Lecture #5! Relaxation through Dipolar Coupling • Topics – Solomon equations – Calculating transition rates – Nuclear Overhauser Effect • Handouts and Reading assignments – Levitt, Chapters 19.1-3, 20.1-3, – Kowalewski, Chapter 3. – Solomon, I, “Relaxation process in a system of two spins”, Physical Review, 99(2):559-565, 1955. 1 Dipolar Coupling • The dominant source of random magnetic field variations is due to dipolar coupling and molecular tumbling. Called “tumbling” rather than “rotating” since molecules constantly bumping into each other. Spins remain aligned with B0 Water with molecule tumbling in a magnetic field ΔB(t) t B = B0 + ΔB(t) € • While Δ B ( t ) = 0 , the instantaneous€ effect is not negligible. • We need to take a close look at the properties€ of ΔB(t). 2 € € Random fields model • Assume the magnetic field seen by a spin is given by ! ! ! ! 2 2 2 2 B0z + Bx (t)x + By (t)y + Bz (t)z where Bx = By = Bz = B isotropic tumbling large main field small perturbation 1 the relaxation of M was given as: = γ 2 B2 J(ω ) z T 0 € 1 • Given the field from a dipole falls off as 1/r3 1 γ 4 1 γ 4 J . J(0) J . ∝ 6 (ω0 ) Similarly ∝ 6 ( + (ω0 )) T1 r € T2 2r Rapid decrease Primarily intramolecular rotational rather than with distance intermolecular translational motion. Bloembergen • The seminal work of Bloembergen, Purcell, and Pound (BPP), Physical Review, 73(7) 1948, recognized the importance of dipolar coupling to NMR relaxation. Nobel Prize Physics 3 1981 Complete Dipolar Effect • A more complete derivation taking into account the correlated perturbations among coupled spins was provided by Solomon Physical Review, 99(1) 1955. That diagram looks an awful lot like the diagram For identical spins, e.g. water, you drew for J-coupling? 2 4 2 Aren’t we dealing with 1 3 ! µ0 $ γ ! dipolar coupling here? = # & 6 (J(ω0 )+ 4J(2ω0 )) T1 20 " 4π % r E 2 4 2 ! $ 3 ! µ0 $ γ ! τ c 4τ c = # & 6 # 2 2 + 2 2 & 10 " 4π % r "1+ω0τ c 1+ 4ω0τ c % 2 4 2 1 3 ! µ0 $ γ ! New term corresponds to both I = # & 3J(0)+ 5J(ω0 )+ 2J(2ω0 ) 6 ( ) and S spins flipping together, i.e. T2 40 " 4π % r double quantum coherence. 2 4 2 ! $ 3 ! µ0 $ γ ! 5τ c 2τ c = # & 6 #3τ c + 2 2 + 2 2 & 20 " 4π % r " 1+ω0τ c 1+ 4ω0τ c % • Let’s look at this derivation more closely… 4 Spin Population Dynamics* • Consider a general dipolar coupled two-spin system. N-- W WI S Note: J=0, but nuclei I and S are close enough W0 N-+ in space that dipolar N+- W coupling is significant. 2 W WS I N++ • At a given point in time, the energy levels are occupied by a certain number of spins, given by N++, N+-, N-+, and N--. • Transition rates: – WI and WS = probability/time spin I or S change energy levels. – W0 and W2 = probability/time of zero of double quantum transition. 5 *see Solomon for details. The Solomon Equations N-- W • Given the transition rates and the populations, WI S W0 let’s compute the dynamics. Namely… N-+ N+- W 2 W WS I dN N++ ++ = −(W + W + W )N + W N + W N + W N dt S I 2 ++ S +− I −+ 2 −− dN +− = −(W + W + W )N + W N + W N + W N dt 0 S I +− 0 −+ I −− S ++ € dN −+ = −(W + W + W )N + W N + W N + W N dt 0 S I −+ 0 +− I ++ S −− € dN −− = −(W + W + W )N + W N + W N + W N dt S I 2 −− S −+ I +− 2 ++ € 6 € Correction for finite temperatures • Before proceeding further, we need to make a small addition, known as the finite temperature correction. • The differential equations on the previous slide assumed equality of the transition probabilities, e.g. just looking at the I spin… N-+ Under this assumption, the system will W evolve until the energy states are E I W I N−+ → N++ = N++ → N−+ = WI equally populated, which, using the Boltzmann distribution, corresponds to N ++ an infinite temperature! • To achieve a finite temperature, we can make an ad hoc correction reflecting the slightly increased probability of a transition that decreases the energy of the system. N-+ This Boltzmann factor can be derived WI(1-b) explicitly if we treat both the spin system E γ!B0 WI(1+b) where b = kT and the lattice as quantum mechanical N Boltzmann factor systems. See Abragam p. 267. ++ 7 Solomon Equations: Mz • Let’s first look at T1 relaxation. ˆ ˆ I z ∝(N++ − N−+ ) + (N+− − N−− ) S z ∝(N++ − N+− ) + (N−+ − N−− ) • Substituting yields a set of coupled differential equations indicating longitudinal magnetization recovers via a combination of two € € exponential terms… direct relaxation cross relaxation ˆ d Iz = −(W + 2W +W ) Iˆ − I eq − (W −W ) Sˆ − Seq dt 0 I 2 ( z z ) 2 0 ( z z ) ˆ d Sz = −(W + 2W +W ) Sˆ − Seq − (W −W ) Iˆ − I eq dt 0 S 2 ( z z ) 2 0 ( z z ) 8 General Solution • The solution can also be written ! $ ! $ eq γ B ˆ ˆ eq I I 0 # Iz & ! $# Iz − Iz & z = d ρI σ IS 2kT # & = −# &# & with dt # σ ρ & eq eq γ B # Sˆ & " IS S %# Sˆ − S & S = S 0 " z % " z z % z 2kT where R is called the ρ = W + 2W +W d ! ! ! eq I 0 I 2 V = −R V −V relaxation matrix with ρ = W + 2W +W dt ( ) S 0 S 2 elements given by: σ IS = W2 −W0 • The general solution is of the form ˆ −λ1t −λ2t Iz = α11e +α12e (not a single exponential) ˆ −λ1t −λ2t Sz = α21e +α22e 9 Identical Spins • For the case of S and I identical (i.e. ωI = ωS and WI = WS = W1) ˆ ˆ d Iz + Sz ( ) = −2(W +W ) Iˆ + Sˆ − I eq − Seq (pure exponential) dt 1 2 ( z z z z ) 1 = 2(W1 + W2) T1 • We now need explicit expressions for the transition probabilities W€1 , and W2. 10 Time-dependent Perturbation Theory • To compute the transition probabilities, we need to use a branch of QM known as time-dependent perturbation theory. Consider the case where: perturbation ˆ ˆ ˆ ˆ ˆ H (t) = H 0 + H1 (t) with H1 << H 0 . • Let m j , j = 1 … N be the eigenkets of the unperturbed Hamiltonian with energies Ej. € ˆ 1 H0 mj = ! E j mj with j =1,…N st € • Assuming the system starts in state m j , then, to 1 order, the probability of being in state m k at time t is given by 2 1 t i t = m Hˆ (t ") m e− ω kj " dt " Pkj 2 ∫o k 1 j (see homework ! € for proof) where€ ωkj = (E k − E j )/! 11 € € Fermi’s Golden Rule • Consider a sinusoidal perturbation ˆ ˆ ˆ V1 iωt −iωt This is actually quite general as we can H1(t) = V1 cosωt = e + e always analyze the Fourier decomposition 2 ( ) of any perturbation. • The transition rate is then given by 2 t Note, we can take the limit of t → ∞ Pkj 1 −iωkjt! ˆ if our measurements are much longer Wkj lim = lim mk H1(t#) mj e dt! = ∫ o than 1 ω . t→∞ t t→∞ t kj € ˆ 2 mk V1 mj This is a famous result known Wkj = δE −E ,−!ω +δE −E ,+!ω as Fermi’s Golden Rule. 4 ( k j k j ) Interaction term Conservation of energy Ek Ek E Absorbing a photon Emitting a photon of of energy Ek-Ej energy Ek-Ej 12 Ej Ej Example: Rf Excitation • For a two spin system (rotating frame, on resonance) … ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ I ˆ S ˆ H = ΩI I z + ΩS S z + 2πJ(I xS x + I yS y + I zS z ) + ω1 I x + ω1 S x ˆ ˆ H 0 H1 ˆ • The interaction term (matrix form with eigenkets of H 0 as the € basis) is given by … + + + − − + − − € + + # 0 1−i 1−i 0 & € + − %1 +i 0 0 1−i( (Simple case of γI = γS) H1 ∝€ € € € € − + %1 +i 0 0 1−i( 0 1 i 1 i 0 € − − $ + + ' € € – No excitation€ if Rf is “off resonance” € – No excitation€ of double or zero quantum coherences 13 Dipolar Coupling • The complete dipolar coupling Hamiltonian is given by ! ! ! ! ! γ Iγ S! µ0 # ˆ ˆ 3 ˆ ! ˆ ! & ˆ where r is the vector Hdipole = − 3 %I ⋅ S − 2 (I ⋅ r)(S ⋅ r)( r 4π $ r ' from spin I to spin S ˆ ˆ ˆ ˆ ˆ ˆ • Using the raising and lowering operators: I+ = Ix +iIy and I− = Ix −iIy the the Hamiltonian can be written in polar coordinates€ as: µ0 γ Iγ S! Hd = − 3 (A + B +C + D + E + F) where 4 r A = Iˆ Sˆ F π z z 0 zero quantum term 1 ˆ ˆ ˆ ˆ 2 B = − I S + I S F F0 (t) =1− 3cos θ 4 ( + − − + ) 0 ˆ ˆ ˆ ˆ 3 −iφ C = I S + I S F F1(t) = 2 sinθ cosθe € ( + z z + ) 1 single quantum terms 3 2 2i * − φ D Iˆ Sˆ Iˆ Sˆ F F2(t) = 4 sin θe = ( − z + z − ) 1 € double quantum terms Note, with molecular E = Iˆ Sˆ F € + + 2 € tumbling, both θ and φ ˆ ˆ * are functions of time. F = I S F2 € − − € € € Example: Calculating W1 2 • Assumption: F(t)F * (t + τ) = F(0) e− τ τ c time average • For a given pair of like spins: 2 t 2 € 1 2 4 2 ∞ 1 γ ! −iω0t! ! µ $ γ ! 2 −τ τ i W = lim F (t!)e dt! 0 c − ω0τ 1 ∫ 2 3 1 = # & F1(0) e e dτ t→∞ t r 6 ∫ 0 " 4π % 4r 0 • For an ensemble of spins: 2 4 2 ! µ0 $ γ ! 2 τ c W1 = # & 6 F1(0) 2 2 " 4π % 4r 1+ω0τ c 2 4 2 ! µ0 $ 3γ ! W1 = # & 6 J(ω0 ) 2π π " 4 % 40r 2 1 π F (0) = 9 sin2 θ cos2 θ sinθdθdφ 1 ∫ ∫ 4 4π 0 0 15 € Transition Probabilities • Using similar equations, the full set of transition rates are: 3 WI = 2 qJ(ωI ) 2τ c 3 J(ω) = 2 2 WS = 2 qJ(ωS ) 1 where + ω τ c 2 € W = 6qJ(ω + ω ) 1 ! µ $ γ 2γ 2!2 2 I S q = # 0 & I S sum of chemical 10 " 4π % r6 shifts IS € € rapid fall W0 = qJ(ωI −ωS ) off with distance difference of € chemical shifts € 16 Identical Spins • For the case of S and I identical (i.e.