Classical Description of NMR Parameters: The Bloch Equations

Transcription

1 Classical Description of NMR Parameters: The Bloch Equations Pascale Legault Département de Biochimie Université de Montréal 1 Outline 1) Classical Behavior of Magnetic Nuclei: The Bloch Equation 2) Precession in a Static Magnetic Field 3) Chemical Shift 4) Laboratory Frame and Rotating Frame 5) The B 1 Field 6) 90 Pulse: Pulse Length 7) T 1 and T 2 Relaxation 2 1

2 The Nobel Prize in Physics 1952 "for their development of new methods for nuclear magnetic precision measurements and discoveries in connection therewith" Felix Bloch Stanford University Stanford, CA, USA b in Zurich, Switzerland d Edward Mills Purcell Harvard University Cambridge, MA, USA b d Ref: 3 1) Classical Behavior of Magnetic Nuclei: The Bloch Equation Bloch formulated a semi-classical model to describe the behavior of a sample of non-interacting spin-1/2 nuclei in a static magnetic field. The evolution of bulk magnetic moment [M(t)] is central to the Bloch equations 4 2

3 1) Classical Behavior of Magnetic Nuclei: The Bloch Equations Evolution of bulk angular momentum J(t): M = γj In the presence of a magnetic field (includes static field B o and radio-frequency (rf) field B 1 ), M(t) experiences a torque (T). T = dj(t)/dt = M(t) x B(t) Evolution of bulk magnetic moment M(t): dm(t)/dt = M(t) x γb(t) M(t) represents the directly observed bulk magnetization. 5 1) Classical Behavior of Magnetic Nuclei: The Bloch Equations Note that M(t) is a vector quantity [also denoted M(t)] The bulk magnetic moment (M) and angular momentum (J) of a macroscopic sample are given by the vector sum of the corresponding quantities for individual nuclei, µ and I. 6 3

4 1) Classical Behavior of Magnetic Nuclei: The Bloch Equations Evolution of M: dm(t)/dt = M(t) x γb(t) M represents the directly observed bulk magnetization. The Bloch equations are solutions of dm(t)/dt. It helps explain: - Isolated spin behaviors - Precession in a Static Field - Chemical Shifts - Lab Frame versus Rotating Frame - The B 1 Field - Effect of Single Pulses - T 1 and T 2 relaxations 7 2) Precession in a Static Field Evolution of M: dm(t)/dt = M(t) x γb(t) Simplified case: B(t) = (0, 0, B o ) and M o = (M x (0), 0, 0) In complex notation: M x (t) = M x (0)cos(ω o t) M y (t) = M x (0)sin(ω o t) M(t) = M x (0)[cosω o t + isinω o t] = M x (0)exp[+iω o t] 8 4

5 Solution to Bloch Equations (time=0 on x axis) General case i j k dm(t)/dt = γ M x M y M z B x B y B o dm x (t)/dt = γ(m y B o - M z B y ); dm y (t)/dt = -γ(m x B o - M z B x ); dm z (t)/dt = γ(m x B y - M y B x ) Simplified case: M o = M x (0); B t = (0, 0, B o ) The static field B o is along the z axis, thus B x = B y = 0 i j k dm/dt = d(m x, M y, M z )/dt = γ M x M y M z 0 0 B o 9 Simplified case (cont d) dm x /dt = γ(m y B o ); dm y /dt = -γ(m x B o ); dm z /dt = 0 dm/dt = γ(m y B o, -M x B o, 0) M x and M y are NOT independent of each other. Mathematically, it is convenient to consider M x and M y as real and imaginary parts of a complex magnetization: M + = M x + im y d(m x + im y )/dt = γ(m y B o - M x B o ) = γb o (M y - im x ) d(m x + im y )/dt = -iγb o (M x + im y ) = -iγb o (M + ) 10 5

6 d(m + )/dt = -iγb o (M + ) d(m + )/dt = -iγb o (M + ) M + (t) = M + (0) e -iγbot M + (t) = M + (0) e +iωot (ω o = -γb o ) M x (t) + im y (t) = [M x (0) + im y (0) ]*[cos(ω o t) + isin(ω o t)] M x (t) = M x (0)cos(ω o t) - M y (0)sin(ω o t) M y (t) = M y (0)cos(ω o t) + M x (0)sin(ω o t) since M o = M x (0) and M y (0) = 0, then: M x (t) = M o cos(ω o t) M y (t) = M o sin(ω o t) 11 3) Chemical Shift Not all nuclei see exactly B o. The local effective field (B o i) is modified by the electronic environment. Different spins have different shielding constants (σ i ) and thus rotate at different angular frequencies (ω i ). ω i = -γb o i = -γb o (1-σ i) The term σ i represents an average or isotropic shielding constant: σ = (σ xx + σ yy + σ zz )/3 The term Δσ i represents the chemical shift anisotropy (CSA): Δσ = σ zz - ( σ xx + σ yy)/2 For simplicity σ i terms are often neglected. 12 6

7 3) Chemical Shift Shielding constants (σ i ) are <<< 1, which means that the variation in ω i (Δω) is much smaller than ω o : ω i = -γb o (1-σ i) = ω o + Δω For 1 H at 500 MHz (B o =11.7 Tesla) ν o = ω o /2π 500 MHz In general : Δω /2π 7500 Hz We choose a reference signal (TMS or DSS) to represent ω o and measure relative frequencies δ =[(ω i - ω DSS )/ ω DSS ] x 10 6 (in ppm) 13 1 H Chemical Shifts in RNA 14 7

8 4) Lab Frame and Rotating Frame Lab Frame: ω i = -γb o i = -γb o (1-σ i) = ω o (1-σ i) M(t) = M x (0)[cosω o t + isinω o t] M(t) = M x (0) e +iωot e +iωot = clockwise rotation At B o = 11.7 tesla, ω o /2π 500 MHz for 1 H 15 4) Lab Frame and Rotating Frame (cont d) Rotating frame: ω eff = ω i - ω rot Mathematically, we rotate the lab frame at an angular frequency ω rot : 16 8

9 4) Lab Frame and Rotating Frame (cont d) General formalism (Rotating frame): [dm(t)/dt] rot = [dm(t)/dt] lab + M(t) X ω rot [dm(t)/dt] rot = M(t) X [γb(t) + ω rot ] B eff = B(t) + ω rot /γ (ω rot = - γb rot ) B eff = B(t) - B rot ω eff = ω(t) - ω rot Special case: B(t) = B o and ω rot = ω o = ω DSS then ω eff = ω i - ω rot ω = ω eff i - ω DSS = Δ and ω eff /2π in Hz-kHz not MHz!!! 17 4) Lab Frame and Rotating Frame (Conclusions) This rotating frame (ω rot = ω o) is very useful when we consider the effect of the radio-frequency field B 1 It allows simplification of time-dependent Bloch equations Also very useful for QM formalism Experimentally, detected signals are subtracted from carrier frequency 18 9

10 5) The B 1 Field B 1 is a radio-frequency field, which creates a linearly polarized oscillating magnetic field. The B 1 field can be decomposed in 2 equal fields of half intensity. For B 1 along x: B rf = 2B 1 cos(ω rf t) B rf = B 1 (e+iωrft + e -iωrft ) where B 1 is the amplitude and ω rf is the frequency related to B rf 19 5) The B 1 Field (cont d) 20 10

11 5) The B 1 Field (cont d) Only the field rotating in the same sense as the magnetic moment interacts significantly with the magnetic moment (resonance effect). B rf = B 1 e +iω rft In the rotating frame: ω rot = ω rf B eff = (B 1, 0, ΔB o ) where B 1 = -ω rf /γ and ΔB o = -Δ/γ; Δ = ω o - ω rf 21 6) 90 Pulse : Pulse Length M o = (0, 0, M z (0)); B eff = (-ω rf /γ,0, -Δ/γ) i j k dm(t)/dt = γ M x M y M z -ω rf /γ 0 -Δ/γ Note: The are there to remind us that we are in the rotating frame. They are often omitted. dm x (t)/dt = -M y Δ dm y (t)/dt = +M x Δ - M z ω rf dm z (t)/dt = +M y ω rf 22 11

12 6) 90 Pulse : Pulse Length (cont d) Lets say that Δ = 0 for now. Thus dm x (t)/dt = 0; dm y (t)/dt = -M z ω rf ; dm z (t)/dt = +M y ω rf. Solution: M x (t) = 0; M y (t) = -M z (0)sinω rf t; M z (t) = M z (0)cosω rf t. The magnetization rotates about x in the y -z plane. 23 6) 90 Pulse : Pulse Length (cont d) length for 90 pulse (θ = π/2 ): θ = ω rf*τ p = π/2 τ p = θ/ω rf = pulse length Example: Typical parameters for hard pulse on 1 H If τ p = 10 µs for 90 pulse (θ = π/2 = 0.25 radian) ω rf = θ/ τ p = 0.25/ 10 µs = 25 khz 24 12

13 6) 90 Pulse : Pulse Length (cont d) Proof: (rotating frame simplifies) dm x (t)/dt = 0; dm y (t)/dt = - M z ω rf ; dm z (t)/dt = +M y ω rf d 2 M y (t)/dt 2 = - ω rf dm z (t)/dt = - ω rf 2 M y Possible solution to differential equation: M y (t) = A cosω rf t + B sinω rf t d 2 M z (t)/dt 2 = ω rf dm y (t)/dt = - ω rf2 M z M z (t) = Acosω rf t + Bsinω rf t 25 6) 90 Pulse : Pulse Length (cont d) M x (t) = M x (0) = 0 M y (0) = A M z (0) = A = 0 dm z (t)/dt = Bω rf cosω rf t = Bω rf t=o t=o dm z (t)/dt = M y ω rf = 0 B = 0 t=o t=o dm y (t)/dt = B ω rf cosω rf t = B ω rf t=o t=o dm y (t)/dt = -M z ω rf = M z (0)ω rf B = M z (0) t=o t=o 26 13

14 7) T 1 and T 2 Relaxation T 1 : longitudinal or spin-lattice relaxation time, returning back to equilibrium (back to +z) T 2 : transverse or spin-spin relaxation time, dephasing in the x-y plane R 1 = 1/T 1 ; R 2 = 1/T 2 R 1 and R 2 are rate constants Lets consider some simplified cases i.e. after pulse (B 1 = 0) and on resonance (ΔB = 0): 27 7) T 1 and T 2 Relaxation T 2 relaxation dm x (t)/dt = -R 2 M x (t) dm y (t)/dt = -R 2 M y (t) Solving for M x : dm x (t)/ M x (t) = -R 2 dt ln M x (t) o = - R 2 t o ln M x (t)/ M x (0) = - R 2 t M x (t) = M x (0)e -R 2t M y (t) = M y (0)e -R 2t t t 28 14

15 7) T 1 and T 2 Relaxation T 1 relaxation dm z (t)/dt = R 1 [M o - M z (t)] Solving for M z : dm z (t)/ [M z (t) - M o ] = -R 1 dt t t ln M z (t) - M o o = - R 1 t o ln [M z (t) - M o ]/ [M z (0) - M o ] = - R 1 t M z (t) = M o - [M o - M z (0)] e -R 1t 29 7) T 1 and T 2 Relaxation: Summary For isolated spins, the bulk magnetization behaves as described by the classical Bloch equation dm(t)/dt = M(t)xγB(t) - [M o - M z (0)] e -R1t - M x,y (0)e -R2t After, 90 x pulse, the bulk magnetization can be detected along y (simple 1D experiment)

16 Exercises (due in a week from now) Solve the Bloch equation [dm(t)/dt = M(t) x γb(t)] for the following case: B(t) = (0, 0, B o ) and M o = (0, M y (0), 0). Explain you results in terms of a vector diagram. Solve the Bloch equation [dm(t)/dt = M(t) x γb(t)] for the following case: M o = (0, 0, M z (0)); B eff = (0,-ω rf /γ,-δ/γ), with Δ = 0. Explain you results in terms of a vector diagram. What is the strength of the B 1 field (in khz) for a 90 pulse of 12,5 µs? 31 16