Spherical box model for London dispersion interactions

Abstract

A simple model for the study of London dispersion interactions is introduced: two spherical boxes, uniformly charged on their surfaces and each containing a freely moving charged particle. The model, also denominated as “interacting quantum balls” model, is applied to calculate the first three dispersion coefficients involving the ground and several excited s states, where dispersion interactions are the only contribution to the interaction energy. Already the first excited symmetric $(1s{-}2s)_+$ state has a negative $C_6$ dispersion coefficient, corresponding to long range repulsion. In general, a drastic state variation of the dispersion coefficients is observed, depending on the occurence of near-resonance excitation/disexcitation coupling.

Appendix: Matrix elements involving radial functions

As most of the required integrals involving the radial functions $R_{nl}(r)$ do not appear to be standard knowledge, they will be derived in the following. Though the normalization factor is known² [4] for the sake of completeness its derivation will also be given.

Proceeding along the lines of the computation of Lommel’s integrals involving products of cylindrical Bessel functions [18] multiplication of the differential equation

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}} r} r^2 \frac{{\mathrm {d}}}{{\mathrm {d}} r} j_l(kr) + (k^2 r^2 -l(l+1)) j_l(kr) = 0 \end{aligned}$$

(26)

with $j_l(k'r)$ and subtraction by a corresponding product with reversed roles of $j_l(kr)$ and $j_l(k'r)$ yields

$$\begin{aligned} (k^2-(k')^2) r^2 j_l(kr) j_l(k'r) &= {} j_l(kr) \frac{{\mathrm {d}}}{{\mathrm {d}} r} r^2 \frac{{\mathrm {d}}}{{\mathrm {d}} r} j_l(k'r) \nonumber \\&\quad -\,j_l(k'r) \frac{{\mathrm {d}}}{{\mathrm {d}}r} r^2 \frac{\mathrm {d}}{\mathrm {d} r} j_l(kr) . \end{aligned}$$

(27)

Partial integration leads to the identity

$$\begin{aligned} \int _0^a {\mathrm {d}} r r^2 j_l(kr) j_l(k'r)= & {} \frac{a^2}{k^2-(k')^2} \left\{ k' j_l(ka) j'_l(k'a) \right. \nonumber \\&\left. -\,k j'_l(ka) j_l(k'a) \right\} , \end{aligned}$$

(28)

where $j'_l(z) = \frac{\mathrm {d} j_l(z) }{{\mathrm {d}} z}$. From this follows orthogonality of the spherical Bessel functions if $ka = z_{nl}$ and $k'a = z_{n'l}$ are different zeros of the spherical Bessel function $j_l(z)$. On the other hand, if $j_l(ka) = 0$ one also gets

$$\begin{aligned} \lim _{k' \rightarrow k} \int _0^a {\mathrm {d}} r r^2 j_l(kr) j_l(k'r)= & {} k a^2 j '_l(ka) \lim _{k' \rightarrow k} \frac{j_l(k'a)}{(k')^2-k^2} \nonumber \\= & {} \frac{a^3}{2} \left( j'_l(ka) \right) ^2 , \end{aligned}$$

(29)

where the rule of l’Hospital has been used in the last equation. From $j'_l(z) = \frac{l}{z} j_l(z) - j_{l+1}(z)$ [3] one finally obtains the normalization factor $N_{nl}$ as given after Eq. (2).

Noting that the last recursion formula may also be written as $r j_{l+1}(kr) = \left( \frac{l}{k} - \frac{\mathrm {d}}{\mathrm {d} k}\right) j_l(kr)$ one obtains

$$\begin{aligned} \int _0^a {\mathrm {d}} r r^3 j_{l+1}(k'r) j_l(kr) = \left( \frac{l}{k} - \frac{{\mathrm {d}}}{\mathrm {d} k}\right) \int _0^a {\mathrm {d}} r r^2 j_l(k'r) j_l(kr) , \end{aligned}$$

(30)

which, employing (28) along with $j_l(ka) = 0$ and appropriate recursion relations, results in

$$\begin{aligned} \int _0^a {\mathrm {d}} r r^3 j_{l+1}(k'r) j_l(kr) = \frac{2 k' k a^2}{(k')^2-k^2} j_l(k'a) j_{l+1}(ka) . \end{aligned}$$

(31)

Taking normalization factors into account, from this one arrives at

$$\begin{aligned} \langle n' (l+1) | r | nl \rangle = \frac{4 a z_{n'(l+1)} z_{nl}}{(z^2_{n'(l+1)} - z^2_{nl})^2} . \end{aligned}$$

(32)

Similarly the remaining integrals needed for the $C_8^{nn',\pm }$ and $C_{10}^{nn',\pm }$ coefficients can be computed as

$$\begin{aligned} \langle n' (l+2) | r^2 | n l \rangle = \frac{8a^2z_{n'(l+2)}z_{nl}}{(z_{n'(l+2)}^2-z_{nl}^2)^2} \left( 1+\frac{2(2l+3)}{z_{n'(l+2)}^2-z_{nl}^2} \right) \end{aligned}$$

(33)

and

$$\begin{aligned} \langle n'(l+3) | r^3 | n l \rangle= & {} \frac{12a^3z_{n'(l+3)}z_{nl}}{(z_{n'(l+3)}^2-z_{nl}^2)^2} \left( 1+\frac{4(2l+5)}{z_{n'3}^2-z_{nl}^2} \right. \nonumber \\&\left. +\frac{8\{(2l+5)(2l+3)-z_{n'(l+3)}^2\}}{(z_{n'3}^2-z_{nl}^2)^2} \right) . \end{aligned}$$

(34)