Linear Stark Effect

Returning to the Stark effect, let us examine the effect of an external electric field on the energy levels of the $n=2$ states of a hydrogen atom. There are four such states: an $l=0$ 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 unperturbed energy, $E_{200} = -e^2/(32\pi \epsilon_0 a_0)$. As before, the perturbing Hamiltonian is

$$H_1 = e \vert{\bf E}\vert z.$$ (956)

According to the previously determined selection rules (i.e., $m'=m$, and $l'=l\pm 1$), this Hamiltonian couples $\psi_{200}$ and $\psi_{210}$. Hence, non-degenerate perturbation theory breaks down when applied to these two states. On the other hand, non-degenerate perturbation theory works fine for the $\psi_{211}$ and $\psi_{21-1}$ states, since these are not coupled to any other $n=2$ states by the perturbing Hamiltonian.

In order to apply perturbation theory to the $\psi_{200}$ and $\psi_{210}$ states, we have to solve the matrix eigenvalue equation

$${\bf U} {\bf x} = \lambda {\bf x},$$ (957)

where ${\bf U}$ is the matrix of the matrix elements of $H_1$ between these states. Thus,

$${\bf U} = e \vert{\bf E}\vert\left(\begin{array}{cc} 0& \la... ...] \langle 2,1,0\vert z\vert 2,0,0\rangle&0 \end{array}\right),$$ (958)

where the rows and columns correspond to $\psi_{200}$ and $\psi_{210}$, respectively. Here, we have again 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 $l$ quantum numbers which differ by unity. It is easily demonstrated, from the exact forms of the 2S and 2P wavefunctions, that

$$\langle 2,0,0\vert z\vert 2,1,0\rangle = \langle 2,1,0\vert z\vert 2,0,0\rangle = 3 a_0.$$ (959)

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

$${\bf x}_1 = \left(\begin{array}{c} 1/\sqrt{2}\ [0.5ex] 1/\sqrt{2}\end{array}\right),$$ (960)

$${\bf x}_2 = \left(\begin{array}{c} 1/\sqrt{2}\ [0.5ex] -1/\sqrt{2}\end{array}\right).$$ (961)

It follows that the simultaneous eigenstates of $H_0$ and $H_1$ take the form

$$\psi_1 = \frac{\psi_{200} + \psi_{210}}{\sqrt{2}},$$ (962)

$$\psi_2 = \frac{\psi_{200} -\psi_{210}}{\sqrt{2}}.$$ (963)

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

$$\Delta E_1 = +3 e a_0 \vert{\bf E}\vert,$$ (964)

$$\Delta E_2 = -3 e a_0 \vert{\bf E}\vert.$$ (965)

Thus, in the presence of an electric field, the energies of states 1 and 2 are shifted upwards and downwards, respectively, by an amount $3 e a_0 \vert{\bf E}\vert$. These states are orthogonal linear combinations of the original $\psi_{200}$ and $\psi_{210}$ states. Note that the energy shifts are linear in the electric field-strength, so this effect--which is known as the linear Stark effect--is much larger than the quadratic effect described in Sect. 12.5. Note, also, that the energies of the $\psi_{211}$ and $\psi_{21-1}$ states are not affected by the electric field 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 (see Sect. 12.5).