Linear Stark Effect

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

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where is the array of the matrix elements of between the degenerate and states. Thus,

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where the rows and columns correspond to the , , , and states, respectively. Here, we have made use of the selection rules, which tell us that the matrix element of between two hydrogen atom states is zero unless the states possess the same quantum number, and quantum numbers that differ by unity. It is easily demonstrated, from the exact forms of the and wavefunctions, that

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It can be seen, by inspection, that the eigenvalues of are , , , and . The corresponding eigenvectors are

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

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In the absence of an electric field, all of these states possess the same energy, . The first-order energy-shifts induced by an electric field are given by

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Thus, the energies of states 1 and 2 are shifted upwards and downwards, respectively, by an amount in the presence of an electric field. States 1 and 2 are orthogonal linear combinations of the original and states. Note that the energy-shifts are

Note that the linear Stark effect depends crucially on the degeneracy of the and 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.