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 energy eigenstates of the hydrogen atom is zero unless the states possess the same magnetic quantum number, , and azimuthal quantum numbers, , that differ by unity. (See Section 7.4.) It is easily demonstrated, from the exact forms of the and wavefunctions, that

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(See Exercise 15.)

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 external electric field, all of these states possess the same energy, . The first-order energy-shifts induced by such a 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 external electric field. States 1 and 2 are orthogonal linear combinations of the original and states. Note that the energy-shifts are linear in the electric field-strength. Consequently, this phenomenon is known as the

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. (See Section 4.6.) Thus, alkali metal atoms do not exhibit the linear Stark effect.