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Suppose that a hydrogen atom is subject to a uniform external electric field, of magnitude , directed along the -axis. The Hamiltonian of the system can be split into two parts. Namely, the unperturbed Hamiltonian,
 (911)

and the perturbing Hamiltonian
 (912)

Note that the electron spin is irrelevant to this problem (since the spin operators all commute with ), so we can ignore the spin degrees of freedom of the system. Hence, the energy eigenstates of the unperturbed Hamiltonian are characterized by three quantum numbers--the radial quantum number , and the two angular quantum numbers and (see Cha. 9). Let us denote these states as the , and let their corresponding energy eigenvalues be the . According to the analysis in the previous section, the change in energy of the eigenstate characterized by the quantum numbers in the presence of a small electric field is given by

 (913)

This energy-shift is known as the Stark effect.

The sum on the right-hand side of the above equation seems very complicated. However, it turns out that most of the terms in this sum are zero. This follows because the matrix elements are zero for virtually all choices of the two sets of quantum number, and . Let us try to find a set of rules which determine when these matrix elements are non-zero. These rules are usually referred to as the selection rules for the problem in hand.

Now, since [see Eq. (529)]

 (914)

it follows that [see Eqs. (481)-(483)]
 (915)

Thus,
 (916)

since is, by definition, an eigenstate of corresponding to the eigenvalue . Hence, it is clear, from the above equation, that one of the selection rules is that the matrix element is zero unless
 (917)

Let us now determine the selection rule for . We have

 (918)

where use has been made of Eqs. (481)-(483), (527)-(529), and (535). Thus,
 (919)

which reduces to
 (920)

However, it is clear from Eqs. (527)-(529) that
 (921)

Hence, we obtain
 (922)

Finally, the above expression expands to give
 (923)

Equation (923) implies that

 (924)

Since, by definition, is an eigenstate of corresponding to the eigenvalue , this expression yields
 (925)

which reduces to
 (926)

According to the above formula, the matrix element vanishes unless or . [Of course, the factor , in the above equation, can never be zero, since and can never be negative.] Recall, however, from Cha. 9, that an wavefunction is spherically symmetric. It, therefore, follows, from symmetry, that the matrix element is zero when . In conclusion, the selection rule for is that the matrix element is zero unless
 (927)

Application of the selection rules (917) and (927) to Eq. (913) yields

 (928)

Note that, according to the selection rules, all of the terms in Eq. (913) which vary linearly with the electric field-strength vanish. Only those terms which vary quadratically with the field-strength survive. Hence, this type of energy-shift of an atomic state in the presence of a small electric field is known as the quadratic Stark effect. Now, the electric polarizability of an atom is defined in terms of the energy-shift of the atomic state as follows:
 (929)

Hence, we can write
 (930)

Unfortunately, there is one fairly obvious problem with Eq. (928). Namely, it predicts an infinite energy-shift if there exists some non-zero matrix element which couples two degenerate unperturbed energy eigenstates: i.e., if and . Clearly, our perturbation method breaks down completely in this situation. Hence, we conclude that Eqs. (928) and (930) are only applicable to cases where the coupled eigenstates are non-degenerate. For this reason, the type of perturbation theory employed here is known as non-degenerate perturbation theory. Now, the unperturbed eigenstates of a hydrogen atom have energies which only depend on the radial quantum number (see Cha. 9). It follows that we can only apply the above results to the eigenstate (since for there will be coupling to degenerate eigenstates with the same value of but different values of ).

Thus, according to non-degenerate perturbation theory, the polarizability of the ground-state (i.e., ) of a hydrogen atom is given by

 (931)

Here, we have made use of the fact that . The sum in the above expression can be evaluated approximately by noting that (see Sect. 9.4)
 (932)

where
 (933)

is the Bohr radius. Hence, we can write
 (934)

which implies that
 (935)

However, [see Eq. (874)]
 (936)

where we have made use of the selection rules, the fact that the form a complete set, and the fact the the ground-state of hydrogen is spherically symmetric. Finally, it follows from Eq. (693) that
 (937)

Hence, we conclude that
 (938)

The exact result (which can be obtained by solving Schrödinger's equation in parabolic coordinates) is
 (939)

Next: Degenerate Perturbation Theory Up: Time-Independent Perturbation Theory Previous: Non-Degenerate Perturbation Theory
Richard Fitzpatrick 2010-07-20