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Quantum Mechanics Chapter 6. Spin §6.1 The Spin Operators d2 2m l (l 1) 2 [r R(r )] 2 E V (r ) [rR (r )] (8.1) dr 2 2mr 2 • Following the method of Section 2.3, we represent the eigenstates of the spin operator as two- dimensional vectors. In the case of electron spin, there are only two eigenfunctions instead of three or more. • The Hilbert spin space is therefore two-dimensional and we can write the eigenfunctions as column vectors just as we did for orbital angular momentum; • the unit vectors that are eigenfunctions of Sz are the two-component vectors • Compare with Eq.(7.18) for the three eigenstates of Lz with l = 1. Following the reasoning given there, we say that any electron-spin function Fs can be a represented by a superposition of │+ >z and │- >z: • Continuing this line of reasoning, we can represent the spin operators Sx , Sy, and Sz as the following 2 × 2 matrices that operate on these vectors: • These may be written • where the dimensionless operators σx, σy, and σz (the Pauli spin matrices) are expressed by the matrices in Eq.(10.3). • These matrices can be derived by using • rules analogous to those of Eqs.(7.17): • It is left as an exercise to verify that these equations are indeed satisfied by the matrices given. You can also verify the anticommutation relation σxσy + σyσx = 0, and the corresponding relations involving σz. • Example Problem 10.1 Find values of the constants a and b that yield normalized eigenfunctions of Sx , by applying this operator to the general expression for a spin state: a│+ >z + b│- >z. – Solution. We know that the eigenvalues of Sx , must be the same as those of Sz . a • The statement that b is an eigenfunction of Sx with eigenvalue / 2 is expressed by the notation a a S x . b 2 b • Substituting the matrix for Sx and performing the a 0 1 a b S indicated matrix operation. we have b 2 1 0 b 2 a . x a b • Therefore b a . or b = a. Normalization requires that a b 1, so that a = b = l/(2)1/2. 2 2 a • Similarly, if is an eigenfunction with the b eigenvalue -ћ/2 , then S a 0 1 a b . x b 2 1 0 b 2 a a b • In this case . b a • or b = -a = -1/(2)1/2. 1 / 2 • In conclusion, the eigenvectors are 1 / 2 1/ 2 • and 1 / 2 respectively. §6.2 Fine Structure in The Hydrogen Spectrum • Effect of Spin on Energy Levels • The intrinsic magnetic moment of the electron, which results from its spin, leads to a splitting of the energy levels of hydrogen into sublevels, provided that there is a magnetic field in the atom, because in each of the states determined by the Schreodinger equation the electron would then have two possible va1ues for its magnetic potential energy, according to the value of the z component of its magnetic moment. • But why should there be a magnetic field in the hydrogen atom? • The only other particle present is the proton. One might expect the proton, like the electron to have internal spin and magnetic moment, and it does, but it turns out that the proton's intrinsic magnetic moment produces a field which is too small to account for the observed fine structure in the hydrogen spectrum. (The proton's magnetic moment leads to hyperfine structure, to be discussed later in this section.) • However, the proton's electric charge leads to the existence of a magnetic field of sufficient strength, • because in the electron's rest frame the proton moves in an orbit around the electron. To put in another way, a body moving through an electric field also “sees” a magnetic field, according to the theory of relativity. • Energy of the Spin-Orbit Interaction • Let us use the preceding idea to estimate the magnitude of the splitting of the hydrogen levels. First we must know the magnetic moment of the electron, which may be deduced from the magnitude of the splitting of the atomic beam in the Stern- Gerlach experiment. • This moment is very close to • μ= -(e/me)S. (or μ= -(e/mec)S in cgs units) (10.6) • where S is the spin angular momentum of the electron, and e is the magnitude of the electronic charge. Since Sz = ±ћ /2, the z component of μ is (in mks units) • μz = ±eћ /2me = ±1 Bohr magneton • [The Bohr magneton is often denoted by the symbol μb and simply called the magnetic moment of the electron. Strictly speaking, μb is not the electron's magnetic moment; it is only one component of that moment. However, one component is all that is observed in an experimental measurement of the magnetic moment of the electron.] • To determine the energy attributable to the existence of the electron's magnetic moment, we write the energy of a magnetic moment μ in a magnetic field B as W = -μ• B, and from Eq.(10.6) we have • W = -(e/me)S•B (10.7) • B may be related to the electron's orbital angular momentum in the following way: • If the velocity of the electron is v, the velocity of the proton relative to the electron is -v. The magnetic field B produced by the proton's charge e is therefore • This may be written in terms of the electron's angular momentum L = r × mv as eL B (10.9) 4r 3m 0c 2 • The interaction energy is then, in the electron's rest frame e2 W B S L (10.10) 4r m 0c 3 2 • Expression (10.10) is not the correct one to use for W, however, because the energy splitting seen in the lab is not the same as that seen in the electron's rest frame. • The reason for this is rather complicated, involving a phenomenon known as the Thomas precession, • which is related to the fact that the time scale is not the same in the laboratory as in the rest frame, of the electron. • The proper relativistic treatment shows that the energy shift seen in the laboratory is just one-half of the value given in Eq.(10.10), or • We can now estimate the value of W. We set ∣L∣≌ћ . For S, we know that the component in any direction, such as the direction of L must be ћ /2 in magnitude. If we then let r ≌ a0, we find W to be of the order of 10-4 eV, • which is of the order of magnitude required by the observed fine structure. We shall see in Chapter 8 that effects of this size may be calculated by the methods of perturbation theory. • It should come as no surprise to find that as a first approximation this energy can be assumed to be the expectation value of S·L, computing this value by using the eigenfunctions already found for the hydrogen atom; that is • where u is an eigenfunction of Eq.(9.10). We should therefore be able to express this energy in terms of the quantum numbers n, l, and m for these eigenfunctions, if we can evaluate the S·L operator; • that is, we need to know the result of the operation (S·L)opu. • Combining Spin and Orbital Angular Momentum • The S·L operator can be included in the operator for the total energy as it appears in the Schreodinger equation. • The energy is then found by finding the eigenvalues of the Schreodinger equation in this form, with the variables S and L represented as operators. Thus • where the function is labeled us to show that it is a function of the spin coordinate as well as the three space coordinates, because S·L must operate on it. • We may account for this spin dependence by assuming that us is some superposition of two wave functions ψ1 and ψ2 that differ in the value of the spin coordinate, as follows: • ψ1 = e-iωt u1(x, y, z)∣+ >s (10.14a) • ψ2 = e-iωt u1 (x, y, z)∣- >s (10.14b) • where the state vectors ∣+ >s and ∣- >s are the two eigenfunctions introduced in Eqs.(10.1). There we saw that the spin operator Sz operates on these functions to yield the two possible eigenvalues of Sz: • Now we must examine the effect of all three components of S, as they appear in the S • L operator. To proceed, we must introduce the total angular momentum vector J = L + S. • Because there are no external torques on the system the total angular momentum is constant during the motion, but neither L nor S is necessarily constant. • The B field caused by the electron's orbital motion exerts a torque on the spin magnetic moment; • when this magnetic moment changes direction, L must also change direction, in order that L + S remain constant. • In quantum mechanical terms, this means that the wave function cannot be either of the functions of Eqs.(10.14), which have fixed values of S. Rather, the wave function must be a superposition of these two functions. The vector S, of course, remains constant in magnitude, because this is a fundamental property of the electron, but the direction of S is not determined. • Having introduced J, we now can express S • L in terms of J, L, and S, using the equation • J2 ≡J·J = (L + S)·(L + S) = S2 + L2 + (S·L) + (L·S) (10.16) • But the L operator must commute with the S operator, because Lop and Sop operate on different coordinates. Therefore (S·L) = (L·S), and we can solve for S·L, obtaining • S·L = (J2 – S2 – L2)/2 (10.17) • The problem of evaluating (S·L) is now reduced to evaluating (J2 – S2 – L2). • We have already found [Eq.(6.58)] that the eigenvalues of L2 can be written in the form • L2 = l(l + 1)ћ2 (10.18) • where l is defined as the maximum value of m, or of Lz / ћ (Section 2.3). • By analog we might suspect that we could also write • S2 = s(s + 1)ћ2 (10.19) • and • J2 =j(j + l)ћ2 (10.20) • where s is defined as the maximum value of Sz /ћ and j is defined as the maximum value of Jz /ћ. • Because Sz =±ћ/2, the value of s must be l/2. This leads to a value for S2 that is consistent with what we know about orbital angular momentum: • From the isotropy of space, Sx2 = Sy2 = Sz2 = ћ2/4, and by definition, • S2 = Sx2 + Sy2 + Sz2 = 3ћ2/4 (10.21) the same result that we find from Eq.(10.19). • Notice the similarity of this discussion to the one that led to Eq.(6.58). • The only difference is that Eq.(6.58) involved expectation values (e.g.,<Lx2>). In the case of spin, Sx2 has only one possible value: <Sx2> = Sx2 in all cases. §6.3 Stepping Operators and Eigenvalues of J2 • The eigenvalues of J2 follow a similar pattern; we only have to determine the maximum value of Jz. We know that Jz = Lz + Sz and that Sz = ±ћ/2. • Thus it would seem clear that the maximum value of Jz must equal the maximum value of Lz plus the maximum value of Sz. • Apparently, therefore, j = (lћ + ћ/2)/ћ = l + 1/2, by definition. This is a possible value for j, but there is more to it. • The value of j is related to the eigenvalue of J2. which in turn is related to the value of L·S. • For a given value of l, there must be two possible values of L·S, because the spin vector always has two possible directions relative to any other direction (including the direction of the L vector). • One of these directions yields a positive value for L·S, the other a negative value of equal magnitude. This tells us that there must be two possible values of j. • Stepping Operators for Angular Momentum Sates • To put the discussion on a firmer basis, we now generate the eigenvalues of J2 by using the stepping- operator method that was introduced in Section 4.2 • for harmonic-oscillator eigenfunctions. In doing this, it is convenient to introduce the symbol [Lx, Ly] to represent the so-called commutator of Lx and Ly: the operator LxLy – LyLx. • It is easy to show (Exercise 6, Chapter 6) that the three operators of this type obey the commutation relations • [Lx, Ly] = ihLz, [Ly, Lz] = ihLx, [Lz, Lx] = ihLy (10.22) • Each relation can be generated from the previous one by the replacement of x by y, y by z, and z by x. • We can now derive the eigenvalues of J2 by making two reasonable assumptions: • 1. Similar commutation relations apply to all forms of angular momentum, so that [Sx, Sy]= iћSz, [Jx, Jy] = iћ Jz, and so on. • 2. We can find a function that is an eigenfunction of both J2 and Jz, with the eigenvalue aћ2 for J2 and eigenvalue bћ for Jz, where a and b are dimensionless. • Denoting this function by the symbol uab, we write • J2uab = aћ2uab (10.23) • Jzuab = bћuab (10.24) • Let us now introduce the operator J+ = Jx + iJy and demonstrate that the function J+uab is also an eigenfunction of Jz. • Applying Jz to J+uab gives us JzJ+uab = Jz(Jx + iJy)uab = (JzJx + iJzJy)uab (10.25) • We can apply Jz directly to uab by using the commutation relations in the form JzJx = JxJz + iћJy and JzJy = JyJz - iћJx (10.26) • Substituting from Eq.(10.26) into Eq.(10.25) produces JzJ+uab = (JxJz + iћJy + iJyJz +ћJx)uabv (10.27) • The substitution of Jzuab = bћuab from Eq.(10.24) now yields JzJ+uab = (Jxbћ + iћJy + iJybћ + ћJx)uab (10.28) • which reduces to • JzJ+uab = (Jx + iJy)(bћ + ћ)uab =(b + 1)ћJ+uab (10.29) • Thus we have proven that the function J+uab is an eigenfunction of Jz, with eigenvalue (b+l)ћ. • Following the same procedure, we could prove that J-uab, where J- = Jx – Jy, is an eigenfunction of Jz with eigenvalue (b - l)ћ. • Therefore we can generate from uab a set of eigenfunctions whose eigenvalues are ћ multiplied by b,b ± 1, b ± 2,…… Each of these functions is also an eigenfunction of J2, as you may verity by applying the J2 operator to the function J+uab, then applying the commutation relations and Eqs.(10.23) and (10.24). • This set of eigenfunctions cannot be infinite, because we know that no eigenvalue of Jz2 can be greater than the eigenvalue of J2, which equals aћ2 for every eigenfunction in the set. • Thus there must be a largest eigenvalue of Jz, which we denote as jћ. We write the eigenfunction corresponding to this eigenvalue as uaj; that means that Jzuaj = jћuaj. • It is clear that • J+uaj = 0 (10.30) • for otherwise J+uaj would be an eigenfunction of Jz with eigenvalue (j + 1)ћ, in contradiction to the definition of jћ as the largest eigenvalue. • Let us now apply the operator (Jx - iJy) to this function to obtain (Jx - iJy)(Jx + iJy)uaj = 0 (10.31) • or [Jx2 + Jy2 +i(JxJy – JyJx)]uaj = 0 (10.32) • The commutation rule for Jx and Jy shows that this can be written as (Jx2 + Jy2 - ћJz)uaj = 0 (10.33) • or, since Jzuaj = jћuaj (Jx2 + Jy2 - jћ2)Uaj = 0 (10.34) • Adding (Jz2 +jћ2)uaj to both sides, we have • (Jx2 + Jy2 + Jz2)uaj = (Jz2 +jћ2)uaj (10.35) • which reduces to J2uaj = j(j + 1)ћ2uaj (10.36) • In summary, we have shown directly from the assumed commutation relations that the eigenvalues of J2 are j(j + 1)ћ2 and that the states with a given value of j form a set for which the eigenvalue of Jz is one of the sequence +jћ, +(j - 1)ћ,..., -jћ. • Thus we may write Jz = mjћ, where mj has a maximum value of j and a minimum value of -j. • Since the difference between any two values of mj must be an integer and the difference between j and -j is 2j, the value of j must be either an integer or a half-integer. • This derivation is based solely upon the commutation relations; nothing in the derivation depends upon the nature of the particle involved. For a single electron, the fact that s = 1/2 makes j a half- integer; for a particle whose spin is an integer, j would also be an integer. • Now we must determine the general relation between the value of j and the value of s. We do this by referring to the z components of the vectors J, L, and S. Since J = L + S we know that Jz = Lz + Sz. • The eigenvalue relation is mjћ = mlћ + msћ, so that mj = ml + ms, or • mj = ml ± 1/2 (10.37) • The maximum value of mj is j. so mj ≤ l + 1/2 (10.38) • and there must be a set of states for which mj takes on the values mj = l + 1/2, l – 1/2, l - 3/2,...,-(l + 1/2) (j = l + ½) (10.39) • and for every state in this set the value of j is l + 1/2, This means that the magnitude of J is greater than the magnitude of L for each of these states, and we deduce that S·L is positive for each of these states. • There must also be states for which S·L is negative; these form a set for which j equals l – 1/2, • and in this set • mj = l – 1/2,l – 3/2,...,-(l – 1/2) (j = l – 1/2) (10.40) • In both of these sets we find values of mj in the range l – 1/2 ≥ mj ≥ -(l – 1/2). Thus there is only one state for which mj = l + 1/2, only one state for which mj = -(l + ½), and two states for each of the other possible values of mj. • Let us now examine in detail the distinction between two states with different j and the same mj. • Consider the case mj = l – 1/2. This value of mj can be produced in two ways; either ml = l - 1 and the spin is positive, or ml = l and the spin is negative. • In Dirac notation (as presented in Section 2.4) we would write these states as ∣l,l – 1>∣+ >s and ∣l,l>∣->s, respectively. • These two states are of course orthogonal to each other. Neither of them is an eigenstate of the J2 operator, but we can form two linear combinations of them, such that one combination is an eigenstate of the J2 with eigenvalue j = l + 1/2, and the other combination is an eigenstate of the J2 with eigenvalue j = l – 1/2. • As illustrated in the following example, S·L is positive when j = l + 1/2, and S·L is negative when j = l – 1/2. • Example Problem 10.2 Find the eigenvalue of J2 for each of the two independent linear combinations of ∣1,0>∣+ >s and ∣1,1>∣- >s that are eigenfunctions of J2. • Solution. Applying the S·L operator to the general expression F = a∣1,0>∣+ >s + b∣1,1>∣- >s gives us the expression ∣SxLx + SyLy + SzLz[a∣1,0>∣+ >s + b∣1,1>∣- >s]. • Using the spin operators [Eqs.(10.3)] combined with the operators for the components of L [Eqs.(7.19)] we find that the individual terms are • SxLx[a∣1,0>∣+ >s + b∣1,1>∣- >s] = ћ2/2(2)1/2[a∣1,1>∣- >s + a∣1,-1>∣- >s + b∣1,0>∣+ >s] • SyLy[a∣1,0>∣+ >s + b∣1,1>∣- >s] = ћ2/2(2)1/2[a∣1,1>∣- >s - a∣1,-1>∣- >s + b∣1,0>∣+ >s] • SzLz[a∣1,0>∣+ >s + b∣1,1>∣- >s = ћ2/2[-b∣1,1>∣- >s] • This becomes, after adding and grouping terms: S·LF = ћ2/(2)1/2[a∣1,1>∣- >s + b∣1,0>∣+ >s] –ћ2/2[b∣1,1>∣- >s] = Aћ2[a∣1,0>∣+ >s + b∣1,1>∣- >s] • where Ah2 is the assumed eigenvalue, A being a dimensionless number. • Equating corresponding coefficients we have a/(2)1/2 – b/2 = Ab and b/(2)1/2 = Aa (10.41) • We can solve these to find two roots for A: A = + 1/2 or A = -l. • Therefore the eigenvalues of S·L are + ћ2/2 and - ћ2. Since L2 = 2ћ2 and S2 = 3ћ2/4, J2 = L2 + S2 + 2S·L= 15ћ2/4 (when S·L = + ћ2/2), and J2 = 3ћ2/4 (when S·L = - ћ2). • Notice that these values agree with the formula J2 = j(j + 1)ћ2 with j = 3/2 in the first case and j = l/2 in the second case. To summarize: • 1. There are two sets of states for a given value of l, when s = 1/2; these have the values of ml given in Eqs.(10.39) and (10.40). • 2. There are two independent states for each value of mj from mj = l - (l/2) down to mj = -[l - (l/2)]. • 3. If one of these states is an eigenstate of J2, that state is not an eigenstate of either Lz or Sz, but it is a linear combination of the two eigenstates ∣l,ml>∣- >s and ∣l,ml – 1>∣+ >s (as illustrated in the preceding example). • 4. In general. the quantities Lz, Sz, and J2 are not simultaneously observable, because the J2 operator does not commute with the Lz and Sz operators. • FIGURE 10.l (a) State in which Lz and Sz are known J2 is unknown. because the direction of L relative to S is unknown. (b) State in which J2 is known. the values of Lz and Sz are not known. because S and L rotate around J. • Figure 10.1 illustrates this point, Figure 10.1a shows the case in which L2, Lz, S2, and Sz are all known. Each of the vectors S and L lies on a cone( whose axis is the z axis, but the position of the vector on its cone is completely undetermined. • Therefore the direction of S relative to L is unknown, S·L is unknown, and J2 must be unknown. • In contrast, Figure 10.1b shows the situation in which J2, L2, S2, and Jz are known. In this case S·L must be known; thus we know that the angle between S and L must be fixed, making S lie on a cone whose axis is the direction of L. • Observe that the triangle formed by S, L, and J is completely determined in size and shape, but this triangle rotates about the direction of J. Thus Lz and Sz are undetermined, but J2 and Jz are known. • Evaluation of the Spin-Orbit Energy • We now have the necessary information to evaluate the spin-orbit energy. Let us generalize Eq.(10.12) to the case of a nuclear charge of Ze in a one-electron ion. Then we have • Using Eq.(1.17) for the operator, we have Ze2 W S L u* ( J 2 S 2 L2 ) op udxdydz 16 0 r 3m 2 c 2 (10.43) • If u is an eigenfunction of J2, S2, and L2, we may write • (J2 - S2 - L2)opu = ћ2[j(j + 1) - l(l + l) - s(s + 1)]u (10.44) • And after setting s equal to 1/2, we have • The angular part of this integral is equal to l, • because of the normalization of the spherical harmonics. • It is a simple matter to evaluate the radial part for any particular state, by plugging in the appropriate radial function (see Table 9.1 or Appendix G). • Evaluating the integral for the general case is more difficult; the result (for l ≠ 0) is • where a’0 is defined as a’0=a0(me+M)/M. • It is instructive to write W in terms of the energy of the unperturbed state [En = -mrc2Z2α2/2n2, where α= e2/4πε0hc ≌ 1/137 (the finestructure constant)]: • Since j equals either l + 1/2 or l - 1/2, we can write this as two expressions: • Thus ∣W∣is proportional to ∣En∣α2, while ∣En∣ itself is proportional to mc2α2. • Thus we could say that we have the first three terms in an expansion of the total energy in a power series in α2. • However, there is another effect, a relativistic contribution to the energy, that is comparable to the effect just computed. • This effect is a consequence of the fact that the kinetic energy of a particle of mass m and momentum p is not accurately given by the expression p2/2m, as had been assumed in constructing the Schreodinger equation. • With both effects included, the energy shift ΔE is given by a single formula • Notice that the total energy depends on n and j only, and not on l. (The calculation involves an approximation that we will discuss in Chapter 7.) • For this reason, the labeling of states includes a subscript giving the value of j. For example, 2s1/2 is the state with n = 2, l = 0, and j = 1/2, and 2p1/2 is the state with n = 2, l = 1, and j = 1/2. • Hyperfine Structure • The proton and neutron also have intrinsic spin and magnetic moment. These are observable because of the energy shift resulting from the interaction of this magnetic moment and the orbital and spin magnetic moments of the electron. • Although the energy involved is tiny, compared to the S·L effect, it has clearly been observed. The magnitude of this “hyperfine” energy splitting can be estimated theoretically by an extension of the methods used above. • In general, an atom has a total angular momentum that can be defined as the vector F = J + I, where I is the internal angular momentum of the nucleus (analogous to S for the electron). • In the 1H atom, I is simply the spin angular momentum of the proton, which is equal to the spin angular momentum of the electron, with two possible values, ±ћ/2, • of the component along any specified direction. • In all other nuclei, I is the vector sum of the angular moment of all of the constituent protons and neutrons. These vectors include spin angular momentum as well as angular momentum resulting from orbital motions of these particles within the nucleus. • The vectors F and I obey the same rules that apply to all angular momentum vectors. • The z component of the vector I is Iћ , and the possible values of I2 are I(I + 1)ћ2. The z component of the vector F is Fћ . and the possible values of F2 are F(F + 1)ћ2. • The quantum number F takes on integrally spaced values between J + I and │J - I│. The magnetic energy resulting from the magnetic moment of the nucleus is proportional to I·J, and this energy has as many possible values as there are values of F— namely, either 2I + 1 or 2J + 1, whichever is smaller. (You may verify this for yourself.) • The energy splitting is so tiny (or hyperfine) because the magnetic moment μp of the proton is very small compared to that of the electron; we find that • μp = 2.79 eћ /2Mp (whereas μB = eћ /2me >> eћ /2Mp) (10.50) • The quantity eћ/2Mp is called the nuclear magneton (μN). • [The magnetic moment of the neutron is smaller than the proton's; its magnitude is 1.93 μN. You might wonder why a neutron, being neutral, would have any magnetic moment at all. • But the neutron is not uniformly neutral; it is composed of other particles (three quarks) that are charged-two of them with a charge of +e/3, one with a charge of -2e/3. It is the motion and spin of these charges that produces the neutron's magnetic moment.] • Because the nuclear magneton is so small relative to the Bohr magneton, hyperfine structure is much more difficult to resolve than fine structure • However, the nuclear magnetic moment is responsible for the splitting of the 1s level of hydrogen into two levels which differ in energy by about 6×10-6eV (considerably smaller than the fine- structure splitting, which is of order 10-4eV). • The wavelength of protons of this energy is 21 cm; this wavelength is quite prominent in radiation from outer space, and it had made it possible to map the distribution of atomic hydrogen in our galaxy. • The agreement between theory and experiment over the energy range from 10 eV to less than 10-5eV is impressive, but even more impressive results are obtained • by including effects of relativity and quantization of the energy levels of the electromagnetic field. • Calculations have shown that the 2s1/2 and 2p1/2 levels are not equal, as Eq.(10.48) implies. Instead, they should be split by 4.3743×10-6eV; this result (the Lamb shift) has been confirmed by experiment..
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