Let us now consider the energy levels of hydrogen-like atoms (i.e., alkali metal atoms) in more detail. The outermost electron moves in a spherically symmetric potential due to the nuclear charge and the charges of the other electrons (which occupy spherically symmetric closed shells). The shielding effect of the inner electrons causes to depart from the pure Coulomb form. This splits the degeneracy of states characterized by the same value of , but different values of . In fact, higher states have higher energies.
Let us examine a phenomenon known as fine structure, which is due to interaction between the spin and orbital angular momenta of the outermost electron. This electron experiences an electric field
However, a non-relativistic charge moving in an electric field also experiences an effective magnetic field
Now, an electron possesses a spin magnetic moment [see Equation (462)]
We, therefore, expect a spin-orbit contribution to the Hamiltonian of the form
where is the orbital angular momentum. Actually, when the above expression is compared to the observed spin-orbit interaction, it is found to be too large by a factor of two. There is a classical explanation for this, due to spin precession, which we need not go into. The correct quantum mechanical explanation requires a relativistically covariant treatment of electron dynamics (this is achieved using the so-called Dirac equation--see Chapter 11).
Let us now apply perturbation theory to a hydrogen-like atom, using as the perturbation (with taking one half of the value given above), and
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 due to the two possible orientations of the electron spin (i.e., ). Thus, we are still dealing with a highly degenerate system. We know, from Section 7.6, that the application of perturbation theory to a degenerate system is greatly simplified if the basis eigenstates of the unperturbed Hamiltonian 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.
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 (568), the allowed values of the total angular momentum are and . We can write
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 . We have also made use of the fact that both the and kets are orthonormal, and have unit lengths. We now need to determine
where the Clebsch-Gordon coefficient is written in form.
Let us now employ the recursion relation for Clebsch-Gordon coefficients, Equation (574), with (lower sign). We obtain
which reduces to
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
It follows from Equation (691) that
We now need to determine the sign of . A careful examination of the recursion relation, Equation (574), shows that the plus sign is appropriate. Thus,
It is convenient to define so called spin-angular functions using the Pauli two-component formalism:
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 wavefunction for an energy eigenstate in a hydrogen-like atom is written
The radial part of the wavefunction, , depends on the radial quantum number and the angular 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 .
The kets are eigenstates of , according to Equation (684). Thus,
It follows that
where the integrals are over all solid angle, .
Let us now apply degenerate perturbation theory to evaluate the shift in energy of a state whose wavefunction is due to the spin-orbit Hamiltonian, . To first order, the energy-shift is given by
where the integral is over all space, . Equations (682) (remembering the factor of two), (698), and (702)-(703) yield
Let us now apply the above result to the case of a sodium atom. In chemist's notation, the ground state is written
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 (707) is positive. The splitting of the energy levels of the sodium atom can be observed using a spectroscope. The well-known sodium D line is associated with transitions between the and states. The fact that there are two slightly different energy levels (note that spin-orbit coupling does not split the energy levels) means that the sodium D line actually consists of two very closely spaced spectroscopic lines. It is easily demonstrated that the ratio of the typical spacing of Balmer lines to the splitting brought about by spin-orbit interaction is about , where
is the fine structure constant. Actually, Equations (705)-(706) are not entirely correct, because we have neglected an effect (namely, the relativistic mass correction of the electron) that is the same order of magnitude as spin-orbit coupling. (See Exercises 3 and 4.)
- Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)