Molecular Spectra and Molecular Structure - Lec 16

Page 1 of 6 pages Lecture #16: 2∏ and 2∑ Matrices Last Time: effect of A2 , Ai , A± on |A α MA〉 basis set |n(L) Λ S ∑〉 |v〉 |ΩJM〉 case (a) basis set L-destroyed, but not Λ and L z , L ± , selection rules H ROT = B( R) R 2 matrix elements n′Λ′S ′ ∑′ v′ ΩJM H Diagonal: ROT nΛS ∑ v ΩJM = δn′nδΛ′ΛδS′Sδ∑′∑δv′vδΩ′ΩδJ′JδM′M × Bv[J(J + 1) – Ω2 + S(S + 1) – ∑2 + L2⊥ ] ∆Ω = ∆∑ = ±1 within Λ–S multiplet state (S-uncoupling): nΛS ∑ ±1 v Ω ± 1JM H ROT nΛS ∑ v ΩJM = –Bv[J(J + 1) – (Ω ± 1)Ω]1/2[S(S + 1) – (∑ ± 1)∑]1/2 **** In my handouts I call J + 1/2 = x. Here, I'll call it y ***** For example: 2 n n n n Λ S 1 1/ 2 1 1/ 2 −1 1 / 2 −1 1 / 2 l ROT ( 2 ∏ ) = H ∏3/ 2 2 Bvπ × ∏1/ 2 2 ∏ −1/ 2 2 ∏ −3/ 2 2 Σ 1/ 2 −1 / 2 −1 / 2 1/ 2 2 ∏ 3/ 2 ∏1/ 2 2 ∏−3/ 2 2 ∏−1/ 2 2 x − 94 + 43 − 14 sym 0 0 −[ x − 43 ]1/ 2 [ 43 + 14 ]1/ 2 x − 14 + 34 − 14 0 0 0 0 ∆Ω = ±1 ∆Λ = ±2 x − 14 + 34 − 14 − [ x − 34 ]1/ 2 sym ⎛ x − 7 (x − 3 )1/ 2 ⎞ 0 0 4 4 ⎜ ⎟ sym x + 41 0 0 ⎟ π⎜ Bv ⎜ ⎟ 1/ 2 0 x + 14 (x − 43 ) ⎟ ⎜ 0 ⎜ 0 0 sym x − 47 ⎟⎠ ⎝ Two identical blocks for Ω > 0 and Ω < 0 - later we will consider parity basis. 0 0 1 9 x − 4 + 43 − 14 Page 2 of 6 pages ∆Ω = ∆Λ = ±1 between Λ-S multiplet states (L-uncoupling) n′Λ + 1S ∑ v′ Ω ± 1JM H ROT nΛS ∑ v ΩJM = –Bv′v[J(J + 1) – (Ω ± 1)Ω]1/2 × n′Λ ± 1 L± nΛ β ↑ a perturbation parameter to be determined by a fit to the spectrum. SO SS SR ⎧effective operators H ,H ,H ⎨ ⎩matrix elements Matrix elements of 2∏, 2∑ effective H. Today: spin-orbit H SO for S = 0 = ∑ a (ri ) i ·si ⎯⎯⎯⎯ → AL·S only 2 i spin-spin H SS spin-rotation H SR 2 2 2 for 2 S = 0 = ⎯⎯⎯⎯ → λ ⎣⎡3S z − S ⎦⎤ + another term 2 ∑ = − 2 Λ = ±2 only 3 = γ R·S usually λ, γ are very small with respect to A and are dominated by second-order spin-orbit effects (thru Van Vleck transformation)—discussed later H SO H is very important SO =∑ i i a(ri )ˆ i ·sˆi electron (not component) , ŝi are vectors with respect to J → ŝi is vector with respect to S → i ⎛ ⎞ ˆ ⎜ not ∑ a i ·sˆ j i, j ⎝ ⎠ L i S ·ŝi is scalar (∆J = ∆M = ∆Ω = 0) with respect to J . ·ŝi is vector with respect to S → ∆S = 0, ±1, ∆∑ = 0, ±1 i·si does no t operate on | ΩJM〉, only on |nΛS 〉; it is therefore NOT INDEPENDENT of Ω because, as vector with respect to L and S, its matrix elements are not independen t of Λ + Σ . Page 3 of 6 pages ∆J = 0 ∆Ω = 0 Selection rules (ASSERTED) +↔ / – (LAB INVERSION I ) / u (body inversion î ) g↔ ∑+ ↔ ∑– (σv) ∆S = 0, ±1 ∆∑ = –∆Λ = 0, ±1 SO H is a one-electron operator, so it has non-zero matrix elements only between electronic configurations differing by a single spin-orbital. Special simplification (due to simple form of Wigner-Eckart Theorem). If B is vector with respect to A , then ∆B = 0 matrix elements of a vector operator ( B ) with respect to angular momentum ( A ) may be evaluated by replacing B by b A (where b is a constant, often called a reduced matrix element)! a ( ri ) i is vector with respect to L ŝi is vector with respect to S For ∆L = 0, ∆S = 0 matrix elements ∑ a (r ) i i ·sˆi → AL·S i H SO 1 ⎡ ⎤ = A ⎢ Lz S z + ( L+ S− + L− S+ ) ⎥ 2 ⎣ ⎦ E.g., for 2∏ ⎛ 1⎞ A ∏ ±3/ 2 = A ( ±1) ⎜ ± ⎟ = ⎝ 2⎠ 2 SO A ⎛ 1⎞ 2 ∏ ±1/ 2 H 2 ∏ ±1/ 2 = A ( ±1) ⎜ ∓ ⎟ = − 2 ⎝ 2⎠ all ∆Ω ≠ 0 matrix elements are = 0. 2 ∏ ±3/ 2 H SO 2 SS 2 2 ⎡ 2 2 2 S S → λ − H ⎯⎯⎯ 3 ⎣ z ⎦⎤ = λ ⎣⎡3 ∑ − S ( S + 1) ⎦⎤ + additional term 2 S =0 3 3 Page 4 of 6 pages Selection rules H SR ∆S = 0 ∆Ω = 0 ∆S = 0 ∆∑ = 0 +↔ / – g↔ / u ∑+ ↔ / ∑– [also ±1, ±2 neglected here] [also ∆∑ = –∆Λ = ±2 (Λ-doubling in 3∏0 neglected here)] = γ R·S = γ ( J – L – S )·S = γ[ J ·S − L·S − S 2 ] we already know how to deal with all three of these! Now we are ready to set up full 2∏, 2∑+ matrix. Start with all of 2∏3/2 and then 2∏1/2 and then 2∑1/2 etc. l =H l elect + H l vib + H l ROT + H l SO + H l SS + H l SR n, 2 ∏ , v = v, n, 2 ∏3/ 2 H 3/ 2 Λ∑ 3∑2 – S(S + 1) Ω ∑ Λ ∑ S(S+1) 2 ⎛ 1 3⎞ 1 3⎞ ⎛3 1 Te ( n 2 ∏ ) + G ( v∏ ) + A∏ (1·1/ 2 ) + λ ⎜ 3· − ⎟ + γ ∏ ⎜ · − 1· − ⎟ 3 Always 2 4⎠ ⎝ 2 4⎠ ⎝2 2 2  0 for S = = 1/2 states! 9 3 1 1 1 7⎞ ⎡ ⎤ ⎛ + Bv∏ ⎢ J ( J + 1) − + − + L2⊥ ⎥ = Ev∏ + A∏ − γ ∏ + Bv∏ ⎜ J ( J + 1) − ⎟ 4 4 4 2 2 4⎠ ⎣ ⎦ ⎝ include with Te + G in E v∏ y2 – 2 Let y = J + 1/2 an integer since J is half-integer for 2∏ and 2∑. Get same results for 2 ∏ −3/ 2 H 2 ∏ −3/ 2 . 2 Get same results for 2 2 ∏1/ 2 H 2 ∏1/ 2 = Ev∏ − 1 1 A∏ − γ ∏ + Bv∏ ⎡ J ( J + 1) + 1/ 4 ⎤ 2 2 ⎢⎣ ⎥⎦ y2 ∏ −1/ 2 H 2 ∏ −1/ 2 . ∑1/ 2 H 2 ∑1/ 2 = 2 1 1 ∑ −1/ 2 H 2 ∑ −1/ 2 = Ev∑ − A∑ 0· − γ ∑ + Bv∑ ⎡ J ( J + 1) − 1/ 4 + 3 / 4 − 1/ 4 ⎤ 2 2 ⎢⎣ ⎥⎦ y2 ↑ always for ∑-states Page 5 of 6 pages [ASIDE: we have two explicit cases where, by evaluation of matrix elements, we see that Λ ∑ Ω H Λ ∑ Ω = −Λ − ∑ −Ω H − Λ − ∑ −Ω . But be careful, this is not true for Λ′ H Λ = −Λ′ H − Λ ! Non-evaluable matrix elements.] Off-Diagonal Matrix Elements 2 2 ∏ 3/ 2 H ∏1/ 2 = 2 2 ∏ −3/ 2 H ∏ −1/ 2 1/ 2 ⎡1 ⎛ 3⎞ 3⎤ = −0 A∏ + ⎢ ⎜ x − ⎟ − ⎥ γ ∏ − Bv∏ 4⎠ 4⎦ ⎣2 ⎝ ∏3/ 2 H 2 ∏ −1/ 2 = 0 ∆Ω = 2 2 ∏3/ 2 H 2 ∏ −3/ 2 = 0 ∆Ω = 3 –BJ–L+ 2 l ∑ ∏3/ 2 H 2 + 1/ 2 1 1/ 2 3 1⎤ ⎡ ( 1) J J + − ⎢⎣ 2 2 ⎥⎦ = − v∏ B(R) v∑ Bv∏ v∑ 1/ 2 = −βv∏ v∑ ⎡⎣ y − 1⎤⎦ 2 2 1/ 2 ⎡ 3 1 ⎛ 1 ⎞⎤ ⎢ 4 − 2 ⎝⎜ − 2 ⎠⎟ ⎥ ⎣ ⎦ [y2 – ∆Ω = 0 2 1/ 2 1 3⎤ ⎡ ⎢⎣ x − 2 2 ⎦⎥ ∏3/ 2 H 2 ∑ −1/ 2 = 0 n ∏ L+ n′ ∑ β(∏,∑) ∆Ω = 2 all done with 2∏3/2 2 ∏1/ 2 H 2 ∏ −3/ 2 = 0 ∆Ω = 2 from L·S 2 from 2L·S ⎡1 ⎤ 2 l 2∑ A + B ( R) ⎥ L+ S− v∑ 2 ∑1/+ 2 ∏1/ 2 H 1/ 2 = v∏ ∏1/ 2 ⎢ ⎣2 ⎦ = [ S ( S + 1) − ∑ ∏ ∑ ∑ ] 1/ 2 = 1 ⎡⎣α v∏ v∑ + βv∏ v∑ ⎤⎦ ⎡ ⎢ v∏ v∑ ⎣ n∏ α ⎤ A L+ n′ ∑ + Bv∏ v∑ n ∏ L+ n′ ∑ ⎥ 2 ⎦ β Page 6 of 6 pages –BJ–L+ 2 ∏1/ 2 H 2 ∑−1/ 2 1/2 ⎡ 1 ⎛ 1 ⎞⎤ = −Bv∏ v∑ ⎢J(J + 1) − ⎜− ⎟⎥ ⎣ 2 ⎝ 2 ⎠⎦ = −βv∏ v∑ y y n ∏ L + n′ ∑ 2 all done with 2∏1/2 –BJ–S+ 2 ∑1/ 2 H 2 ∑−1/ 2 1/2 1/ 2 ⎡ 1 ⎛ 1 ⎞⎤ ⎡ 3 1 ⎛ 1 ⎞⎤ = −Bv∑ ⎢J(J + 1) − ⎜− ⎟⎥ ⎢ − ⎜− ⎟⎥ ⎣ 2 ⎝ 2 ⎠⎦ ⎣ 4 2 ⎝ 2 ⎠⎦ = −β v∑ y all done with 2∑1/2. Are we done? Not quite. Must worry about 2∑+ ~ 2∏–3/2 and 2∑+ ~ 2∏–1/2 matrix elements. What happens to the 〈∏|L+|∑〉 unevaluable factor? Need to consider effects of σv(xz) reflections and ∑+, ∑– symmetry.