Molecular Spectra and Molecular Structure - Lec 17

Page 1 of 6 pages Lecture #17: Parity and e/f Basis for 2∏, 2∑± Last Time: H=H ROT +H SO SS +H +H SR for 2∏, 2∑ Ω > 0 and Ω < 0 * two identical blocks for 2∏ 2 2 * ∏~ ∑ matrix elements involve L± and cannot be evaluated explicitly. Define: βv ∏ v∑ αv ∏ v∑ = Bv ∏ v∑ n ∏ L + n′ ∑ = v∏ v∑ n∏ A L + n′ ∑ 2 Caution: we do not know relationship between two similar appearing symbols. E.g. ? nΛ = +1 L + n′Λ = 0 = nΛ = −1 L − n′Λ = 0 . Another caution: The 2∑ block of the matrix does not factor into Ω > 0 and Ω < 0 subblocks. 2 2 2 ∑+1/2 ∑−1/2 ∑ +1/ 2 H 2 ∑ −1/ 2 = 2 E v − γ∑ 2 + Bv y 2 ∑ (γ sym ∑ ∑ 2 − Bv )y ∑ E v − γ∑ 2 + Bv y 2 ∑ ∑ ⎛ 1 ⎞ ∑ +1/ 2 Bv ( − J − S+ ) + γ ⎜ + J − S+ ⎟ 2 ∑ −1/ 2 ⎝ 2 ⎠ ∑ = ( γ ∑ 2 − Bv ∑ )[ J ( Jy+ 1) − (1/ 2)(−1/ 2)] [ S ( S + 1)1− (1/ 2)(−1/ 2)] 1/ 2 1/ 2 So are we wrong about being able to factor H into two identical blocks? Not quite. Today: Parity σ̂ v ( xz ) * apply σ̂ v to 3 parts of | 〉 and to operators * mysterious phases for α, β matrix elements * ± parity and e/f basis functions e Derive full 2∏, 2∑+, 2∑– ⎛⎝ f ⎞⎠ Effective Hamiltonian expressed in terms of E, B, A, γ, α, β What happens when we apply σ̂ v (xz ) [reflection in body frame] to | HLB-RWF pages 74-75, 138-145 JTH, pages 15-22, 23-25 〉? Trivial. What happens when we apply σ̂ v to| ΩJM〉? Need to look at form of wavefunction. Page 2 of 6 pages Assertion σ̂ v (xz) | ΩJM〉 = (–1)J–Ω | ΩM〉 known as Condon and Shortley phase convention. This leads to Ĵ x and Ĵ± matrix elements real and positive. Problem, what do we do about | nΛ〉? We have no L quantum number so what do we do to ensure the Condon and Shortley phase choice? Assertion σ̂ v (xz) | nΛsS∑〉 = (–1)Λ+s + S–∑ | n–ΛsS–∑〉 is s? A special index for ∑ states. Necessary because σ̂ v n ∏ = (−1)1 n – ∏ Λ = +1 Λ = –1 but σ̂ v n ∑ = (−1)s n ∑ Λ=0 Λ=0 When Λ ≠ 0 σ̂ v takes one basis state into another. When Λ = 0 σ̂ v takes a basis state into itself. Such a situation requires that a | n∑〉 state be an eigenfunction of σ̂ v (xz) belonging to either the +1 (∑+) or the –1 (∑–) eigenvalue! s = 0 for and s = 1 for – + Putting it all together: J−Ω+ Λ+s+S−∑ σ v v ΩJM nΛ sS∑ = (−1) v −ΩJM n − Λ s S − ∑ J–S+s [because (–1)S– = (–1) – S + 2(S – ) ] an integer (–1)2n = +1 So now we are ready to figure out the relationship between + ∏ L + ∑ s and − ∏ L − ∑ s . Page 3 of 6 pages ⎡⎣σv , H ⎤⎦ = 0 σˆ v ( xz ) ⎡⎣ 2 ∏Ω L + σˆ v [ a number ] = [same number ] OR 2 s ⎤ ∑ Ω− 1 ⎦ = σˆ v but also 2 ∏Ω L + 2 s ∑Ω− 1 ( −1) n ∏Ω = 2 Λ s S ∑ Λ + 0 +1/ 2 − ( Ω−Λ ) n 2 ∏ −Ω σ̂ v (xz) x → x y → –y z→z σˆ v L + = σˆ v ⎣⎡ L x + iL y ⎦⎤ = σˆ v ⎡⎣ ypz − zp y + i ( xpz − zpx ) ⎤⎦ = − L x + iL y = − L − σ̂ v L± = –L∓ turns out to be basis for ∑+~∑– H SO s ( −1) σv 2 ∑ Ω− = 1 N selection rule. ∑ 0 + s +1/ 2 − ( Ω −1) 2 ∑ −s Ω+1 ∑ Putting it all together: 1/ 2 −Ω s ( ) n 2 ∏ Ω L + n′ 2 ∑ Ω− 1 n 2 ∏ −Ω (−1)1 L − (−1) s +1/ 2−Ω+1 n′ 2 ∑ −s Ω+1 = − 1 s +1/ 2 +1/ 2 +1+1−2 Ω = (−1) odd integer even integer s n 2 ∏ −Ω L − n′ 2 ∑ −Ω+ 1 (−1) s More generally: ± 2 S +1 ∏ −Ω L − n′ 2 S +1 ∑ −s Ω+1 n 2 S +1 ∏ Ω L + n′ 2 S +1 ∑ Ω− 1 = ± n Now return to 2∏, 2∑+ matrix and go to either +/– or e/f basis in which every basis function is eigenfunction of σ̂ v (xz). J – S+ s Since σ̂ v nΛ sS ∑ ΩJM v = (−1) n − ΛS − ∑ −ΩJM v Page 4 of 6 pages try ψ = 2 −1/2 ⎡⎣ nΛ sS∑ ΩJM v ± (−1) J – S+s n − ΛS − ∑ −ΩJM v ⎤⎦ . So what do we get? σˆ v ψ = 2−1/ 2 ⎡( −1) ⎢ ⎣ = ±ψ J −S + s n − Λ s S − ∑ – ΩJM v ± (−1) J − S + s (−1) J − S + s nΛS ∑ ΩJM v ⎤ ⎥ +1 ⎦ So these are the E.g. 2 ∏ 3/2 ± ≡ 2 −1/2 ⎡⎣ 2 ∏+3/2 ± (–1)J−S 2 ∏−3/2 ⎤⎦ Or more generally n Λ s S∑± Ω JM ≡ 2 −1/2 ⎡⎣ nΛ s ≥ 0S∑ Ω = Λ + ∑ JM ± (−1)J−S+s nΛ s ≤ 0S − ∑ Ω = − Λ − ∑ JM ⎤⎦ note |absolute value| on Λ and ž only need careful notation to deal with case where S > |Λ| e.g. 4∏ where |Ω| = 5/2, 3/2, 1/2, –1/2 NOTE that the phase factor in front of the second term alternates with J (and S). This turns out to be inconvenient for labeling spectroscopic transitions and spectroscopic perturbations. Alternate form of parity: e/f labels. Want e: σv(xz)ψe = +(–1)J–1/2ψe f: σv(xz)ψf = –(–1)J–1/2ψf use J–1/2 for even multiplicity systems S = 1/2, 3/2, etc. use J for odd multiplicity systems S = 0, 1, 2, etc. For our problem we will use e/f labeled basis functions. Check that the following conform to the above definition. 2 ∏ 3/2 e = 2 −1/2 ⎡⎣ 2 ∏+3/2 ± 2 ∏−3/2 ⎤⎦ f 2 ∏1/ 2 e ∑1/± 2 e = 2 −1/2 ⎡⎣ 2 ∑±+1/2 ± (−1)s 2 ∑±−1/2 ⎤⎦ f 2 f = 2 −1/2 ⎡⎣ 2 ∏1/2 ± 2 ∏−1/2 ⎤⎦ (Recall s = 0 for ∑ + and s = 1 for ∑ – ) Page 5 of 6 pages Now we can write down the full 2∏, 2∑+, 2∑– matrix in the case (a) e/f basis. 2 e/f 2 ∏3/2 2 ∏3/2 ( E v∏ + 12 A∏ + Bv∏ × 1 2 (y 2 − 2)− 12 γ∏ 2 2 2 ∏1/2 ) γ∏ − Bv∏ × E v∏ − 12 A∏ + Bv∏ y 2 − 12 γ∏ ∑+ 2 1/2 –β+v∏ v + ⎡⎣y 2 −1⎤⎦ ∑ ⎡⎣y 2 − 1⎤⎦1/2 ∏1/2 ∑+ α+v∏ v ∑+ Ev ∑+ + β+v∏ v ∑+ + Bv ∑+ [1 my ] ∑– 1/2 –β−v∏ v − ⎡⎣y 2 −1⎤⎦ ∑ α−v∏ v ∑− (y 2 my) + β−v∏ v ∑− αv ∑+ [1± y ] v ∑− − 12 γ∑+ (1 my ) 2 ∑– Ev ∑− + Bv ∑− (y 2 ± y) − 12 γ∑− (1± y ) y = J + 1/2 See derivations below for details regarding ∑ states. There are some surprising terms here. Consider 1 2 ∑+ e H f 2 ∑+ e 1 = [ H1/ 21 / 2 + H −1/ 2 −1/ 2 ± H1/ 2 −1/ 2 ± H −1/ 2 1/ 2 ] = B ( y 2 ∓ y ) + γ / 2 ( ± y − 1) f 2 [–γ/2 + By2] 2 ∑+ e f H 2 ∑+ f e = ±(γ/2 – B)y 1 [ H1/ 2 1/ 2 − H −1/ 2 −1/ 2 + H −1/ 21/ 2 − H1/ 2 −1/ 2 ] = 0 2 Factorization into separate e/f blocks! 2 2 ∑− e H f 2 ∑− e 1 = [ H1/ 2 1/ 2 + H −1/ 2 −1/ 2 ∓ H1/ 2 −1/ 2 ∓ H −1/ 2 1/ 2 ] = B ( y 2 ± y ) + γ / 2 ( ∓ y − 1) f 2 Also 3 2 ∑+ e H f 2 ∏1/ 2 L+S– J+L– J–L+ e 1 L–S+ = ⎡⎣ H ∑1/ 2 ∏1/ 2 + H ∑ −1/ 2 ∏−1/ 2 ± H ∑−1/ 2 ∏1/ 2 ± H ∑1/ 2 ∏ −1/ 2 ⎤⎦ f 2 1⎡ ( α + β ) + ( α + β ) ± ( y 2 )1/ 2 β ± yβ⎤⎦ ⎣ 2 = α + β(1 ± y ) = Page 6 of 6 pages 4 2 ∑− e e 1 = ⎡⎣ H ∑1 / 2 ∏1/ 2 − (1) s H ∑ −1/ 2 ∏−1/ 2 ± H ∑1/ 2 ∏−1/ 2 ∓ H ∑ −1/ 2 ∏1/ 2 ⎤⎦ H 2 ∏1/ 2 f f 2 1⎡ ( α + β ) − (−1)1 ( α + β ) ± yβ(−1)1 ∓ yβ⎤⎦ 2⎣ = α + β(1 ∓ y ) =