5.73 F2018 Lecture 35: Spin-Orbit

5.73 Lecture #35 35 - 1 SPIN - ORBIT: Many-Electron ς(N, L, S) ↔︎ Single-Orbital ςnℓ Coupling Constants LAST TIME: ∑ e 2 rij i≥j death of orbital picture expansion of 1/rij: multipoles, integrals over AOs in nucleus-centered coordinates SELECTION RULES: orbital and many-e– basis sets k k k Gaunt Coefficients: a ,b , c [products of 3-j coefficients] Slater-Condon (Fk, Gk)⟶(Fk, Gk)Parameters Sum Rule Method - avoid necessity to derive: *eigenvectors * off-diagonal elements in Slater basis Hund’s 1st and 2nd Rules → predict lowest L-S term of configuration based on rapid decrease in size of Fk as k increases A single-configuration pattern of assigned L-S terms. For nf2 there are 91 Slater determinants. To use the sum rule method to get the relative energies of all 7 L-S terms (1I, 3H, 1G, 3F, 1D, 3P, 1S), it turns out it is necessary to compute 28 diagonal matrix elements in the f2 Slater determinantal basis set. This is wonderful because it is unnecessary to compute any offdiagonal matrix elements (see page 34-8). These L-S term energies are expressed in terms of F0, F2, F4, and F6 (and Gk) Slater-Condon parameters. TODAY: A. General Importance of spin-orbit term HSO = ∑ a(ri )ℓ̂ 1ŝ1 1 − e − operator i B. Trick: replace 1–e– operator by more convenient �(N,L,S)L · S for ∆S = 0, ∆L = 0 special case matrix elements C. Pattern: Landé Interval Rule (Patterns are for breaking! Broken patterns provide information about “dark” states) 1 5.73 Lecture #35 D. 35 - 2 HSO matrix elements in Slater Determinantal Basis Set * another operator replacement * A single-orbital integral is the most fundamental control parameter: n,ℓ-scaling of �nℓ * �nℓ⟷�(N, L, S) relationship between a single-orbital coupling constant and that for a specific L-S state * off-diagonal spin-orbit matrix elements: ∆L ≠ 0, ∆S ≠ 0, ∆J = 0. PATTERN BREAKING next time → Hund’s 3rd Rule and Lande gJ–factors A. Importance of spin-orbit 1. HSO produces diagnostically significant “fine structure” CONFIGURATIONAL ASSIGNMENTS (based on which L-S terms are present and the ± signs of spin-orbit splittings) L,S term assignments are based on PATTERNS:* # components * sign of pattern (largest splitting on top or on bottom) * statistical weight (2J + 1) of lowest vs. highest energy component * overall magnitude of splitting SO 2. for heavy atoms, H is responsible for such large splittings and off-diagonal interactions that L-S terms “vanish”, ∆S selection rules are violated. “Inter-System Crossing (ISC)”. Need to “deperturb” to recover Fk, Gk inter-electronic (1/rij )parameters which should vary smoothly from atom to atom (isoelectronic series) (shielding rules). Periodicity! The PERIODIC TABLE 2 5.73 Lecture #35 35 - 3 3. Spin-forbidden transitions provide energy linkages between manifolds of states with different values of S. “InterSystem Crossing (ISC)”, e.g. Hg 3P1← 1S0 254nm 4. Non-textbook Zeeman tuning coefficients (clues about unobserved “dark” states) finer detail to be used after S-O patterns reduce the possibilities that must be considered. Atoms, Molecules, Quantum Dots, solids: in an electronic transition, light acts on a single e– and operates exclusively on the spatial part SO of ψ → spin-flips are forbidden except when H mixes states of different S — forbidden transitions “borrow” intensity from allowed transitions. In the time-domain: a short pulse prepares, at t = 0, a non-eigenstate that is a pure ∆S = 0 excitation (and ∆ℓ = ±1) basis state. The “pluck”! Language: the name of each eigenstate is based on its dominant (i.e., “nominal”) character. It is the name of the dominant basis state. Use of the same name for both eigenstate and basis state is a source of confusion. 3 5.73 Lecture #35 35 - 4 B. Operator Replacement for HSO HSO = ∑ a(ri )ℓ i i s i a one-e– operator i Wigner-Eckart Theorem for a vector operator — operator replacement for special cases where only ∆J = 0 matrix elements are considered. ˆ J ˆ JM ′ = J A JM J A J proportionality constant reduced matrix element ∆J = 0 CTDL p. 1054 use projection theorem: V! = JM J Jˆ JM J′ matrix element of corresponding component of Ĵ J•V J. J2 Especially useful when V is an angular momentum that is included in J. useful trick SO for H a(ri)ℓi si vector with respect to L vector with respect to S Special case of ∆L = 0, ∆S = 0 matrix elements in |NLMLSMS〉 basis set configuration label HSO = ∑ a(ri )ℓ i ⋅s i → ζ(N , L,S)L ⋅S i * ζ(N , L,S) ≡ ∑ operator replacement! L a(ri )ℓ i L S s i S i * a different spin-orbit coupling constant for EACH L-S term of the N configuration (loss of simplicity) * convenient because it is easy to evaluate matrix elements of L∙S without having to resort to use of the Slater determinantal basis set ⎡ ASIDE: a(r )ℓ and s are both vectors with respect to J, thus HSO is scalar i i i ⎢ ⎢ with respect to J, hence matrix elements in the NJLSM J basis set have 3 ⎢ ⎢⎣special characteristics: ∆ J = 0,∆ M J = 0, and independent of M J . ⎤ ⎥ ⎥ ⎥ ⎥⎦ CAUTION: the L∙S operator seems to imply a ∆S = 0 selection rule, but we assumed ∆S = 0 in deriving the simplified form of HSO: �L∙S 4 5.73 Lecture #35 35 - 5 C. Landé Interval Rule (useful for recognizing and assigning an isolated L-S term) J 2 = L2 + S2 + 2L ⋅S J=L+S J2 − L2 − S2 L⋅S= 2 NJLSM J H SO NJLSM J !2 = ζ(N , L,S) ⎣⎡ J (J + 1) − L(L + 1) − S(S + 1)⎤⎦ 2 can be positive, zero, or negative So, within an L-S term, HSO causes splitting into 2S+1 (or 2L+1 if S > L) components. J higher J of the 2 levels (ℏ2/2)�(N,L,S)[J(J + 1) – (J – 1)J] = ℏ�(N,L,S)J Landé interval rule J–1 3 P2 2 P1 spacing is ∝ to larger of 2 J’s largest spacing locates largest J large spacing on top: “regular” L-S term bottom: “inverted” L-S term 3 3 [Convince yourself that you are able to tell D from P…] * * * 1 3 P0 regular 3P [2 intervals: 2:1] 3D 3F 3:2 4P 5/2:3/2 4D (3 components) Easy to show that the degeneracy–weighted average (each J-component has degeneracy 2J + 1) spin-orbit energy of a multiplet = 0 (easiest to show from the SO trace of the H in ∣LMLSMS〉basis). Off-diagonal elements (between same-J components of different L-S states) do not affect the trace of HSO. The interval rule plus the number of J-components of a multiplet determine the values of both L and S. [4P 5:3, 2 intervals; 2P 1 interval, 4D 7:5:3, 3 intervals] 5 4:3 7/2:5/2:3/2 (4 components) L+S ∑ J = |L−S| < latexit 5.73 Lecture #35 GOALS: HSO = ∑∑ i '' * * * i replace a(ri )ℓ i by ζ nℓ ℓ i using completeness: X so, for n′ = n, 35 - 6 D. Matrix Elements of HSO in Slater Determinantal Basis Set ∆S ≠ 0 matrix elements, ∆L ≠ 0 matrix elements relationships between �(N,L,S) and �nℓ L-S term ℓi can't change ℓ in ℓmℓ → ℓ′′ = ℓ ℓ i ⋅s i does not operate on radial part of ψ → n′′ = n ⎛ n′ ⎞ n′ℓ a(ri ) nℓ = (n′n )−3/2 ζ°ℓ = ⎜ ⎟ ⎝ n⎠ Rydberg scaling for inner part of orbital 6 orbital excluding interconfigurational H SO matrix elements NONLECTURE: alternative operator replacement for HSO that is appropriate for orbital matrix elements a(ri )`i · si n′ℓ′mℓ′sms′ HSO nℓmℓ sms = n′ℓ′mℓ′sms′ a(ri ) n′′ℓ′′mℓ′′sms′′ n′′ℓ′′mℓ′′sms′′ ℓ i i si nℓmℓ sms completeness a ( ri ) is scalar with respect to s i → m′s = m′′s and value is m′-independent s a ( ri ) is scalar with respect to ℓ i → ℓ′ = ℓ′′,mℓ′ = mℓ′′, and value is mℓ′-independent thus n′ℓ′mℓ′sms′ HSO nℓmℓ sms = δ ℓ′ℓ n′ℓ′ a(ri ) nℓ ℓmℓ′sms′ ℓ ⋅s ℓmℓ sms −3/2 ζ nℓ ∝n nℓ a ri nℓ = ζnℓ = n−3ζ°ℓ () Spin-orbit scaling for all members of a Rydberg series. −3 5.73 Lecture #35 35 - 7 This reduction of HSO shows that, for atoms, HSO acts exclusively within a configuration except for interconfigurational matrix elements where the two configurations differ by a single spin-orbital of the same value of ℓ: "$$$$$ nℓ ↔ n′ℓ # same ℓ Examples A is a single Slater determinant ( ) A HSO A = ∑ uk a rk ℓ k ⋅sk uk k spin-orbitals = ∑ ζn ℓ ℓ k mℓ sk ms ℓ ⋅s ℓ k mℓ sk ms k k k k k k k diagonal element picks out ℓzsz = "2 ∑ ζn ℓ mℓ ms k k k k k spin-orbitals if A is also an eigenfunction of L2 , Lz , S 2 , and S z then NLM L SM S H SO NLM L SM S = ζ ( N, L,S ) ! 2 M L M S . ζ ( N, L,S ) = ∑ζ nk ℓ k mℓ k m s k k MLMS The matrix element is evaluated 2 ways in order to reduce a many-e– spin-orbit coupling constant (ς(N,L,S)) to a sum of one-e– orbital coupling constants (ςnℓ)! This reveals the “periodicity” for which the periodic table is famous. 7 5.73 Lecture #35 35 - 8 Example 1. nf2 3H uncoupled representation uncoupled nf 2 3 H M L = 5 M S = 1 = 3α2α ζ (nf , H ) = 2 3 3 × 1/2 2 × 1/2 ζ nf [3(1 / 2) + 2(1 / 2)] (fill in the steps!) 5 ⋅1 Single Slater determinant! = ζ nf 2 MLMS Example 2. nf2 in coupled representation coupled nf 2 3 H6 MJ = 6 Landé : = 3α2α nf 2 3 H 6 M J = 6 H SO nf 2 3 H 6 6 !2 = ⎡⎣ J ( J + 1) − L ( L + 1) − S ( S + 1) ⎤⎦ ζ ( nf 2 , 3 H ) 6∙7 2 5∙6 1∙2 2 2 3 – = ! 5ζ ( nf , H ) from many-e form = ! 2ζ nf ⎡⎣ 3(1 / 2 ) + 2 (1 / 2 ) ⎤⎦ from orbital form ∴ζ ( nf 2 , 3 H ) = ζ nf 2 what you measure Example 3: ζ(nf2, 3F) what you want to know 3F is never expressed as a single Slater Evaluate in either of determinant for any value of (ML,MS) 2 ways: a. Obtain explicit linear combination of Slater determinants using ladders and orthogonality or using L2 ,S2 to get nf 2 3 F M L = 3 MS = 1 [laborious]. b. Slater sum rule method [simple]. M L = 3, M S = 1 box: contains only 3α0α , 2α1α ( ) ( trace 3α0α + 2α1α = E 1st Matrix Element ⎡3 ⎤ 3α0α = 3α0α H SO 3α0α = ! 2ζ nf ⎢ + 0 ⎥ ⎣2 ⎦ 3 H M L = 3, M S = 1 + E 8 3 ) F M L = 3, M S = 1 5.73 Lecture #35 second matrix element 35 - 9 ⎡ 1⎤ 2α1α = 2α1α HSO 2α1α = ! 2ζ nf ⎢1 + ⎥ ⎣ 2⎦ trace of ML = 3, M S = 1 box is 3!2ζnf exploit sum rule E ( 3 ) H ML = 3 MS = 1 = 3 ( ) H M L = 3 M S = 1 H SO 3 H 31 = ζ nf 2 , 3 H ! 2 3⋅1 but we already showed that ζ(nf2, 3H) = ζnℓ/2 ( ) E 3 H M L = 3, M S = 1 = ! 2ζ nf (3 2 ) ∴E ( FM 3 L ) ( ) ( ) = ζ (nf , F ) (3 ⋅1) ! = 3, M S = 1 = 3! 2ζnf − 3 2 ! 2ζnf = 3 2 ! 2ζnf sum for the ML = 3, MS = 1 box = 3 F 31 HSO 3 F 31 ( 2 3 ) 1 ∴ζ nf 2 , 3 F = ζnf 2 ⎛ ⎞ 1 ⎜ actually would find, for nf ,ζ (nf , L) = ζ for all L ⎟ ⎝ ⎠ 2 2 3 2 [not true for all configurations] nf We are not done. There are some ∆J = 0 off-diagonal in L,S matrix elements among the L-S-J terms of the same configuration. 2 1 nf , I, 6 3 H, 1 G, 3 F, 1 D, 3 P, 1 S, 6 5 HSO is diagonal in J must diagonalize two 3 × 3 and two 2 × 2 matrices. 4 4 4 3 2 2 2 1 0 9 0 2 5.73 Lecture #35 35 - 10 set up the J = 6 matrix because it is simplest 1 3 1 I6 6 = 3α3β H 6 6 = 3α2α I6 6 HSO 3 H6 6 = 3α3β HSO 3α 2α Mismatch is in 2nd spin-orbital. Needs 1/2 ℓ+s– operator to give nonzero spin-orbit matrix element. 1 ℓ s 2α 2 + − 1/2 1 = "2ζnf [3⋅ 4 − 2⋅3] = "2ζnf 3 2 2 = 3β ( ) 1/2 zero for all singlet states SO J= 6 H I6 ⎛ 0 =3 ⎜ 1/2 H 6 ⎜⎝ (3 / 2) 1 3 SO 3 H6 6 H (3 / 2)1/2 ⎞ ⎟⎟ h"2ςnf 5/2 ⎠ ( !2 H6 6 = ⎣⎡ J( J + 1) − L(L + 1) − S(S + 1)⎤⎦ ζ nf 2 , 3H 2 = !25ζ nf 2 , 3H = !2 (5 / 2) ζnf ( ) ) for more complex configurations such as (nℓ)a(n′ℓ′〉)b → ςnℓ and ςn′ℓ′: two ς parameters needed, 1 for each of the two open subshell orbitals. But can use the value of ςnℓ determined from some other configuration: e.g. ς3d from 3d64s2 can be used to predict the 3d part of HSO in 3d6 4s4p. Unexpected inter-relationships between superficially unrelated observables. Small number of control parameters. Magic decoder! Hund’s 3rd Rule: lowest energy J level of lowest energy L – S term is J = |L – S| if subshell is less than ½ full, but it is J = L + S if subshell is more than ½ full, and J = S (no spin-orbit splitting) because L = 0 for a half-filled subshell. Sign of �(n,L,S) as diagnostic! NEXT TIME! 10