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

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

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

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

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

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

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

$${\bf x}_1 = \left( \begin{array}{c} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \\ 0 \end{array} \right),$$ (667)

$${\bf x}_2 = \left( \begin{array}{c} 1/\sqrt{2} \\ - 1/\sqrt{2} \\ 0 \\ 0 \end{array} \right),$$ (668)

$${\bf x}_3 = \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right),$$ (669)

$${\bf x}_4 = \left( \begin{array}{c} 0 \\ 0\\ 0 \\ 1\end{array} \right).$$ (670)

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

$$\vert 1\rangle = \frac{\vert 2,0,0\rangle + \vert 2,1,0\rangle}{\sqrt{2}},$$ (671)

$$\vert 2\rangle = \frac{\vert 2,0,0\rangle - \vert 2,1,0\rangle}{\sqrt{2}},$$ (672)

$$\vert 3\rangle = \vert 2,1,1\rangle,$$ (673)

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

$${\mit\Delta} E_1 = +3\,e\,a_0\, \vert{\bf E}\vert,$$ (675)

$${\mit\Delta} E_2 = -3\,e\,a_0 \,\vert{\bf E}\vert,$$ (676)

$${\mit\Delta} E_3 = 0,$$ (677)

$${\mit\Delta} E_4 = 0.$$ (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.