MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 5.80 Lecture #15 Fall, 2008 2 Page 1 of 6 pages 2 Lecture #15: ∏ and ∑ Matrices Last Time: 2 , A ,A ± on |A α M 〉 basis set effect of A A i case (a) basis set |n(L) Λ S ∑〉 |v〉 |ΩJM〉 z, L ± , selection rules L-destroyed, but not Λ: L 2 matrix elements ROT = B(R)R H ROT nΛS∑ v ΩJM = δ δ δ δ δ δ δ δ n′Λ′S′ ∑′ v′ ΩJM H n′n Λ′Λ S′S ∑′∑ v′v Ω′Ω J′J M′M Diagonal: × Bv[J(J + 1) – Ω2 + S(S + 1) – ∑2 + L2⊥ ] ∆Ω = ∆∑ = ±1 within Λ–S multiplet state (S-uncoupling): ROT nΛS∑ v ΩJM = –B [J(J + 1) – (Ω ± 1)Ω]1/2[S(S + 1) – nΛS∑±1 v Ω ±1JM H v (∑ ± 1)∑]1/2 **** In some of my handouts I call J + 1/2 = x. Here, I'll call it y ***** Here x = J(J + 1), y = J + 1/2 For example: Start by listing all relevant basis states. Λ S Σ 2 n 1 1/ 2 1/ 2 ∏ 3/2 2 ∏ 2 n 1 1 / 2 −1 / 2 ∏1/2 2 n −1 1 / 2 −1 / 2 ∏−3/2 2 n −1 1 / 2 1 / 2 ∏−1/2 ROT ( 2 ∏) = H ∏ 3/2 2 ∏1/2 Bvπ × 2 ∏ −1/2 2 ∏−3/2 ⎛ x − 7 4 ⎜ ⎜ sym Bvπ ⎜ ⎜ 0 ⎜ 0 ⎝ 2 x − 94 + 43 − 14 sym 0 0 −[x − 43 ]1/2 [ 43 + 14 ]1/2 x − 14 + 43 − 14 0 0 0 0 ∆Ω = ±1 ∆Λ = ±2 x − 14 + 43 − 14 − [x − 43 ]1/2 sym ⎞ ⎟ x + 41 0 0 ⎟ ⎟ 1/2 0 x + 41 ( x − 43 ) ⎟ 0 sym x − 47 ⎟⎠ Two identical blocks for Ω > 0 and Ω < 0 - later we will consider parity basis. What about 2∑+? Class should do this. (x − 43 ) 1/2 0 0 0 0 1 9 x − 4 + 43 − 14 5.80 Lecture #15 Fall, 2008 Page 2 of 6 pages ∆Ω = ∆Λ = ±1 between Λ-S multiplet states (L-uncoupling) ROT nΛS∑ v ΩJM = –B [J(J + 1) – (Ω ± 1)Ω]1/2 × n′Λ ±1 L nΛ n′Λ +1S∑ v′ Ω ±1JM H v′v ± β a perturbation parameter to be determined by a fit to the spectrum. ↑ Today: spin-orbit SO , H SS , H SR ⎧⎨effective operators H ⎩ matrix elements Matrix elements of 2∏, 2∑ effective H. ∆ S=0 SO = ·S H → AL ∑ a(ri ) i ·si ⎯for⎯⎯⎯ only (restricted validity operator replacement) i spin-spin 2 ⎡⎣ 2 2 ⎤⎦ ∆ S=0 SS = ⎯for H ⎯⎯⎯ → λ 3Sz − S + another term ∆ ∑ = − ∆ Λ = ±2 only 3 spin-rotation SR = γR ·S H 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 SO is very important H SO H = ∑ a(ri )̂ i ·ŝ i i electron (not component) ⎛ ⎞ ⎜⎜ not ∑ â i ·ŝ j ⎟⎟ ⎝ ⎠ i, j L S i ,ŝ i are vectors with respect to J → i ·ŝ i is scalar (∆J = ∆M = ∆Ω = 0) with respect to J . ŝ i is vector with respect to S → i ·ŝ i is vector with respect to S → ∆S = 0, ±1, ∆∑ = 0, ±1 i·si does not 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 independent of Λ and ∑. 5.80 Lecture #15 Fall, 2008 Selection rules (ASSERTED) Page 3 of 6 pages ∆J = 0 ∆Ω = 0 +↔ / – (LAB INVERSION I ) (parity) g↔ / u (body inversion î ) + ∑ ↔ ∑– (σv) ∆S = 0, ±1 ∆∑ = –∆Λ = 0, ±1 SO is a one-electron operator, so it has non-zero matrix elements only between electronic H configurations differing by a single spin-orbital. (e.g. π orbital = 1α, 1β, –1α, –1β spin-orbitals) is vector with respect to Special simplification (due to simple form of Wigner-Eckart Theorem). If B ) with respect to angular momentum ( A , then ∆B = 0 matrix elements of a vector operator ( B ) may A by b A (where b is a constant, often called a reduced matrix element)! be evaluated by replacing B a ( ri ) i is vector with respect to L ŝ is vector with respect to S i For ∆L = 0, ∆S = 0 matrix elements ∑ a (r ) ·ŝ i i i ·S → AL (limited validity operator replacement) i SO ⎡ ⎤ 1 H = A ⎢L zSz + ( L +S− + L −S+ )⎥ ⎣ ⎦ 2 E.g., for 2∏ ⎛ ⎞ A SO 2 ∏ ( ) 1 ∏±3/2 H ±3/2 = A ±1 ⎜± ⎟ = ⎝ 2⎠ 2 ⎛ ⎞ A 2 SO 2 ∏ ( ) 1 ∏±1/2 H ±1/2 = A ±1 ⎜ ⎟ = − ⎝ 2⎠ 2 all ∆Ω ≠ 0 matrix elements are = 0. 2 SS 2 ⎡⎣ 2 2 ⎤⎦ 2 ⎡ 2 H ⎯⎯⎯ → λ 3Sz − S = λ ⎣3∑ −S(S +1)⎤⎦ + additional term ∆ S=0 3 3 5.80 Lecture #15 Fall, 2008 Selection rules ∆S = 0 ∆Ω = 0 ∆S = 0 ∆∑ = 0 +↔ / – g↔ / u + ∑ ↔ / ∑– Page 4 of 6 pages [also ±1, ±2 neglected here] [also ∆∑ = –∆Λ = ±2 (Λ-doubling in 3∏0 neglected here)] SR ·S = γ ( J – L – S )·S = γ[J·S − L·S − S2 ] H = γR 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 matrix elements of 2∏3/2 and then 2∏1/2 and then 2∑1/2 etc. v,n, ∏ 3/2 2 elect vib ROT SO SS SR H = H + H + H + H + H + H n, 2 ∏ 3/2 , v = Λ∑ 3∑2 – S(S + 1) Ω ∑ Λ ∑ S(S+1) 2 2 ⎛ ⎛ 1⎞ 3⎞ 1 ⎛31 2 Te ( n ∏ ) + G (v∏ ) + A ∏ (1·1 / 2 ) + λ ⎜ 3·⎜ ⎟ − ⎟ + γ ∏ ⎜ · − 1· − ⎝2 2 3 ⎝ ⎝ 2⎠ 4⎠ 2 2 –Ω 2 2 +S – ∑ 3⎞ ⎟ 4⎠ always = 0 for S = 1/2 states! 9 3 1 1 1 ⎛ ⎡ ⎤ +Bv∏ ⎢ J(J + 1) − + − + L2⊥ ⎥ = E v∏ + A ∏ − γ ∏ + Bv∏ ⎜ J(J + 1) − ⎝ ⎣ 4 4 4 ⎦ 2 2 include with Te + G in E v∏ y ≡ J + 1/2, thus y is an integer since J is half-integer for 2∏ and 2∑. Get same results for 2 2∏ ∏−3/2 H −3/2 . 2 1 1 2∏ +1) +1 /4 ⎤ ∏1/2 H A∏ − γ∏ + Bv∏ ⎡J(J 1/2 = E v∏ − 2 2 ⎢⎣ ⎥⎦ y2 y2 – 2 7⎞ ⎟ 4⎠ 5.80 Lecture #15 Fall, 2008 2 Get same results for 2 Page 5 of 6 pages 2∏ ∏−1/2 H −1/2 . 2∑ ∑1/2 H 1/2 = 2 1 1 2∑ ∑−1/2 H γ∑ + Bv∑ ⎡J(J +1) −1 / 4 + 3 / 4 −1 / 4⎤ −1/2 = E v∑ − A ∑ 0· − 2 2 ⎢⎣ ⎥⎦ 2 y ↑ always for ∑-states [ASIDE: we have two explicit cases where, by evaluation of matrix elements, we see that Λ ∑Ω = −Λ − ∑−Ω H − Λ − ∑−Ω . But be careful, this is not true for Λ ∑Ω H Λ = −Λ′ H − Λ ! Non-automatically-evaluable matrix elements.] Λ′ H Off-Diagonal Matrix Elements Always ask what operator do we need to get non-zero matrix element between specified basis states? 2 ∏ 3/2 2∏ H 1/2 = 2 ∏ −3/2 2∏ H −1/2 1 ( J + S− + J − S + ) 2 1/2 ⎡1 ⎛ 3⎞ ⎤ = −0A∏ + ⎢ ⎜ x − ⎟ ⎥ γ ∏ − Bv∏ 4⎠ ⎦ ⎣2 ⎝ ∆Ω = 0 2 2∏ ∏ 3/2 H −1/2 = 0 ∆Ω = 2 2 2∏ ∏ 3/2 H −3/2 = 0 ∆Ω = 3 –BJ–L+ 2 2 ∑+ ∏ 3/2 H 1/2 J+ ⎡ 3 1⎤ = − v∏ B(R) v∑ ⎢ J(J +1) − ⎣ 2 2 ⎥⎦ B 1/2 v∏ v∑ = −βv∏ v∑ ⎡⎣y −1⎤⎦ 2 2 2∑ ∏ 3/2 H −1/2 = 0 ∆Ω = 2 all done with 2∏3/2 2 2∏ ∏1/2 H −3/2 = 0 1/2 1/2 1 3⎤ ⎡ 3 1 ⎛ 1⎞ ⎤ ⎡ ⎢⎣ x − 2 2 ⎥⎦ ⎢⎣ 4 − 2 ⎜⎝ − 2 ⎟⎠ ⎥⎦ [y2 – 1]1/2 1 ∆Ω = 2 1/ 2 n∏ L + n′∑ β(∏,∑) 5.80 Lecture #15 Fall, 2008 from AL·S 2 Page 6 of 6 pages from 2BL·S ⎡1 ⎤ 2 2 + 2∑ ∏1/2 H 1/2 = v∏ ∏1/2 ⎢ A + B(R)⎥ L +S− v∑ ∑1/2 ⎣2 ⎦ 1/ 2 ⎡ = ⎣⎡S(S +1) − ∑∏ ∑∑ ⎤⎦ ⎢ v∏ v∑ ⎣ = 1⎡⎣α v∏ v∑ + β v∏ v∑ ⎤⎦ –BJ–L+ 2 2∑ ∏1/ 2 H −1/ 2 n∏ ⎤ A L + n′∑ + Bv∏ v∑ n∏ L + n′∑ ⎥ 2 ⎦ β α ⎡ 1 ⎛ 1 ⎞⎤ = −Bv∏ v∑ ⎢J(J +1) − ⎜− ⎟⎥ ⎣ 2 ⎝ 2 ⎠⎦ = −βv∏ v∑ y y 1/2 n ∏ L + n′∑ 2 all done with 2∏1/2 –BJ–S+ 2 2∑ ∑1/2 H −1/2 J+ 1/2 1 ⎛ 1⎞ ⎤ ⎡ = −Bv∑ ⎢ J(J + 1) − ⎜ − ⎟ ⎥ 2 ⎝ 2⎠ ⎢ ⎥ 2 y ⎣ ⎦ = −Bv∑ y S+ ⎡3 ⎢4 − ⎣ 1/2 1 ⎛ 1⎞ ⎤ ⎜− ⎟ 2 ⎝ 2 ⎠ ⎥⎦ 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 in order to get the correct relative signs of off-diagonal matrix elements.