Solution Set 3. Hand out : i d dt. Ψ(t) = Ĥ Ψ(t) + and

Transcription

1 Physikalische Chemie IV Magnetische Resonanz HS Solution Set 3 Hand out : 5.. Repetition. The Schrödinger equation describes the time evolution of a closed quantum system: i d dt Ψt Ĥ Ψt Here the state of the system is described by a state function Ψt. The state of a quantum system can also be described by a density operator ˆρt, which for pure states contains the same information as the state function. The time evolution of the density operator is described by the Liouville-von-Neumann-equation: d dt i ˆρt [Ĥ, ˆρt] Dening the density operator as Ψt Ψt : ˆρt one can derive the Liouville-von-Neumann-equation out of the Schrödinger equation as shown in chapter 4.9 in the script.. The Hilbert space dimension for a spin system with I is D I 3, where I is the spin quantum number. An example of an appropriate vector basis is given by, and which are the common eigenvectors of the operators Î with the eigenvalue and Î z with the eigenvalues,, -. By the matrix representation of these vectors,,, we can easily show that, i.e. the three vectors form an orthogonal basis. This basis is not unique. For example, the vectors x i, y i, and, full the conditions x x y y z and x y x y and form another vector basis of this 3-dimensional Hilbert space. 3. The eigenvalues of a Hamiltonian operator Ĥ are the solutions E k of the time-independent Schrödinger equation. H Ψk Ek Ψk The expectation value of an observable A represented by an operator Â, in a quantum system characterized by the state function Ψt, is dened by  t Ψt  Ψt Ψt Ψt

2 Physikalische Chemie IV Magnetische Resonanz HS 4. The spin Hamiltonian ĤS, is derived from the full system Hamiltonian Ĥ, by applying the Born-Oppenheimer approximation and integrating over the space coordinates of all electrons and over the spin coordinates of the paired electrons. This process is represented by Ĥ S Ψ q, s e, s k Ĥ Ψ q, se, s k In this simplied Hamiltonian all the information about molecular structure is found in numerical constants and the only variables of the spin Hamiltonian ĤS, are the spin coordinates in the form of nuclear spin operators s k. 5. An ensemble of spin systems in which all spins are in the same state is called a pure state. In such a state only one element of the density matrix is unequal zero. In the case, where all spins are in the α-state, only the rst element in the density matrix is populated: ρ Problem : Equilibrium spin density operator. Only the interaction of the spin system with the external magnetic eld is considered: H ˆ H ˆS ˆ H Z γ Î B γ B Îz rad s for H 3.5 rad s for free e As shown in the lecture the equilibrium spin denstiy operator can be simplied by an exponential expansion. Keeping terms of. order one gets: ˆσ ˆ T r{ˆ} H ˆ k B T T r{ˆ}.5 for.4999 H ˆσ.4664 for free e In the st part of this exercise one saw, that NMR is a quite insensitive method as only.4 % of all spins contribute to the overall magnetization. In contrast, using electron spins, one get a value of about.6 %. In optics, the population of a two spin system is given by the Boltzmann distribution too. For the given transition the energy dierence between Js m s the two levels is given by E h ν hc λ 53 9 nm J. For the Boltzmann distribution one gets f E e E kt e J.38 3 K J K e Hence, only the ground state is populated an one get's nearly % contribution. Remark : Optical transitions are much easier to measure than spin transitions < I x > Tr{ˆσ I x } for H and e. < I z > Tr{ˆσ I z } for e and for H. 4. ˆσ ˆσ ˆ T r{ˆ} 4.5 for.5 H.336 for free e.336

3 Physikalische Chemie IV Magnetische Resonanz HS 5. The spin-up population is represented in ˆσ, the spin-down population in ˆσ. Therefore the ratio is calculated as { ˆσ.9998 for H ˆσ.44 for free e For a positive negative gyromagnetic constant one gets a spin-up to spin-down ratio smaller larger than, as the α β-spin states lie at lower energies and are therefore more populated. Problem : Pauli matrices. Ix i i i i i i Iy i i i i i i Iz i i i. I x I y ; I i i y I x i i I i x I y I y I x i i I z 3. Tr {I x } ; Tr {I y } ; Tr {I z } 4. M a a I x a I y a 3 I z a a a a a a 3 a i a a i a a a 3 a m m [ 5. I I i i i ; a m m ] i a i a a3 a 3 ; a i m m ; a 3 m m I z I 6. Remark : In the exercise the commutator [I, I z ] is not equal to zero and hence the example is not correct. Instead, the solution is given for I z and I. As I I x I y I z 3 according to., the product I I z is equal to I z I in mathematical terms the two operators commute whereas I x I y i I z according to. Hence one can assume, that the two operators have to commute to make the special case of the Baker-Campbell-Hausdor formula be valid. 3

4 Physikalische Chemie IV Magnetische Resonanz HS Problem 3: D NOESY Nuclear Overhauser Enhancement SpectroscopY. The dierential equation is give by d MAz R σ AB dt M Bz σ AB R MAz M Bz where R T. The characteristical polynom of the matrix is: R λ σ det AB pλ! σ AB R λ R λ σ AB λ, R ± σ AB To diagonalize the matrix one need furthermore the eigenfunctions of the Matrix which can be calculated out of the eigenequation see exercise : R λ, σ AB c, σ AB R λ, c, ±σab σ AB c, σ AB ±σ AB c, ±c, c, ±c, c, c Resulting in the two eigenvectors c c and c. The new magnetization vectors expressed in the new basis can be derived from the original one by R σ writing the relaxation matrix R AB as R T σ AB R Λ T T M Λ T M where Λ is the diagonal matrix with the eigenvalues on the diagonal and T is a matrix consisting of the eigenvectors and is given by T. The new magnetisation vectors can now be derived from ṀAz R σ AB MAz Ṁ Bz R σ AB M Bz ṀAz ṀBz R σ AB MAz M Bz R σ AB M Az M Bz ṀAz ṀBz The solution is now quit simple: MAz t M Bz t M Az t M Bz t e R σ AB t e R σ ABt which results in the solution for M Az and M Bz in the form: M Az M Bz M Az M Bz M Az t M Az MBz e λ t M Az MBz e λ t M Bz t M Az MBz e λ t M Az MBz e λ t As the pulse sequence for a D NOESY spectrum is identical to that for a D exchange spectroscopy, we obtain the NMR signal st, τ, t M {a τ cos Ω t cos Ω t a τ cos Ω t cos Ω t a τ cos Ω t cos Ω t a τ cos Ω t cos Ω t } 4

5 Physikalische Chemie IV Magnetische Resonanz HS with the relative diagonal- and cross- peak intensities eλt e λt a a eλt e λt a a cosh σ AB t e t T sinh σ AB t e t T One derives four signals with intensities a, a, a and a, respectively and frequency Ω and Ω each damped with the factor e t T. M A M e z π/ M e x t M cos Ω t e x sin Ω t e y π/ M cos Ω t e z sin Ω t e y τ mixing according to the differential equation π/ s t, τ, t M z t, τ cos Ω t π/ π/ π/ Η t τ t Figure : D NOESY experiment with magnetisation analysis. 5

6 Physikalische Chemie IV Magnetische Resonanz HS. For the homonuclear case, the cross-relaxation rate is given by σ AB µ γ 4 4π rab 6 τ c 6 τ c 4ω τ c with γ the gyromagnetic ratio of the involved type of nucleus here protons with γ Proton,675 8 s T and τ c the correlation time we assume τ c 3. ns. By inserting one point except τ s of the buildup curve given in Table into the equation above, we are able to calculate both r AB and T. The cross-relaxation rate σ AB can be derived by division of the two formulas derived above: a a tanh σ AB t. The values calculated from this are given by τ [s] σ AB [ s ] r AB [Å] T [s] mean Table : List of distances and relaxation time T derived from the intensities given on the exercise sheet. 6