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0 s —> 00, as we did in Section 78 for the magnitude of the orbital angular momentum
L by letting its quantum number 1—» 00. An equivalent statement is that in the
classical limit the magnitude of S is completely negligible because it is so small, so
spin is essentially nonclassical. This being the case, it is sometimes more harmful than
helpful to think of spin in terms of a classical model like a small spinning sphere; but
it must be admitted that it is difﬁcult to avoid thinking in such terms. 84 THE SPINORBIT INTERACTION Although spin itself is subtle, there is nothing subtle about many of the effects it
produces. Perhaps the most important is that it doubles the number of electrons
which the “exclusion principle” allows to populate the quantum states of multi
electron atoms. When we study this effect in Chapter 10, we shall see that the ground
states of atoms would be very much altered if electrons did not have spin. This would
have profound consequences on the periodic properties of atoms, and therefore on all
of chemistry and solid state physics. In the present section we shall study the interaction between an electron’s spin
magnetic dipole moment and the internal magnetic ﬁeld of a oneelectron atom. Since
the internal magnetic ﬁeld is related to the electron’s orbital angular momentum, this
is called the spinorbit interaction. It is a relatively weak interaction which is respon
sible, in part, for the ﬁne structure of the excited states of oneelectron atoms. The spinorbit interaction also occurs in multielectron atoms, but in such atoms it
is reasonably strong because the internal magnetic ﬁelds are very strong. Further
more, an effect completely analogous to the spinorbit interaction occurs in nuclei.
The nuclear spinorbit interaction is so strong that it governs the periodic properties
of nuclei. . The origin of the internal magnetic ﬁeld experienced by an electron moving in a
oneelectron atom is easy to understand if we consider the motion of the nucleus
from the point of view of the electron. In a frame of reference ﬁxed on the electron,
the charged nucleus moves around the electron and the electron is, in effect, located
inside a current loop which produces the magnetic ﬁeld. The argument is illustrated
qualitatively in Figure 87. To make the argument quantitative, we note that the
charged nucleus moving with velocity —v constitutes a current element j, where j= —Zev According to Ampere’s law, this produces a magnetic ﬁeld B which, at the position
of the electron, is ojxr Zeuovxr
B: =_ 3 7tr3 41: r _v‘ Figure 87 Left: An electron moves in a circular Bohr orbit, the motion as seen by the
nucleus. Right: The same motion. but as seen by the electron. From the point of view 0‘
the electron. the nucleus moves around it. The magnetic field B experienced by the
electron is in the direction out of the page at the electron’s location. It is convenient to express this in terms of the electric ﬁeld E acting on the electron.
According to Coulomb’s law From the last two equations, we have
B = *50140" x E 1
B=——c—2vxE (824)
since c = l/‘leouo. The quantity B is the magnetic ﬁeld strength experienced by the
electron when it is moving with velocity v relative to the nucleus, and therefore
through the electric ﬁeld of strength E which the nucleus exerts on it. Equation (824)
is actually of very general validity, and it can be derived from relativistic considera
tions. The electron and its spin magnetic dipole moment can assume diﬂ‘erent orienta
tions in the internal magnetic ﬁeld of the atom, and its potential energy is different
for each of these orientations. If we evaluate the orientational potential energy of the
magnetic dipole moment in this magnetic ﬁeld, from an equation analogous to (813),
we have AE= —u,B Using (819), this can be written in terms of the electron’s spin angular momen
tum S as AE=9”%sB
h
But this energy has been evaluated in a frame of reference in which the electron
is at rest, whereas we are interested in the energy as measured in the normal frame
of reference in which the nucleus is at rest. Because of an effect of the relativistic trans
formation of velocities, called the Thomas precession, the transformation back to the
nuclear rest frame results in a reduction of the orientational potential energy by a
factor of 2. Thus, the spinorbit interaction energy is Mm
AE = — S  B 
2 h (8 25)
The transformation leading to the factor of 2 is interesting, but rather complicated,
so we shall not carry it out here. (It is carried out in Appendix 0.)
We shalt ﬁnd it convenient to express (825) in terms of S  L, the scalar, or dot, product of the spin and orbital angular momentum vectors. To this end, we use, in
(824), the relation 4m=F between the electric ﬁeld E and the force F acting on the electron of charge —e.
We also use the relation _ dV(r) r
dr r between the force and the potential. (The term r/r is a unit vector in the radial direc
tion which gives F its proper direction.) With these relations, (824) becomes Imwn
«ac2 r dr F: B: vxr NOllOVHEILNI ilSHO—NldS 3H1 613 o
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o Multiplying and dividing by the electron mass m allows us to write this in terms of
the orbital angular momentum, L = r x mv = —mv x r, as 1 ldV(r)
emc2 r dr Note that the strength of the magnetic ﬁeld B, experienced by the electron because
it is moving about the nucleus with orbital angular momentum L, is proportional
to the magnitude of L, and also that the magnetic ﬁeld vector is in the same direction
as the angular momentum vector. With this result, we can express the spinorbit
interaction energy, (825), as B = L (826) AE _ an, 1dV<r) _ 2emc2h ? dr SiL Evaluating g5 and a,” we obtain _ 1 1dV(r)
AE— 2m2c2? dr S‘L (8—27) This equation was ﬁrst derived in 1926 by Thomas, using as we have a combination
of the Bohr model, Schroedinger quantum mechanics, and relativistic kinematics.
However, it is in complete agreement with the results of the relativistic quantum
mechanics of Dirac. It is important in the theory of multielectron atoms as well as of oneelectron atoms. Furthermore, a similar equation is central to the understand
ing of the theory of the structure of nuclei, as we shall see later in the book. Example 83. Estimate the magnitude of the orientational potential energy AE'for the n = 2,
I = 1 state of the hydrogen atom, to check whether it is of the same order of magnitude as the
observed ﬁnestructure splitting of the corresponding energy level. (There is no spinorbit en ergy in the n = 1 state, since for n = 1 the only possible value for lis l = 0, which means L = 0.)
>The potential is 82 V = — ‘1
(r) 41t€0 r dV(r) _ e2 _2
dr _ 47:60 r e2 1
AE = — — S  L
47:602m2c2 r3
The magnitude of S  L is approximately hz since each of these angular momentum vectors
has a magnitude of approximately in. The expectation value of l/r3 for the n = 2 state 15
approximately 1/(3a0)3. Thus . e2 1 m3e6 2 me8 47:602m2c2 3—3(41r€0)3h6 = 54 x (47teo)4c2h4
(9 x 109 nt—mz/coulz)4 x 9 x 10~31 kg x (1.6 x 10‘19 coul)a
54 x (3 x 108 m/sec)2 x (1.1 x 10‘34joule—sec)4
~10‘23jou1e ~10'4 eV
Since S  L can be either positive or negative, depending on the relative orientation of the two
vectors, the energy level is split by roughly 2 x 10‘4 eV. Comparing this with the energy of the n = 2, l = 1 level of hydrogen, E2 = — 3.4 eV. we see
that the ratio of the predicted energy splitting to the energy itself, AE/E, is about one part
in 104. This is in reasonable agreement with the splitting required to explain the ﬁne structure
of the lines of the hydrogen spectrum associated with this level, as discussed in Section 410. and therefore it provides some conﬁrmation of the theory we have developed. A more details?
comparison of the theory with experiment will be made shortly. IAEI ; ...
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 Winter '07
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 Physics

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