QCexercises10

<span class='text_page_counter'>(1)</span>10‒1. Express the Hamiltonian operator for a hydrogen molecule in atomic units. 10‒2. The vibrational energy levels of [H2]+ (in cm‒1) are given by Chase, M. W., Jr., et al., J. Phys. Chem. Ref. Data 1985, vol. 14, supplement no. 1, known as the JANAF Thermochemical Tables. These tables were updated in 1998 by M. W. Chase, Jr. As the NIST‒JANAF Thermochemical Tables, 4th ed., monograph no. 9 (parts 1 and 2), available from the American Institute of Physics. 2. 3. 1 1 1 1     G (υ ) = 2323.23υ +  − 67.39υ +  + 0.93υ +  − 0.029υ +  2 2 2 2    . 4. The molecule dissociates in the limit that ∆G(υ + 1) ‒ ∆G(υ + 1) → 0, and so υmax, the maximum vibrational quantum number, is given by ∆G(υmax) = 0. Plot ∆G(υ) = G(υ + 1) ‒ G(υ) against υ and show that υmax = 18. 10‒4. The value of De for [H2]+ is 0.10264 Eh, yet the minimum energy in Figure 10.2 is ‒0.60264 Eh. Why is there a difference?. Figure 10.2. 10‒28. Show that ψ given by Equation 10.38 is an eigenfunction of Sˆ z = Sˆ z1 + Sˆ z 2 with Sz = 0.. ψ=. 1 σ bα (1) σ b β (1)  1  = σ b (1)σ b ( 2)  [α (1) β ( 2) − α (2) β (1)] 2! σ bα ( 2) σ b β ( 2)  2 . (Equation 10.38). 10‒30. The six wave functions given by Equation 10.49 are actually linear combinations of the determinantal wave functions given by Equation 10.48. Show that ψ1 and ψ2 are the spatial parts of D1 and D2, respectively; that two of ψ3, ψ4 or ψ5 are the spatial parts of D3 and D6; that the remaining one of ψ3, ψ4 or ψ5 is the spatial part of ( D4 + D5 ) take. linear. 2 ; and that ψ6 is the spatial part of ( D4 − D5 ). combinations ( D4 ± D5 ). 2. is. because. 2 . The reason that we. these. combinations. are.

<span class='text_page_counter'>(2)</span> eigenfunctions of Sˆ 2 .. D1 =| σ bα (1)σ b β (2)⟩. D2 =| σ aα (1)σ a β (2)⟩. D3 =| σ bα (1)σ aα (2)⟩. D4 =| σ bα (1)σ a β (2)⟩. D5 =| σ b β (1)σ aα (2)⟩. D6 =| σ b β (1)σ a β (2)⟩. ψ 1 = σ b (1)σ b (2) ψ 2 = σ a (1)σ a (2) ψ 3 , ψ 4 , ψ 5 = σ b (1)σ a (2) − σ a (1)σ b (2) ψ 6 = σ b (1)σ a (2) + σ a (1)σ b (2). (Equation 10.48) (Equation 10.49). 10‒31. Use the result of the previous problem that the complete wave functions (spin functions included) given by Equation 10.49 are. ψ1 =. 1 σ b (1)σ b (2)[α (1) β (2) − α (2) β (1)] 2. ψ2 =. 1 σ a (1)σ a (2)[α (1) β (2) − α (2) β (1)] 2. α (1)α (2)  1 [σ b (1)σ a (2) − σ a (1)σ b (2)]1 2[α (1) β (2) + α (2) β (1)] ψ 3 ,ψ 4 ,ψ 5 = 2 β (1) β (2)  1 ψ 6 = [σ b (1)σ a (2) + σ b (2)σ a (1)][α (1) β (2) − α (2) β (1)] 2 Show that Sˆ ψ = ( Sˆ + Sˆ )ψ = 0 z. 1. z1. z2. 1. Sˆ zψ 2 = ( Sˆ z1 + Sˆ z 2 )ψ 2 = 0 Sˆ ψ = ( Sˆ + Sˆ )ψ = ψ z. 3. z1. z2. 3. 3. Sˆ zψ 4 = ( Sˆ z1 + Sˆ z 2 )ψ 4 = 0 Sˆ zψ 5 = ( Sˆ z1 + Sˆ z 2 )ψ 5 = ψ 5 Sˆ ψ = ( Sˆ + Sˆ )ψ = 0 z. 6. z1. z2. 6. It turns out that ψ3, ψ4 and ψ5 are wave functions of a triplet state and the others are wave functions of singlet states.. 10‒32. Show that all six wave functions in the previous problem are antisymmetric..

<span class='text_page_counter'>(3)</span> 10‒34. In this problem, we shall prove that if an operator F̂ commutes with Ĥ , then matrix elements of Ĥ , H ij = ⟨ψ i | Hˆ | ψ j ⟩ , between states with different eigenvalues of F̂ vanish. For simplicity, we prove this only for nondegenerate state. Let F̂ be an operator that commutes with Ĥ , and let its eigenvalues and eigenfunctions be denoted by and , respectively. Show that. [ Hˆ , Fˆ ]λλ ′ = ⟨ψ λ ′ | [ Hˆ , Fˆ ] |ψ λ ⟩ = (λ − λ ′) Hˆ λ ′λ Now argue that Hˆ λ ′λ = 0 unless λ = λ ′ . For degenerate states, it is possible to take linear combinations of the degenerate eigenvalues and eigenfunctions to carry out a similar proof. 10‒40. Show that the normalization condition for Equation 10.64 is c12 + 2c1c2 S AB + c22 = 1 where SAB is the overlap integral involving ϕA and ϕB.. ψ = c1φA + c2φB. (Equation 10.64). 10‒48. In the B‒O approximation, we assume that because the nuclei are so much more massive than the electrons, the electrons can adjust essentially instantaneously to wany nuclear motion, and hence we have a unique and well‒defined energy, E(R), at each internuclear separation R. Under this same approximation, E(R) is the internuclear potential and so is the potential field in which the nuclei vibrate. Argue, then that under the B‒O approximation, the force constant is independent of isotopic substitution. Using the above ideas, and given that the dissociation energy for H2 is D0 = 430.3 kJ·mol‒1 and the fundamental vibrational frequency υ is 1.32 × 1014 s‒1, calculate D0 and for deuterium, D2. Realize that then observed dissociation energy is given by D0 = D0 ‒ 0.5hυ where De is the value of E(R) at Req..

<span class='text_page_counter'>(4)</span>