grzesiek_relaxation

-1- Notes on relaxation and dynamics EMBO Practical Course on NMR, Basel, July 6-13, 2007 Stephan Grzesiek Biozentrum der Universität Basel Klingelbergstr 70 CH-4056 Basel, Switzerland e-mail: [email protected] 1. RELAXATIO N AND DYNAM ICS...................................................................................................... ..............2 1.1 G ENERAL REMARK ............................................................................................................................... ..............2 1.2 T HE DIPOLAR INTERACTION ............................................................................................................................... 4 1.3 T2 MEASURED BY THE CARR-PURCELL -MEIBOOM -GILL EXPERIMENT .............................................................5 1.4 T HE LINEWIDTH INCREASES WITH INCREASING MOLECULAR WEIGHT .............................................................6 1.5 A SIMPLE EXAMPLE OF FLUCTUATION -DISSIPATION : WHY IS T2 FOR LARGER MOLECULES INVERSELY PROPORTIONAL TO THE MOLECULAR TUMBLING TIME ? ...................................................................................................7 1.6 T1 RELAXATION ............................................................................................................................... .................13 1.7 S PIN-FLIPS INDUCED BY BROWNIAN MOTION , THE NUCLEAR OVERHAUSER EFFECT .....................................15 1.8 T HE SOLOMON EQUATION ............................................................................................................................... .17 1.9 H OMONUCLEAR NOE ............................................................................................................................ ..........19 1.10 H ETERONUCLEAR NOE............................................................................................................................ ........21 1.11 D YNAMICAL INFORMATION FROM RELAXATION EXPERIMENTS .....................................................................23 2. A QUICK DERIVATION OF TH E MASTER EQ UATION .......................................................................26 2.1 T HE TOTAL HAMILTONIAN ..............................................................................................................................2 6 2.2 L IOUVILLE -VON NEUMANN EQUATION ...........................................................................................................26 2.3 T RANSFORM OPERATORS TO INTERACTION PICTURE ......................................................................................26 2.4 I NTEGRATE ONCE ............................................................................................................................... ..............27 2.5 T AKE ENSEMBLE AVERAGE (––––) OF ALL TERMS .............................................................................................27 2.6 I NTRODUCE RETURN TO EQUILIBRIUM AD HOC ...............................................................................................27 2.7 E XPAND H1 IN SPIN AND TIME DEPENDENT PART ............................................................................................28 2.8 U SE ASSUMPTIONS ON RANDOM MOTIONS ......................................................................................................28 2.9 U SE ONE MORE ASSUMPTIONS ON RANDOM MOTIONS ....................................................................................28 2.10 S ECULAR APPROXIMATION ..............................................................................................................................3 0 3. REFERENCES ................................................................................................................... ................................31 4. APPE NDIX ..................................................................................................................... .....................................32 4.1 E FFICIENCY OF SCALAR MAGNETIZATION TRANSFER IN THE PRESENCE OF RELAXATION .............................32 4.2 T HE SIZE OF THE SCALAR COUPLINGS ..............................................................................................................33 4.3 S OME PRODUCT OPERATOR GYMNASTICS ........................................................................................................34 4.4 I RREDUCIBLE REPRESENTATION OF THE DIPOLAR INTERACTION ....................................................................34 4.5 E XAMPLE : HETERONUCLEAR T2 FROM REDFIELD THEORY .............................................................................35 1/7/07 16:04 -2- 1. Relaxation and dynamics 1.1 General Remark Application of RF-frequency pulses changes the density matrix and moves the spins away from their thermal equilibrium distribution. E.g. for protons in a 14T magnetic field the density matrix at thermal equilibrium (T = 293K) is given by eq=exp(H/kT) trace exp(H/kT) ()
1 2(1H/kT)=1 2(1(hIzB) /kT)
1 2(1+ 104Iz) (1.1) where H=
hIzB, exp(H/kT)
1H/kT (since H<<kT), 1 is the unity matrix, trace 1 =2, trace H = 0. A 180 ˚ pulse around the x-axis (NOE-experiment) transfers this density matrix to 1 2(1+ 10
4Iz)180 ˚Ix     1 2(1
10
4Iz) (1.2) Relaxation is the phenomenon that the spins have the tendency to come back to their thermal equilibrium distribution
eq. The time constants for the return to thermal equilibrium are T1 for the longitudinal magnetization (z-magnetization) and T2 for the transverse magnetization (x,y- magnetization). For somewhat historical reasons T1 is also called spin-lattice and T2 is called spin-spin relaxation time. For the relaxation phenomenon to happen, internal forces must act on the individual spins. In solution, the main source for these forces are random fields caused by the random (Brownian)motion of the molecule. Individual spins experience different random field fluctuations. Theaverage of these fluctuations leads to the return to thermal equilibrium of the entire macroscopic ensemble of all spins. -3- The macroscopic relaxation rates can be calculated from a statistical description of the individual microscopic fluctuations. Thus the macroscopic relaxation rates containinformation on the average dynamical behavior of individual spins and their surroundings The approach of thermal equilibrium is mainly induced by the following internal forces (interactions): 1. dipolar interactions between the magnetic moments of the spins (typical energies in biomacromolecules < 10 4-105 Hz * h) 2. the anisotropy of the chemical shift (CSA). Different orientations of the molecule in the solution lead to different shieldings, i.e. different Larmor frequencies. This results in internal forces on the spins (Energies/h < 104-105 Hz). 3. electric quadrupolar interaction s of the nucleus with the inhomogeneous electric field produced by the electrons. This interaction is very large, but it is only observed for spinswith I > 1/2. (quadrupolar energy/h ~ 200 kHz for 2H and ~ 3 MHz for 14N). Remark 1: In NMR, the forces resulting from these random fields are rather weak compared to the force that the external magnetic field exerts on the spins. These weak forces lead torather long relaxation times, e.g. for biological macromolecules in solution T 2 values are in the range of tens of milliseconds and T1 values are on the order of seconds. Therefore we can observe the oscillations of the magnetic moments on the order of 600 MHz * 10 ms =6
000
000 times. This is the reason why despite of the small magnitude of the equilibrium polarisation (10 -4), NMR can still be observed with reasonable sensitivity. Remark 2: The correlation between macroscopic relaxation phenomena and microscopic fluctuations is described in a very general way by the fluctuation-dissipation theorem : the macroscopic relaxation rates are proportional to (the Fourier transform of) the autocorrelation function of the fluctuating forces (see e.g. F. Reif). This is for example the case for the viscosity
of a medium (proportional to the autocorrelation of Brownian force fluctuations on individual particles) or the electrical resistance R (proportional to the autocorrelation of thermal voltage fluctuations in a conductor). The NMR theory of relaxation (Redfield theory)is the equivalent of the fluctuation-dissipation theorem applied to an ensemble of spins described by quantum statistics: the NMR relaxation rates are proportional to the autocorrelation function of the fluctuating microscopic fields. We shall see this in detailbelow. -4- 1.2 The dipolar interaction The dipolar interaction is the most important contribution to relaxation for biological macromolecules. A magnetic nucleus with gyromagnetic ratio
and spin I is surrounded by a magnetic dipolar field Bd which is given by the following equation: Bd=hµ0 4
r3I-3r(Ir) r2    (1.3) where r is the distance vector to the nucleus. hI/2
Bdr Fig.: dipolar field around nucleus with magnetic moment µ=
hI The dipolar energy Hd of a second nucleus with gyromagnetic ratio
2 and spin I2 in the field of the first nucleus with gyromagnetic ratio
1 and spin I2 is then given as Hd=µ2Bd1=(2hI2)1hµ0 4
r3I1-3r(I1r) r2    =12h2µ0 4
r3 I1I2-3(I1r)(I2r) r2    (1.4) We notice that depending on the position of the second nucleus (distance vector r), the dipolar field and therefore the dipolar energy can have different values. If the two nuclei are part of the same molecule, then during the thermal movement of the molecule (Brownian rotation) the direction of r changes, whereas the length of r stays usually more or less constant. If the molecule has isotropic orientation in solution then the average of the dipolar interaction energy over all angles vanishes. The Brownian rotation, however, leads to fluctuations in the dipolar field and dipolar interaction energy. These fluctuating fields can induce rotations of the spins. As the fluctuations are driven by the Brownian motion, this provides a mechanism to exchange energy between the spins and also between the spins and the mechanical motion of themolecule. -5- 1.3 T2 measured by the Carr-Purcell-Meiboom-Gill experiment A classical experiment to visualize T2 is the spin-echo experiment with the Carr-Purcell- Meiboom-Gill (CPMG) sequence: 90˚x- [
180 ˚y

echo]n Fig.: Harris p 83. An initial 90 ˚ pulse turns the equilibrium magnetization into the xy-plane. The magnetization vectors from different nuclei then begin to fan out (b) and the signal decays. A 180 ˚y pulse is applied after time  (c) which has the effect of rotating all the magnetization vectors about the y-axis, or in other words reflecting them in the zy-plane, they continue to move in the same direction (d), and after a further time  (e) they are again in phase in the y-direction, and the signal in the receiver coil is maximal. The process of refocusing by the 180 ˚ pulse can be repeated many times. The amplitudes of successive echoes decay exponentially and the truevalue of T 2 can be found from the envelope of the echoes: Fig.: Harris p. 83. -6- 1.4 The linewidth increases with increasing molecular weight A puzzle for many students of NMR is the fact that the line widths of the nuclear resonance lines are becoming larger when the molecular weight of the molecules in solution is increased.Remember that the linewidth is given by =/2
=1/
T2() (1.5) An example for this increase can be seen for the three molecules magainin (3 kDa), BPTI (7 kDa), and staphyloccocal nuclease (18 kDa): Fig.: one-dimensional proton spectra of magainin, BPTI, and staphyloccocal nuclease. This increase in linewidth has very important consequences for high resolution NMR in solution: 1. It leads to unsolvable overlap problems 2. As the lines are becoming wider, their amplitudes decrease. Remember that the integral of the line stays constant. The smaller amplitude corresponds to a loss in the signal to noiseratio. 3. The speed of magnetization transfer in COSY experiments is given by the size of the scalar coupling constant J. The time required for this transfer is given by ~1/J. Only when themagnetization lives long enough, i.e. T 2 > ~1/J, the transfer can be carried out in an effective way. This means that for very large molecules, COSY experiments become completely inefficient. -7- 1.5 A simple example of fluctuation-dissipation: why is T2 for larger molecules inversely proportional to the molecular tumbling time? A simple explanation for this behavior will be attempted in the following section. Consider a system of two spins A and B, which are close to each other in a molecule. The magnetic moment of spin B produces a dipolar magnetic field at the location of spin A. The Brownian motion of the molecule in solution has the effect that this dipolar magnetic field fluctuates. Inprinciple there can be x,y, and z-components of the dipolar field at the location of Adepending on the relative orientation of the two nuclei. Fig.: Derome p.102 If the dipolar field at the location of A is parallel to the external field, the magnetic moment of A will rotate a little faster, if the dipolar filed is antiparallel, A will rotate a little slower thanits normal Larmor frequency  which is given by B o. Now, let’s make the simplification that the dipolar field has a constant size Bd and that the field can only have two orientations, i.e. parallel or antiparallel to z. At the same time we assume that the field stays constant in one direction for a characteristic time c and that after this time it can either invert its direction or continue in the same orientation. This change is completely random and occurs with equal probability. Now let’s also assume that the magnetic moment of A is in the xy-plane. In the absence of the dipolar field, after a time c, the magnetic moment would have rotated by an angle =c=
Boc (1.6) Because of the presence of the random dipolar field this angle  is modified by an addition
 which is given by -8-
=±Bdc=±dc (1.7) where the plus and minus sign depends on whether the dipolar field had been parallel or antiparallel to Bo and where we have use the abbreviation d =
Bd. Now, lets wait for a longer period t: t=n
c (1.8) In this case, the dipolar field had had n times the possibility to change its direction. The problem is identical to the random walk (or drunken man’s walk) in one dimension. How fardoes the drunken man get if he only steps left and right by a length L of 1 m with equalprobability and he repeats this process n times? Fig.: Reif pg.5 If this experiment is repeated many times (by many drunkards) the answer is the binomial probability distribution: Pn(m)=n! n+m 2    !n
m 2    !1 2    n (1.9) This is the probability of finding the man at position m*L after n steps with equal probability to left or right. The binomial distribution leads to the Gaussian distribution if the number ofsteps becomes very large. After their walk, some men will end up a little bit to the left, some men a little to the right, but the average position of all men after their walk is the spot where they started (x=0). -9- -20-18-16-14-12-10-8-6-4-20246810121416182000.020.040.060.080.10.120.140.160.18 mP(m)binomial distribution n = 20 F ig.: Binomial probability distribution for n = 20 steps with equal probability (p = q = 0.5) in left and right direction. The graph shows the probability Pn(m) of a net displacement of m units to the right (adapted from Reif p. 11). However, the width of the probability distribution or the root mean square displacement increases with the number of steps. For the binomial distribution the width is given by n1/2. This means for 100 steps the mean square displacement is 10. Now consider the situation for the dipolar coupling of the two spins A and B in a molecule in solution: of course we are not just dealing with a single molecule but with many moleculesand we detect the nuclear moments of the spins A of all the molecules as a sum. After the timet, i.e. after doing n times the random addition of  dc to the phase angle of the single spins A, the phase angles of all the spins A of all the molecules will have a random distribution. The width  of this random distribution of phases is given by
=dcn1 2=dct c     1 2 =dct()12 (1.10) Now, when  reaches the value 2
the spins A of all the molecules are more or less completely spread out in the xy plane. The result is that the sum of all the magnetic moments becomes 0 and the signal becomes zero. -10- Fig.: the distribution of phase angles over the whole circle (pizza distribution) gives a net magnetic moment of zero for the whole ensemble of spins A. To a first approximation, the relaxation time T2 is the time when  has reached the value 2
. 2
=dcT2()1 21 T2=d 2
    2 c=Hd h    2 c (1.11) Thereby one can see that T2 is inversely proportional to c and it is directly proportional to the square of the (dipolar) interaction energy Hd. This is a very general principle resulting from the fluctuation-dissipation theorem, which would also hold if the system had been disturbed by a different interaction. In reality, the dipolar field does not only jump between the +z and -z direction. The dipolar field moves continuously as the molecules rotate. The characteristic time for this molecularBrownian rotation is called the molecular tumbling time or rotational correlation time  c. For a spherical molecule of radius a rotating in a liquid of viscosity , c is given by c=4
a3 3kT=V kT(1.12) where V is the volume of the molecule. As the volume is proportional to the molecular weight, the rotational correlation time is proportional to the molecular weight. Therefore, T2 is inversely and the line width is directly proportional to the molecular weight. As a rule of thumb, the rotational correlation time [given in ns] of a molecule in aqueous solution at room temperature is about half its molecular weight in kDa. -11- 1.5.1 Example: Estimate the dipolar T2 for a proton which is 1.01 Å away from a 15N nucleus with a rotational correlation time of 8 ns. Answer: the dipolar frequency is derived from the dipolar coupling: 1Bd2=d=Hd/h=12hµ0 4
r3 15N=2.71107(Ts)1;1H=2.68108(Ts)1 µo=4
107VsA-1m-1;T=Vsm-2 h=1.0541034Js r=1.011010m;c=8109s d/2
=1.18104Hz 1 T2=d/2
()2c=1 (2
)2µ0 4
    2I2S2h2 r6c 1 T2=1.1 Hz(1.13) The above simplifications have neglected a number of things: 1. the movement of the molecules is continuous and not a two-site random jump 2. there are also contributions of the dipolar field along the x- and y-axis which remove magnetization from the xy-plane 3. the assumption that T2 corresponds to the time when the width of the phase distribution has reached a full circle overestimates T2. An exact calculation for two (dipolarly) interacting spins I and S at a distance r shows that 1 T2I=1 20µ0 4
   2I2S2h2 r64J(0,c)+J(IS,c)+ 3J(I,c)+6J(S,c)+6J(I+S,c)     (1.14) and by symmetry: 1 T2I(
I,
S)=1 T2S(
S,
I)(1.15) -12- The function J(,c) is called spectral density. It gives the probability of finding the frequency (energy)  within the thermal motions of the molecule. For spherical molecules, J(,c) has the following form: J(,
c)=
c 1+2
c2 (1.16) where c is the rotational correlation time of the molecule. Remark 1: in the slow motion limit ( c >> 1), the terms J(,c) in Eq. 1.14 are very small compared to J(0,c) = c. Thus 1/T2 is simply proportional to c. This holds also for other relaxation mechanism such as quadrupolar relaxation and CSA relaxation. Therefore T2s for different nuclei and from different relaxation mechanisms within a single molecule should be proportional to each other (provided there is no variation in local mobility and there is noadditional line broadening from chemical exchange). Remark 2: In the slow motion limit, the crude estimate for the heteronuclear1/T 2 that we derived above is off from the exact value by a factor of 5/(2
)2  0.13. Remark 3: Redfield theory (see below, Eq. 2.16) shows that the spectral density is given by the Fourier transform of the autocorrelation function of the field fluctuations F(t) resulting from the thermal motions: J()=F(0)F*( t ) 0
 exp(i t )d t (1.17) where the overbar presents the ensemble average over all spins. Thus the relaxation rates, e.g. Eq. 1.14 are proportional to the square (autocorrelation) of the field fluctuations. This is (again) the central point of the fluctuation-dissipation theorem. In an equilibrium situation, the autocorrelation of the field fluctuations K(s)=F(t)F*(t+s) does not depend on the absolute value of the time t, but only on the difference s. K(s) is a decaying function with a maximum at s = 0. In the case of rotational Brownian motion, the decay is monoexponential and the decay time is the rotational correlation time c. The Fourier transform of the monoexponential decay gives Eq. 1.16. -13- Fig.: Illustration of the correlation function K(s)=F(t)F*(t+s) of a random function F(t). (From Reif, pg. 571) 1.6 T1 relaxation The return to equilibrium along the z-axis of the spins involves transitions of the spins from the down to the up state and vice versa. Therefore in the random magnetic fields at thelocation of the nucleus I, there must be components that correspond to this transition, which happens at the frequency  o. It can be shown that for dipolar relaxation the T1 relaxation rate is proportional to the square of the dipolar field strength times the spectral density of the field fluctuations at frequency o. The spectral density J(,
c)=
c 1+2
c2has the following appearance: 106107108109101000.20.40.60.81x 10-8 Larmor freq.  [Hz]spectral density [s]
c = 10 ns MWT ~ 20 KDa
c = 5 ns MWT ~ 10 KDa
c = 1 ns MWT ~ 2 KDafast motion 
c << 1 J ~
c slow motion 
c >> 1 J ~ 1/
c Fig.: the spectral density or the efficiency of relaxation for different Larmor frequencies and rotational correlation times
c. -14- What the graphic shows is that the spectral density is more or less constant from zero frequency up to a limiting frequency, which is given by 1/
c. For larger frequencies the spectral density goes to zero. The area under the spectral density function stays the same independent of
c. For small molecules (short correlation time, 
c << 1) the function is spread out such that there is not much density at the position of the Larmor frequency. For large molecules (long correlation time, 
c >> 1) the spectral density is very small at the position of . Therefore both for small and large molecules T1 relaxation is inefficient. Between the two limits, there is the most efficient T1 relaxation, the “T1 minimum”. The exact calculation of T1 (non-selective T1) for two identical spins I (two protons) by dipolar coupling yields: 1 T1=3 10µ0 4
    2I4h2 r6J(I,c)+4J(2I,c) {} (1.18) The behavior of the T1 and T2 as a function of the correlation time is shown in the following figure: 10-1210-1110-1010-910-810-710-310-210-1100101102103 tau_c [s] [s]T1 T2400, 600, 800 MH z400, 600, 800 MH z Fig.: T1 and T2 for two-spin system consisting of two protons with identical Larmor frequencies (400, 600 or 800 MHz) at a distance of 2 Å as a function of the correlation time. -15- 1.7 Spin-flips induced by Brownian motion, the nuclear Overhauser effect Consider now a system of two spins I and S, which are in dipolar interaction. Spin I is moved away from thermal equilibrium by a 180 ˚ Ix-pulse as described above. The dipolar field of the second nucleus S at the position of nucleus I fluctuates because of the Brownian motion of themolecule. These fluctuations generate transitions between the states of the IS spin system.The dominant transition is the so-called spin flip process where spin I changes from the up- state to the down-state and spin S from the down-state to the up-state and vice versa (   ). We look at the spin system for the following sequence of events: (1) – 180 ˚ I x – (2) – I-S spin-flip –(3): IS IS IS IS spin states at equilibrium (1)    probability1 4+
P(I) 2+
P(S) 214+
P(I) 2
P(S) 214
P(I) 2+
P(S) 214
P(I) 2
P(S) 2 spin states after 180 ˚-Ix pulse (2)    spin states after    spin flip (3)    Consider the table above and the distribution of states at equilibrium. We calculate for the ensemble average values of the observables (time point 1): << I>>(1)=1 4+
P(I) 2+
P(S) 2+1 4+
P(I) 2
P(S) 2=1 2+
P(I) << I>>(1)=1 4
P(I) 2+
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(I) << S>>(1)=1 4+
P(I) 2+
P(S) 2+1 4
P(I) 2+
P(S) 2=1 2+
P(S) << S>>(1)=1 4+
P(I) 2
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(S)(1.19) where the
Ps are given from the Boltzmann distribution. We also notice that 1 2<< I>>
<< I>>    =<< Iz>> 1 2<< S>>
<< S>>    =<< Sz>>(1.20) and therefore -16- << Iz>>(1)=
P(I) << Sz>>(1)=
P(S)(1.21) After the 180 ˚ Ix-pulse (time point 2), the following ensemble averages are calculated: << I>>(2)=1 4
P(I) 2+
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(I) << I>>(2)=1 4+
P(I) 2+
P(S) 2+1 4+
P(I) 2
P(S) 2=1 2+
P(I) << S>>(2)=1 4+
P(I) 2+
P(S) 2+1 4
P(I) 2+
P(S) 2=1 2+
P(S) << S>>(2)=1 4+
P(I) 2
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(S)(1.22) and therefore << Iz>>(2)=
P(I) << Sz>>(2)=
P(S)(1.23) If the    spin flips between point (2) and (3) happened with a 100% probability the distribution of states would be the following: << I>>(3)=1 4+
P(I) 2+
P(S) 2+1 4
P(I) 2+
P(S) 2=1 2+
P(S) << I>>(3)=1 4+
P(I) 2
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(S) << S>>(3)=1 4
P(I) 2+
P(S) 2+1 4
P(I) 2
P(S) 2=1 2
P(I) << S>>(3)=1 4+
P(I) 2+
P(S) 2+1 4+
P(I) 2
P(S) 2=1 2+
P(I)(1.24) and therefore << Iz>>(3)=
P(S)=<< Sz>>(1) << Sz>>(3)=
P(I)=<< Iz>>(1)(1.25) Clearly the spin flips exchange the z-magnetizations between the two spins (trivial). If the probablities
P(I) and
P(S) were equal, then the net effect from time point (1) to (3) would be that spin I is back to equilibrium whereas spin S has its equilibrium distribution inverted. The effect that z-magnetization can be exchanged between spins on the way to thermalequilibrium by means of the dipolar interaction is called the nuclear Overhauser effect . -17- 1.8 The Solomon equation The probability for the spin-flips is not 100%, but is given as a rate constant (per unit time). This constant is the NOE build-up rate kNOE. kNOE is proportional to the square of the dipolar interaction times a function that depends on the rotational correlation time
c (spectral density). The total return to equilibrium for the z-magnetizations of a two-spin system is described by the Solomon equation [Solomon, I. (1955). Phys. Rev. 99, 559]: d dt<<Iz>> <<Sz>>     =
k1(I)
kNOE
kNOE
k1(S)     <<Iz>>
<<Iz>>eq <<Sz>>
<<Sz>>eq      (1.26) In the case of an isolated, heteronuclear two-spin system, the rate constants k1(I), k1(S), kNOE are given by k1(I)=1 10µ0 4
   2I2S2h2 r63J(I,c)+J(IS,c)+6J(I+S,c) {} k1(S)=1 10µ0 4
   2I2S2h2 r6 3J(S,c)+J(SI,c)+6J(s+I,c) {} kNOE=1 10µ0 4
   2I2S2h2 r66J(I+S,c)J(IS,c) {} (1.27) Remark: the approach to thermal equilibrium described by the Solomon equantion, i.e by the NOE, i.e. by the spin-flips resembles very much the propagation of a heat wave. At time zero spin I is heated, then the heat dissipates to the next neighbor to the next neighbor to the next neighbor ... -18- inversion of spin I: deviation from thermal equilibriumt = 0 some spin flips with next neigborst = t some spin flips with next neigborst = 2t some spin flips with next neigborst = 3t equal distribution at t =
Fig.: approach to thermal equilibrium by NOESY spin-flips 1.8.1 Initial rate approximation For NOE mixing times tm, which are short compared to the NOE buildup rate kNOE, the increase in z-magnetization on spin S due to NOE transfer from spin I can be approximated from the Solomon equation (dt 
t = tm): <<Iz>> <<Sz>>    
k1(I)kNOE kNOEk1(S)     <<Iz>><<Iz>>eq <<Sz>><<Sz>>eq     t (1.28) -19- The amount of magnetization, which has started on Iz and was transferred to Sz during the mixing time, is given by: <<Sz>>NOE
kNOE<<Iz>>(t=0)<<Iz>>eq () tm (1.29) In the NOESY experiment, the amount of Iz-magnetization at the beginning of the mixing period ( <<Iz>>(t=0)) equals
<<Iz>>eqcos(It). Therefore
<<Sz>>NOE=kNOEcos(It1)+1 () tm<<Iz>>eq (1.30) This means that the NOESY (I->S) cross-peak intensity (value
in the original equation of the NOESY experiment) is given by 
kNOEtm<<Iz>>eq (1.31) Therefore the NOESY crosspeak intensity is proportional to the NOESY build-up rate kNOE in the initial rate approximation. From the Solomon equation we realize that kNOE is proportional to
I2
S2r6. It follows that the NOE is strongest for short distances and between protons (because protons have the largest gyromagnetic ratios). 1.9 Homonuclear NOE For the homonuclear case (e.g. NOE between protons) we can set I
S
. kNOE is then given as kNOE=1 10µ0 4
   24h2 r66J(2,c)J(0,c) {} (1.32) For spherical Brownian motion, this function has a zero crossing at c=5/4
1.12 (1.33) This means that for B = 14T, no NOE can be observed between protons ifc1.12 /(2
600MHz) =0.3 ns . At correlation times larger than 0.3 ns, kNOE becomes negative (but the cross peak in the NOESY spectrum will have the same sign as the diagonal peak). For 
c>>1 (slow tumbling limit), the contribution of J(2,
c) can be neglected and kNOE is simply given by kNOE1 10µ0 4
   24h2 r6c (1.34) -20- 10-1210-1110-1010-9-0.8-0.6-0.4-0.200.20.4 tau_c [s]k_noe [Hz]400 MHz 800 MHz600 MHz Fig.: NOE buildup rate kNOE for two protons at a distance of 2 Å for Larmor frequencies of 400, 600, 800 MHz. 0 0.5 1 1.5 2 2.5 3 x 10-8-30-25-20-15-10-50 tau_c [s]k_noe [Hz] Fig.: NOE buildup rate kNOE for two protons at a distance of 2 Å for Larmor frequencies of 400, 600, 800 MHz. -21- 1.10 Heteronuclear NOE From the homonuclear case, we have seen that kNOE depends on the rotational correlation time of the internuclear vector of the two nuclei, which undergo dipolar relaxation. In principle, the rotational correlation time could be extracted from the intensities of the homonuclear proton- proton NOESY experiment. In practice, this is very difficult, because often the exact distancesare not known and secondly many protons interact at equal strength since their distances aresimilar. The heteronuclear NOE between a proton and a directly bonded heteronucleus is much easier to interpret, because the one-bond distance is rather well determined and because this NOE is the dominant relaxation effect on the heteronucleus. { 1H}-15N NOEs and to a lesser extent {1H}-13C NOEs are used to obtain information on the correlation times of the respective internuclear vectors in biological macromolecules. Usually the steady-state NOE effect is measured. The proton spin is often called I and the heteronuclear spin S. The steadystate NOE enhancement compares the z-magnetization of the S-spin in thermal equilibrium tothe z-magnetization of the S-spin at equilibrium when the I-spin is saturated: NOE({I}- S}) =<<Sz>>(I-sat) <<Sz>>eq(1.35) Saturation of the I-spin means that its magnetization vanishes. This can be obtained by applying a large number or random RF-pulses on I. The z-magnetization of the S-spin for this case can be calculated as a steady state solution of the Solomon equation. For the steady state, the time-derivatives are simply set to 0. Because of the saturation of spin I, we use <<Iz>> = 0: 0=
k1(I)
kNOE
kNOE
k1(S)    
<<Iz>>eq <<Sz>>I-sat
<<Sz>>eq      (1.36) Solving for <<Sz>>I-sat yields: 0=kNOE<<Iz>>eq
k1(S)<<Sz>>I-sat
<<Sz>>eq () <<Sz>>I-sat=kNOE k1<<Iz>>eq+< < Sz>>eq(1.37) And therefore: NOE({I}- S}) =kNOE k1<<Iz>>eq <<Sz>>eq+1=kNOE k1
I
S+1 (1.38) k1 is the longitudinal relaxation rate (1/T1). It can also contain contributions from other relaxation mechanisms besides the dipolar 1-bond relaxation. A significant contribution to k1 for 15N amide nuclei is the relaxation stemming from chemical shift anisotropy. The figure shows the behavior of the amide {1H}-15N-NOE as a function of the correlation time when this mechanism is taken into account. The effect clearly distinguishes between motions faster orslower than ~ 0.5 ns. -22- 10-1110-1010-910-810-7-4-3-2-101
c [s]{1H}-15N NOE 800 400600 Fig: values of the steady state heteronuclear {1H}-15N NOE as a function of correlation time for magnetic field strengths corresponding to 400, 600, and 800 MHz proton Larmorfrequency. The effect of 160 ppm chemical shift anisotropy (
 -
) on k1 has been taken into account. -23- 1.11 Dynamical information from relaxation experiments The relaxing way to study motion (adapted from G. Wagner, 1995, nature structural biology, 2, 255-257) 15N NMR relaxation studies have emerged as a powerful approach for the determination of the motional properties of molecules in solution. In an 15N relaxation experiment, one creates non- equilibrium spin order and records how this relaxes back to equilibrium. At equilibrium, themagnetization of the 15N spins is aligned along the external field, and this alignment can be inverted by radio frequency pulses. The spins will relax back to equilibrium along the direction of the magnetic field with a longitudinal relaxation time, T 1, or the longitudinal relaxation rate RN(Nz). The magnetization can also be oriented perpendicular to the external magnetic field. The relaxation time of this spin order back to equilibrium is called thetransverse relaxation time, T 2, and the rate can be denoted as R N(Nx). A third relaxation parameter is the so-called heteronuclear nuclear Overhauser effect (NOE). This is measured by saturating the proton ( 1H) signal and observing changes in the 15N signal. The rate at which this occurs is the heteronuclear cross relaxation rate, RHN(HZ<->NZ); for long proton saturation, it reaches the steady state NOE(HZ<->NZ) value. These are essentially the three experiments Nicholson et al1 perform to learn about mobility in HIV protease. Other spin orders could be created and the relaxation analyzed. The relaxation of the spin orders is due to rotational diffusive motions of the nitrogen atom and the orientation of its chemical bonds (N-H bond) relative to the external field. The molecular motions cause the 15N nucleus to experience energy fluctuations inducing transitions between the Zeeman energy levels - and result in relaxation. Thus, the relaxation of spin orders is related to rotational motions of theamide group; this includes the overall, and internal motions of the HIV protease. The effect ofthe motions on relaxation rates is described with a spectral density function J(
). This function describes the relative distribution of the frequencies of the rotational diffusive motions. For the three parameters measured by Nicholson and colleagues, these relations are: RN(Nz)=d2J(N
H)+3J(N)+6J(N+H) [] +c2J(N) RN(Nx)=0.5d24J(0)+J(N
H)+3J(N)+6J(H)+6J(N+H) [] + c2 64J(0)+3J(N) [] +Rex NOE (HzNz)=1+Hd2 NRN(Nz)6J(N+H)
J(N
H) [](1.39) In these relations, d and c represent the strengths of the dipolar interaction between the proton and nitrogen, and the chemical shift anisotropy: they also represent the main relaxationmechanisms for 15N in proteins.
N and
H are the resonance frequencies of 15N and 1H. In a typical spectrometer, these are 50 and 500 MHz, respectively. The equations indicate that 15N 1 L. Nicholson, T. Yamazaki, D.A. Torchia, S. Grzesiek, A. Bax, S.J. Stahl, J.D. Kaufman, P.T. Wi ngfield, P.Y.S. Lam, P.K. Jadhav, C.N. Hodge, P.J. Domaille, and C.-H. Chang: Flexibility and function in HIV-1 protease. Nature Structural Biology 2, 274-279, 1995. -24- relaxation is primarily sensitive to very fast motions in the order of the frequencies of protons and nitrogens as well as the sum and difference thereof. However, the transverse relaxationrates can also be influenced by slow conformational exchange, represented by R ex. Typically, motions on the order of ms to µs would have an effect on the transverse relaxation rates. To measure dynamics from such relaxation experiments one has to perform further ex-periments, or alternatively one can make assumptions about an analytical form of the spectraldensity function that contains a sufficiently small number of parameters can be fitted to theexperiments. The most commonly used analytical form for a spectral density function hasbeen proposed by Lipari and Szabo which is a sum of two Lorentzian shapes: J()=S2m 1+(m)2+1
S2() 1+()2 with 
1=m
1+e
1(1.40) Accepting that this is a reasonable assumption, the relaxation rates depend on only one global and two local parameters, the overall correlation time
m which is the same for all residues, a correlation time for internal motions,
e and an order parameter S 2 which is a measure of the amplitude of the internal motions, With the measurement of three relaxation parameters, thelongitudinal relaxation rate, R N(Nz), the transverse relaxation rate, RN(Nx), and the heteronuclear NOE(HZ<->NZ), the three parameters,
m,
e, and S2, can readily be fitted. The following figure shows the measured T1, T2, and NOE values for the HIV-1 protease in complex with two different inhibitors: -25- A fit of the parameters S2 and Rex is shown in the next figure (Fig. 2). Clearly visible is the enhanced mobility around residues 38-42. These residues are at the hinges of the two enzymeflaps that have to open in order to allow the binding of an inhibitor or substrate to the proteaseactive site. Fig.: The mobility of HIV-1 protease as determined by NMR relaxation experiments. Themobility of the two enzyme flaps makes it possible for the substrate to enter into the active siteof the protease. All three figures from Nicholson et al. (1995) Flexibility and function in HIV- 1 protease. Nature Structural Biology 2, 274-279. -26- 2. A quick derivation of the master equation according to M. Goldman (1993) with special thanks to Bloembergen, Purcell, Pound (1948), Wangsness and Bloch (1953), Bloch (1956), Solomon (1955), Redfield (1957), Abragam (1961), Ernst, Bodenhausen, Wokaun (1987). 2.1 The total Hamiltonian H=H0+H1(t) (2.1) H0: main Hamiltonian, discrete levels, time-independent H1: random, spin-lattice coupling 2.2 Liouville-von Neumann equation d dt=
iH,[](2.2) 2.3 Transform operators to interaction picture i.e. into the rotating frame ˜ Q (t)=exp(iH0t)Qexp(
iH0t) d˜  dt=
i(˜ H
H0),˜  []  d˜  dt=
i˜ H 1(t),˜  [](2.3) H0 disappears! -27- 2.4 Integrate once ˜  (t)=˜  (0)i˜ H 1( t ),˜  ( t ) []0t
d t (2.4) put into original Liouville-von Neumann equation: d˜  (t) dt=i˜ H 1(t),˜  (0) [] ˜ H 1(t),˜ H 1( t ),˜  ( t ) [] []0t
d t (2.5) 2.5 Take ensemble average (––––) of all terms d˜  (t) dt=˜ H 1(t),˜ H 1( t ),˜  ( t ) [] []0t
d t (2.6) (˜ H 1(t) from random inner forces
˜ H 1(t)=0). This equation is exact. It shows nicely the second order nature of relaxation theory, i.e. time derivatives are proportional to the square of the disturbing Hamiltonian. Remark: Taking the ensemble average “seems strange, since the very concept of density matrix is intimately linked to the ensemble average. That a further averaging is necessary stems from the fact that the operator ˜ H 1(t) is a random function of time; in different parts of the system otherwise identical, simulated by different members of a Gibbs ensemble, it may have a different history, resulting in a different density matrix ˜
.” (M. Goldman, 1993). 2.6 Introduce return to equilibrium ad hoc ˜  (t)˜  (t)
˜  eq (2.7) This forces ˜
(t)to end up at ˜
eq. This hat trick is necessary because the lattice is not treated quantum-mechanically. d˜  (t) dt=˜ H 1(t),˜ H 1( t ),˜  ( t )˜  eq [] []0t
d t (2.8) -28- 2.7 Expand H1 in spin and time dependent part H1(t)=VF 
(t)=V+F* 
(t)(2.9) V
spin part F
(t) random function of time ˜ V (t)=exp(iH0t)Vexp(
iH0t)=exp(it)V (2.10) such a decomposition is always possible => d˜  (t) dt=˜ V (t),˜ V +( t ),˜  ( t )˜  eq [][] F(t)F*( t ) 0td t ,
(2.11) 2.8 Use assumptions on random motions 1. F(t)F*( t )
0, when t -  t >>c 2. ˜
( t ) varies slowly on time scale of
c This has the following consequences: 1. ˜
( t ) can be replaced by ˜
(t) in the integral 2. all members of the ensemble experience many random events , when |t-t'| >>
c => all ˜
(t) are nearly the same and they are nearly equal to ˜
(t) => ––– can be omitted on ˜
(t) => d˜  (t) dt=˜ V (t),˜ V +( t ),˜  (t)˜  eq [][] F(t)F*( t ) 0td t ,
(2.12) 2.9 Use one more assumptions on random motions F(t)F*( t )=G(t
 t ) (2.13) -29- d˜  (t) dt= V,V+,˜  (t)˜  eq [][]exp(iti t )G0t(t- t )d t ,
K 0td t =exp(i()t) exp( i(t t ))G0t(t- t )d t (2.14) The integral can be extended to infinity since G(t)
0, for t>>c exp(i(t t ))G0t(t- t )d t =exp(i)G0t()d  exp(i)G0
()d=J() (2.15) where the definition J()G() 0
exp(i)d(2.16) has been used. => d˜  (t) dt= V,V+,˜  (t)˜  eq [][] exp( i()t)J() ,
(2.17) This is the principal result of Redfield's theory! -30- 2.10 Secular approximation The secular approximation retains only terms with 
=  in the Redfield equation, since other terms are assumed to oscillate rapidly on the time scale of the changes of the density matrix. This approximation leads to a system of coupled linear differential equations, which is easily solved by standard methods of linear algebra. The relaxation supermatrix ˆ ˆ R adopts the Redfield kite structure. d˜  (t) dt V,V+,˜  (t)˜  eq [][] J() , =
ˆ ˆ R (˜  (t)˜  eq) (2.18) From Ernst, Bodenhausen, Wokaun 1987, pg. 54 -31- 3. References Abragam, A. (1961) The Principles of Nuclear Magnetism (Clarendon Press, Oxford) Bloch, F. (1956) Phys. Rev. 102, 1956. Bloembergen, N.; Purcell, E.M.; Pound, R.V. (1948) Phys. Rev. 73, 679. Cavanagh, W. J. Fairbrother, A.G. Palmer, N.J. Skelton, Protein NMR Spectroscopy , Academic Press, San Diego, 1996. Ernst, G. Bodenhausen, A. Wokaun, “Principles of Nuclear Magnetic Resonance in One and Two Dimensions”, Clarendon Press, Oxford, 1987. M. Goldman, Quantum Description of High-Resolution NMR in Liquids , Oxford University Press, Oxford, 1988. M. Goldman, Introduction to Some Basic Aspects of NMR , appeared in Rendiconti della Scuola Internazionale di Fisica Enrico Fermi (S.I.F.) - corso CXXIII (Ravenna, 13.-21.10. 1992), pg. 1-66. Title of course: Nuclear Magnetic Double Resonance. English seriestitle: Proc. Intl. School of Physics Enrico Fermi, course 123. Maraviglia, B. (Ed.), Amsterdam, 1993. My all time favourite!!! Redfield, A.G. (1957) IBM J. Res. Develop., 1, 19; Theory of relaxation processes in Advances in Magnetic Resonance , edited by J.S. Waugh, Vol. 1 (Academic Press, New York, N.Y., 1965), p. 1. F. Reif, Fundamentals of Statistical and Thermal Physics, McGraw-Hill Kogakusha, Tokyo, 1965. Solomon, I. (1955) Phys. Rev. 99, 559. Wangsness, R.K.; Bloch, F. (1953) Phys. Rev. 89, 278. -32- 4. Appendix 4.1 Efficiency of scalar magnetization transfer in the presence of relaxation In the presence of relaxation the efficiency of the scalar magnetization transfer depends strongly on the ratio of the transfer speed J and the decay rate 1/T2. If the magnetization has decayed too fast at the end of the transfer internal, no efficient transfer is possible. 1H 15N90˚x 90˚x90˚y180 ˚y 180 ˚y Hz-Hy2
JHzNz*2 -Hycos(2
J)+ 2HxNzsin(2
J)-Hycos(2
J)+ 2HzNysin(2
J) exp(-2/T2(H)) Relaxation: 0 0.25 0.500.10.20.30.40.50.60.70.80.91 *Jtransfer"INEPT" sin(2
J) exp(-2/T2(H)) opt < 0.25*J sin(2
J)*exp(-2/T2(H))transfer term opt=1 2
Jatan
JT2;JT2<<1 no efficient transfer Fig.: efficiency of scalar magnetization transfer by INEPT -33- 4.2 The size of the scalar couplings The size of the coupling constants between protons that are connected via three bonds varies in the range from ~ 3 to 10 Hz. This means that it takes between 0.1 to 0.3 s to transfer magnetization completely via the proton-proton three-bond J-couplings. The size of the heteronuclear one-bond coupling constants is usually much larger, e.g. ~ 140 Hz for 1H-13C and ~ 93 Hz for 1H-15N. This means that the magnetization transfer takes ~ 7 ms for 1H -> 13C and ~ 11 ms for 1H -> 15N N HC O H C CHC
6 8 N H O HC H
C C N HC
14 852 42 136 H Hslow tumbling limit: 1/T2 ~ J(0) ~ c ~ MWT Typical T2-values (ms) for a ( c ~ 15 ns) protein (MWT ~ 30 kDa) 140 35 119255 15140-7 11 3 - 10 J-values (Hz) of a proteinJ T2 << 0.5 => no efficient transfer 1JNC
T2(C
) = 0.15 !! for 30 kDaH Fig.: T2-values for a 30 kDa protein and J-couplings of a protein. -34- 4.3 Some product operator gymnastics The following commutator rules hold for any spin I: Ix,Iy[] =IxIy
IyIx=iIz [and cylcic permutations thereof (x->y->z->x)] (4.3.1) For a spin-1/2 particle the angular momentum operators can be represented by 2 x 2 matrices, which are called Pauli matrices: Ix=1 201 10     , Iy=1 20
i i0     ,Iz=1 210 0
1     (4.3.2) In addition to the canonical commutator rules, the following relations hold for spin-1/2: Ix
Ix=Ix2=Iy2=Iz2=1 410 01     (4.3.3) IxIy=
IyIx=1 2iIz [and cylcic permutations thereof (x->y->z->x)] (4.3.4) 4.4 Irreducible representation of the dipolar interaction The dipolar Hamiltonian (in angular momentum units) Hd=12hµ0 4
r3 I1I2-3(I1r)(I 2r) r2     (4.4.1) can be written as Hd=d4 5Y2m(,) m=2,2
Vm,i+ i
=d D0m2(,,) m=2,2
Vm,i+ i
with d=12hµ0 4r3(4.4.2) and Y2m(
,) is the second order spherical harmonic with angular arguments that correspond to the polar angles of the internuclear distance vector r, and D0m2(,,
) is a Wigner rotation matrix. The spin part of the Hamiltonian and its characteristic frequencies are given by -35- V0,1=2IzSz;0,1=0 V0,2=1 2I+S;0, 2=IS V0,3=1 2IS+;0,3=SI V1, 1=6 2I+Sz;1, 1=I V1, 2=6 2IzS+;1, 2=S V2,1=6 2I+S+;2, 1=I+S for m
0:Vm,i=(1)mVm,i+;m,i=m,i(4.4.3) 4.5 Example: heteronuclear T2 from Redfield theory We want to use the Redfield equation and calculate an expression for the T2 for a heteronuclear two spin-system IS due to dipolar interaction under the assumption of spherical diffusion. We further assume the slow motion limit ( c >> 1), such that all spectral densities besides J(0) can be neglected. An inspection of the operators for the dipolar interaction shows that only the term V0,1 = -2 IzSz will give rise to a zero frequency spectral density. Therefore, the summation in the Redfield sum over the indices
and  reduces to the case where
=  = (0,1) d˜  (t) dt V,V+,˜  (t)˜  eq [][] J() , ==0
= =V(0,1 ),V(0, 1)+,˜  (t)˜  eq [][] J(0, 1)( 0, 1) ((0, 1))= =2IzSz,2IzSz,˜  (t)˜  eq [] [] J(0, 1)( 0,1 ) (0)(4.5.1) Note that the indices
and  are used to label different parts of the random Hamiltonian corresponding to distinct rotational frequencies in the rotating frame. If we look at the expansion of the dipolar Hamiltonian, we see that every pair of indices (m,i) corresponds to adistinct frequency  m,i and vice versa. Therefore we can just make the identification
= (m
, i
) and  = (m, i). We need to find an expression for J(0 ,1)(0, 1) (0). From the expansion of the dipolar interaction, we see that the rotational diffusion of the molecule leads to a time-dependent variation of the spherical harmonical functions Y2m(
,). The random functions F
(t) are therefore given by F(t)=d4
5Y2m((t),(t)) (4.5.2) -36- For spherical diffusion, the correlation function F(t)F*( t )=G(t
 t )is given by F(t)F*( t )=d24
5Y2m( (t), (t))Y2m*( ( t ), ( t ))= =d2 5mmexp(t t /c)(4.5.3) Such that J
()is given by J( )=F()F*(0)exp(i )d 0
=d2 5mmexp(/c)exp( i )d 0
 =d2 5mm1 1/ci =d2 5mmc 1+ 2c2+i c 1+ 2c2     (4.5.4) And we realize that J(0,1 ) (0,1 ) (0)=d2 5
c (4.5.5) Therefore d˜  (t) dt
4IzSz,IzSz,˜  (t)˜  eq [][]d2 5c (4.5.6) The equilibrium magnetization ˜
eq contains only longitudinal or unity operators which commute with IzSz and can be omitted in this equation. d˜  (t) dt
4IzSz,IzSz,˜  (t) [][]d2 5c (4.5.7) From the commutator algebra (Eqs. 4.3.1, 4.3.3), it is easy to see that a density matrix which proportional to Ix at time 0, will always be proportional to Ix (I and S are assumed to be spin 1/2). ˜
(0)=a(0)Ix˜
(t)=a(t)Ix (4.5.8) Putting this “Ansatz” into Eq. 4.5.7 yields Ixda(t) dt
4IzSz,IzSz,a(t)Ix [] []d2 5c= =4IzSz,IzSz,Ix [] [] a(t)d2 5c= =4IzSz,iIySz [] a(t)d2 5c=Ixa(t)d2 5c(4.5.9) -37- or da(t) dtd2 5ca(t)=1 T2a(t) a(t)a(0)exp(t/T2) 1 T2d2 5c=12hµ0 4
r3   2c 5(4.5.10) This expression for T2 corresponds to textbook expressions (Eq. 1.14) in the slow tumbling limit. The complete expression for spectral density terms at non-zero frequencies is derived in an analogous manner.