find the probability to successfully observe the system without any unwanted decay process (also known as the fidelity $$f$$) to be
 * $$f = (1-p_p)(1-p_r),$$

where $$p_p$$ and $$p_r$$ are the probability for decay events from the intermediate and from the Rydberg state, respectively. In the limit, where these probabilities are small, we can express them as
 * $$p_p = |\psi_p|^2 \gamma_p t\approx \frac{\Omega^2}{4\Delta^2}\,\gamma_p$$
 * $$p_r \approx \gamma t,$$

where $$t$$ is the time the system is evolving. Now assume that we want to evolve the system until it has reached a certain fraction $$\phi$$ of a Rabi cycle, which requires the time $$t = 4\phi\Delta/\Omega^2$$. In order to maximize the fidelity $$f$$ we solve the equation $$\partial_\Delta f = 0$$, which has the solution
 * $$\Delta = \pm \sqrt{\frac{\gamma_p}{\gamma}}\,\frac{\Omega}{2}.$$

Consequently, the optimal choice for the detuning depends on the ratio of the decay rates. It occurs at the point where the errors from the two processes occur with equal probability, i.e., $$p_p=p_r$$.