12 Understanding Solution State NMR Experiments: Product Operator Formalism

Transcription

1 12 Understanding Solution State NMR Experiments: Product Operator Formalism As we saw in the brief introduction in the previous chapter, solution state NMR pulse sequences consist of building blocks. Product operator formalism (POF) allows us to walk through a pulse sequence and determine which interactions are present at any point in time Density Matrix and POF In product operator formalism, the density operator of the spin system is expressed as a linear combination of base operator products. Cartesian base operators (I x, I y, I z ) are used to describe pulse effects and time evolution of the spin system. For a weakly coupled two spin system (I = 1/2), there are 16 product operators: 1. zero-spin operator: unit operator 1/2 2. one-spin operators: I 1x, I 1y, I 1z, I 2x, I 2y, I 2z

2 3. two-spin operators: 2I 1x I 2x, 2I 1x I 2y, 2I 1x I 2z, 2I 1y I 2x, 2I 1y I 2y, 2I 1y I 2z, 2I 1z I 2x, 2I 1z I 2y, 2I 1z I 2z where the operators in red are observable, or represented graphically:

3 12.2 Product Operator Formalism

4 As we saw in the Quantum chapter, we can use density matrices and Hamiltonians to describe the effect of RF pulses or the evolution of a spin with time. Recall: ˆρ(t) = e iĥt/ h ˆρ(0)e iĥt/ h (12.1) If the density operators and Hamiltonians are expressed in terms of cartesian base operators (I x, I y, I z ), then the solution to these equations can be worked out. The results are general rules governing how the spin operators are affected by RF fields, the chemical shift and the scalar coupling. 1) Rules for RF irradiation i - Every experiment starts off from the thermal equilibrium state of I z. ii - If applying an RF pulse from the y-direction to a two spin system, the following rotations of the spin operators occur:

5 I x βi y I x cosβ I z sinβ I y βi y I y (12.2) I z βi y I z cosβ + I x sinβ where β is the tilt angle (e.g. 90 degrees). z z β y βi y y x x For a three spin system, the same rules apply. 2) Rules for chemical shift evolution i - If no pulses are applied, the chemical shifts will evolve. For a two-spin system, evolving for a time t, I x ΩtI z Ix cosωt + I y sinωt I y ΩtI z Iy cosωt I x sinωt (12.3) I z ΩtI z Iz.

6 y ti Ω z y Ωt x x For a three spin system, the same rules apply. 3) Rules for scalar coupling i - If two spins are not decoupled from one another, the scalar interaction between them J 12 will be active and will evolve over time t such that I 1x I 1y 2I 1x I 2z 2πJ 12 ti 1z I 2z I1x cosπj 12 t + 2I 1y I 2z sinπj 12 t 2πJ 12 ti 1z I 2z 2πJ 12 ti 1z I 2z I1y cosπj 12 t 2I 1x I 2z sinπj 12 t 2I1x I 2z cosπj 12 t + I 1y sinπj 12 t

7 2I 1y I 2z 2πJ 12 ti 1z I 2z 2I1y I 2z cosπj 12 t I 1x sinπj 12 t (12.4) y 2πJ 12 ti 1z I 2z y x x For a three spin system with scalar coupling between spin 1 and 2 and between spin 1 and 3, i.e. with J 12 and J 13, the relationships above become more complex since both couplings must be taken into account. For example, 2I 1x I 2z 2πJ 13 ti 1z I 3z 2I1x I 2z cosπj 13 t + 4I 1y I 2z I 3z sinπj 13 t (12.5) The last relationships above involve the evolution of

8 double quantum terms. These spin operators cannot be directly observed in an NMR experiment - only single quantum operators such as (I x, I y, I z ) can be directly observed. They can, however, be affected by RF and the chemical shift evolution. For instance 2I 1x I 2z 2I 1x I 2z 2I 1x I 2y 90 y Ω 1 ti 1z +Ω 2 ti 2z 2I 1z I 2x 2I1x I 2z cosω 1 t + 2I 1y I 2z sinω 1 t Ω 1 ti 1z +Ω 2 ti 2z 2 (I 1x cosω 1 t + I 1y sinω 1 t) (I 2y cosω 2 t I 2x sinω 2 t) (12.6) See also H. Kessler et al., Angewandte Chemie,27, 490, Example: Polarization transfer In solution state, the INEPT sequence is used to transfer magnetization from the I z spins (at equilibrium) to the S spins. This is how

9 cross-polarization takes place in solution state. We will see later than in solids, a very different approach is used. NOTE: solid state as well? The pulse sequence is: Can an INEPT sequence be used in

10 I S (90) y (180) x (90) x (180) y A B C D E (180) x (90) y (180) y (90) x F G H I J K 1/2J 1/2J At point A, the magnetization is given by

11 A : I z. (12.7) After each successive pulse applied to the I spins, we have C: 90 y B : I z I z cos90 + I x sin90 = I x (12.8) I x I x cosω I t + I y sinω I t I x I x cosπj IS t + 2I y S z sinπj IS t (12.9) D: I x cosω I t + I y sinω I t I x cosω I t I y sinω I t I x cosπj IS t + 2I y S z sinπj IS t I x cosπj IS t + 2I y S z sinπj IS t (12.10) E: I x cosω I t I y sinω I t I x cosω I t I y sinω I t (12.11)

12 which for the total time τ results in I x cosω I τ τ=t t I x (12.12) I x cosπj IS τ +2I y S z sinπj IS τ τ=1/2j IS 2I y S z (12.13) After the 90 x pulse applied on the I spins: which at F becomes: 90 2I y S x z 2Iz S z (12.14) G: 90 y F : 2I z S z 2I z S x (12.15) H: 2I z S x 2I z S x cosω S t + 2I z S y sinω S t 2I z S x 2I z S x cosπj IS t + S y sinπj IS t (12.16) 2I z S x cosω S t + 2I z S y sinω S t

13 2I z S x cosω S t 2I z S y sinω S t I: 2I z S x cosπj IS t + S y sinπj IS t 2I z S x cosπj IS t + S y sinπj IS t (12.17) 2I z S x cosω S t 2I z S x cosπj IS t + S y sinπj IS t τ=t t 2I z S x which finally results in J: τ=1/2j IS S y (12.18) S y 90 x Sz. (12.19) Thus the overall INEPT sequence transforms I z S z, i.e. transfering magnetization from the abundant I spins to the S spins. -Using the product operator formalism in the solid state- Although the product operator formalism, as given

14 above, is of limited use in the solid state, it can be useful to understand the different components of a pulse sequence and to derive an appropriate phase cycle for a given sequence. An example of this will be given below. The principle reason why the rules given above are not generally applicable in the solid state is that they cannot take into account the dipolar interaction and quadrupolar interaction and cannot describe any averaging processes such as magic angle sample spinning. Other approaches which take into account the full Hamiltonian and density operators must be used instead (e.g. coherence averaging theory, Floquet theory, etc.). Nonetheless, product operator formalism can provide insight into the workings of a pulse sequence when it is assumed that the dipolar interaction is averaged and there is not quadrupolar interaction present. For instance, consider the pulse sequence below, applied under fast MAS conditions:

15 I S π/2 A B CP CP DEC DEC π π π/2 π/2 C D F G H Receiver E selective Once again, we start with A being I z. The first 90 x pulse will convert that into I y. By applying two

16 simultaneous pulses on the I and S spins (CP), this I y gets converted into S y. Thus at C: after CP S y (12.20) D: evolution during time t = t 1 2 S 1y S 1y cosω 1 t S 1x sinω 1 t S 1y S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t S 2y S 2y cosω 2 t S 2x sinω 2 t S 2y S 2y cosπj 12 t 2S 2x S 1z sinπj 12 t (12.21) Note that a distinction is made between spins S 1 and S 2 since the selective pulse will only work on one set of spins, say S 1. Thus, at E: selective π x pulse S 1y cosω 1 t S 1x sinω 1 t π x S 1y cosω 1 t S 1x sinω 1 t S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t π x

17 S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t (12.22) whereas the S 2 spins remain unaffected. If in the following scan, a selective π y pulse is applied, then: E : selective π y pulse S 1y cosω 1 t S 1x sinω 1 t π y S 1y cosω 1 t + S 1x sinω 1 t At S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t π y S 1y cosπj 12 t + 2S 1x S 2z sinπj 12 t (12.23) F: non-selective π x pulse S 1y cosω 1 t S 1x sinω 1 t π x S 1y cosω 1 t S 1x sinω 1 t S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t π x S 1y cosπj 12 t + 2S 1x S 2z sinπj 12 t S 2y cosω 2 t S 2x sinω 2 t π x S 2y cosω 2 t S 2x sinω 2 t

18 S 2y cosπj 12 t 2S 2x S 1z sinπj 12 t π x S 2y cosπj 12 t + 2S 2x S 1z sinπj 12 t (12.24) and F : non-selective π x pulse S 1y cosω 1 t + S 1x sinω 1 t π x S 1y cosω 1 t + S 1x sinω 1 t S 1y cosπj 12 t 2S 1x S 2z sinπj 12 t π x S 1y cosπj 12 t + 2S 1x S 2z sinπj 12 t (12.25) with the S 2 components being the same as in F. G: evolution during time t = t 1 2 so that the total evolution during t 1 is S 1y cosω 1 t 1 S 1x sinω 1 t 1 S 1y cosπj 12 t 1 + 2S 1x S 2z sinπj 12 t 1 (12.26) S 2y cosω 2 t 1 S 2x sinω 2 t 1 S 2y cosπj 12 t 1 + 2S 2x S 1z sinπj 12 t 1 (12.27) and G : evolution during time t = t 1 2 so that the

19 total evolution during t 1 is S 1y cosω 1 t 1 + S 1x sinω 1 t 1 S 1y cosπj 12 t 1 + 2S 1x S 2z sinπj 12 t 1 S 2y cosω 2 t 1 S 2x sinω 2 t 1 S 2y cosπj 12 t 1 + 2S 2x S 1z sinπj 12 t 1 (12.28) By picking the delays t appropriately, one can make the contribution from the scalar interaction vanish. Suppose that during t 1, we would like however to maintain the chemical shift evolution of the S 1 spins. In order to select these spins from the S 2, we have to phase cycle the receiver such that Receiver: phase = +x for the first scan and phase = -x for the second scan S 1y cosω 1 t 1 S 1x sinω 1 t 1 [ S 1y cosω 1 t 1 + S 1x sinω 1 t 1 ] = 2[S 1y cosω 1 t 1 S 1x sinω 1 t 1 ] S 2y cosω 2 t 1 S 2x sinω 2 t 1

20 [ S 2y cosω 2 t 1 S 2x sinω 2 t 1 ] = 0 (12.29) Thus by analysing the pulse sequence using product operator formalism, we can design a pulse sequence which refocuses the scalar coupling between two homonuclear spins S 1 and S 2, while maintaining the chemical shift evolution of the S 1 spins. With the chemical shift evolution of the S 2 spins being subtracted with the phase cycle, one can expect no artifacts in the detected spectra coming from S 2. QUESTION: 1) Given the pulse sequence below, determine what is being detected at point 10, taking into account the phases of the pulses as drawn in.

21 π π/2 +x x π I WALTZ16 WALTZ S π π/2 +x x π π/2 +y y