Rydberg Atoms/Light-matter interactions

So far, we have only looked into the dynamics of the atomic states, ignoring the effects on the light field. However, there are situations where the interplay between Rydberg interactions and coupling between the atoms and the radiation field results in some very interesting consequences. This is particularly true when the light field is so weak that its quantization in terms of single photons becomes relevant.

Open quantum systems

To understand the dynamics of an atom coupled to the radiation field, one needs to adopt a description for an open quantum system as atoms in excited states can spontaneously emit photons via an interaction with the vacuum of the radiation field. The most straight-forward way to describe the state of an open quantum system is to use a statistical sum (mixture) of Hilbert space vectors, giving rise to a density matrix of the form

$\rho =\sum \limits _{i}p_{i}|\psi _{i}\rangle \langle \psi _{i}|,$

where $0\leq p_{i}\leq 1$ denotes the probability to find the system in the (pure) quantum state described by the Hilbert space vector $|\psi _{i}\rangle$. Being probabilities, they are subject to the constraint $\sum _{i}p_{i}=1$, resulting in $\mathrm {Tr} \{\rho \}=1$ while the non-existence of negative probabilities requires the density matrix to be positive-semidefinite.

As the state of the quantum system is no longer described by a Hilbert space vector, the dynamics is no longer described by the Schrödinger equation. Instead, we have

${\begin{aligned}{\frac {d}{dt}}\rho &={\frac {d}{dt}}\sum \limits _{i}p_{i}|\psi _{i}\rangle \langle \psi _{i}|=\sum \limits _{i}p_{i}\left[\left({\frac {d}{dt}}|\psi _{i}\rangle \right)\langle \psi _{i}|+|\psi _{i}\rangle \left({\frac {d}{dt}}\langle \psi _{i}|\right)\right]\\&=-\sum \limits _{i}p_{i}{\frac {i}{\hbar }}\left(H|\psi _{i}\rangle \langle \psi _{i}|-|\psi _{i}\rangle \langle \psi _{i}|H\right)=-{\frac {i}{\hbar }}\left[H,\rho \right],\end{aligned}}$

which is called the Liouville-von Neumann equation. Note that this equation is analogous to the Liouville for the classical phase space density, where the commutator is replaced by Poisson brackets.

If we also want to introduce irreversible processes such as spontaneous emission, we have to go beyond the Liouville-von Neumann equation and describe the dynamics in terms of a quantum master equation. In the simplest case, the environment (i.e., the radiation field) is static and does not have a memory of previous irreversible events, such an environment is called to be Markovian. Then, one can show that the dynamics of the atom coupled to the radiation field is given by a Lindblad master equation

${\frac {d}{dt}}\rho =-i[H,\rho ]+\sum \limits _{i=1}^{d^{2}-1}\gamma _{i}\left(c_{i}\rho c_{i}^{\dagger }-{\frac {1}{2}}\left\{c_{i}^{\dagger }c_{i},\rho \right\}\right),$

where $d$ is the dimension of the Hilbert space of the atom, $\gamma _{i}$ are decay rates that describe the frequency of the irreversible events, and the quantum jump operators $c_{i}$ describe the action of the irreversible event . Note that the jump operators can be non-Hermitian. For a better understanding of the meaning of the Lindblad master equation, it is instructive to look into an alternative formulation in terms of a stochastic process . In this approach, each state $|\psi _{j}\rangle$ of the statistical ensemble is propagated under the non-Hermitian Hamiltonian

$H'=H-{\frac {i}{2}}\sum \limits _{i}\gamma _{i}c_{i}^{\dagger }c_{i}.$

For each infinitesimally small timestep $\tau$, the probability of a quantum jump in the $i$th channel is calculated as

$p_{i}=\tau \langle \psi _{j}|c_{i}^{\dagger }c_{i}|\psi _{j}\rangle .$

If such a quantum jump occurs, the system is brought into the state $|\psi '_{j}\rangle =c_{i}|\psi _{j}\rangle$ and the final result is normalized. Repeating this process over all channels and states in the ensemble then reproduces the Lindblad form of the quantum master equation. Remarkably, this method converges rapidly with the number of states in the ensemble, $n$. Essentially, the stochastic process performs a Monte-Carlo sampling of the quantum master equation, yielding an error for any observable $\langle O\rangle ={\text{Tr}}\{O\rho \}$ proportional to $1/{\sqrt {n}}$. In many cases, this allows the number of samples to be chosen much smaller than the Hilbert space dimension of the problem, $d$, reducing the computational complexity of simulating the quantum master equation.

Resonance fluorescence

As a first example for an open quantum system, let us consider a single two-level system interacting with the vacuum of the radiation field. Additionally, we will assume that the two-level system consisting of the states $|g\rangle$ and $|e\rangle$ is resonantly driven by an external laser. The Hamiltonian is then given bymation we neglect the fast oscillating term. Then, we can write the laser Hamiltonian as

$H={\frac {\Omega }{2}}(\sigma _{+}+\sigma _{-}),$

where we have introduced the spin-flip operators $\sigma _{+}=|e\rangle \langle g|$ and $\sigma _{-}=\sigma _{+}^{\dagger }=|g\rangle \langle e|$. The interaction with the radiation field will lead to spontaneous emission with a rate

$\gamma ={\frac {4}{3}}\omega ^{3}d^{2},$

where $\omega$ is the frequency of the transition and $d$ is the dipole matrix element . Spontaneous emission results in the atom changing the state from $|e\rangle$ to $|g\rangle$, so the corresponding quantum jump operator is simply given by $\sigma _{-}$. The total quantum master equation then reads

${\frac {d}{dt}}\rho =-i\left[H,\rho \right]+\gamma \left(\sigma _{-}\rho \sigma ^{+}-{\frac {1}{2}}\left\{\sigma _{+}\sigma _{-},\rho \right\}\right).$

The interesting aspect about this master equation is that it exhibits a competition between the coherent dynamics generated by the laser and the dissipative dynamics arising from the decay into the vacuum of the radiation field. The solution to this master equation can be found by introducing the Bloch vector $\langle {\vec {\sigma }}\rangle$ given in terms of the expectation values of the Pauli matrices as

$\langle {\vec {\sigma }}\rangle =\left({\begin{array}{c}\langle \sigma _{x}\rangle \\\langle \sigma _{y}\rangle \\\langle \sigma _{z}\rangle \end{array}}\right).$

The equations of motion for the Bloch vector are given by

${\frac {d}{dt}}\langle {\vec {\sigma }}\rangle =G\langle {\vec {\sigma }}\rangle +{\vec {b}},$

using the matrix $G$ given by

$G=\left({\begin{array}{ccc}-\gamma /2&0&0\\0&-\gamma /2&-\Omega \\0&\Omega &-\gamma \end{array}}\right)$

and the vector ${\vec {b}}$,

${\vec {b}}=\left({\begin{array}{c}0\\0\\-\gamma \end{array}}\right).$

This equation of motion is called the optical Bloch equation. Its stationary state $\langle {\vec {\sigma }}\rangle _{s}$ can be found from the condition $d/dt\langle {\vec {\sigma }}\rangle =0$ and is given by

${\begin{aligned}\langle \sigma _{z}\rangle _{s}&=-{\frac {\gamma ^{2}}{\gamma ^{2}+2\Omega ^{2}}}\\\langle \sigma _{+}\rangle _{s}=\langle \sigma _{-}\rangle _{s}^{*}&=-i{\frac {\Omega \gamma }{\gamma ^{2}+2\Omega ^{2}}}.\end{aligned}}$

Note that the population of the excited state,

$p_{e}={\frac {1}{2}}(1+\langle \sigma _{z}\rangle _{s})={\frac {\Omega ^{2}}{\gamma ^{2}+2\Omega ^{2}}}$

is always less than $1/2$ even in the limit of strong driving, i.e., $\Omega \gg \gamma$. Thus, it is not possible to create population inversion in a two level system in the stationary state by coherent driving. The population will merely saturate at $p_{e}=1/2$. The off-diagonal elements are related to the linear susceptibility $\chi$, which is defined in terms of the polarization $P$, according to

$\langle P\rangle ={\frac {\chi E}{4\pi }}={\frac {N}{V}}d\langle \sigma _{+}\rangle ,$

where $N/V$ is the atomic density. Using the relation $\Omega =dE$ and solving for $\chi$, we obtain

$\chi =8\pi i{\frac {N}{V}}d^{2}{\frac {\gamma }{\gamma ^{2}+2\Omega ^{2}}}.$

As the susceptibility is purely imaginary, the medium formed by the atoms is purely absorptive, with an absorption coefficient that is proportional to the decay rate $\gamma$.

Electromagnetically Induced Transparency

The steady state of the driven two-level atom is not particularly exciting, but this changes drastically when a third level (e.g., a Rydberg state) is introduced. The presence of the third state enables interference effects between different excitation and deexcitation paths and leads to a much richer structure in the susceptibility. In the following, we consider a ladder-type three-level system, consisting of a ground state $|g\rangle$, an intermediate excited state $|e\rangle$, and a Rydberg state $|r\rangle$. Furthermore, we assume laser driving of the transitions between the ground state and the intermediate state and the intermediate state and the Rydberg state. Then, the Hamiltonian is given by

$H=\left({\begin{array}{ccc}0&&{\frac {\Omega _{p}}{2}}\\0&\delta &{\frac {\Omega _{c}}{2}}\\{\frac {\Omega _{p}}{2}}&{\frac {\Omega _{c}}{2}}&\Delta \end{array}}\right),$

where $\Delta$ is the detuning between $|g\rangle$ and $|e\rangle$, and $\delta$ is the two-photon detuning between $|g\rangle$ and $|r\rangle$.

On two-photon-resonance, the eigenstates of this Hamiltonian can be expressed by introducing two angles $\theta$ and $\phi$, which are defined as

${\begin{aligned}\tan \theta &={\frac {\Omega _{p}}{\Omega _{c}}}\\\tan 2\phi &={\frac {\sqrt {\Omega _{p}^{2}+\Omega _{c}^{2}}}{\Delta }}.\end{aligned}}$

Then, the eigenstates of the Hamiltonian are given by

${\begin{aligned}|a^{+}\rangle &=\sin \theta \sin \phi |g\rangle +\cos \phi |e\rangle +\cos \theta \sin \phi |r\rangle \\|a^{0}\rangle &=\cos \theta |g\rangle -\sin \theta |r\rangle \\|a^{-}\rangle &=\sin \theta \cos \phi |g\rangle -\sin \phi |e\rangle +\cos \theta \cos \phi |2\rangle .\end{aligned}}$

Remarkably, the state $|a^{0}\rangle$ does not contain the intermediate state $|e\rangle$. This means that once the system is in the state $|a_{0}\rangle$, it remains there as spontaneous emission is no longer effective. On the other hand, if we wait long enough, the system will eventually end up in $|a_{0}\rangle$ as the other two eigenstates of the Hamiltonian can decay via spontaneous emission events. Such a behavior is called coherent population trapping. Moreover, the once the system is in $|a_{0}\rangle$, its susceptibility vanishes since we have $\rho _{ge}=\langle a_{0}|g\rangle \langle e|a_{0}\rangle =0$. Consequently, $|a_{0}\rangle$ is called a dark state of the dynamics, as the system does not absorb any photons any more. As all incoming photons simply pass through such a medium, this effect is known as electromagnetically induced transparancy (EIT).

Away from the two-photon resonance ($\delta \neq 0$), we can calculate the susceptibility in the limit of a weak probe laser, i.e., $\Omega _{p}\ll \Omega _{c},\gamma$ , obtaining

$\chi =4\pi {\frac {N}{V}}d^{2}\left[{\frac {4\delta (\Omega _{c}^{2}-4\delta \Delta )}{|\Omega _{c}^{2}+i2\delta (\gamma +i2\Delta )|^{2}}}+i{\frac {8\delta ^{2}\gamma }{|\Omega _{c}^{2}+i2\delta (\gamma +i2\Delta )|^{2}}}\right].$

EIT and interacting Rydberg atoms

We are now interested in what happens to the EIT phenomenon in the presence of strong Rydberg interactions. First, let us consider the case of two atoms. Suppose we place the system into the EIT condition for non-interacting atoms. If one of the atoms is found to be in the Rydberg state, then the Rydberg interaction will appear to the second atom as a two-photon detuning, which will destroy the EIT resonance. Therefore, we can expect a reduction of the EIT transmission from the interaction between the Rydberg atoms.

We can calculate this effect in more detail by making a few assumptions. First, we assume that the single photon detuning $\Delta$ is large, allowing us to adiabatically eliminate the $|e\rangle$. At first, it might seem to be a bad idea to get rid of the state since we are interested in the absorption coefficient, which involves the density matrix element $\rho _{ge}$. However, we can still obtain information about the absorption coefficient. Notice that the real and imaginary part of the susceptibility $\chi =\chi '+i\chi ''$ are not independent from each other, as they are related through the Kramers-Kronig relations

${\begin{aligned}\chi '(\omega )&={\frac {1}{\pi }}{\text{PV}}\int {\frac {\chi ''(\omega ')}{\omega '-\omega }}d\omega '\\\chi ''(\omega )&=-{\frac {1}{\pi }}{\text{PV}}\int {\frac {\chi '(\omega ')}{\omega '-\omega }}d\omega ',\end{aligned}}$

where ${\text{PV}}$ denotes the Cauchy principal value. For the case of an EIT resonance, we observe that close to the two-photon resonance, the dispersion $\chi '$ scales linearly with the two-photon detuning, while the absorption $\chi ''$ scales quadratically. Therefore, we can express the absorption coefficient in terms of the energy shift that the (no longer zero) energy eigenstate $|a^{0}\rangle$ obtains as a consequence of the Rydberg interaction. Furthermore, we assume that the probe laser field $\Omega _{p}$ is much weaker than the coupling laser field $\Omega _{c}$, allowing to expand the mixing angle as $\theta \approx \Omega _{p}/\Omega _{c}$ and the state $|a^{0}\rangle \approx |g\rangle -\theta |r\rangle$. Finally, we make the simplification that we take all Rydberg interaction matrix elements as being equal to $V$. For a system of $N$ atoms, we can again go to collective basis of symmetric basis, now consisting of the three states

${\begin{aligned}|S_{0}\rangle &=|g_{1}g_{2}g_{3}\ldots g_{N}\rangle \\|S_{1}\rangle &={\frac {1}{\sqrt {N}}}\sum \limits _{i}|g_{1}\ldots r_{i}\ldots g_{N}\rangle \\|S_{2}\rangle &={\sqrt {\frac {2}{N(N-1)}}}\sum \limits _{i=1}^{N}\sum \limits _{j=1}^{i-1}|g_{1}\ldots r_{i}\ldots r_{j}\ldots g_{N}\rangle .\end{aligned}}$

We also need the matrix elements involving the doubly excited $|S_{2}\rangle$ state, which we find to be

${\begin{aligned}\langle S_{1}|H|S_{2}\rangle &={\sqrt {2(N-1)}}{\frac {\Omega _{p}\Omega _{c}}{4\Delta }}\\\langle S_{2}|H|S_{2}\rangle &=V.\end{aligned}}$

In the basis consisting of the three collective states, the Hamiltonian for $\delta =0$ can then be written in the form

$H={\frac {\Omega _{c}^{2}}{4\Delta }}\left({\begin{array}{ccc}N\theta ^{2}&{\sqrt {N}}\theta &\\{\sqrt {N}}\theta &1+(N-1)\theta ^{2}&{\sqrt {2(N-1)}}\theta \\&{\sqrt {2(N-1)}}\theta &v+2+(N-2)\theta ^{2}\end{array}}\right),$

where we have introduced the reduced interaction strength $v=V/(\Omega _{c}^{2}/4\Delta )$. To see an effect of the interaction, we have to expand up to fourth order perturbation theory in $\theta$. Then, we find for the energy shift

$\Delta E=-{\frac {\Omega _{c}^{2}}{4\Delta }}2N(N-1)\left({\frac {1}{2+v}}-{\frac {1}{2}}\right)\theta ^{4}={\frac {N(N-1)V}{v+2}}\theta ^{4}.$

As expected, the energy shift reproduces the limit $\langle a^{0}a^{0}|V|rr\rangle \langle rr|a^{0}a^{0}\rangle =V\theta ^{4}$ for weak interactions and $N=2$. Interestingly, the EIT resonance is not completely destroyed in the limit of very strong interactions. However, a large number of blockaded atoms, which is roughly equivalent to $N$, quickly deteriorates the EIT condition. This behavior has also been observed experimentally .

Rydberg polaritons

So far, we have treated the electromagnetic field as a purely classical field. However, this is only an approximation and in the limit of very weak fields, it is possible to see effects arising from the quantization of the radiation field in terms of single photons. As a first step, we will again neglect the consequences of the Rydberg interactions. The radiation field can be quantized in terms of a sum of harmonic oscillators, leading to the Hamiltonian

$H=\sum \limits _{{\vec {k}}\lambda }\omega _{\vec {k}}a_{{\vec {k}}\lambda }^{\dagger }a_{{\vec {k}}\lambda },$

where $\lambda$ is the polarization index . If the wavelength of the radiation field is much larger than the spatial extent of the atomic wavefunction, the interaction between the atoms and the radiation field is given by the electric dipole term,

$H_{I}=-{\vec {d}}{\vec {E}},$

where ${\vec {d}}$ is the dipole operator of the atom and ${\vec {E}}$ is the quantized electric field,

${\vec {E}}=i\sum \limits _{{\vec {k}}\lambda }{\sqrt {\frac {2\pi \omega _{k}}{V}}}{\vec {e}}_{\lambda }\left(a_{{\vec {k}}\lambda }-a_{{\vec {k}}\lambda }^{\dagger }\right),$

where $V$ is the quantization volume. In the following, we will focus on the case where only a single mode of the field is relevant, e.g., a single mode driven by a weak laser, propagating along the $x$ axis. The quantized field takes over the role of the probe field $\Omega _{p}$, while the coupling field $\Omega _{c}$ remains being described in terms of a classical field. In this case, it is possible to define a polariton field that describes quasiparticles of the combined light-matter system . In particular, one can express the system in terms of dark and bright state polaritions, respectively, according to

$\Psi (x,t)=\cos \theta {\mathcal {E}}(x,t)-\sin \theta {\sqrt {n}}\rho _{ge}(x,t)e^{i\Delta k}$

$\Phi (x,t)=\sin \theta {\mathcal {E}}(x,t)+\cos \theta {\sqrt {n}}\rho _{ge}(x,t)e^{i\Delta k}.$

Here, ${\mathcal {E}}=E/{\sqrt {2\pi \omega }}$ is an operator describing the reduced field strength, $n$ is the atomic density, and $\Delta k$ is a phase factor accounting for the difference in wave vectors between the pump and probe fields. The mixing angle is now given by

$\tan ^{2}\theta ={\frac {3n\lambda ^{2}c\gamma }{2\pi \Omega _{c}^{2}}}.$

Under an EIT condition, only the dark state polaritons contribute, i.e., the bright state polariton field $\Phi$ vanishes. One can show that the polariton field obeys a modified wave equation given by

$\left({\frac {\partial }{\partial t}}+c\cos ^{2}\theta {\frac {\partial }{\partial z}}\right)\psi (x,t)=0.$

Its solution corresponds to a wave propagation with a group velocity of

$v_{gr}=c\cos ^{2}\theta .$

For small mixing angles $\theta$, the propagation speed is roughly equivalent to the speed of light in the vacuum. However, for $\theta \to \pi /2$, the propagation speed in the medium actually vanishes! In practice, this means to make the coupling laser field $\Omega _{c}$ very weak. Many experiments have successfully demonstrated such slowing and stopping of light beams in an EIT medium.

Let us now return to the question on what happens when we include interactions between the Rydberg atoms. As we have seen that Rydberg interactions lead to absorption, two spatially close Rydberg excitations will lead to coupling of the dark state polaritons to the bright state polariton, leading to strong losses from absorption . On the other hand, if the Rydberg excitations are further separated than a critical distance given by

$r_{c}={\sqrt[{6}]{\frac {C_{6}|\Delta |}{2\pi nd^{2}\omega _{p}}}},$

then the interaction is weak enough such that the coupling to the bright state polaritons can be treated as a small perturbation . In this regime, the interaction between the Rydberg atoms leads to an effective interaction between the dark state polaritons according to

$H_{\text{int}}={\frac {C_{6}}{2}}\cos ^{4}\theta \int drdr'{\frac {\Psi ^{\dagger }(r)\Psi ^{\dagger }(r')\Psi (r')\Psi (r)}{(r-r')^{6}}}.$

This repulsive interaction between the polaritons can lead to ordered polariton structures, as well as to the generation of strongly correlated light pulses.