Quadratic Stark Effect

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

This energy-shift is known as the

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) |

(915) |

(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

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) |

(922) |

Equation (923) implies that

(924) |

(925) |

which reduces to

(926) |

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

(929) |

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) |

(932) |

(933) |

(934) |

(935) |

(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) |

(938) |

(939) |