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Next: Fine Structure Up: Time-Independent Perturbation Theory Previous: Degenerate Perturbation Theory

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 Section 7.4, the perturbing Hamiltonian is

$\displaystyle H_1= e\, \vert{\bf E}\vert\, z.$ (663)

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

$\displaystyle {\bf U} \,{\bf x} = \lambda \,{\bf x},$ (664)

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

$\displaystyle {\bf U} = e \,\vert{\bf E}\vert \left( \begin{array}{cccc} 0&\lan...
...2,1,0\vert\,z\,\vert 2,0,0\rangle&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right),$ (665)

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 that differ by unity. It is easily demonstrated, from the exact forms of the $ 2s$ and $ 2p$ wavefunctions, that

$\displaystyle \langle 2,0,0\vert\,z\,\vert 2,1,0\rangle = \langle 2,1,0\vert\,z\,\vert 2,0,0\rangle = 3\,a_0.$ (666)

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

$\displaystyle {\bf x}_1$ $\displaystyle = \left( \begin{array}{c} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \\ 0 \end{array} \right),$ (667)
$\displaystyle {\bf x}_2$ $\displaystyle = \left( \begin{array}{c} 1/\sqrt{2} \\ - 1/\sqrt{2} \\ 0 \\ 0 \end{array} \right),$ (668)
$\displaystyle {\bf x}_3$ $\displaystyle = \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right),$ (669)
$\displaystyle {\bf x}_4$ $\displaystyle = \left( \begin{array}{c} 0 \\ 0\\ 0 \\ 1\end{array} \right).$ (670)

It follows from Section 7.5 that the simultaneous eigenstates of the unperturbed Hamiltonian and the perturbing Hamiltonian take the form

$\displaystyle \vert 1\rangle$ $\displaystyle = \frac{\vert 2,0,0\rangle + \vert 2,1,0\rangle}{\sqrt{2}},$ (671)
$\displaystyle \vert 2\rangle$ $\displaystyle = \frac{\vert 2,0,0\rangle - \vert 2,1,0\rangle}{\sqrt{2}},$ (672)
$\displaystyle \vert 3\rangle$ $\displaystyle = \vert 2,1,1\rangle,$ (673)
$\displaystyle \vert 4\rangle$ $\displaystyle = \vert 2,1,-1\rangle.$ (674)

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

$\displaystyle {\mit\Delta} E_1$ $\displaystyle = +3\,e\,a_0\, \vert{\bf E}\vert,$ (675)
$\displaystyle {\mit\Delta} E_2$ $\displaystyle = -3\,e\,a_0 \,\vert{\bf E}\vert,$ (676)
$\displaystyle {\mit\Delta} E_3$ $\displaystyle = 0,$ (677)
$\displaystyle {\mit\Delta} E_4$ $\displaystyle = 0.$ (678)

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 Section 7.4. 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 that 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.

next up previous
Next: Fine Structure Up: Time-Independent Perturbation Theory Previous: Degenerate Perturbation Theory
Richard Fitzpatrick 2013-04-08