Basic Theory of Solid-State NMR

Transcript

1 Basic Theory of Solid-State NMR Mei Hong, Department of Chemistry, MIT ˆρ ( t) = ˆρ cosωt ω i Ĥ, ˆρ sinωt 5 th Winter School on Biomolecular Solid-State NMR, Stowe, VT, Jan. 7-12, 218

2 Magnetic Dipole Moment From Nuclear Spins µ = γ S γ N = g N e 2m p Energy of a magnetic moment in a B field: Torque on a magnetic moment in a B field: E = µ i B τ = µ B 2

3 Precession of the Magnetic Moment around B For rotational motion: τ = µ B τ = d S dt = 1 d µ γ dt ( Recall linear motion : F = dp dt)!!! µ B! i j k! = µ x µ y µ z B =! i µ y B! j µ x B +! k d µ dt = γ µ B dµ x dt dµ y dt dµ z dt Bloch eqn = γ µ y B µ x B Solution for µ() in the x-z plane: µ x t µ y t µ z t = µ x cosωt = µ x sinωt = µ z = cnst Larmor frequency 3

4 Fields, Frequencies, Energies Hamiltonians B B = 18.8 Tesla ω = γ B = 2πν 1 H Larmor frequency: ω = -2π 8 MHz E = µ B ΔE = ω Energy splitting Ĥ E Ĥ = ˆµ B = γ Ŝ B ˆρ ( t) = e iĥt ˆρ e +iĥt Precession frequency = transition frequency Resonance: apply electromagnetic radiation near the Larmor frequency ω. 4

5 Resonance: Maximizing Transition Probability W << H () Complete transition (P ab =1) can be achieved even when the applied field strength V ba =ħω 1 << ħω (~1 khz vs ~1 MHz). P ab t = V ba ω=ω ab = sin 2 ω r t sin 2 ω r t ω r 2 ω r = 1 2 Rabi frequency: ( ω ba ω) V ab 2 5

6 Resonance, the Rotating Frame, and RF Pulses Lab frame Rotating frame, On resonance Rotating frame, Off resonance ω rf 2π 5 1 MHz, ω 1 2π 5 1 khz 6

7 Phase of Radiofrequency Pulses Change in phase of B 1 (t) by 9 in the lab frame corresponds to a change in direction of B 1 in the rotating frame (e.g. from x to y) B 1 Lab t = B 1 cosω rf t = B 1 2 cosω rf t sinω rf t + B 1 2 cosω rf t sinω rf t B 1 Lab t = B 1 2 cosω rf t sinω rf t B 1 rot t = B B 1 Lab t = B 1 sinω rf t = B 1 2 sinω rf t cosω rf t + B 1 2 sinω rf t cosω rf t B 1 Lab t = B 1 2 sinω rf t cosω rf t B 1 rot t = B 1 2 1

8 Analyzing Pulses: Spin Echo the Vector Model Precession at chemical shift offsets in the rotating frame; Pulse effects: left-hand rule; No quantum mechanics needed. 8

9 A Central Equation: Local Fields Frequencies ω L = γ! B Resonance frequency Magnetic field at the nucleus Local fields from electrons other nuclei are much weaker than the Zeeman field 9

10 Truncation of Weak Interactions: Keep B // B tot = B + B loc = ( B + B loc,// ) B loc, = ( B + B loc,// ) 1+ = Taylor expansion 2 B loc, ( B + B loc,// ) 2 2 B ( B + B loc,// ) 1+ loc, 2( B + B loc,// ) 2 << 1 B + B loc,// ω L γ ( B + B loc,// ) = ω γ B loc,// 1

11 Local Field Interactions: Hamiltonians Ĥ Zeeman = γ Îz B chemical shielding tensor Ĥ CS = γ! Î σ! B Ĥ D = µ 4π j k γ j γ k 3 ( I!ˆ ) r Î! k r! jk jk r jk r 3 jk! Î j! Î k Ĥ Q = electric quadrupole moment e Q 2I ( 2I 1) ˆ I V Î electric field gradient tensor Hamiltonians expressed in frequency units. 11

12 Truncated Hamiltonians: Keep the Operators that Commute with the Zeeman Interaction Ĥ Zeeman = ω Îz H CS,D,Q Ĥ CS = γσ Lab zz B Îz = σ isoω δ ( 3cos2 θ 1 ηsin 2 θ cos2φ) Îz Ĥ D,IS = µ 4π! γ Iγ S r cos2 θ 1 2ÎzŜz Ĥ D,II = µ 4π! γ 2 Ĥ Q = j k r jk cos2 θ jk 1 eq 2I ( 2I 1)! V zz Lab 3ÎzÎz Î Î ( 3Îz j Ŝ k z - Î j Î k ), where V Lab zz = 1 2 eq 3cos2 θ 1 η sin 2 θ cos 2φ Ĥ Local = ω iso +ω aniso θ,φ spatial part, affected by MAS sample alignment Spin Part Affected by rf pulses 12

13 Understanding Pulse Sequences for Coupled Spins The vector model cannot conveniently describe how dipolar J interactions change the magnetization; complex multiple-pulse sequences. Need statistical and quantum mechanics: density operators. 14

14 Spin States Represented by Vectors Spin angular momentum is quantized, as shown by the Stern-Gerlach expt. Spin-1/2 nuclei have 2 basis states: and In a suitably chosen basis, these 2 states are represented by vectors: = 1, = 1 A pure spin state is a linear superposition of the basis states: ψ = C + + C = C + ( C ) + * 15

15 Observables Represented by Matrices Observables are QM (Hermitian) operators with eigenvalue eqns: Î z ±z = ± 1 2 ±z, Î x ±x = ± 1 2 ±x, Î y ±y = ± 1 2 ±y Spin operators follow commutation relations:! Îx, Îy = iîz,! Î y, Îz = iîx,! Î z, Îx = iîy Commutator Â, ˆB Â ˆB ˆBÂ Spin-1/2 operators are represented by 2 x 2 Pauli matrices: Î x ˆ= , Îy ˆ= 1 2 i i, Îz ˆ= Î 2 x = Îy 2 = Îz 2 = = 1 4 ˆ1 16

16 Our Samples: an Ensemble of Spins Expectation (average) value of an observable A: For pure states: A ψ  ψ * = C + * C A ++ A + A + A C + C = C + * C + A ++ + C * C A +C + * C A + + C * C + A + A statistical ensemble of spins is described by a density operator: C k+ = p k k C k ˆρ p k ψ k ψ k k * * C k+ C k = For mixtures: A = p k ψ k  ψ k = Tr( ˆρ ) C + C + * C C + * C + C * C C * At thermal equilibrium: Ĥ kt e ˆρ eq = ˆ1 Ĥ Ĥ kt Tr e 2I +1 kt 1 ) ( ˆρ eq 1 γ B 2I +1 kt Î z ) ( Îz Trace, sum of diagonal ω << kt Dropped: commutes with all operators The observable in NMR is the magnetization... 17

17 Magnetization: the sum of all magnetic dipole moments: M = γ ˆ I = Î x Î y Î z = γ Tr ( ˆρÎx ) Tr ( ˆρÎy ) Tr ( ˆρÎz ) The only non-zero traces are: Tr ( Îx 2 ), Tr ( Îy 2 ), Tr ( Îz 2 ) Î x ˆ= , Îy ˆ= 1 2 i i, Îz ˆ= , Îx 2 = Îy 2 = Îz 2 = = 1 4 ˆ1 So the x, y magnetization are non- only when ρ contains I x and I y terms. e.g. for ˆρ = cosωt Îx + sinωt Îy, (( Îx ± )) iîy = cosωt ± isinωt = e ±iωt = f t Î x ± iîy = Tr cosωt Îx + sinωt Îy 18

18 Time Evolution of the Density Operator ψ Ĥ ψ ( t) ρ rf pulses, chem shifts dipolar couplings ρ ( t) Schrödinger eqn i d ψ dt = Ĥ ψ von Neumann eqn d ˆρ dt = i Ĥ, ˆρ If H and ρ commute, then ρ does not change in time. e.g. S z magnetization is not affected by I z S z dipolar coupling. If H is time-independent, then the formal solution to the von Neumann eqn is: ˆρ ( t) = e iĥt ˆρ e iĥt = Û ( t) ˆρ Û 1 ( t), where U ( t) = e iĥt. More useful solution: (by Taylor-expansion of the exponential operators) ˆρ ( t) = ˆρ cosωt + Ĥ, ˆρ iω sinωt, provided Ĥ, Ĥ, ˆρ = ω2 ˆρ 19

19 Evolution of ρ Under 2-Spin Dipolar Coupling Let ˆρ = Îx, Ĥ IS = ω IS 2ÎzŜz H IS ˆρ ˆρ ( t)? ˆρ ( t) = ˆρ cosωt + Ĥ, ˆρ Ĥ, ˆρ iω = ω IS 2ÎzŜz, Îx Ĥ, Ĥ, ˆρ = ω IS sinωt, provided Ĥ, Ĥ, ˆρ 2 Îz, Îx = ω IS 2ÎzŜz,ω IS 2iÎyŜz = ω IS Thus, Ĥ, Ĥ, ˆρ Ŝz = ω IS 2iÎyŜz 2 4i Î z, Îy = ω IS 2 ˆρ = ω 2 ˆρ Ŝz 2 = ω 2 IS i( iîx ) = ω 2 IS Î x ˆρ ( t) = Îx cosωt + 2ÎyŜz sinωt Detected as x-magn. unobservable. 2

20 The Spatial Part of Nuclear Spin Hamiltonians Ĥ Local = ω iso +ω aniso θ,φ spatial part, affected by MAS sample alignment Spin Part Affected by rf pulses ( Ĥ CS = σ iso ω δ 3cos2 θ 1 ηsin 2 + ( θ cos2φ) ) *, - Îz Ĥ D,IS = µ 4π γ Iγ S r cos2 θ 1 2ÎzŜz Ĥ D,II = µ 4π γ 2 Ĥ Q = eqeq 2I 2I 1 j k 1 r jk cos2 θ jk 1 2 3cos2 θ 1 η sin 2 θ cos2φ ( 3Îz j Ŝ k z Î j Î k ) 3ÎzÎz Î Î 21

21 Chemical Shift Tensor the Principal Axis System Ĥ CS = γ Î! σ B! =... σ iso ω δ ( 3cos2 θ 1 ηsin 2 θ cos2φ) Îz Ĥ CS = γ! Î σ! B = γ I x I y I z = γ I z B σ zz Lab Lab σ xx Lab σ yx σ zx Lab Choose the principal axis system (PAS), where the tensor is diagonal σ PAS PAS σ xx PAS σ yy σ zz PAS 1 Lab σ xy Lab σ yy Lab σ yz bilinear, invariant with coordinate change ( ( ( ( Lab σ xz Lab σ yz σ zz Lab = γ I z B B! b T σ! b PAS = γ I zσ Lab zz B truncation 22

22 Chemical Shift Anisotropy Asymmetry T Ĥ CS = γ I z B ( b σ b ) PAS = ( cosφ sinθ, sinφ sinθ, cosθ) Principal values: ω ii PAS = ω σ ii PAS PAS ω xx PAS ω yy ω zz PAS = ( ω PAS xx cos 2 φ sin 2 θ +ω PAS yy sin 2 φ sin 2 θ +ω PAS zz cos 2 θ) I z Isotropic shift : ω iso 1 ( 3 ω xx + ω yy + ω zz ) Anisotropy parameter : δ ω zz PAS ω iso Asymmetry parameter : η ω yy ω xx ω zz ω iso b PAS ) cosφ sinθ ) ) sinφ sinθ ) ) ) cosθ ) ( ( (θ, ϕ): reflect orientations of the molecules in the sample: powder, oriented, single crystal, etc. (δ, η): reflect electronic environment at the nuclear spin > structure info. I z ω ( θ,φ;δ,η) = δ ( 2 3cos2 θ 1 ηsin 2 θ cos2φ ) + ω iso 23

23 NMR Frequencies are Orientation-Dependent ω ( θ,φ;δ,η) = δ ( 2 3cos2 θ 1 ηsin 2 θ cos2φ ) + ω iso Orientation dependence allows the measurement of: Helix orientation in lipid bilayers; Changes in bond orientation due to motion; Torsion angles, i.e. relative orientation of molecular segments. 24

24 Static Powder Spectra Schmidt-Rohr Spiess, Spectra S(ω) reflect orientation distribution P(θ, ϕ) of the molecules. For η=, ω( θ) = 1 2 δ 3cos2 θ 1 + ω iso P( θ) dθ = S ( ω) dω, and P( θ) = sinθ S ( ω) = P θ dθ dω = 1 6δ ω + δ principal values: directly read off from maximum and step positions. 25

25 Lab frame Spinning: Ĥ Local ( t) = ω ( t) ( Spin Part) Rotor frame b Rotor = sinθ m cosω r t sinθ m sinω r t cosθ m ( ( ( = ω ω CS θ,φ ( t) = sinθ m cosω r t, sinθ m sinω r t, cosθ m σ b ( T b ) Rotor rotor ω xx rotor ω yx rotor ω zx rotor ω xy rotor ω yy rotor ω zy rotor ω xz rotor ω yz rotor ω zz sinθ m cosω r t sinθ m sinω r t cosθ m = ω iso + 1 ( 2 3cos2 θ m 1) ( ω rotor zz ω ) iso ( ω rotor rotor yy ω xx )sin 2 θ m cos2ω r t + ω rotor xy sin 2 θ m sin2ω r t + ω rotor xz sin2θ m cosω r t + ω rotor yz sin2θ m cosω r t θ m = 54.7 : 1 ( 2 3cos2 θ m 1) = θ m 54.7 : OMAS, SAS, DAS... 26

26 Periodic Time Signal Under MAS = ω iso + ω yy ω CS θ,ω r t ( rotor rotor ω xx )sin 2 θ m cos2ω r t +ω xy At t = nt r = n 2π ω r, ω CS ( θ,φ) = ω iso f ( t r ) = e iω isot rotor sin 2 θ m sin2ω r t +... Rotation echoes At all times: f ( t) = e t i ω CS (t )dt = e iω isot e t i ω CSA (t )dt At 6 1 khz MAS: ω r >> ω ij rotor t ω CS (t)dt = ω iso t + c ij i, j ω ij rotor ω r cosn ij ω r t f ( t) e iω isot 27

27 Dipolar Coupling Tensor! D PAS = PAS D xx PAS D yy D zz PAS! = δ 2 δ 2 δ Uniaxial (η = ) along the internuclear vector. Isotropic component =. ω IS ( θ;δ) = 1 2 δ ( 3cos2 θ 1), where δ = µ 4π γ Iγ S r 3 Common dipolar coupling constants (δ): 1 H- 1 H in CH 2 groups: ~1.8 Å, ~32 khz (includes 1.5 x for homonuclear) 13 C- 1 H single bond: 1.1 Å, ~23 khz 15 N- 1 H single bond: 1.5 Å, ~1 khz 13 C- 13 C single bond: 1.54 Å (aliphatic), ~3.1 khz; 1.4 Å (aromatic), ~4.2 khz 13 C- 15 N bonds in protein backbones: 1.5 Å,.9 1. khz 28

28 Detecting the NMR Time Signal The precessing M induces a voltage in the rf coil, which is detected. B x ( t) = µ M x ( t) = µ M sinω L t Magnetic flux: φ t = B t da Voltage: U = dφ dt = µ = µ ω L M cosω L t da dm x ( t) dt da NMR signal ω L M cosω L t ΔE γ, µ γ S N = ω L M cosω L t ω L M = N ( γ) 2 1 B 4kT S N γ 5 2 B T 3 2 NMR signals are measured in the rotating frame: = f cos ω L ω rf f t t t + isin( ω L ω rf )t = f e i ω L ω rf 29

29 From Time Signals to Frequency Spectra: FT The spectrum S(ω) is the weighting factors of e iωt for a given ω in the total f(t): f ( t) = 1 2π S( ω)e iωt dω The spectrum is the Fourier transform of the time signal. S( ω) = 1 2π f ( t)e iωt dt Many useful Fourier theorems relate the shapes of f(t) and S(ω): Inverse width: dwell time = 1/spectral width, Δν 1/2 = 1/πT 2 Convolution theorem Integral theorem Symmetry theorem Sampling theorem 3