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
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. 1905 in Zurich, Switzerland d. 1983 Edward Mills Purcell Harvard University Cambridge, MA, USA b. 1912 d. 1997 Ref: http://nobelprize.org/nobel_prizes/physics/laureates/1952/press.html 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
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
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
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
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
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
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
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
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
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
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
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
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
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). 30 15
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