Rydberg Atoms/Quantum Defect Theory

The Hydrogen Atom

Rydberg atoms are excited states of atoms with a large principle quantum number, where the Rydberg electron is only weakly bound to the ionic core. This weak binding makes Rydberg atoms very sensitive to external perturbations and results in a wide range of unique features. To understand the basic properties of Rydberg atoms, it is instructive to first look at the solution of the hydrogen atom. In atomic units, the Hamiltonian is given by

$H=\frac{{\mathbf{\text{p}}}^2}{2\mu }-\frac{1}{r},$

where position and momentum operators are canonically conjugated and $\mu =1-1/m_p$ is the reduced mass according to the proton mass $m_p\approx 1836$. To solve the hydrogen problem, we first introduce the angular momentum operators

$L_i={\epsilon }_{ijk}{x}_j{p}_k$

and its square $L^2=L_x^2+L_y^2+L_z^2$. Then, we can write the Hamiltonian as

$H=\frac{p_r^2}{2\mu }+\frac{L^2}{2\mu r^2}-\frac{1}{r}.$

The angular momentum part can be solved independently from the radial part, using the angular momentum algebra. The eigenvalues of $L^2$ are given by $l(l+1)$, with $l$ being a non-negative integer. Each eigenvalue of $L^2$ is $2l+1$-fold degenerate. This degenerate manifold can be expressed in terms of eigenvectors of $L_z$, having integer eigenvalues $m$ with $|m|\le l$.

The radial part can be solved by expressing the radial momentum term in its coordinate representation, resulting in

$H=-\frac{1}{2\mu r}\frac{{\partial }^2}{\partial r^2}r+\frac{l(l+1)}{2\mu r}-\frac{1}{r}.$

The resulting Schrödinger equation can then be solved using standard techniques . We then obtain the solution

$\psi (r)=\sqrt{{\left(\frac{2}{n}\right)}^3\frac{(n-l-1)!}{2n[(n+l)!{]}^3}}\exp (-r/n{a}_0^{*})(2r/n{a}_0^{*}{)}^l{L}_{n-l-1}^{2l+1}(2r/n{a}_0^{*}),$

where $n$ is the principal quantum number and $a_0^{*}=m_p/(m_p-1)$ is the reduced Bohr radius. The associated Laguerre polynomials are normalized according to

$L_{n-1}^1(0)=(n-1)!.$

The eigenvalues do not depend on $l$, a fact that follows from the conservation of the Runge-Lenz vector, and are given by

$E_n=-\frac{1}{2\mu n^2}$.

Fig. 1: Probability distribution of a Rydberg atom in the 35s state of hydrogen.

Rydberg states are atomic excitation with a large principal quantum number $n$. We can call $n$ being large if the properties of the Rydberg state are drastically different from the ground state. For practical purposes, this usually means $n\gtrsim 10$. For example, consider the expectation value of the radius $r$ ,

$⟨r⟩=\frac{a_0^{*}}{2}[3n^2-l(l+1)].$

Already at relatively modest values of $n$, the spatial extension of the Rydberg state is already orders of magnitude larger than that of the ground state, see Fig. 1. Such an enhanced scaling with the principal quantum number $n$ is characteristic for Rydberg states and leads to strongly exaggerated properties of Rydberg atoms. We will see many examples of such a scaling in the following.

Quantum defect

Tab. 1: Quantum defect in Rb.

$l$	${\delta }_l$
0	3.13
1	2.64
2	1.35
3	0.016
>3	0

Of course, we are not only interested in the properties of hydrogen atoms. However, for Rydberg states with a single excited electron, the eigenenergies can be well described by a simple phenomenological extension of the expression for hydrogen, as was first noted by Rydberg himself . This can be done by introducting a quantum defect ${\delta }_l$ that depends (in leading order) only on the angular momentum quantum number, yielding

$E_n=-\frac{1}{2\mu (n-{\delta }_l{)}^2}.$

The quantum defect accounts for the corrections to the Coulomb potential by the core electrons. We can introduce an effective quantum number $n^{*}=n-{\delta }_l$. Already on the basis of the hydrogen wave functions we see that the probability to find the Rydberg electron within the core decreases with $l$, therefore the quantum defect should also decrease with $l$. Using laser spectroscopy, the eigenenergies of atoms can be measured very accurately, and as an example, the quantum defect for rubidium is shown in Tab. 1. For $l>3$, the eigenstates are essentially hydrogenic. Experimentally, the quantum defect can be measured with much higher accuracy up to a relative uncertainty of $10^{-7}$ , but requires the treatment of the electron spin and introducing an $n$-dependence of the quantum defect.

However, it is evident that the Hamiltonian has to be modified to yield the desired eigenvalues. The easiest way to achieve this is to add an additional term $V_{eff}(r)$ to the radial part such that

$\frac{l(l+1)}{2\mu r}+V_{eff}(r)=\frac{l^{*}(l^{*}+1)}{2\mu r},$

where the effective quantum number $l^{*}$ is given by

$l^{*}=l-\delta (l)+I(l)$

with $I(l)$ being an integer . The radial eigenfunction can then be expressed using an extension of the associated Laguerre polynomials for non-integer $\alpha =2l^{*}-1$,

$L_n^{\alpha }(\rho )=\sum _{p=0}^n(-\rho {)}^p\frac{\mathrm{\Gamma }(n+\alpha +1)}{p!\mathrm{\Gamma }(p+\alpha +1)\mathrm{\Gamma }(n-p-1)}.$

Then, we obtain for the radial eigenfunctions

${\psi }_{n^{*}{l}^{*}}(r)=\sqrt{{\left(\frac{2}{n^{*}}\right)}^3\frac{\mathrm{\Gamma }(n-l-I)}{2n\mathrm{\Gamma }(n^{*}+l^{*}+1)}}\exp (-r/n^{*}{a}_0^{*}){(r/n^{*}{a}_0^{*})}^{l^{*}}{L}_{n-l-I-1}^{2l^{*}+1}(r/n^{*}{a}_0^{*}).$

Note the different normalization factor compared to the hydrogen case due to the different definition of the associated Laguerre polynomials. The value of the integer $I$ can be fixed by an appropriate choice of the number of nodes, $n-l-I-1$, e.g., the ground state should not have any nodes. Alternatively, it is possible to improve the wave functions by choosing $I$ such that transition matrix elements $⟨{n^{*}}^{\prime },{l^{*}}^{\prime }|x|n^{*},l^{*}⟩$ to match experimentally observed values .

Polarizability of Rydberg states

One important property of quantum defect theory is that the scaling relations we already saw in the hydrogen case remain valid, with the principal quantum number $n$ being replaced by $n^{*}$. The key difference is that the $l$ degeneracy found in the hydrogen atom is lifted. This means that we no longer have atoms with a permanent electric dipole moment (linear Stark effect) , but induced dipole moments (quadratic Stark effect). The strength of the quadractic Stark shift is captured in the polarizability $\alpha$, according to

$\mathrm{\Delta }E=-\frac{1}{2}\alpha E^2,$

where $\mathrm{\Delta }E$ is the energy shift from the Stark effect and $E$ is the applied electric field. Within the electic dipole approximation, the perturbation from the electric field is given by the Hamiltonian $H^{\prime }=-dE$, where $d$ is the electric dipole operator. In second order perturbation theory, we obtain for the Stark shift

$\mathrm{\Delta }E=\sum _{n^{\prime }{l}^{\prime }{m}^{\prime }}\frac{⟨{\psi }_{nlm}|d|{\psi }_{n^{\prime }{l}^{\prime }{m}^{\prime }}⟩⟨{\psi }_{n^{\prime }{l}^{\prime }{m}^{\prime }}|d|{\psi }_{nlm}⟩}{E_{nl}-E_{n^{\prime }{l}^{\prime }}}{E}^2.$

Using this expression, we can identify the polarizability as

$\alpha =2\sum _{n^{\prime }{l}^{\prime }{m}^{\prime }}\frac{|⟨{\psi }_{n^{\prime }{l}^{\prime }{m}^{\prime }}|d|{\psi }_{nlm}⟩{|}^2}{E_{n^{\prime }{l}^{\prime }}-E_{nl}}.$

Note that while the polarizability is always positive for the ground state (since $E_{n^{\prime }{l}^{\prime }}>E_{nl}$), the polarizability of Rydberg states can actually be negative.

We are now interested in the scaling behavior of the polarizability with the effective quantum number $n^{*}$. For this, we first assume that the main contribution to the polarizability comes from the coupling to a single state, namely the one which is closest in energy for any dipole-allowed transition (i.e., $l^{\prime }=l\pm 1$). The matrix element $⟨{\psi }_{n^{\prime }{l}^{\prime }{m}^{\prime }}|d|{\psi }_{nlm}⟩$ is proportional to a length and therefore has the same scaling as the expectation value for the orbital radius, $⟨r⟩\sim {n^{*}}^2$. The scaling for the energy difference in the denominator can be calculated as

$\lim {\mathrm{\backslash limits}}_{n\to \mathrm{\infty }}{E}_{n,l+1}-E_{nl}=\lim {\mathrm{\backslash limits}}_{n^{*}\to \mathrm{\infty }}\frac{1}{2{n^{*}}^2}-\frac{1}{2(n^{*}+\epsilon {)}^2}=\epsilon {n^{*}}^{-3}.$

If we combine these two scaling relations, we obtain for the polarizability an asymptotic behavior according to $\alpha \sim {n^{*}}^7$. This dramatic scaling with the seventh power of the principal quantum number makes Rydberg atoms very sensitive to external electric fields and therefore very good candidates for the realization of electric field sensors.

If one wants to go beyond simple scaling relations, it is often appropriate to use semiclassical approximations for the dipole matrix element . This allows to express the dipole matrix elements in an analytical form,

$⟨{\psi }_{n^{\prime }{l}^{\prime }{m}^{\prime }}|d|{\psi }_{nlm}⟩=\frac{(-1{)}^{n^{\prime }-n}}{{n^{*}}^{\prime }-n^{*}}(n^{*}{n^{*}}^{\prime }{)}^{11/6}{\left(\frac{n^{*}+{n^{*}}^{\prime }}{2}\right)}^{-5/3}[{𝐉}_{{n^{*}}^{\prime }-n^{*}-1}({n^{*}}^{\prime }-n^{*})+{𝐉}_{n^{*}-{n^{*}}^{\prime }+1}({n^{*}}^{\prime }-n^{*})]I_{ml}^{m^{\prime }{l}^{\prime }},$

where ${𝐉}_{\nu }(z)$ is the Anger function and $I_{ml}^{m^{\prime }{l}^{\prime }},$ are integrals of spherical harmonics representing the angular part . For example, this semiclassical expression predicts for the dipole moment between the 43s and the 43p state in rubidium a value of $d=1041$, whereas the exact value is $d=1069$ , i.e., less than 3% discrepancy.