Fine Structure

Let us examine a phenomenon known as *fine structure*, which caused by the
interaction between the spin and orbital angular momenta of the outermost
electron. This electron experiences an electric field [49]

(7.95) |

However, a non-relativistic charge moving in an electric field also experiences an effective magnetic field [49]

(7.96) |

Now, an electron possesses a spin magnetic moment

[See Equation (5.47).] We, therefore, expect a contribution to the Hamiltonian of the form [49]

where is the orbital angular momentum. This phenomenon is known as

Let us now apply perturbation theory to a hydrogen-like atom, using

as the perturbation (noting that takes one half of the value given previously), and

(7.100) |

as the unperturbed Hamiltonian. We have two choices for the energy eigenstates of . We can adopt the simultaneous eigenstates of and , or the simultaneous eigenstates of and , where is the total angular momentum. Although the departure of from a pure form splits the degeneracy of same , different , states, those states characterized by the same values of and , but different values of , are still degenerate. (Here, and are the quantum numbers corresponding to and , respectively.) Moreover, with the addition of spin degrees of freedom, each state is doubly degenerate because of the two possible orientations of the electron spin (i.e., ). Thus, we are still dealing with a highly degenerate system. However, we know, from Section 7.5, that there is no danger of singular terms appearing to second order in the perturbation expansion if the degenerate eigenstates of the unperturbed Hamiltonian (and the set of commuting operators needed to uniquely label the degenerate eigenstates) are also eigenstates of the perturbing Hamiltonian. Now, the perturbing Hamiltonian, , is proportional to , where

It is fairly obvious that the first group of operators ( and ) does not commute with , whereas the second group ( and ) does. In fact, is just a combination of operators appearing in the second group. Thus, it is advantageous to work in terms of the eigenstates of the second group of operators, rather than those of the first group (because the former eigenstates are also eigenstates of the perturbing Hamiltonian).

We now need to find the simultaneous eigenstates of and . This is equivalent to finding the eigenstates of the total angular momentum resulting from the addition of two angular momenta: , and . According to Equation (6.26), the allowed values of the total angular momentum are and . We can write

(7.102) | ||

(7.103) |

Here, the kets on the left-hand side are kets, whereas those on the right-hand side are kets (the labels have been dropped, for the sake of clarity). We have made use of the fact that the Clebsch-Gordon coefficients are automatically zero unless . (See Section 6.3.) We have also made use of the fact that both the and kets are orthonormal. We now need to determine

(7.104) |

where the Clebsch-Gordon coefficient is written in form.

Let us now employ the recursion relation for Clebsch-Gordon coefficients, Equation (6.32), with , , , , and , choosing the lower sign. We obtain

which reduces to

(7.105) |

We can use this formula to successively increase the value of . For instance,

This procedure can be continued until attains its maximum possible value, . Thus,

Consider the situation in which and both take their maximum values, and , respectively. The corresponding value of is . This value is possible when , but not when . Thus, the ket must be equal to the ket , up to an arbitrary phase-factor. By convention, this factor is taken to be unity, giving

(7.107) |

It follows from Equation (7.109) that

(7.108) |

Hence,

(7.109) |

We now need to determine the sign of . A careful examination of the recursion relation, Equation (6.32), shows that the plus sign is appropriate. Thus,

It is convenient to define so-called *spin-angular functions* using the
Pauli two-component formalism:

(See Section 5.7.) These functions are eigenfunctions of the total angular momentum for spin one-half particles, just as the spherical harmonics are eigenfunctions of the orbital angular momentum. A general spinor-wavefunction for an energy eigenstate in a hydrogen-like atom is written

The radial part of the wavefunction, , depends on the principal quantum number , and the azimuthal quantum number . The wavefunction is also labeled by , which is the quantum number associated with . For a given choice of , the quantum number (i.e., the quantum number associated with ) can take the values . (However, for the special case .)

The kets are eigenstates of , according to Equation (7.102). Thus,

(7.114) |

giving

(7.115) | ||

(7.116) |

It follows that

where the integrals are over all solid angle, .

Let us now apply degenerate perturbation theory to evaluate the energy-shift of a state whose spinor-wavefunction is caused by the spin-orbit Hamiltonian, . To first order, the energy-shift is given by

(7.119) |

where the integral is over all space, . Equations (7.100), (7.116), and (7.120)-(7.121) yield

where

Incidentally, for the special case of an state, , and there is no state with , so Equation (7.124) is redundant. Thus, it is clear that for an state, which is not surprising, given that such a state possesses zero orbital angular momentum (i.e., it is characterized by .)

Let us now apply the previous result to the case of a sodium atom. In chemist's notation [53], the ground state is written

(7.123) |

[Here, implies that there are two electrons in the state, et cetera.] The inner ten electrons effectively form a spherically symmetric electron cloud. We are interested in the excitation of the eleventh electron from to some higher energy state. The closest (in energy) unoccupied state is . This state has a higher energy than due to the deviations of the potential from the pure Coulomb form. In the absence of spin-orbit interaction, there are six degenerate states. The spin-orbit interaction breaks the degeneracy of these states. The modified states are labeled and , where the subscript refers to the value of . The four states lie at a slightly higher energy level than the two states, because the radial integral (7.125) is positive. (See Exercise 14.) The splitting of the energy levels of the sodium atom can be observed using a spectroscope (which measures the frequency of spectral lines caused by transitions between quantum states of different energy--this frequency is, of course, --see Section 8.9). The well-known