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
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
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
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
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 4 2 2 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) 2 2 + V ab 2 5
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
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 1 2 1 7 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
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
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
Truncation of Weak Interactions: Keep B // B tot = B + B loc = ( B + B loc,// ) 2 2 + 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
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
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 + 1 2 δ ( 3cos2 θ 1 ηsin 2 θ cos2φ) Îz Ĥ D,IS = µ 4π! γ Iγ S r 3 1 2 3cos2 θ 1 2ÎzŜz Ĥ D,II = µ 4π! γ 2 Ĥ Q = j k r jk 3 1 2 3cos2 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
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
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
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= 1 2 1 1, Îy ˆ= 1 2 i i, Îz ˆ= 1 2 1 1 Î 2 x = Îy 2 = Îz 2 = 1 4 1 1 = 1 4 ˆ1 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
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= 1 2 1 1, Îy ˆ= 1 2 i i, Îz ˆ= 1 2 1 1, Îx 2 = Îy 2 = Îz 2 = 1 4 1 1 = 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
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
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
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 + 1 2 δ 3cos2 θ 1 ηsin 2 + ( θ cos2φ) ) *, - Îz Ĥ D,IS = µ 4π γ Iγ S r 3 1 2 3cos2 θ 1 2ÎzŜz Ĥ D,II = µ 4π γ 2 Ĥ Q = eqeq 2I 2I 1 j k 1 r jk 3 1 2 3cos2 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
Chemical Shift Tensor the Principal Axis System Ĥ CS = γ Î! σ B! =... σ iso ω + 1 2 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 1........................ 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
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
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
Static Powder Spectra Schmidt-Rohr Spiess, 1995. 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 2 1 2 3 principal values: directly read off from maximum and step positions. 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
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
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
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
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