# Principles of Quantum Mechanics as Applied to Chemistry and Chemical Physics

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6 0 (7:34) [S^z, L^ : S^] [S^z, S^x]L^x [S^z, S^ y]L^y i"(L^xS^ y ÿ L^ yS^x) 6 0 (7:35) where equations (5.10) and (7.2) have been used. Similar expressions apply to the other components of L^ and S^. Thus, the vectors L and S are no longer 7.4 Spin\u2013orbit interaction 203 constants of motion. However, the operators L^2 and S^2 do commute with L^ : S^, which follows from equations (5.15), so that the quantities L2 and S2 are still constants of motion. We now introduce the total angular momentum J, which is the sum of L and S J L S (7:36) The operators J^ and J^ 2 commute with H^0. The addition of equations (7.34) and (7.35) gives [J^z, L^ : S^] [L^z, L^ : S^] [S^z, L^ : S^] 0 The addition of similar relations for the x- and y-components of these angular momentum vectors leads to the result that [J^, L^ : S^] 0, so that J^ and L^ : S^ commute. Furthermore, we may easily show that J^ 2 commutes with L^ : S^ because each term in J^ 2 L^2 S^2 2L^ : S^ commutes with L^ : S^. Thus, J^ and J^ 2 commute with H^ in equation (7.33) and J and J2 are constants of motion. That the quantities L2, S2, J 2, and J are constants of motion, but L and S are not, is illustrated in Figure 7.1. The spin magnetic moment Ms, which is antiparallel to S, exerts a torque on the orbital magnetic moment M, which is antiparallel to L, and alters its direction, but not its magnitude. Thus, the orbital angular momentum vector L precesses about J and L is not a constant of motion. However, since the magnitude of L does not change, the quantity L2 is a constant of motion. Likewise, the orbital magnetic moment M exerts a torque on Ms, causing S to precess about J. The vector S is, then, not a constant of S J L Figure 7.1 Precession of the orbital angular momentum vector L and the spin angular momentum vector S about their vector sum J. 204 Spin motion, but S2 is. Since J is fixed in direction and magnitude, both J and J 2 are constants of motion. If we form the cross product J^ 3 J^ and substitute equations (7.36), (5.11), and (7.3), we obtain J^ 3 J^ (L^ S^) 3 (L^ S^) (L^ 3 L^) (S^ 3 S^) i"L^ i"S^ i"J^ where the cross terms (L^ 3 S^) and (S^ 3 L^) cancel each other. Thus, the operator J^ obeys equation (5.12) and the quantum-mechanical treatment of Section 5.2 applies to the total angular momentum. Since J^x, J^ y, and J^z each commute with J^ 2 but do not commute with one another, we select J^z and seek the simultaneous eigenfunctions jnlsjm ji of the set of mutually commuting operators H^ , L2, S2, J2, and J^z H^ jnlsjmji Enjnlsjmji (7:37a) L^2jnlsjmji l(l 1)"2jnlsjmji (7:37b) S^2jnlsjmji s(s 1)"2jnlsjmji (7:37c) J^ 2jnlsjmji j( j 1)"2jnlsjmji (7:37d) J^zjnlsjmji mj"jnlsjmji, mj ÿ j, ÿ j 1, . . . , jÿ 1, j (7:37e) From the expression J^zjnlsjmji (L^z S^z)jnlsjmji (m ms)"jnlsjmji obtained from (7.36), (5.28b), and (7.5), we see that mj m ms (7:38) The quantum number j takes on the values l s, l sÿ 1, l sÿ 2, . . . , jl ÿ sj The argument leading to this conclusion is somewhat complicated and may be found elsewhere.3 In the application being considered here, the spin s equals 1 2 and the quantum number j can have only two values j l \ufffd 1 2 (7:39) The resulting vectors J are shown in Figure 7.2. The scalar product L^ : S^ in equation (7.33) may be expressed in terms of operators that commute with H^ by L^ : S^ 1 2 (L^ S^) : (L^ S^)ÿ 1 2 L^ : L^ÿ 1 2 S^ : S^ 1 2 (J^ 2 ÿ L^2 ÿ S^2) (7:40) 3 B. H. Brandsen and C. J. Joachain (1989) Introduction to Quantum Mechanics (Addison Wesley Longman, Harlow, Essex), pp. 299, 301; R. N. Zare (1988) Angular Momentum (John Wiley & Sons, New York), pp. 45\u20138. 7.4 Spin\u2013orbit interaction 205 so that H^ becomes H^ H^0 12î(r)(J^ 2 ÿ L^2 ÿ S^2) (7:41) Equation (7.37a) then takes the form f H^0 12"2î(r)[ j( j 1)ÿ l(l 1)ÿ s(s 1)]gjnlsjmji Enjnlsjmji (7:42) or H^0 l" 2 2 î(r) \ufffd \ufffd jn, l, 1 2 , l 1 2 , mji Enjn, l, 12, l 12, mji if j l 12 (7:43a) H^0 ÿ (l 1)" 2 2 î(r) \ufffd \ufffd jn, l, ÿ1 2 , l ÿ 1 2 , mji Enjn, l, ÿ12, l ÿ 12, mji if j l ÿ 1 2 (7:43b) where equations (7.37b), (7.37c), (7.37d), and (7.39) have also been intro- duced. Since the spin\u2013orbit interaction energy is small, the solution of equations (7.43) to obtain En is most easily accomplished by means of perturbation theory, a technique which is presented in Chapter 9. The evaluation of En is left as a problem at the end of Chapter 9. Problems 7.1 Determine the angle between the spin vector S and the z-axis for an electron in spin state jÆi. 7.2 Prove equation (7.19) from equations (7.15) and (7.17). S L J S L J j 5 l 1 j 5 l 2 1 2 1 2 Figure 7.2 The total angular momentum vectors J obtained from the sum of L and S for s 1 2 and s ÿ1 2 . 206 Spin 7.3 Show that the pair of operators ó y, ó z anticommute. 7.4 Using the Pauli spin matrices in equation (7.25) and the spinors in (7.13), (a) construct the operators ó and óÿ corresponding to S^ and S^ÿ (b) operate on jÆi and on jâi with ó 2, ó z, ó, óÿ, ó x, and ó y and compare the results with equations (7.14), (7.15), (7.16), and (7.17). 7.5 Using the Pauli spin matrices in equation (7.25), verify the relationships in (7.19) and (7.22). Problems 207 8 Systems of identical particles The postulates 1 to 6 of quantum mechanics as stated in Sections 3.7 and 7.2 apply to multi-particle systems provided that each of the particles is distin- guishable from the others. For example, the nucleus and the electron in a hydrogen-like atom are readily distinguishable by their differing masses and charges. When a system contains two or more identical particles, however, postulates 1 to 6 are not sufficient to predict the properties of the system. These postulates must be augmented by an additional postulate. This chapter intro- duces this new postulate and discusses its consequences. 8.1 Permutations of identical particles Particles are identical if they cannot be distinguished one from another by any intrinsic property, such as mass, charge, or spin. There does not exist, in fact and in principle, any experimental procedure which can identify any one of the particles. In classical mechanics, even though all particles in the system may have the same intrinsic properties, each may be identified, at least in principle, by its precise trajectory as governed by Newton\u2019s laws of motion. This identification is not possible in quantum theory because each particle does not possess a trajectory; instead, the wave function gives the probability density for finding the particle at each point in space. When a particle is found to be in some small region, there is no way of determining either theoretically or experimentally which particle it is. Thus, all electrons are identical and there- fore indistinguishable, as are all protons, all neutrons, all hydrogen atoms with 1H nuclei, all hydrogen atoms with 2H nuclei, all helium atoms with 4He nuclei, all helium atoms with 3He nuclei, etc. Two-particle systems For simplicity, we first consider a system composed of two identical particles 208 of mass m. If we label one of the particles as particle 1 and the other as particle 2, then the Hamiltonian operator H^(1, 2) for the system is H^(1, 2) p^ 2 1 2m p^ 2 2 2m V (q1, q2) (8:1) where qi (i 1, 2) represents the three-dimensional (continuous) spatial coordinates ri and the (discrete) spin coordinate ó i of particle i. In order for these two identical particles to be indistinguishable from each other, the Hamiltonian operator must be symmetric with respect to particle interchange,