Kramer's Relation and the Circular State

An atom which has a valance electron in an extremely high principal quantum number state is in a Rydberg state. These states are called circular states for the following reason. (We use the simpler hydrogen case due to the fact that it is not a terrible approximation to hydrogenic atoms.)

The radial equation (see the compendium on the hydrogen atom) was found to be simplifiable to,

$$-\frac{\hbar^2}{2m}u'' + \left(-\frac{e^2}{4\pi \epsilon_0}\frac{1}{r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right)u = Eu$$ (2.1)

With,

$$E=-\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0}\right)^2\frac{1}{n^2}$$ (2.2)

and $a=4\pi \epsilon_0 \hbar^@ / me^2$ , we can rewrite this equation as,

$$u''=\left(\frac{l(l+1)}{r^2} - \frac{2}{ar} + \frac{1}{n^2 a^2}\right)u$$ (2.3)

We then consider that (using integration by parts and knowing that the infinite limit of the radial component must be zero)

$$\int ur^su'' dr = \int u r^s\left(\frac{l(l+1)}{r^2} - \frac{2}{ar} + \frac{1}{n^2 a^2}\right)u \ dr$$

$$= \langle r^{s-2} \rangle l(l+1)-\frac{2}{a}\langle r^{s-1}\rangle + \frac{1}{n^2a^2}\langle r^s \rangle$$

$$= -\int (ur^s)'u' dr$$

$$= -\int u'r^su'dr - s\int ur^{s-1}u'dr$$

$$= -\left(\frac{2}{s+1}\int u''r^{s+1}u'dr\right) - s\left(-\frac{s-1}{2}\int ur^{s-2}u \ dr\right)$$

$$= -\left(\frac{2}{s+1}\int \left(\frac{l(l+1)}{r^2} - \frac{2}{ar} ... ...{1}{n^2 a^2}\right)ur^{s+1}u'dr\right) + \frac{s(s-1)}{2}\langle r^{s-2}\rangle$$

$$= -\frac{l(l+1)(s-1)}{s+1}\langle r^{s-2}\rangle +\frac{2s}{a(s+1)}... ... - \frac{1}{n^2a^2}\langle r^s \rangle + \frac{s(s-1)}{2}\langle r^{s-2}\rangle$$ (2.4)

Algebraic exercise yields Kramer's relation,

$$\frac{s+1}{n^2}\langle r^s \rangle - (2s+1)a\langle r^{s-1}\rangle + \frac{s}{4}\left((2l+1)^2 - s^2\right)a^2\langle r^{s-2}\rangle =0$$ (2.5)

When we set $s=0$ we find,

$$\frac{1}{n^2}\langle 1 \rangle - a \langle \frac{1}{r}\rangle = 0 \ \ni \ \langle \frac{1}{r}\rangle = \frac{1}{n^2 a}$$ (2.6)

Thus setting $s=1$ yields,

$$\frac{2}{n^2}\langle r \rangle - 3 a\langle 1 \rangle + \frac{1}{... ...r}\rangle = 0 \ \ni \ \langle r \rangle = \frac{a}{2}\left(3n^2 - l(l+1)\right)$$ (2.7)

When we use the states such that $l=n-1$ and $\vert m\vert=n-1$ , i.e. the maximum allowed, we have

$$\langle r \rangle = an^2 + \frac{an}{2}$$ (2.8)

For very large n, this approximates the classical result for circular orbits, $\langle r \rangle = n^2 a$ , thus the name circular Rydberg states.