Rydberg Atoms/Rydberg blockade

Dipole-dipole and van der Waals interactions

In the previous chapter, we have seen that Rydberg atoms are very sensitive to external electric fields, with their polarizability scaling with the principal quantum number like ${n^{*}}^{7}$. This is also the case when the electric field is generated by the charge distribution of the Rydberg electron of another atom, therefore we can expect Rydberg atoms to exhibit very strong interactions. To understand the interactions between Rydberg atoms, we will assume in the following that they are separated far enough so that their electron wave functions do not overlap. As the Rydberg wave functions decay exponentially at large distances, it is possible to express this in terms of a single quantity, the Le Roy radius $R_{LR}$ , which is given by

$R_{LR}=2\left({\sqrt {\langle n_{1},l_{1}|r^{2}|n_{1},l_{1}\rangle }}+{\sqrt {\langle n_{2},l_{2}|r^{2}|n_{2},l_{2}\rangle }}\right),$

where $|n_{i},l_{i}\rangle$ refers to the electron eigenstate of the $i$th atom. Treating the Rydberg electrons as hydrogenic, we obtain for the expectation value (see )

$\langle r^{2}\rangle ={\frac {n^{2}}{2}}[5n^{2}+1-3l(l+1)].$.

For example, the Le Roy radius of two rubidium atoms in the $43S$ state is $R_{LR}=532\,{\text{nm}}$. In this regime, the interaction potential between two atoms separated by a distance $R$ can be expressed as an Laurent series in $R$,

$V(\mathbf {r} _{1},\mathbf {r} _{2})={\frac {1}{|\mathbf {r} _{1}-\mathbf {r_{2}} |}}=-\sum \limits _{n=1}^{\infty }{\frac {C_{n}}{R^{n}}}.$

The first two terms of the series correspond to the Coulomb and charge-dipole interaction, respectively, and therefore vanish for neutral atoms. The first contribution therefore comes from the dipole-dipole interaction, which is given by

$V(\mathbf {r} _{1},\mathbf {r} _{2})=(1-3\cos ^{2}\vartheta _{ij}){\frac {d_{i}d_{j}}{R^{3}}},$

where $d_{i}$ and $d_{j}$ are the electric dipole operators and $\vartheta _{ij}$ is the angle between the interatomic axis and the quantization axis of the atoms. The higher order terms can be expressed in terms of a series expansion involving spherical harmonics ,

$V(\mathbf {r} _{1},\mathbf {r} _{2})=\sum \limits _{l,L=1}{\frac {V_{lL}(\mathbf {r} _{1},\mathbf {r} _{2})}{R^{l+L+1}}}$

$V_{lL}(\mathbf {r} _{1},\mathbf {r} _{2})={\frac {(-1)^{L}4\pi }{\sqrt {(2l+1)(2L+1)}}}\sum \limits _{m}{\sqrt {{\binom {l+L}{l+m}}{\binom {l+L}{L+m}}}}r_{1}^{l}r_{2}^{l}Y_{lm}({\textbf {r}}_{1})Y_{L-m}({\textbf {r}}_{2}).$

In the following we will concentrate on two atoms in the same $s$ state $|r\rangle$, i.e., $n_{1}=n_{2}\equiv n$. In this case, the main contribution comes from a single combination of $p$ states $|r'r''\rangle =|n_{1}',p_{1},n_{2}',p_{2}\rangle$. In the case of rubidium, we have $n_{1}'=n$ and $n_{2}'=n-1$, as the difference in the quantum defect for $s$ and $p$ states is close to 0.5. The energy difference between the states $|rr\rangle$ and $|r'r''\rangle$ is the Förster defect $\delta _{F}$. Then, we can write the interaction Hamiltonian in the basis consisting of $|rr\rangle$ and $|r'r''\rangle$ as

$H=\left({\begin{array}{cc}\delta _{F}&{\frac {d_{r'r'}d_{rr''}}{R^{3}}}{\sqrt {D_{\varphi }}}\\{\frac {d_{r'r'}d_{rr''}}{R^{3}}}{\sqrt {D_{\varphi }}}&0\end{array}}\right),$

where the coefficient $D_{\varphi }=3$ follows from the angular part of the dipole operators . The eigenvalues of the interaction Hamiltonian are

$V_{int}(R)={\frac {\delta _{F}}{2}}\pm {\frac {1}{2}}{\sqrt {\delta _{F}^{2}+4{\frac {(d_{1}d_{2})^{2}D\varphi }{R^{6}}}}}.$

Let us now consider two regimes of this interaction Hamiltonian, depending on the strength of the interaction $V_{dd}=d_{1}d_{2}/R^{3}$ compared to the Förster defect $\delta _{F}$. At short distances, we have $V_{dd}\gg \delta _{F}$, therefore the eigenvalues of $H$ are given by

$V_{int}(R)=\pm {\frac {d_{r'r'}d_{rr''}}{R^{3}}}{\sqrt {D_{\varphi }}}.$

As the dipole-dipole interaction is so strong that it mixes the electronic eigenstates, the interaction decays like $1/R^{3}$ even though the unperturbed eigenstates do not have a finite electric dipole moment. The other regime is where the atoms are so far apart that $V_{dd}\ll \delta _{F}$. Then we can perform a Taylor expansion of $V_{int}(R)$, obtaining

$V_{int}(R)=\pm {\frac {(d_{1}d_{2})^{2}D_{\varphi }}{\delta _{F}R^{6}}}.$

This interaction is a van der Waals interaction decaying like $1/R^{6}$. In the limit $R\to \infty$, the negative eigenvalue connects to the unperturbed state $|rr\rangle$. From this, we can read off the van der Waals coeffcient $C_{6}$ to be

$C_{6}={\frac {(d_{1}d_{2})^{2}D_{\varphi }}{\delta _{F}}}.$

Note that $\delta _{F}$ may be negative for certain combination of states, in this case the van der Waals interaction is repulsive. Consequently there is is a crossover from a resonant dipole interaction at short distances to a van der Waals interaction at large distances taking place at a critical radius $r_{c}$, which is given by

$r_{c}={\sqrt[{6}]{\frac {4(d_{1}d_{2})^{2}}{\delta _{F}^{2}}}}.$

In the case of two Rb atoms in the $43s$ state, the van der Waals coefficient is given by $C_{6}=1.7\times 10^{19}$. The behavior of the interaction energy depending on the interatomic separation is shown in Fig. 1. For comparison, the van der Waals coefficient for the ground state of Rb is $C_{6}=4707$ .

This huge difference suggests a dramatic scaling of the van der Waals coefficient with the principal quantum number $n^{*}$. Each transition dipole moment scales as $d_{i}\sim {n^{*}}^{2}$, resulting in a ${n^{*}}^{8}$ dependence from the dipole matrix elements. However, the Förster defect $\delta _{F}$ has the same scaling as the energy splitting between neighboring Rydberg states, $\delta \sim {n^{*}}^{-3}$. Overall, this results in a scaling of the van der Waals coefficient like $C_{6}\sim {n^{*}}^{11}$.

Finally, Rydberg atoms in states with nonzero angular momentum quantum number $l$ have a finite quadrupole moment. The quadrupole-quadrupole interaction decays like $1/R^{5}$, therefore it will become larger than the van der Waals interaction at very large distances. The distance where both interactions are equal can essentially be calculated as

${\frac {\langle r^{2}\rangle ^{2}}{R^{5}}}={\frac {\langle r\rangle ^{4}}{\delta _{F}R^{6}}}.$

The expectation values are comparable, resulting in $R=1/\delta _{F}$. For typical Förster defects on the order of a few GHz, this translates to a distance of $R\approx 100\,\mu {\text{m}}$, where both interaction are already vanishingly small. Therefore, one can simply neglect the quadrupole-quadrupole interaction in most cases.

The Rydberg blockade mechanism

Let us now have a look how the presence of interactions modifies the excitation dynamics. As before, we will assume resonant laser excitation between the ground state $|g\rangle$ and a single Rydberg state $|r\rangle$. Furthermore, we will assume that the atoms are separated by more than $r_{c}$ so that the treatment in terms of a van der Waals interaction is appropriate. Therefore, the system is fully described by the states $|gg\rangle$, $|gr\rangle$, $|rg\rangle$, and $|rr\rangle$. The Hamiltonian in this basis is of the form

$H={\frac {\Omega }{2}}\left(|g\rangle \langle r|\otimes 1+1\otimes |g\rangle \langle r|+{\text{H.c.}}\right)\,-{\frac {C_{6}}{R^{6}}}|rr\rangle \langle rr|={\frac {\Omega }{2}}\left(|gg\rangle \langle gr|+|gg\rangle \langle rg|+|gr\rangle \langle rr|+|rg\rangle \langle rr|+{\text{H.c}}\right)\,-{\frac {C_{6}}{R^{6}}}|rr\rangle \langle rr|.$

We can simplify the problem by noting that the state $|-\rangle =(|gr\rangle -|rg\rangle )/{\sqrt {2}}$ does not take part in the dynamics, i.e., it is an eigenstate of the Hamiltonian with an eigenvalue of zero. We can therefore introduce an effective three-level system consisting of the states $|gg\rangle$, $|+\rangle =(|gr\rangle +|rg\rangle )/{\sqrt {2}}$, and $|rr\rangle$. In this new basis, the Hamiltonian becomes

$H={\frac {{\sqrt {2}}\Omega }{2}}\left(|gg\rangle \langle +|+|+\rangle \langle rr|+{\text{H.c.}}\right)\,-{\frac {C_{6}}{R^{6}}}|rr\rangle \langle rr|.$

Note the enhancement of the Rabi frequency by a factor of ${\sqrt {2}}$ in this basis. The resulting dynamics depends on the strength of the interaction compared to the Rabi frequency. In the weakly interacting regime given by $|C_{6}|/R^{6}\ll \Omega$, the system will undergo slightly perturbed Rabi oscillations with Rabi frequency $\Omega$, but the qualitative picture is hardly modified compared to the single atom case. However, in the regime of strong interactions denoted by $|C_{6}|/R^{6}\gg \Omega$, the system behaves differently. The first excitation from $|gg\rangle$ to $|+\rangle$ is unaffected by the interaction. The second excitation from $|+\rangle$ to $|rr\rangle$ is off-resonant because of the strong interaction, see Fig. 2. Effectively, the $|rr\rangle$ state is decoupled from the dynamics, as it can never be reached. This decoupling of the doubly excited state is called the "Rydberg blockade". We can reduce the description to a two level system consisting only of $|gg\rangle$ and $|+\rangle$, governed by the Hamiltonian

$H={\frac {{\sqrt {2}}\Omega }{2}}\left(|gg\rangle \langle +|{\text{H.c.}}\right).$

The dynamics of this Hamiltonian again produces Rabi oscillations, however with two important differences to the non-interacting case. First, the maximum probability to find an atom in the Rydberg state, $p_{r}$ is $1/2$, as the $|+\rangle$ state has only one of the two atoms in the Rydberg state. Second, the Rabi frequency is enhanced by a factor of ${\sqrt {2}}$, resulting in

$p_{r}(t)={\frac {1}{2}}\sin ^{2}\left({\sqrt {2}}\Omega t\right).$

The distance at which the blockade sets in can be determined by setting the interaction strength equal to the Rabi frequency. This results in a blockade radius $r_{b}$ given by

$r_{b}={\sqrt[{6}]{\frac {|C_{6}|}{\Omega }}}.$

In typical experiments, $r_{b}$ is on the order of $5{\text{--}}10\,\mu {\text{m}}$.

If we stop the laser excitation process at the time $t=\pi /{\sqrt {8}}\Omega$, the system will be in the state $|+\rangle =(|gr\rangle +|rg\rangle )/{\sqrt {2}}$. This state has some very interesting properties. Remarkably, this state cannot be written as a product state of the form

$|\psi \rangle =|\phi \rangle \otimes |\chi \rangle$.

This can be proven by noting that for each product state $|\psi \rangle$, there exists a unitary transformation $U=U^{(1)}\otimes U^{(2)}$ that transforms the product state $|gg\rangle$, i.e.,

$|\psi \rangle =U^{(1)}\otimes U^{(2)}|gg\rangle .$

$U^{(1)}$ and $U^{(2)}$ are unitary $2\times 2$ matrices, and we can carry out the matrix multiplication explicitly using their matrix elements $U_{\alpha \beta }^{(i)}$,

$U^{(1)}\otimes U^{(2)}|gg\rangle =U^{(1)}\left[U_{gg}^{(2)}|gg\rangle +U_{rg}|gr\rangle \right]=U_{gg}^{(1)}U_{gg}^{(2)}|gg\rangle +U_{gg}^{(1)}U_{rg}^{(2)}|gr\rangle +U_{rg}^{(1)}U_{gg}^{(2)}|rg\rangle +U_{rg}^{(1)}U_{rg}^{(2)}|rr\rangle .$

The $|+\rangle$ state does not have any $|gg\rangle$ component, so either $U_{gg}^{(1)}$ or $U_{gg}^{(2)}$ would have to be zero. But any of these choices will also cause the $|gr\rangle$ or $|rg\rangle$ component to vanish, making it impossible to write $|+\rangle$ as a product state. Such quantum states that cannot be written as product states are entangled. Entanglement is a type of quantum correlation that is not found in classical system and is a key ingredient for a quantum computer. The state $|+\rangle$ is actually a maximally entangled state because the two atoms are perfectly anticorrelated. This possiblity to create entangled quantum states using the Rydberg blockade has resulted in proposals to use Rydberg atoms as the building blocks of a quantum computer , and the first proof-of-principle experiments have already been carried out .

As the blockade radius $r_{b}$ can be much larger than the typical interparticle distance that can be achieved with laser-cooled atoms, it is natural to ask what happens when there are more than two atoms located inside the blockade radius. For $N$ atoms that can be either in the $|g\rangle$ or in the $|r\rangle$ state, the tensor product Hilbert space contains $2^{N}$ basis states. This exponential scaling of the complexity of the problem cannot be underestimated: For $N=300$, the number of basis states already exceeds the number of atoms in the universe! Nevertheless, we can expand each state according to

$|\psi \rangle =\sum \limits _{k=1}^{2^{N}}c_{k}|k\rangle$

$|k\rangle =\prod _{i=1}^{N}|\alpha _{i}\rangle _{i}$,

where $\alpha _{i}$ is either $g$ or $r$. The Hamiltonian for this quantum many-body system can be written as

$H={\frac {\Omega }{2}}\sum \limits _{i}\left(|g\rangle _{i}\langle r|_{i}+{\text{H.c.}}\right)+\sum \limits _{i<j}V_{ij}|r\rangle _{i}|r\rangle _{j}\langle r|_{i}\langle r|_{j},$

where $V_{ij}$ is the strength of the van der Waals interaction between atoms $i$ and $j$. Now, we are interested in the blockaded regime, where all the interaction strengths are much larger than the Rabi frequency, i.e., $V_{ij}\gg \Omega$. Then, the dynamics of the system is restricted two the manifold of zero or one Rydberg excitation. We can simplify the problem further by going to a basis of collective states given by:

$|G\rangle =\prod _{i=1}^{N}|g\rangle _{i}=|g_{1}g_{2}g_{3}\ldots g_{N}\rangle$

$|R\rangle ={\frac {1}{\sqrt {N}}}\sum \limits _{i}|g_{1}\ldots r_{i}\ldots g_{N}\rangle .$

The $|R\rangle$ state is a symmetric superposition of a single Rydberg excitation. In order to express the Hamiltonian in the new basis, we need to calculate the transition matrix element $\langle G|H|R\rangle$, which is given by

$\langle G|H|R\rangle ={\frac {\Omega }{2}}\langle G|\sum \limits _{i=1}^{N}|g_{i}\rangle \langle r_{i}|R\rangle ={\frac {\Omega }{2}}\sum _{i=1}^{N}{\frac {1}{\sqrt {N}}}={\frac {{\sqrt {N}}\Omega }{2}}.$

In the collective basis, the Hamiltonian is given by

$H={\frac {{\sqrt {N}}\Omega }{2}}\left(|G\rangle \langle R|+{\text{H.c.}}\right),$

as the matrix elements to all other states in the single excitation manifold are zero. Consequently, we obtain collectively enhanced Rabi oscillation by a factor of ${\sqrt {N}}$, leading to a probability to find an atom in the Rydberg state that is given by

$p_{r}(t)={\frac {1}{N}}\sin ^{2}\left({\sqrt {N}}\Omega t\right).$