Density Matrix Second Order Spectra BCMB/CHEM 8190

Transcription

1 Density Matrix Second Order Spectra BCMB/CHEM 819

2 Operators in Matrix Notation If we stay with one basis set, properties vary only because of changes in the coefficients weighting each basis set function μ x = γ(h/2π)<ψ Ix ψ> ψ= c1 αα + c2 αβ + c3 βα + c4 ββ = j c j φ j <ψ Ix ψ> = j,k c j * c k <φ j Ix φ k > We need calculate <φ j Ix φ k > only once if we stay with this basis set these can be put in a n x n matrix. Matrix equivalent: <ψ Ix ψ> = (c1, c2, )* c1 Ix c2

3 Special Case: Pauli Spin Matrices Ix = ½ ½ Note: < α Ix α> = ½ < α β > = < α Ix β > = ½ < α α > = ½ Iy = i½ -i½ Iz = ½ ½ How do they work? Try something we know: Ix α > = ½ β ½ ½ 1 = = ½ = ½ β ½ 1 Operators are a matrix of numbers, Spin functions a vector of numbers

4 Larger Collections of Spin ½ Nuclei =??????? αα I Ax = βα???? αβ???? ββ <αα I Ax βα> = <αα ½ αα> = ½ 1/2 1/2 1/2? αα βα αβ ββ 1/2 I Xx = 1/2 1/2 1/2 1/2 Easier way: direct products: E I Ax with 2X2 matrices

5 Density Matrices and Observables for an Ensemble of Spins Expectation values pertain to single spin properties; we observe net behavior of an ensemble of spins. For a particular system (two spin ½ nuclei) operator representations are the same; all variations are in basis set coefficients. We need to average over these coefficients. <ψ Ix ψ> = (c1, c2, )* c1 Ix c2 = j,k c j * c k Ix jk Products of averaged coefficients can also be collected and used in matrix form. This is called a density matrix, ρ. <ψ Ix ψ> = Tr c j c k * Ix = Tr ρ Ix

6 Solving for the Time Dependence of ρ Our observables are time dependent (magnetization precesses). All time dependence can also be put in basis set coefficients, or in coefficient products of the density matrix. Schrodinger s time dependent equation (H ψ(t) = - i (h/2π) d(ψ(t))/dt) Allows us to solve for dc j /dt, d/dt (c j c k *) = c j (dc k */dt) + (dc j */dt) c k Result in terms of a density matrix is Lioville von Neuman eq. d/dt ρ = i / (h/2π) { ρ H - H ρ } Implications: If we know ρ at any time (equilibrium at time ) and know the Hamiltonian (H), we can solve for ρ at any time, and calculate any observable as the Tr ρ O (O is any operator).

7 Test: Does X Magnetization Precess in B? What elements of ρ dictate x magnetization? <I x > = Tr [ρ] [I x ] - examine spin ½ case = Therefore, consider ρ() = What is H?, H = -(γhb /(2π))I z =-(γhb /(2π)) d / dt ρ21 i / h( γhb ρ11 Tr ρ21 ρ12 = ) ρ21 ρ12 ρ22 1/ 2 ρ12 1/ 2 1/ 2 ρ21 1/ 2 = 1/ 2ρ12 + 1/ 2ρ21 ρ12 1/ 2 1/ 2 1/ 2 ρ21 1/ 2 ρ12

8 Consider evolution of one element (d/dt)ρ 12 (t) = iγb (-½ ρ 12 (t) ½ ρ 12 (t)) = iω ρ 12 (t) Solution: ρ 12 (t) = ρ 12 () exp(-iω t) or: = ρ 12 () (cos(ω t) i sin(ω t) Same happens for ρ 21 (t) hence elements corresponding to x magnetization precess t = t = 1

9 Density Matrix Simulation of 2nd Order Spectra For two spins basis set is: αα, αβ, βα, ββ ρ = ρ11 ρ21 ρ31. ρ12 ρ22.. ρ ρ 12 ρ 34 ρ 13 ρ 24 ρ 12,ρ 21, are associated with lines in an AX spectrum Don t have to have a 1:1 association if ψ is not an eigen function of H. c i φ I + c i φ I may evolve coherently as one line ie ρ 12,ρ 21, are mixed Calculation of Mx still works!

10 Second Order Spectrum Example: Acetonitrile 1 H C 1 H A C B A N C 1 H B AMX spectrum Twelve lines, all equal intensities J AB = -2 Hz, δ C = 5.6 ppm J AC = 11 Hz, δ B = 6.1 ppm J BC = 16 Hz, δ A = 6.3 ppm Actual spectrum is more complex, field dependent

11 Field dependence of second order spectra 6 MHz 9 MHz 25 MHz 49 MHz

12 Line intensities: density matrix elements and transition probabilities ρ kl = c* k c l ; c* k c l is the probability of being in state k c* k c k c* l c l is the probability of starting in k and ending in l If we know we start in k, c* k c k = 1, and just need c l from Schrodinger equation: dc l /dt <φ l H φ k > Integrating from t = and combining with c* l P k->l <φ l H φ k > 2 gives line intensities

13 Understanding second order spectra A 2 vs AX or H 2 O vs HF For HF.. P ββ->αβ <ββ γb 1 (I x1 +I x2 ) αβ> 2 ; I x = (I + +I - )/2 this is finite and same for all P ij implies same line intensities ββ αβ βα αα E

14 For an A 2 system: αα, αβ, βα, ββ are not good wave functions Imply distinguishability Use: αα, (αβ+βα)/ 2, (αβ-βα)/ 2, ββ <ββ γb 1 (I x1 +I x2 ) (αβ+βα)/ 2 > = (½ + ½)/ 2 <ββ γb 1 (I x1 +I x2 ) (αβ-βα)/ 2 > = (½ -½ )/ 2 = Outer transitions are not allowed ββ (αβ+βα)/ 2 αα (αβ-βα)/ 2 E

15 Intermediate behavior of 2 nd order spectra H 2 O AB Outer lines are not Observed Couplings still exist HF

16 Second order effects can be important Base O φ H 5 C H 5 ' OH J 45? H 4 X HO OH Coupling of H4 and H5, H5 protons in ribose rings an ABX system Karplus equation might be used to deduce torsion angle, but can couplings be measured from multiplet? For degenerate H5, H5, see a triplet, but splitting is (J 45 +J 45 )/2 Cannot conclude that J 45 = J 45 Answer: simulate spectra See: S.A Smith, J. Magn Res, 166, 75 (1994); M.Veshtort, R.G.Griffin, J Magn. Res., 178, (26)