Lecture #5 Relaxation Through Dipolar Coupling

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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 = 1 … N be the eigenkets of the unperturbed j 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 ∫ k 1 " j " ! o (see homework € for proof) where€ ω = (E − E )/ kj k 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 γ µ # ! ! 3 ! ! ! ! & ! ˆ I S! 0 ˆ 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. ωI = ωS and WI = WS = W1)

2 4 2 1 3 ! µ0 $ γ ! = 2(WI + W2)= # & 6 (J(ω0 )+ 4J(2ω0 )) T1 20 " 4π % r 2 2 4 2 ! $ ! $ 3 4 2 3 ! µ0 $ γ ! τ c 4τ c µ0 γ ! τ c = # & # + & # & 6 6 2 2 2 2 Extreme " 4π % 2r 10 " 4π % r "1+ω0τ c 1+ 4ω0τ c % narrowing € • If we crunch through the numbers…

2 4 2 1 3 ! µ0 $ γ ! = # & 6 (3J(0)+ 5J(ω0 )+ 2J(2ω0 )) T2 40 " 4π % r 2 4 2 ! $ 2 4 2 3 ! µ0 $ γ ! 5τ c 2τ c ! µ $ 3γ τ = 3τ + + 0 ! c # & 6 # c 2 2 2 2 & Extreme # & 6 20 " 4π % r " 1+ω0τ c 1+ 4ω0τ c % narrowing " 4π % 2r

• Later in the course, we’ll develop Redﬁeld theory and not have to

“crunch the numbers” every time. 17 T1 and T2 of water • Some numbers for pure water….

−12 water τ c = 5.0 ×10

2 4 2 ! µ $ 3 γ ! 10 K = # 0 & =1.02 ×10 " 4π % 10 r6

! $ T1 ≈ T2 ≈ 3.92 s 1 τ c 4τ c = K# 2 2 + 2 2 & T1 "1+ω0τ c 1+ 4ω0τ c % (extreme narrowing condition)

! $ 1 K 5τ c 2τ c = #3τ c + 2 2 + 2 2 & T2 2 " 1+ω0τ c 1+ 4ω0τ c %

18 Cross Relaxation • Let’s try to gain some more physical insight/intuition into the phenomenon of cross relaxation. The Solomon Equations ! $ ! $ ˆ ˆ eq Iz ! $ Iz − Iz d # & ρI σ IS # & # & = −# &# & dt eq # Sˆ & "# σ IS ρS %&# Sˆ − S & " z % " z z %

• We’ll start by examining the results of a series of saturation recovery experiments in which the z magnetization from the I or S spins (or both) are saturated, and we then watch the recovery of the longitudinal magnetization over time.

19 Saturation Recovery

• Case (a) Initial conditions Experiment Saturate M for I spins Observe recovery of Mz z for I spins Leave S spins untouched

90o t RFI Saturation pulse

RFS Recovery curve is not a signal single exponential!

20 Saturation Recovery

• Case (a*) Initial conditions Experiment Saturate M for I spins Observe recovery of Mz z for I spins Leave S spins untouched Observe recovery of Mz for S spins

90o t RFI Saturation pulse 90o RFS Shapes of the curves signal depends on the strength of the dipolar coupling.

21 Saturation Recovery

• Case (b) Initial conditions Experiment Saturate M for I spins Observe recovery of Mz z for I spins Saturate Mz for S spins

90o t RFI Saturation pulse

RFS Saturation pulse signal curve not a single exponential.

22 Saturation Recovery

• Case (c) Initial conditions Experiment Saturate M for I spins Continue keeping Mz z for S spins saturated Saturate Mz for S spins Observe recovery of Mz for I spins

90o t RFI Saturation pulse

RFS Saturation pulse signal This curve does recover exponential with the true T1 of spin I. However it eq does not recover to Iz ! 23 Nuclear Overhauser Effect (NOE) • The NOE is the change in the equilibrium magnetization of one nuclei with the RF irradiation of a nearby nuclei (nearby deﬁned in terms of dipole coupling) • The change in magnetization can be positive (generally with small rapidly tumbling molecules) or negative (as with slower tumbling molecules) • The effect was ﬁrst proposed by Albert Overhauser in 1953. • We will describe NOE… - mathematically - graphically (via energy diagrams) Albert Overhauser - with in vivo examples

24 Calculating the NOE ˆ d Iz • Start: = −(W + 2W +W ) Iˆ − I eq + (W −W ) Sˆ − S dt 0 I 2 ( z z ) 2 0 ( z 0 ) ˆ • Saturate Sz S z = 0 d Iˆ ˆ z Iz S " W −W % • At steady state… 0 0 2 0 = eq =1+ eq $ ' dt Iz Iz #W0 + 2WI +W2 & € • Rewriting in a more convenient form and letting Ie be the steady state magnetization… % ( € eq γ S W2 −W0 Ie = (1+η)Iz where η = ' * γ I & W0 + 2WI + W2 ) This is often just expressed as: γ $ W −W ' enhancement factor NOE =1+ S & 2 0 ) =1+ η (can be positive or negative) €γ I % W0 + 2WI + W2 ( 25

€ Energy Diagram Formulation • Using an energy diagram notation… ˆ ˆ I z + S z Starting conditions Population difference ˆ − − I z − + + −€ € Decreased I

+ + polarization € € €

ˆ − − €S z − + + − €

+ + € € € Increased I Population excess polarization Population deﬁcit € (ignores WI and WS relaxation pathways)

26 P. J. Hore, NMR, Oxford University Press, p 62. NOE versus τc γ $ W −W ' NOE =1+ S & 2 0 ) γ I % W0 + 2WI + W2 ( W 6qJ( ) 2 = ωI + ωS η

W0 = qJ(ωI −ωS ) € solid S spin on large € immobile liquid molecule € € S spin on small mobile molecule

2 2 γ S ω τ c <<1⇒ NOE =1+ 2γ I

€ 31P Muscle NOE Example

31P saturate o 1H nuclei 90 RF off-resonance RF G z w/ decoupling signal and NOE

w/ decoupling Typical in vivo NOE enhancement factors for 31P-1H and 13C-1H interactions are 1.4-1.8 and 1.3-2.9 conventional respectively.

28 Transverse Cross Relaxation • NOE based on cross relaxation of longitudinal magnetization, but can cross relaxation of transverse magnetization be observed? • Answer: usually no, but sometimes yes • No effects for identical spins. But consider a dipolar-coupled system with the two spins having different chemical shifts.

Transferred magnetization components cancel

• Hence, cross relaxation of transverse magnetization is not observed between spins with different chemical shifts. 29 Spin Locking • Consider the following pulse sequence:

# π & % ( $ 2 ' x 1H Spin lock acquire t € The spin-lock Rf pulse inhibits chemical shift evolution. • Cross relaxation of transverse magnetization can now occur

T K T K Rauto = 5J (0) + 9J (ω0 ) + 6J (2ω0 ) Rcross = − 2J (0) + 3J (ω0 ) 20 ( ) 10 ( )

• Longitudinal relaxation during a spin-lock pulse is characterized

by a time constant Τ1ρ, (more in upcoming lecture on cartilage).

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