We have seen that the total magnetic moment or magnetization, M, of a sample of nuclear spins is the sum of the nuclear moments and is given by:

Transcription

1 Bloch Equations We have seen that the total magnetic moment or magnetization, M, of a sample of nuclear spins is the sum of the nuclear moments and is given by: M = [] µ i i In terms of the total spin angular momentum, P, it is M = γp [] The interaction of the applied field and magnetic moment produces a torque on the system and the motion of the nuclear magnetic moment in a magnetic field, B, is given by the following equation: dp = ΜxB [3] Since the angular momentum and total nuclear magnetic moment is related by equation [], equation [3] can be rewritten as: dμ = γμxb [4] The vector product can be expanded as MxB = i j k B x B y B z = ( B z B y i +( B x B z j +( B y B x k This can be written as individual components as follows: d = γ ( B z B y d d = γ ( B x B z = γ ( B y B x [6] By convention, the static magnetic field is along the z-direction. This means that B z has non-zero value. Whereas, B x and B y are zero and as a consequence the spins precess around B z. will be converted to and into. will remain constant. This is the result that we see when a sample is placed in a magnet. In order to get any changes in under these conditions we have to add radiofrequency (RF field terms. [5]

2 d d d = 0 = γ B z = γ B z [7] For free precession in the absence of relaxation and RF, these equations have to be solved. The solutions to these equations are as follows (with ω=γb z : (t = (0cos(ωt (0sin(ωt (t = (0cos(ωt (0sin(ωt [8] Relaxation terms: Bloch assumed that spin-lattice and spin-spin relaxation could be treated as first order processes with characteristic times and respectively (details of these processes will be discussed later in Relaxation lecture. This assumption is true for liquids but not for solids as far as is concerned. d d d = γ ( B z B y = γ ( B x B z = γ ( B y M ( y B x [9] The including relaxation terms and treating that the B o is applied along z-axis (B x =B y =0, B o =B z are given by: d d d = γm B o x = γ B o = z

3 Solving for yields [0] (t = equ + [ (0 equ ]exp( t /T [] Since the Bloch equation for z-component is based on first order kinetics with rate /, the behavior of would be a simple exponential approach to equilibrium as shown in Figure : eq (t Figure (0 t = null T ln( (0 eq Figure shows the recovery of longitudinal magnetization,, following its inversion (for initial condition Mz(0 = - equ The equations for and can be decoupled by defining M± as M + = + i M = i [] The solution for resulting equations are: M ± (t = M ± (0exp( t / iγbt [3] (t and (t can be found by inverting the above equations (t = [ (0cos(γBt (0sin(γBt]exp( t / [4] (t = [ (0sin(γBt + (0cos(γBt]exp( t / If the magnitude of the transverse magnetization at time zero is M o and the angle between M o and the x-axis at time zero is Φ o, then (t = M 0 (0cos(γBt Φ o exp( t / (t = M 0 (0sin(γBt Φ o exp( t / [5] 3

4 Representing the decay of x and y components of the magnetization in the laboratory frame is difficult since γb o is millions of times larger than π/. In practice, we never detect these oscillations in lab frame. Experimentally, the x- and y- components of M are detected in a frame rotating about z- with an angular frequency near ω o = γb o such that the differences between the rotating frame of detection and ω o are generally between 0 and 00 khz. On this time scale the decay of and with first-order rate constant / is easily observed as shown in Figure 3 (see the pulse sequence analysis section. Effect of RF field: In NMR experiments, a small oscillating field of amplitude, cosωt, is applied, in the x-direction. This field can be resolved into two components rotating in opposite directions with angular velocities ± ω. These can be considered in the limit of small and the component rotating in the opposite direction to the Larmor precession (the one with +ω can be neglected. If has arbitrary phase, φ, w.r.t. the laboratory x, and y axes, then the form of (t will be (t = B [icos(ωt + φ + jsin(ωt + φ] As mentioned earlier, the static field, Bo=B z ; an RF field rotating at ω, with φ=0, is applied in the transverse plane, i.e perpendicular to Bo (non zero RF phase is included in expressed in the Matrix form (see equation 9. Now the field components including the RF are: B x = cos(ωt,b y = sin(ωt,b z = B o Including relaxation and RF field components, the are given by: d d d = γ ( B o + M sin(ωt x = γ ( cos(ωt B o = γ ( sin(ωt + cos(ωt z Additional time dependence introduced in the form of time dependent RF field complicates the analysis. All that one wants to detect in NMR is the differences in Larmor precession frequencies caused by the immediate environment of the spins, not the absolute Larmor frequency. This can be accomplished by working in a frame that is rotating at Larmor precession frequency which essentially removes the time dependence of the RF field. [6] 4

5 Rotating frame Transformation into coordinate system (x,y,z rotating around Bo, with an angular frequency -ω: In the new coordinate system, u and v represent component in the direction of x and y respectively. Let us define the new variables u and v in terms of and as: u = cos(ωt sin(ωt v = ( sin(ωt + cos(ωt [7] which then can be used to recast and as = ucos(ωt vsin(ωt = (ucos(ωt + vsin(ωt [8] du dm dm = x y cos(ωt sin(ωt ω[m x sin(ωt + cos(ωt] = γb 0 v u + ωv [9] Similarly, dv/ can be written. Rewriting the by substituting the new equations for and. du = (ω o ωv u dv = (ω o ωu ω v d = ω v z [0] Usually u and v are written as Mx and My, respectively to represent the terms in the rotating frame. However, we will drop the primes here and now on we will work in the rotating frame with the following equations: d d d = (ω o ω = (ω o ω ω = ω z [] 5

6 Steady state solutions: If we assume that the system has been allowed to soak in this combination of static and time varying fields, a steady state will ultimately will be reached in which none of the components change with time. Therefore, we will have the following equations: (ω o ωm x = 0 (ω o ω ω = 0 [] ω z = 0 These equations (three equations and three unknowns can be easily solved to obtain three components of the magnetic moment in the rotating frame under the influence of a rotating field: ω T δω = + ω T T + (T δω M o = ω + ω T + ( δω M o + T δω = + ω T T + (T δω M o with δω = (ω o ω ω = γ Small RF limit valid for CW experiments: In the limiting case of small RF limit, ω << The above equations become T δω (ω M o + (T δω = ω M o G(δω T (ω M o + (T δω = ω M o F(δω [3] [4] [5] 6

7 with T δω G(δω = + (T δω F(δω = + ( δω [6] (π - Signal amplitude (a.u Absorption mode Dispersion mode Frequency (Hz-- Figure : Lorentzian absorption and dispersion lineshapes given in the above equations. The functional forms F and G are the absorption and dispersion, respectively, for a line known as a Lorentzian line. Signal due to and components are known as absorption and dispersion, respectively. These are plotted in Figure : Using full, we can analyze the effect of simple pulse sequences on spin / systems without additional couplings. 7

8 Single Pulse response: During a RF pulse (of amplitude ω and duration t w applied along the x-axis and ignoring relaxation and chemical shift during the pulse, the results of the above differential equations are: ((+ represents the magnetization just after and (- represents the magnetization just after (+ = ( (+ = ( cos(ω t w + ( sin(ω t w (+ = ( cos(ω t w ( sin(ω t w [7] Following the pulse, the spin system is allowed to evolve in the presence of chemical shift Δω during the detection period, t and the component of the magnetization is given by (+ = [ ( cos(δωt ( sin(δωt]exp( t / (+ = [ ( cos(δωt + ( sin(δωt]exp( t / [8] (+ = M ( exp( t /T 0 (t CosΦ o or CosΔω (t SinΦ o or SinΔω Figure 3: Decay of transverse magnetization that is evolving under a chemical shift (or frequency offset of Φ ο or Δω. The oscillation shows the chemical shift frequency and the decay is due to relaxation. In equation [8], M(- terms are the same as M(+ terms in equation [7]. Similarly, the response of a nuclear spin system to two pulse response and formation of spin echo can be analyzed. in the matrix form: 8

9 The in the rotating frame can be arranged in a matrix form as: d M x 0 δ ω M sinφ M x y = δ 0 ω cosφ M M y z ω sinφ ω cosφ 0 M z / / ( equ / [9] Where δ is the resonance offset, ω (=γ is the nutation frequency of the pulse and φ is the phase of the RF. As described above, are convenient for treating the response of an isolated spin / system to simultaneous effects of relaxation, RF (continuous, and resonance offsets or chemical shifts. are well suited for describing continuous wave NMR where steady state magnetization is detected as a function of the static field. However, after introduction of FT NMR which is more sensitive and versatile and where majority of the experiments are performed in the pulsed mode, NMR theory and pulse responses are analyzed using quantum mechanical approaches. 9