Polarizability of the hydrogen atom

In order to calculate the polarizability of the hydrogen atom we need the transition moments for the allowed transitions. In the presence of a Z-polarized applied electric field we have the induced dipole moment

Since m = aE₀ we have for the polarizability

obtained from Cohen-Tannoudji (page 1280). It appears from the selection rules that Dn = any value, Dl = ±1, and Dm = 0. We have the following matrix elements:

making the substitution x = cosq, and collecting constants we have

Let u = 3r/2a₀, then r = (2a₀/3)u and dr = (2a₀/3)du so

Evaluating the integrals we obtain

The energy denominator is

so

The polarizability can be expressed

Thus, for example for the 210 state we have

The constant in front of the polarizability expression is:

which works out to:

and it has the correct units. For the 310 state everything will be the same except the radial integral and the energy denominator.

The radial integrals are

where we have included the factor of 4p/3 that comes from the angular integrals. Using the substitution u = 4r/3a₀ we have r = (3a₀/4)u and dr = (3a₀/4)du. The result is 0.298a₀. The contribution to the polarizability is:

Note that from the ground quantum state with l = 0 transitions will be allowed only to states with l = 1. However, from the first excited state (e.g. the 210 state) transitions can occur to 300, 320, and 100.

For example,

Here again, the angular integrals are done in advance and give 4p/3.

Upon making the substitution u = 5r/6a₀ we have

= -0.624a₀. The contribution to the polarizability is:

We consider the contribution from the transition moment for 210 à 320:

which can be written as

or

We have used the fact that

If this is correct then we have