The linear Stark effect

Let us examine the effect of an electric field on the excited energy levels of a hydrogen atom. For instance, consider the $n=2$ states. There is a single $l=0$ state, usually referred to as $2s$, and three $l=1$ states (with $m=-1, 0, 1$), usually referred to as $2p$. All of these states possess the same energy, $E_{200} = -e^2/(32\pi \epsilon_0 a_0)$. As in Sect. 6.4, the perturbing Hamiltonian is

$$H_1= e \vert{\bf E}\vert z.$$ (667)

In order to apply perturbation theory, we have to solve the matrix eigenvalue equation

$${\bf U} {\bf x} = \lambda {\bf x},$$ (668)

where ${\bf U}$ is the array of the matrix elements of $H_1$ between the degenerate $2s$ and $2p$ states. Thus,

$${\bf U} = e \vert{\bf E}\vert \left( \begin{array}{cccc} 0... ...t 2,0,0\rangle&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right),$$ (669)

where the rows and columns correspond to the $\vert 2,0,0\rangle$, $\vert 2,1,0\rangle$, $\vert 2,1,1\rangle$, and $\vert 2,1,-1\rangle$ states, respectively. Here, we have made use of the selection rules, which tell us that the matrix element of $z$ between two hydrogen atom states is zero unless the states possess the same $m$ quantum number, and $l$ quantum numbers which differ by unity. It is easily demonstrated, from the exact forms of the $2s$ and $2p$ wave-functions, that

$$\langle 2,0,0\vert z\vert 2,1,0\rangle = \langle 2,1,0\vert z\vert 2,0,0\rangle = 3 a_0.$$ (670)

It can be seen, by inspection, that the eigenvalues of ${\bf U}$ are $\lambda_1= 3 e a_0 \vert{\bf E}\vert$, $\lambda_2 = - 3 e a_0 \vert{\bf E}\vert$, $\lambda_3=0$, and $\lambda_4 =0$. The corresponding eigenvectors are

$${\bf x}_1 = \left( \begin{array}{c} 1/\sqrt{2} \ 1/\sqrt{2} \ 0 \ 0 \end{array}\right),$$ (671)

$${\bf x}_2 = \left( \begin{array}{c} 1/\sqrt{2} \ - 1/\sqrt{2} \ 0 \ 0 \end{array}\right),$$ (672)

$${\bf x}_3 = \left( \begin{array}{c} 0 \ 0 \ 1 \ 0 \end{array}\right),$$ (673)

$${\bf x}_4 = \left( \begin{array}{c} 0 \ 0\ 0 \ 1\end{array}\right).$$ (674)

It follows from Sect. 6.5 that the simultaneous eigenstates of the unperturbed Hamiltonian and the perturbing Hamiltonian take the form

$$\vert 1\rangle = \frac{\vert 2,0,0\rangle + \vert 2,1,0\rangle}{\sqrt{2}},$$ (675)

$$\vert 2\rangle = \frac{\vert 2,0,0\rangle - \vert 2,1,0\rangle}{\sqrt{2}},$$ (676)

$$\vert 3\rangle = \vert 2,1,1\rangle,$$ (677)

$$\vert 4\rangle = \vert 2,1,-1\rangle.$$ (678)

In the absence of an electric field, all of these states possess the same energy, $E_{200}$. The first-order energy shifts induced by an electric field are given by

$$\Delta E_1 = +3 e a_0 \vert{\bf E}\vert,$$ (679)

$$\Delta E_2 = -3 e a_0 \vert{\bf E}\vert,$$ (680)

$$\Delta E_3 = 0,$$ (681)

$$\Delta E_4 = 0.$$ (682)

Thus, the energies of states 1 and 2 are shifted upwards and downwards, respectively, by an amount $3 e a_0 \vert{\bf E}\vert$ in the presence of an electric field. States 1 and 2 are orthogonal linear combinations of the original $2s$ and $2p(m=0)$ states. Note that the energy shifts are linear in the electric field-strength, so this is a much larger effect that the quadratic effect described in Sect. 6.4. The energies of states 3 and 4 (which are equivalent to the original $2p(m=1)$ and $2p(m=-1)$ states, respectively) are not affected to first-order. Of course, to second-order the energies of these states are shifted by an amount which depends on the square of the electric field-strength.

Note that the linear Stark effect depends crucially on the degeneracy of the $2s$ and $2p$ states. This degeneracy is a special property of a pure Coulomb potential, and, therefore, only applies to a hydrogen atom. Thus, alkali metal atoms do not exhibit the linear Stark effect.