ntroduction to NMR Product Operato C. Griesinger Max Planck nstitute for Biophysical Chemistry Am Faßberg 11 D-3777 Göttingen Germany cigr@nmr.mpibpc.mpg.de http://goenmr.de EMBO Coue Heidelberg Sept. 23
63 Density Matrices: We have two levels of description for the status of a spin system, the density matrix expressed as cartesian operato, or single element operato or the energy level diagram. Cartesian Basis Single element basis ρ 1:, x; y; z ; ρ :, ; + ; α β ; Energy level diagram α β + Time evolution of Density Matrices: The density matrix transforms under the Liouville von Neumann equation in the following form: i = [ H ρ ] ρ, Hamiltonian Operato under which the density matrix transforms are: Chemical Shift: H CS = Ω z + Ω S S z Pulses: H pulse the Hamiltonian: H = γ B1 = ω. After time τ the pulse has aquired the phase: ω τ = β. A pulse with pulse x x = γ B1 = ω of duration τ is therefore called a β ( ) pulse. x x x Scalar Coupling between spin and S: J H = 2πJ S z S z Transformation of Density Matrices under pulses, chemical shift and coupling for a two spin system of and S in the cartesian operator basis: Chemical Shift: CS CS CS H H H x x cos Ω t + y sin Ω t ; y y cos Ω t x sin Ωt ; z z CS CS CS H H H Sx S x cos ΩSt + S y sin ΩSt ; Sy y cos Ω St Sx sin ΩSt ; S z S z
64 Pulses: S β β x ( ) β x( ) ; cos β + sin β ; cos β sin β x ( ) x x ( S ) x x S x y β β x ( S) ; S S cos β + S sin β y y y z z z β x ( S) ; S S cos β S sin β z z z z z Scalar Coupling: x J J H H cos πj t + 2 S sin πj t ; cos πj t 2 S sin πj t x J H S y z S z J H y y S x z S ; z J H S t 2 xs z 2 xs z cosπj S t + y sin πj S 2 S 2 S cosπj t sin πj ; t y z y z S x Transformation of Density Matrices under pulses, chemical shift and coupling for a two spin system of and S in the single element operator basis: Chemical Shift: + + CS H S+ S + e CS H iωt e ; iω St ; CS H S S CS H + iωt e ; H CS α α iω St e + ; S H CS α Sα CS ; β H β CS ; S β H Sβ Pulses: β β β φ ( ) 2 2 i iφ iφ α α cos + β sin + ( + e + e ) sin β 2 2 2 β β β φ ( ) 2 2 i iφ iφ β β cos + α sin ( + e + e ) sin β 2 2 2 β φ ( ) 2 β i2φ 2 β i iφ iφ + + cos + e sin + αe + βe sin 2 2 2 β φ ( ) 2 β i 2φ 2 β i iφ iφ cos + + e sin α e + β e sin 2 2 2 Scalar Coupling: + Sα J H iπj St + Sα e ; + S J H + iπj St β + Sβ e ; S S ( ) β ( ) β J H iπj St α Sαe ; H J α α β J H S β e iπj St J ; β H β
65 Transformation of Density Matrices under pulses, chemical shift and coupling for a two spin system of and S in the energy level picture: S S S + S α S S The evolution of coherences between the levels is given by the energy difference of the levels connected. i Et The evolution goes like: e : E.g. J CS H + H + Sα + S α e i( Ω + πj S) t Populations have no energy difference, therefore they are invariant under time evolution. Pulses can be calculated by looking at the action of a pulse on each of the functions involved in a coherence. For that we need the transformation of the functions under pulses: β β β α > x α > cos + i β > sin 2 2 β β β β > x β > cos + i α > sin 2 2 Thus a β x ( ) pulse onto the operator S + α = αα >< βα effects the following transformation: e e i β x + i( β / 2) S α e i β x = ( αα>< βα + βα>< αα + αβ>< ββ + ββ>< αβ ) i( β / 2) ( αα>< βα + βα>< αα + αβ>< ββ + ββ>< αβ ) αα >< βα e n order to simplify this expression, we fit notice that the the two pai of operato contained in each of the exponentials commute: [ αα >< βα + βα >< αα, αβ >< ββ + ββ >< αβ ] = Due to this fact, they can be applied sequentially:
66 e e i( β i( β / 2) / 2) ( αα >< βα + βα >< αα + αβ >< ββ + ββ >< αβ ) ( αα >< βα + βα >< αα ) i( β / 2) ( αβ >< ββ + ββ >< αβ ) We calculate now one of the two expressions: e i( β / 2) n ( αα >< βα + βα>< αα ) e [ i( β / 2) ( αα >< βα + βα >< αα )] cos( β / 2) or in matrix form: cos( β / 2) cos( β / 2) isin( β / 2) = n! n = ( αα >< αα + βα >< βα ) + isin( β / 2) ( αα >< βα + βα >< αα ) ( αα >< αα + βα >< βα ) + i sin( β / 2) ( αα >< βα + βα >< αα ) i sin( β cos( β / 2) / 2) 1 1 The ordering of the basis functions is: αα, βα, αβ, ββ. The second rotation matrix can be constructed in the same way and we find: cos( β / 2) 1 1 ( αβ >< αβ + ββ >< ββ ) + i sin( β / 2) ( αβ >< ββ + ββ >< αβ ) cos( β i sin( β / 2) / 2) isin( β / 2) cos( β / 2) Application to S + α = αα >< βα can be obtained now in the following way: e i β x + S α e i β x = ( αα>< βα + βα>< αα + αβ>< ββ + ββ>< αβ ) i( β / 2) ( αα>< βα + βα>< αα + αβ >< ββ + ββ>< αβ ) i( β / 2) e αα >< βα e = { c β ( αβ >< αβ + ββ >< ββ ) + is β ( αβ >< ββ + ββ >< αβ )}{ c β ( αα >< αα + βα >< βα ) + is β ( αα >< βα + βα >< αα )} αα >< βα { c ( αβ >< αβ + ββ >< ββ ) is ( αβ >< ββ + ββ >< αβ )}{ c ( αα >< αα + βα >< βα ) is ( αα >< βα + βα αα )} β β β β >< = = =
67 Of all the indicated transitions, only the ones that carry αα> as ket and <βα as bra Calculation of the transformation leads to: S S + S α β x S α β S α + S α α S α Fictitious Two Level Operato: How do we transfer population on the αβ state into the βα state? We can look how we do this in the simplest spin system, namely a single spin system and we ask the question, how do we transfer population on the α state into population on the β state. This can be done by a π pulse as we know. A iπ π x pulse is given by: π = = i( π / 2)( α>< β + β>< α ) x e x e. Thus if we want to apply a pulse iθ i( / 2)( r s s r ) across a certain transition, we simply apply a x e x θ >< + >< θ = = e pulse. Here the flip angle is θ. The effect of this pulse will be as known for the single spin operator transformations: z x y θ x θ y θ z cos θ y sinθ; z z cosθ + x sinθ; z z z θ θ θ ; y x cosθ sin θ; cosθ x x x z z x x + y sin θ Eq. [5] θ x cosθ y + z sinθ; y θ y y ; y θ z y cosθ x sinθ Optimization of Pulse Sequences: When optimizing pulse sequences one has to distinguish the following levels:
68 a) Optimal experimental implementation of the pulse sequence? This addresses the question whether the pulse sequence does what it is supposed to do. E.g. off-resonance effects of pulses or polarization transfer segements, decoupling bandwidth etc. There is a vast body of literature addressing this problem: e.g. broad band decoupling (WALTZ, GARP, WURST), TOCSY transfer: (CW, MLEV, DPS, FLOPSY etc.). nveion pulses (normal, composite, shaped). This shall be mentioned only shortly in this lecture. b) Optimal pulse sequence? Every pulse sequence consists of one or several transfe of coherences. The question addresses the problem whether the coherence transfe are optimal with respect to sensitivity. This question can be answered by analysing the bounds of coherence transfer ignoring relaxation. These bounds will be introduced and examples will be given. t will also be discussed on several examples how to find the optimal pulse sequence. c) Optimal coherences? This is perhaps the most fundamental question out of those a)-c). t addresses the question whether the right coherences are chosen to create the spectrum. There are several coherences that provide the same spectroscopic information but they may give rise to considerably varying quality of spectra even after optimization according to b) and a). Bounds and optimal pulse sequences for Hermitian A and C: The optimization of a desired coherence transfer ignoring relaxation requires to find the unitary transformation U that maximizes a transfer that goes from a coherence A to a coherence C: A U ac + B Eq. [1] a is given by: 1 Tr{ UAU C } a = Eq. [2] Tr{ C C} Let's assume that A and C are both hermitian matrices and that we have chosen the eigenbasis of C with the eigenvalues in descending order. Then the trace is maximized if A is also diagonalized and the eigenvalues of A are also sorted in descending order.
69 = 44 33 22 11 44 33 22 11 C C C C A A A A Tr max. Eq. [3] Example: NEPT transfer in S system: For the NEPT transfer it is possible to transfer completely x -> S x in an S spin system, however, this is not possible for a 2 S spin system where one can achieve 1x + 2x -> S x. The question is, whether this is a fundamental limitation or whether the NEPT is not optimal. The matrices for x and S x in an S spin system are both diagonalized by a 9 y pulse. We look therefore for z and S z : = = = =.5.5.5.5 ;.5.5.5.5 z S z C A Eq. [4] The eigenvalues are still not yet ordered correctly for S z. f we do this by exchanging the αβ and the βα populations we obtain: = =.5.5.5.5 ;.5.5.5.5 ' z S z Eq. [5]
7 Obviously z and S' z are the same. Therefore the transfer from z to S z is possible with an a=1, thus full transfer. Fictitious Two Level Operato: How do we transfer population on the αβ state into the βα state? We can look how we do this in the simplest spin system, namely a single spin system and we ask the question, how do we transfer population on the α state into population on the β state. This can be done by a π pulse as we know. A iπ π x pulse is given by: π = = i( π / 2)( α>< β + β>< α ) x e x e. Thus if we want to apply a pulse iθ i( / 2)( r s s r ) across a certain transition, we simply apply a x e x θ >< + >< θ = = e pulse. Here the flip angle is θ. The effect of this pulse will be as known for the single spin operator transformations: z x y θ x θ y θ z cos θ y sinθ; z z cosθ + x sinθ; z z z θ θ θ ; y x cosθ sin θ; cosθ x x x z z x x + y sin θ Eq. [5] θ x cosθ y + z sinθ; y θ y y ; y θ z y cosθ x sinθ With this information, we can now directly write down the pulse sequence for this transfer z -> S z. Application of a π αβ> βα> x pulse will effect the desired transfer. The popagator that represents this pulse is: e i( π / 2)( αβ>< βα + βα>< αβ ). We now have to translate ( αβ >< βα + βα >< αβ ) into the cartesian product operato. This yields: ( αβ >< βα + βα >< αβ ) 1 = = 2( xsx + ysy ) 1 Eq. [6] Thus we have to apply the following propagator: π αβ> βα> x = e i( π / 2)( αβ>< βα + βα>< αβ ) = Eq. [7a] i( π / 2)(2 xsx + 2 ys y ) e = Eq. [7a] i( π / 2)(2 xs x) i( π / 2)2 ys y ) Eq. [7c] e e
71 The fit transformation Eq. 7a is a planar coupling Hamiltonian πj ( x Sx + ys y )(1/ J ) applied during duration (1/J). This is indeed an implementation of the heteronuclear polarization transfer. The planar Hamiltonian π J ( xsx + ys y) is generated from the weak coupling Hamiltonian 2 πj z S z by a multipulse sequence that applies pulses only along x (e.g. DPS-2) and flanking 9 y pulses on and S. The multipulse sequence generates: π J ( zsz + ys y ) and the flanking 9 y pulses on and S rotate this to π J ( xsx + ysy ) : e e = i( π / 2)(2 xsx + 2 ys y ) = i( π / 2) y i( π / 2) S y i( π i( π / 2) e e (2 zs z + 2 y S y ) e e y y z z y y / 2) y i( π e i( π / 2) i( π / 2) S i( π / 2)(2 S + 2 S ) i( π / 2) i( π / 2) S e e e y e / 2) S y y Eq. [8] This yields the implementation given in Fig. 2a). x 2τ -x a) 1 H 13 C x -x b) x -x y -y c) 1 H τ1 τ2 1 H x τ1 -x y τ2 13 C x y -x -y 13 C -x y -y d) 1 H 13 C x τ1 -y z -y τ2 z-y-z e) x τ1 -y 1 H 13 C -y τ2 x
72 The second (Eq. 7c) consists of two terms. Looking at the fit of the two we find the transformation of Eq. 8: i ( / 2)(2 x S x ) e π i( π / 2) y i( π / 2) S y i( π i( π / 2) e e (2 = zs z ) e e i( π / 2) y i( π / 2) S y e e i( π / 2) y i( π / 2) S y i( π / 2)(2 i y i = zs ( / 2) ( e e e z ) π π e e i( π / 2)(2 S ) 9 y(, S) e z z 9 y (, S) / 2) / 2) S y i( π / 2) S e y y = Eq. [9] The middle term is free evolution of heteronuclear coupling during a delay (2J) -1 : 2π J ( zsz )(1/ 2J ). Thus the pulse sequence that implements the propagator of Eq. 7c is given by: 9 y(, S)(1/ 2J )9 y (, S)9x(, S)(1/ 2J )9x (, S) This can be transformed from Fig. 2b) to e) by using the fact that e.g. 9 y = 9 -z 9 x 9 z. Furthermore, z rotations that are before or after the pulse sequence can be introduced and skipped as well. Example: NEPT transfer in 2 S system: Let's now look at the 2 S case: Here the matrices are a twice as big: 1 1 1z + 2z = ; Sz 1 1.5 =.5.5.5.5.5.5.5 Application of Eq. [2] to this transfer yields after reordering of the matrix elements: a=1. The problem contains symmetry and we can use this symmetry to achieve a simpler notation. The composite spin of 1 and 2 is called F. There is a spin 1 and a spin multiplicity. The corresponding matrices then become:
73 = = +.5.5.5.5.5.5.5.5 ; 1 1 1 1 2 1 z z z S The ordering of the levels is: (1,1)α, (1,)α, (1,-1)α, (,)α, (1,1)β, (1,)β, (1,-1)β, (,)β, where (1,1) is the m=1 state of the spin 1 combination of the two spins 1 and 2. The (,) means the m= state of the spin combination of the two spins. The second polarization state refe to S. The states of the composite spin = can be ignored since they have no population in the initial operator. Also except for the application of selective pulses, the spins will always be affected in the same way. This however means that throghout the whole pulse sequence the total spin does not change. We then are left with two 6x6 matrices: = = +.5.5.5.5.5.5 ; 1 1 1 1 2 1 z z z S We can obtain now total transfer if we apply the following π pulses: > > > > β α β α π π 1 1 x x. Thus we have to rotate by: { } 1 1 2) / ( α β β α π >< + >< and { } 1 1 2) / ( α β β α π >< + ><. The two rotations affect different levels, therefore they commute. We note that for a spin 1 the operator F x looks in the following way: = 2 / 2 2 / 2 2 / 2 2 / 2 x F Thus it connects the and -1 levels as well as the 1 and levels. Therefore from analogy with the previous S spin system, we arrive at the operator expression that implements the desired pulses:
74 ( π / 2) 2( Fx Sx + FyS y ) = 2 / 2 2 / 2 2 / 2 2 / 2 ; This Hamiltonian is again a planar heteronuclear mixing operator. t can be implemented by two flanking 9(S) pulses and a pulse sequence that generates the Hamiltonian πj ( FxSx + FyS y ) during a time ( 2J ) 1. This is the same sequence as for the S spin system, however, the delay τ = ( 2J ) 1 instead 1 of τ = J as in the case of the S spin system. t should also be noted that this pulse sequence is a little bit shorter than the familiar implementation of the NEPT with delays τ 1 = (2J) -1 and τ 2 = (4J) -1. Bounds and optimal pulse sequences for non-hermitian A and C: n the hermitian case we have obtained the bounds by calculating the eigenvalues of the involved matrices A and C. These eigenvalues are, however, always for transfe that interchange operato that would be selected by echo gradient selection. Therefore another bound, the so called gradient bound was developed. So far there are no analytical expressions for this bound. We just give their values and then try to find implementations of pulse sequences that achieve a certain transfer. Let's look at the antiphase transfer in a two spin system: 2S z shall be transfered to. The gradient bound tells that this transformation should be possible with an a of 1. ndeed, it is possible to accomplish the following transfer: 2S x z -> y by a (π/2)s x x rotation and 2S y z -> - x by a (π/2)s y y rotation. The two operato commute and therefore one can either apply them consecutively or simultaneously. You can see that the required transfe are exactly the same as for the NEPT in the S spin system. Therefore the implementation is the same as in Fig. 2b. However, now, only the last 9(S) pulse can be ommitted yielding the pulse sequences in Fig. 2f or g. Now, let us consider the 2 S spin system. The maximum achievable transfer is like in the case of the hermitian operato given by: a = 1. We can again look at the necessary matrices:
75 = = 2 2 2 2 ; 2 2 2 F F z S Now, let us consider the 2 S spin system. We recognize that by a rotation about the transitions: 2,4 and 3,5 we accomplish the desired transfer. This rotation can again be expressed as: ; 2 2 / 2 / 2 2 / 2 2 / 2 ) 2( 2) / ( = + y y x x S F S F π Thus we find that the heteronuclear Hartmann Hahn transfer indeed achieves in a two spin system the desired optimal transfer. Optimal coherences? We shall now discuss the question of choosing optimal coherences for pulse sequences. This question has become very interesting for large molecules where certain coherences relax much slower than other coherences. There are essentially two examples of this approach. a) The use of heteronuclear multiple quantum coherences that relax slower than single quantum coherences. b) The use of single multiplet components that relax much slower than other multiplet components. Multiple Quantum Coherences: We consider only molecules that are in the slow tumbling regime. Then we can neglect all spectral densities to the relaxation except for J(). For the dipolar relaxation double commutator, we find: dρ/dt = 2 3 4 S S r π γ γ µ h [2S z z,[2s z z,ρ] J() = if [2S z z,ρ]=.
76 t is obvious that the only coherences that commute with this double commutator are also those that do not evolve heteronuclear coupling. Thus, z-magnetization or zero or double quantum coherences. This approach has been successfully used for proteins, especially with partial deuteration and RNA. A selection of respective pape can be found in the pape accompanying this lecture. Differential Line Widths in Submultiplets (TROSY): The concept that a multiplet has the same linewidth is no longer true as soon as molecules become larger. Let's look again at an S spins system. There will be relaxation due to the dipolar interaction and there will be relaxation due to the anisotropy of the transvee spin. Thus there will be autocorrelated γ relaxation due to the dipolar interaction: ( ) Sγh µ 4 πr 3 S z z = b D S z z and due to the anisotropy S of the S spin: 3 1 ( σ σ ) γ S B S z = b a S z. n addition there will be the cross term due to the cross correlation of the two interactions. f we look at the individual multiplet components: S a and S b we find the following expressions: ( S α ) [ b S [ b S S α ] [ b S [ b S S α 2τ = D z z, D z z, + a z, a z, ] c 5 ( + [ b S [ b S S α ] [ b S [ b S S α D z z, a z, + a z, D z z, ]) 1 ( 3cos 2 θ 1) 2τ c 5 α ( 2 2 2 2τ = S b / 4 (3cos 1) / 2) c D + ba + bdba θ 5 ( S β ) [ b S [ b S S β ] [ b S [ b S S β 2τ = D z z, D z z, + a z, a z, ] c 5 ( + [ b S [ b S S β ] [ b S [ b S S β D z z, a z, + a z, D z z, ]) 1 ( 3cos 2 θ 1) 2τ c 5 β ( 2 2 2 2τ = S b / 4 (3cos 1) / 2) c D + ba bdba θ 5 Obviously, if ( 2 / 4 + b 2 ± b b (3cos 2 θ 1) / 2) b D a D a is, the linewidth can be very small. For a NH bond, the nitrogen as well as the hydrogen CSA tensor are almost exactly aligned along the NH bond. Therefore (3cos 2 θ 1) / 2 is close to 1. This means that ( 2 / 4 + b 2 b b (3cos 2 θ 1) / 2) = b D a D a. S β and Thus the optimal coherences are: S β between those coherences in an optimal way are desirable.. Therefore transfer sequences that transfer
77 There are two implementations that can achieve this transfer so far published in the literature. The fit is again the planar heteronuclear Hartmann Hahn mixing: t implements a rotation about the 2,3 transition. This rotation achieves the following transfer: β S + β S β α S ; S + β + α S ; S α S + α S These transformations will create a spectrum that contains in the antiecho and echo part the following peaks: Antiecho Echo α β Ω(S) Ω(S) Ω() Ω() Of coue either one of the peaks will be broad and it is not optimal to have it in the spectrum. This problem is solved in the TROSY sequence. Here one can show that the following transformations are achieved: β S + β S β α S ; S + β + α S ; S + α S α S
78 This produces the following spectrum: 1 H 13 C x τ1 -y -y τ2 x Antiecho Echo α Ω(S) Ω(S) β Ω() Ω() Ω() This is optimal in combination with gradients to select exactly one line only.