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# 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.)
