Lecture #9 Redfield theory of NMR relaxation

Transcription

1 Lecture 9 Redfield theory of NMR relaxation Topics Redfield theory recap Relaxation supermatrix Dipolar coupling revisited Scalar relaxation of the st kind Handouts and Reading assignments van de Ven, Chapters 6.2. Kowalewski, Chapter 4. Abragam Chapter VIII.C, pp , 955.

2 Redfield theory We ended the last lecture with the following master equation dσ dt = iĥ 0 σ Γ σ σ B ( ) and we were happy! relaxation superoperator 2

3 Redfield theory The relaxation superoperator was defined as: - spectral density functions: J q e q - correlation functions: G q 0 ( ) = G q τ ( τ ) = F q ( t )F q t τ Γ = J ( q e ) q  q  q q ( )e ie qτ dτ. ( ) H 0 eigenoperators Recipe:. Given 3. Plug and chug. Ĥ ( t) = Ĥ0 + ( Ĥ t). Random functions of time 2. Express t as a linear combination of eigenoperators of Ĥ Ĥ ( ) Ĥ 0, ( t) = F q ( t)  q. q But how do we compute T or T 2? 3

4 The Relaxation Supermatrix Rather than directly solving the master equation, we often just want to calculate the time dependence of particular coherences, e.g. Î x, Î y, or Î z. Express the density operator in the product operator basis σ = j C j C j C = Tr( σ C )! σ =! I x S x! 2I z S z C j { 2 Two-spin case E, I x, S x, I y, S y,,2 I z S z } vector in a 6-D coherence (Liouville) space Rewrite the master equation as a vector/matrix equation: 4

5 The Relaxation Supermatrix Rewrite the master equation as a matrix equation: Γ R Supermatrix with elements R jk = Tr Ĉ j ΓĈ k! If we reorder σ to first list populations, then singlequantum terms, then doublequantum terms,... - Relaxation supermatrix, R, is block diagonal (secular approx). - Cross relaxation only occurs for coherences with degenerate eigenvalues of Ĥ 0. trace ( ) = Ĉ j Γ Ĉk Notation used in van de Ven. Don t confuse with expected value. H 0 eigenvalue 0 ±(ω I -ω S ) ±ω I, ±ω S From Problem Set, these eigenvalues are the transition frequencies, i.e. sums and differences of the system energy levels ±(ω I +ω S ) 5

6 Calculating Relaxation Times Rewrite the master equation in terms of the operator coefficients: d dt Ĉ j = i Ĉ j Ĥ 0 Ĉ k Ĉ k Ĉ k Ĉk k ( ) Ĉ j Γ Ĉ k B ( Rotations Relaxation Examples T,I = Îz Γ Îz T,S = Ŝz Γ Ŝz T,cross = Îz Γ Ŝz T 2,I = Î x Γ Î x = Î y Γ Î y we just need to compute some (a bunch) of commutators. 6

7 Relaxation due to a random field (or hints for Problem Set 5) Consider a Hamiltonian of the form Ĥ = Ĥ0 + Ĥ t ( ) = γb 0 Î z γδb( t)îz with ΔB t ( ) = 0 ΔB( t)δb t τ ( ) = B 2 e τ τ c Noting that Ĥ 0 Î z = 0 Îz Eigenvalue Eigenoperator ( ) = γδb ( t) J 0 ( ω) = γ 2 B 2 τ c then  0 = Îz, F 0 t, and T 2 = Î x Γ = J ( q e ) q  q  q q Γ Î x = J 0 ( 0)B 2 Î z Î z +ω 2 τ c 2 Assumed to be normalized = γ 2 B 2 τ c Tr( Î Î x z Î z Î ) = γ 2 x B 2 τ c Tr( iî Î x z Î ) = γ 2 y B 2 τ c Tr( Î xî ) x Hence: T 2 = γ 2 B 2 τ C T =? 7

8 Dipolar Coupling Revisited The complete dipolar coupling Hamiltonian is given by Ĥ dipole = γ Iγ S! µ 0! I Ŝ! 3 r 3 4π r (! Î r)(! Ŝ! r)! ( 2 This can be written as: Ĥ D ( t) = γ Iγ S! µ 0 r 3 4π q F q ( t) Âq where! r where vector from spin I to spin S With tumbling, both θ and φ are functions of time.  0 = ( 6 2ÎzŜz 2 Î+Ŝ ) 2 Î Ŝ+  ± = ± 2  ±2 = 2 αŜ± ( Î ± Ŝ z + ) ÎzŜ± and F 0 (t) = 2 3 ( 3cos 2 θ ) F ± (t) = ±3sinθ cosθe iφ F ±2 (t) = 3 2 sin2 θe 2iφ Hey! These look like rank 2 spherical harmonics.

9 Dipolar Coupling Revisited Noting the following are eigenoperators of (see Problem Set ) H 0 Eigenoperator Eigenvalue Î z Ŝ z Î + Ŝ + 0 ( ω I +ω S ) Î Ŝ Î + Ŝ ω I +ω S ( ω I ω S ) Together with F q * F q = 6 5 Î Ŝ + ω I ω S Î + Ŝ z ω I Î Ŝ z ω I Î z Ŝ + ω S Î z Ŝ ω S We can now compute Γ.

10 Dipolar coupling superoperator Case : unlike spins, after much algebra Γ = 4 γ 2 ( Iγ 2 S! 2 * ) 2J 0 0r 6 + * + ( ) ÎzŜz Î z Ŝ z 4 J ( ω ω I S ) J ( ω +ω I S ) J ω I ( ) Î xŝz Î x Ŝ z+ Î y Ŝ z Î y Ŝ z 4 J ( ω ω I S ) 3 2 J ( ω +ω I S ) Î xŝx J ω S Î xŝx Î x Ŝ x+ Î y Ŝ y ( ) ÎzŜx Î y Ŝ y+ Î y Ŝ y Î y Ŝ y+ Î x Ŝ y Î z Ŝ x+ Î z Ŝ y Î x Ŝ x Î x Ŝ y Î x Ŝ y+ Î y Ŝ x Î z Ŝ y Î y Ŝ x Î y Ŝ x Î y Ŝ x Î x Ŝ y Note: Î z Ŝ z Î z Ŝ z

11 Dipolar coupling superoperator Before calculating a bunch of commutators, we should note that there are multiple terms of the form C q Ĉ q, and this can make things easier C q Ĉ q Ĉ p = 0 if Ĉ q Ĉ p = 0 Ĉ p if Ĉ q Ĉ p 0 Remember all product operators cyclically commute. Terms of the form C q Ĉ r give rise to cross relaxation Example: Î x Ŝ y Î y Ŝ x Î z = 4 Ŝz

12 Dipolar coupling unlike Spins Calculating T 2. Let q = µ 2 0 γ 2 I γ 2 S! 2 6π 2 0r 6 ΓÎ = q 2J 0 x ( ( ) + J ω 2 ( I ω S ) + 3J ( ω I +ω S ) J ( ω I ) + 3J ( ω S ))Î x T 2,I = Î x Γ Î x = q 4J 0 2 ( ( ) + J ( ω I ω S ) + 6J ( ω I +ω S ) + 3J ( ω I ) + 6J ( ω S )) Let s calculate T. ΓÎz = q ( J ( ω I ω S ) + 6J ( ω I +ω S ) + 3J ( ω I ))Îz + J ω I ω S T,I = Îz Γ Îz ( ( ) + 6J ( ω I +ω S ) + 3J ( ω I )) = q J ω I ω S And the cross relaxation term is: T,IS = Ŝz Γ Îz ( ( ) J ( ω I ω S )) = q 6J ω I +ω S ( ( ) 6J ( ω I +ω S ))Ŝz

13 Dipolar coupling like spins Case 2: The equation for spins with the same (or nearly the same) chemical shift, ~ω 0, is even longer due to cross terms between Î z Ŝ z and Î ± Ŝ. Γ = see van de Ven p. 355 T = q 3J(ω 0 )+2J(2ω 0 ) ( ) But now there is also transverse cross relaxation between spins I and S. T 2,I = Î x Γ Î x = q 5J 0 2 ( ( ) + 9J ( ω 0 ) + 6J ( 2ω 0 )) T 2,IS = Ŝx Γ Î x = q 2J 0 ( ( ) + 3J ( ω 0 )) This effect is exploited in some spin lock experiments.

14 Summary of Redfield theory dσ dt = iĥ 0 σ Γ σ σ B ( ) Relaxation arises from perturbations having energy at the transition frequencies. That is, if the eigenvalues of Ĥ 0 (= energies of the system/!) are e, e 2, etc. Then, the spectral density function is probed as frequencies ±(e i e j ). Cross relaxation only occurs between coherences with the same transition frequencies. Although Redfield theory may seem much more complicated than the Solomon equations for dipolar relaxation, it is actually very useful. For example, T and T 2 due to chemical shift anisotropy or scalar relaxation of the st and 2 nd kind are readily calculated. 4

15 Example: Scalar relaxation of the st kind Consider a J-coupled spin pair with the following Hamiltonian: ( ) Ĥ = Ĥ0 + Ĥ = ω I Î z ω S Ŝ z + 2π J ÎzŜz + Î xŝx + Î yŝy We would normally expect a doublet from the I spin, however chemical exchange by the S spin can become a relaxation mechanism. Under exchange, with an exchange time of τ ex, the coupling constant between the I spin and a spin S i becomes a random function of time. Rewriting the perturbing Hamiltonian:! where A 2 i = A i ( t) A i t +τ Ĥ A 2 = 4π 2 J 2 0 otherwise ( t) = A i ( t)!! I S if I and S i are on the same molecule ( ) = A 2 e τ τ ex = probability the I and S i spins are on the same molecule at time t + τ, given that they are at time t.

16 Example: Scalar relaxation of the st kind Thus we have: Ĥ 0 = ω I Î z ω S Ŝ z and Ĥ t Noting the eigenoperators and corresponding eigenvalues of Ĥ 0 Eigenoperator Eigenvalue Î z Ŝ z 0 Î + Ŝ ( ω I ω S ) Î Ŝ + ω I ω S Written as a sum of eigenoperators of, the perturbing Hamiltonian becomes Ĥ H 0 ( ) = A i ( t) ÎzŜz + Î xŝx + Î yŝy ( t) = A i ( t)îzŝz + A 2 i ( t)î+ŝ + A 2 i ( t)î Ŝ+ ( ). All we need now is the spectral density function, which we ll denote J ex ( ω). Let P i be the probability spins I and S i are on the same molecule, then J ex ( ω) = P i A i ( t) A i ( t +τ ) cosωτ dτ. i 0 Assume the I spin is always coupled to some S spin, i.e. P i = J ex ( ω) = A 2 τ ex +ω 2 2 τ ex i

17 Example: Scalar relaxation of the st kind Hence: Γ = A 2 J ex ( 0)ÎzŜz Î z Ŝ z+ 4 A 2 J ex ( ω I ω S )Î+Ŝ Î + Ŝ + 4 A 2 J ex ( ω I ω S )Î Ŝ+ Î Ŝ + From which it follows T,I = Îz ( ) Γ Îz = 2A S S τ ex + ( ω I ω S ) 2 2 τ ex = 8π 2 J 2 S( S +) 3 τ ex + ( ω I ω S ) 2 2 τ ex T 2,I = Î x Γ Î x = Î y ( ) = Note: the S(S+)/3 factor comes from Tr Ŝp 2 S(S +), p = product operator 3 where S = spin of the unpaired electron system or nucleus. ( ) Γ Î y = Γ Î+ Î + = 4π 2 J 2 S S + 3 τ ex + τ ex + ( ω I ω S ) 2 2 τ ex For those who complete the homework, we note that these equations have the same form as scalar relaxation of the 2 nd kind with the correlation time given by τ ex instead of T,S and T 2,S.

18 Next Lecture: Redfield theory- Examples 8