Selection Rules and Available Transitions

A change in energy levels for a Rydberg atom in a circular state must obey selection rules so that $\Delta m = \pm 1$ or 0, and $\Delta l = \pm 1$ . Thus an atom in a Rydberg state under guarded environmental condtions can only transition as

$$n \ \rightarrow \ n-1, \ l \ \rightarrow \ l-1, \vert m\vert \ \rightarrow \ \vert m\vert -1$$

Thus a Rydberg atom approximates a two-level system.

We demonstrate the theory behind elementary selection rules with a simple example. A two-state system in the presence of puturbing Hamiltonian can be described as,

$$\Psi(t)=c_a(t)\psi_ae^{-iE_at/\hbar} + c_b(t)\psi_b e^{-iE_bt/\hbar}$$ (3.1)

Its evolution is described by

$$(H+H'(t))\Psi = i\hbar \frac{\partial \Psi}{\partial t}$$

from which we have,

$$c_a(t)\langle\psi_a\vert H' \vert\psi_a\rangle e^{-iE_at/\hbar} +... ...\vert H' \vert\psi_b\rangle e^{-iE_bt/\hbar}=i\hbar \dot{c_a(t)}e^{-iE_at/hbar}$$ (3.2)

Rejoining convention and writing $H'_{ij}=\langle \psi_i\vert H'\vert\psi_j\rangle$ , $\omega_0 = (E_b - E_a)/\hbar$ , and assuming (as is warented in the experiments we discuss) that the diagonal of the perturbing part of the Hamiltonian is zero, we can obtain a set of equations for the prefactors,

$$\dot{c_a} = -\frac{i}{\hbar}H'_{ab}e^{-i\omega_0t}c_b$$ (3.3)

$$\dot{c_b} = -\frac{i}{\hbar}H'_{ba}e^{i\omega_0t}c_a$$ (3.4)

Following [4] we consider a basic hydrogen atom in the $n=2$ state in an electric field so that $H'=-eEr\cos\theta$ . As we showed in a previous compendium, the wave functions for hydrogen, $n=1, n=2$ , are

$$\psi_{100} = \frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$

$$\psi_{200} = \frac{1}{\sqrt{8\pi a^3}}\left(1-\frac{r}{2a}\right)e^{-r/2a}$$

$$\psi_{210} = \frac{1}{\sqrt{32 \pi a^3}}\frac{r}{a}e^{r/a}\cos \theta$$

$$\psi_{211} = -\frac{1}{\sqrt{64 \pi a^3}}\frac{r}{a}e^{-r/2a}\sin \theta e^{i\phi}$$

$$\psi_{21-1} = \frac{1}{\sqrt{64 \pi a^3}}\frac{r}{a}e^{-r/2a}\sin \theta e^{-i\phi}$$

The perturbation matrix is simple in that all but one of these will be even in $r\cos \theta$ , so the perturbation matrix ( $\langle n' l' m'\vert H'\vert nlm\rangle$ ) will have all zero elements except

$$\langle 100\vert H'\vert 210 \rangle = - \frac{eE}{4\sqrt{2}\pi a... ...2 \cos^2\theta r^2 \sin\theta dr d\theta \ d\phi = -\frac{2^8}{\sqrt{2} 3^5}eEa$$ (3.5)

	$\psi_{100}$	$\psi_{200}$	$\psi_{210}$	$\psi_{211}$	$\psi_{21-1}$
$\psi_{100}$	0	0	$-\frac{2^8}{\sqrt{2} 3^5}eEa$	0	0
$\psi_{200}$	0	0	0	0	0
$\psi_{210}$	$-\frac{2^8}{\sqrt{2} 3^5}eEa$	0	0	0	0
$\psi_{211}$	0	0	0	0	0
$\psi_{21-1}$	0	0	0	0	0

Thus we see that under this perturbation, the $100 \leftrightarrow 210$ levels are ``selected' and we have an approximate two-state system under the right conditions (ideal).

In general, we can derive selection rules for m and l transitions. In the case of m, we consider that $[L_z,z]=0$ so that

$$0=\langle n'l'm'\vert[L_z,z]\vert nlm\rangle=(m'-m)\langle n'l'm'\vert z\vert nlm\rangle$$ (3.6)

That $[L_z,x]=i\hbar y$ gives,

$$(m'-m)\langle n'l'm'\vert x\vert nlm\rangle=i\langle n'l'm'\vert y\vert nlm\rangle$$ (3.7)

Finally, $[L_z,y]=-i\hbar x$ gives,

$$(m'-m)\langle n'l'm'\vert y\vert nlm\rangle=-i\langle n'l'm'\vert x\vert nlm\rangle$$ (3.8)

Thus $(m'-m)^2 =1$ or $\langle\vert x,y\vert\rangle=0$ . For the $l,l'$ case, it can be shown that

$$L^2,[L^2,r]]=2\hbar^2(rL^2 + L^2r)$$

From this we can show that

$$2\hbar^4(l(l+1)+l'(l'+1))\langle n' l' m'\vert r\vert nlm\rangle = \hbar^4(l'(l'_1)-l(l+1))^2)\langle n' l' m'\vert r\vert nlm\rangle$$ (3.9)

Rewriting

$$l'(l'+1)-l(l+1)=(l'+l+1)(l'-l), \$$   and$$\ 2(l(l+1)+l'(l'+1))=(l'+l+1)^2+(l'-l)^2-1$$

We thus conclude that $l'=l\pm1$ and $m'=m\pm1 \$   or$\ m$