11.7: 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

$$H_1 = e\,|{\bf E}|\,z.$$ According to the previously determined selection rules (i.e., $m'=m$, and $l'=l\pm1$), 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, because 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

$${\bf U}\,{\bf x} = \lambda\,{\bf x},$$ where ${\bf U}$ is the matrix of the matrix elements of $H_1$ between these states. Thus, $${\bf U} = e\,|{\bf E}|\left(\begin{array}{cc} 0,& \langle 2,0,0|z|2,1,0\rangle\\[0.5ex] \langle 2,1,0|z|2,0,0\rangle,&0 \end{array}\right),$$ 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 that differ by unity. It is easily demonstrated, from the exact forms of the 2S and 2P wavefunctions, that $$\langle 2,0,0|z|2,1,0\rangle = \langle 2,1,0|z|2,0,0\rangle = 3\,a_0.$$

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

$$\begin{aligned} {\bf x}_1&=\left(\begin{array}{c} 1/\sqrt{2}\\[0.5ex] 1/\sqrt{2}\end{array}\right),\\[0.5ex] {\bf x}_2&=\left(\begin{array}{c} 1/\sqrt{2}\\[0.5ex] -1/\sqrt{2}\end{array}\right).\end{aligned}$$ It follows that the simultaneous eigenstates of $H_0$ and $H_1$ take the form $$\begin{aligned} \psi_1 &= \frac{\psi_{200} + \psi_{210}}{\sqrt{2}},\\[0.5ex] \psi_2 &= \frac{\psi_{200} -\psi_{210}}{\sqrt{2}}.\end{aligned}$$ 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 $$\begin{aligned} {\mit\Delta} E_1 &=+3\,e\,a_0\,|{\bf E}|,\\[0.5ex] {\mit\Delta} E_2 &= -3\,e\,a_0\,|{\bf E}|.\end{aligned}$$ Thus, in the presence of an electric field, the energies of states 1 and 2 are shifted upwards and downwards, respectively, by an amount $3\,e\,a_0\,|{\bf E}|$. 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 Section 1.5. 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 Section 1.5.)

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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