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Nuclear magneticresonance

Trond Saue

Trond Saue (LCPQ, Toulouse) Nuclear magnetic resonance Virginia Tech 2015 1 / 51

Nuclear spin

The atomic nucleus is acomposite particle built from Zprotons and N neutrons, bothspin-1/2 fermions, which are inturn built from quarks.

Nuclei with even (odd) massnumber A = Z + N have integer(half-integer) spin I .

Nuclei with Z and N both evenhave I = 0, e.g. 12C, 16O, 56Fe.

Nuclei with Z and N both oddhave I > 0, e.g. 2H(1),10B(3),14N(1), 40K(4).


Gyromagnetic ratio: classical model

The gyromagnetic ratio is theratio between magnetic dipolemoment and angular moment.

m[1] = γL

In a simple classical model weobtain

L = r × p = rmvn

m[1] =1

2r × qv =

1

2rqvn

such that

γ =m[1]

L=

q

2m


Gyromagnetic ratio: electrons and nuclei

Electron: γe = − e2mge ; ge = 2.0023193043617(15)

Nuclei: γ = e2mp

g ; g − nucleus g-factor; e2mp

= 4.789 s−1T−1

Spin Abundance γ(107rad s−1T−1) Spin Abundance γ(107rad s−1T−1)

1H 1/2 ∼100% 26.7522 23Na 3/2 ∼100% 7.0808493

2H 1 0.015% 4.10662791 27Al 5/2 ∼100% 6.9762715

3H 1/2 0 28.5349779 29Si 1/2 4.7% -5.3190

10B 3 19.9% 2.8746786 31P 1/2 ∼100% 10.8394

11B 3/2 80.1% 8.5847044 35Cl 3/2 75.77% 2.624198

12C 0 98.9% 37Cl 3/2 24.23% 2.184368

13C 1/2 1.1% 6.728284 63Cu 3/2 69.17% 7.1117890

14N 1 99.6% 1.9337792 65Cu 3/2 30.83% 7.60435

15N 1/2 0.37% -2.71261804 107Ag 1/2 51.84% -1.0889181

16O 0 ∼100% 109Ag 1/2 48.16% -1.2518634

17O 5/2 0.04% -3.62808 129Xe 1/2 24.4% -7.452103

19F 1/2 ∼100% 25.18148 207Pb 1/2 22.1% 5.58046


The NMR experiment


Larmor presession

The NMR experiment probes transitions between levels split bynuclear spin Zeeman interaction,

We place a nucleus of magnetic moment m = γIin a magnetic field B0 = B0n

The interaction spin Hamiltonian is given by

H0 = −m · B0 = −γI · B0

The time-evolution of the nuclear spin function is

ψ(t) = e−iHtψ(0) = e+iγB0t(n·I)ψ (0) = R I (−γB0t,n)ψ(0)

...corresponding to a rotation about the magnetic field vector n withangular frequency ω0 = −γB0.


Larmor precession

The time-evolution of the nuclear spin I can be followed using theEhrenfest theorem

d 〈Ii 〉dt

= −i 〈[Ii ,H0]〉 = +iγ 〈[Ii , Ij ]〉B0;j = −γεijk 〈Ik〉B0;j

which can be re-arranged to

d 〈I〉dt

= 〈m〉 × B0

...showing that the nuclear spin precesses aboutthe magnetic field vector with the Larmorfrequency ω0 = −γB0.

Most nuclei have positive γ such that theLarmor precession is clockwise.

In the following we will assume that the magnetic field is along the z-axis B0 = B0ez .


A poetic moment

I remember, in the winter of our firstexperiments,...,looking on snow with new eyes.

There the snow lay around my doorstep — greatheaps of protons quietly precessing in the earth’smagnetic field.

Edward Mills Purcell

(1912-1997)


Rotating-frame Hamiltonian

We have seen that the time-evolution of the nuclear spin function in themagnetic field is

ψ(t) = R Iz (ω0t)ψ(0); ω0 = −γB0

We now transform to a frame rotating with angular frequency ωref aroundthe magnetic field.In this frame the nuclear spin function is

ψ(t) = R Iz (−Φ)ψ(t); Φ(t) = ωref t + φref

The time-dependent Schrodinger equation can now be expressed as

H0R Iz (Φ) ψ(t) = i

d

dtR Iz (Φ) ψ(t) = R I

z (Φ)

(ωref Iz + i

d

dt

)ψ(t)

Pre-multiplication with R I (−Φ) gives

H0ψ(t) = id

dtψ(t)

...where we have introduced the rotating-frame spin Hamiltonian

H0 = R Iz (−Φ) H0R I

z (Φ)− ωref Iz = (ω0 − ωref ) Iz


Radiofrequency probe

The system is now probed by a weak radiofrequency pulseperpendicular to the magnetic field (along the x-axis)

B1 = B1ex cos (Φp) ; Φp(t) = ωref t + φp

For computational convenience it will be split into two parts ofopposite circular polarization

B1 = B+1 + B−1

with

B±1 =1

2√

2B1 (e± exp [iΦp] + e∓ exp [−iΦp]) ; e± =

1√2

(ex ± iey )

The component B−1 describes circular polarization following the Larmorprecession, whereas B+

1 describes circular polarization in the opposite senseand can in most situations be discarded, as we will do in the following.The Hamiltonian describing the radio pulse is therefore given by

H−1 = −γI · B−1 = −1

4γB1

(I−e+iΦp + I+e−iΦp

)Trond Saue (LCPQ, Toulouse) Nuclear magnetic resonance Virginia Tech 2015 8 / 51

Radiofrequency pulse in the rotating frame

Using the Baker-Campbell-Hausdorff expansion

exp (A) B exp (−A) = A + [B,A] +1

2[B, [B,A]] + . . .

and the commutation relation

[Iz , I±] = ±I± : I± = Ix ± iIy

we find that

I−e+iΦp = R Iz (Φp) I−R I

z (−Φp) ; I+e−iΦp = R Iz (Φp) I+R I

z (−Φp)

such that the total spin Hamiltonian in the rotating frame becomes

H = (ω0 − ωref ) Iz + R Iz (−Φ) H1R I

z (Φ)

= (ω0 − ωref ) Iz −1

4γB1R I

z (−Φ + Φp) (I− + I+) R Iz (−Φp + Φ)

= (ω0 − ωref ) Iz −1

2γB1R I

z (−φref + φp) IxR Iz (−φp + φref )


Radiofrequency pulse in the rotating frame

As a final step we employ our freedom in choosing the phase constantφref to bring the total spin Hamiltonian into identical form for bothpositive and negative gyromagnetic ratio γ:

H = (ω0 − ωref ) Iz + ωnutRIz (φp) IxR

Iz (−φp) ; φref =

π for γ > 0

0 for γ < 0

where we have introduced the nutation frequency ωnut = 12 |γ|B1.

The presence of

Ix =1

2(I+ + I−)

in the Hamiltonian adds a nutation to theprecession generated by Iz .


Spin-1/2 functions in the rotating frame

Let us introduce the notation

|+z〉 = |α〉 and |−z〉 = |β〉

for the eigenfunctions of the sz operator in the rotating frame.

Starting from |+z〉 we can now use the spin rotation operator

R12 (φ,n) = cos

1

2φ− i sin

1

2φ

...to generate

|+x〉 = R12y

(π2

)|+z〉 = 1√

2

[1 −11 1

] [10

]= 1√

2

[11

]|−x〉 = R

12y

(−π

2

)|+z〉 = 1√

2

[1 1−1 1

] [10

]= 1√

2

[1−1

]|+y〉 = R

12x

(−π

2

)|+z〉 = 1√

2

[1 ii 1

] [10

]= 1√

2

[1i

]|−y〉 = R

12x

(π2

)|+z〉 = 1√

2

[1 −i−i 1

] [10

]= 1√

2

[1−i

]Trond Saue (LCPQ, Toulouse) Nuclear magnetic resonance Virginia Tech 2015 11 / 51

Spin-1/2 functions in the rotating frame

We may check our results as follows:

sx |+x〉 = 12σx |+x〉 = 1

2

[0 11 0

]1√2

[11

]= + 1

2|+x〉

sx |−x〉 = 12σx |−x〉 = 1

2

[0 11 0

]1√2

[1−1

]= − 1

2|+x〉

sy |+y〉 = 12σy |+y〉 = 1

2

[0 −ii 0

]1√2

[1i

]= + 1

2|+y〉

sy |−y〉 = 12σy |−y〉 = 1

2

[0 −ii 0

]1√2

[1−i

]= − 1

2|−y〉


Nuclear spin-1/2 Zeeman splitting

The spin Hamiltonian describing theinteraction of a nuclear spin I = 1/2in a magnetic field along the z-axis is

H0 = −1

2γσzB0 =

[− 1

2γB0 00 + 1

2γB0

]with the ordering of eigenfunctionsdepending on the sign of γ.


Radiofrequency pulse at resonance

H = (ω0 − ωref ) Iz + ωnutRIz (φp) IxR

Iz (−φp) ; φref =

π for γ > 0

0 for γ < 0

Consider now the application of a radiofrequency pulse of phaseφp = 0 at resonance, that is ωref = ω0.The spin Hamiltonian in the rotating frame is accordingly

H = ωnut Ix =1

2ωnutσx ; ωnut =

1

2|γ|B1

...and the time development of the spin function in the rotating frameis

ψ(t) = e−i Ht ψ(0) = e−i 12ωnutσx t ψ(0) = R

12x (ωnutt) ψ(0)

Let ψ(0) = |+z〉. We then see that

ψ(t) = R12x (ωnutt) |+z〉 = cos

1

2ωnutt |+z〉 − i sin

1

2ωnutt |−z〉

...corresponding to Rabi oscillations of frequency 12ωnut = 1

4 B1 |γ|.


Radiofrequency pulse at resonance

In general, however, the radiofrequency pulse has a specific durationτp which allows the selection of final state

ψ(τp) = R12x (βp) |+z〉 = cos

1

2βp |+z〉 − i sin

1

2βp |−z〉

where we have introduced the flip angle βp = ωnutτp

Examples are:

I a (π/2)x - pulse:

ψ(τp) =1√2|+z〉 − i√

2|−z〉 = |−y〉

I a (π)x -pulse:

ψ(τp) = −i |−z〉

...and so we see that the pulses have a straightforward geometricalinterpretation.


The NMR experiment

1 The sample is placed in a strong magnetic field which induces a longitudinalmagnetization

Mnucz (t) = Mnuc

eq (1− exp {− (t − ton) /T1})

where T1 is the longitudinal (or spin-lattice) relaxation time constant.

2 Transverse nuclear magnetization is induced by a weak radiofrequency pulseat resonance ω0

Mnucx (t) = Mnuc

eq sin (ω0t) exp {−t/T2}Mnuc

y (t) = −Mnuceq cos (ω0t) exp {−t/T2}

where T2 is the transverse relaxation time constant.

3 The NMR signal, that is the oscillating electric current [free-induction decay(FID)] induced by the precessing nuclear transverse magnetization isdetected.


Continuous-wave NMR

1 frequency-sweep method: the magnetic field B0 is held constantand the RF signal swept to determine resonances

2 field-sweep method: the RF signal is held constant, and themagnetic field B0 to bring the energy splittings into resonance


Fourier-transform NMR

The magnetic field B0 is heldfixed and a short (1-10µs)square RF pulse is applied toexcite all nuclei in their localenvironments.

The free-induction decay isdetected andFourier-transformed to give thefinal spectrum.


NMR parameters


Nuclear shielding

The principal chemical interest of NMR resides in the fact that thelocal magnetic field experienced by a nuclear spin is sensitive to itschemical environment. The nuclear spin Zeeman operator isaccordingly modified to

H0 = −mK · Bloc0

The local magnetic field can to a very good approximation beassumed linear in the applied external field

Bloc0 =

(I3 − σK

)B0; σK =

σKxx σK

xy σKxz

σKyx σK

yy σKyz

σKzx σK

zy σKzz

where we have introduced the shielding tensor σK of nucleus K .

Experiments in isotropic liquids give the isotropic shielding:

σiso =1

3

(σKxx + σK

yy + σKzz

)due to rotational averaging.

In general, the more electronegative the nucleus is, the smaller theshielding.Other factors, such as ring currents and bond strain, also play a role.


Chemical shift: 1H

The chemical shift is defined as the relative resonance frequency shiftwith respect to a reference compound

δ =ω − ωref

ωref=σref − σ1− σref

≈ σref − σ

and is generally reported in parts-per-million (ppm).

1H chemical shift ranges:(R = alkyl or H, Ar = aryl)Gyromagnetic ratio:26.7522·107rad s−1T−1

Reference compound:Tetramethylsilane(TMS) < 1% in CDCl3 = 0 ppmChemical shift range:13 ppm, from -1 to 12


Chemical shift: 13C

Gyromagnetic ratio:6.728284·107rad s−1T−1

Reference compound:Tetramethylsilane(TMS) < 1% in CDCl3 = 0 ppmChemical shift range:200 ppm, from 0 to 200


Chemical shift: 129Xe

Gyromagnetic ratio:-7.452103·107rad s−1T−1

Reference compound:Xenon oxytetrafluorideXeOF4 (neat)Chemical shift range:5300 ppm, from -5200 to 100


Chemical shift: 195Pt

Gyromagnetic ratio:5.8385·107rad s−1T−1

Reference compound:1.2 M Na2PtCl6 in D2OChemical shift range:6700 ppm, from -6500 to 200


Direct dipole-dipole coupling

The vector potential of the magnetic dipole moment mL of nucleus Lat position RL is conveniently given by

AL(RK ) =1

c2

mK × RKL

R3KL

The corresponding magnetic field is

BL (RK ) =1

c2

{3nKL (mL · nKL)−mL

R3KL

+8π

3mLδ (RKL)

}where we have used

∇i∇j1

r= 3ri rj r

−5 − δij r−3 − 4π

3δ(r).

We can, however, ignore the second contact term since RKL > 0.The direct dipole-dipole coupling is accordingly given by

HDD = −1

2

∑K 6=L

{3 (mK · nKL) (mL · nKL)−mK ·mL

c2R3KL

}In an isotropic liquid the direct dipole-dipole interaction averages tozero.


Indirect dipole-dipole coupling

Spin-spin couplings mediated by electrons do survive in isotropicliquids. The corresponding general spin Hamiltonian is

HJ =1

2

∑K 6=L

2πIK · JKL · IL

...although in isotropic liquids only the isotropic part is accessed

J isoKL =

1

3

(σKxx + σK

yy + σKzz

)For the study of trends in the indirect spin-spin coupling it is generallybetter to use the reduced spin-spin coupling defined via

2πJKL = γKγLKKL


NMR spin Hamiltonian for isotropic liquids

Combining terms we obtain(keeping the full tensors of the non-zero contributions):

HNMR = −∑K

mK ·(I3 − σK

)B0 +

1

2

∑K 6=L

mK · KKL ·mL

The electrons contribute through the shielding tensor σK and thereduced indirect spin-spin coupling constant KKL.

These NMR parameters can be calculated by mapping thephenomenological NMR Hamiltonian onto the expression for theelectronic energy in the presence of nuclear spins and uniformmagnetic moments, notably

σKµν =

d2E

dmK ;µdB0;ν

∣∣∣∣B0={mL}=0

+ δµν

KKL:µν =d2E

dmK ;µdmL;µ

∣∣∣∣B0={mL}=0


Perturbed Hamiltonian

Introducing the vector potentials associated with nuclear magneticmoments and an external magnetic field

AB(ri ) =1

2B× riO AK (ri ) =

1

c2

mK × riKr 3iK

; riK = ri − RK

into the one-electron Hamiltonian

h =p2

2m+

e

mA · p +

e2

2mA2 +

e

2mσ · B− eφ

gives first- and second-order Hamiltonians

h1(ri ) = B ·(

horb

B + hspin

B

)+∑K

mK ·(

hpso

K + hsd

K + hfc

K

)h2(ri ) =

1

2B · hdia

BB · B +∑K

B · hdiaBK ·mK +

1

2

∑K 6=L

mK · hdsoKL ·mL


First-order Hamiltonians

Orbital Zeeman horbB =

(1

2m

)(riO × pi ) S

Spin Zeeman hspnB =

(1

2m

)σi T

Paramagnetic spin-orbit hpsoK =

(1

mc2r 3iK

)(riK × pi ) S

Spin-dipole hsdK = −

(1

2mc2

)σr 2

iK − 3 (σi · riK ) riKr 5iK

T

Fermi contact hfcK =

(4π3c2

)δ (riK )σi T


Second-order Hamiltonians

B B hdiaBB =

1

8m

[I3r 2

i − ri · rTi]

B MK hdiaBK =

1

2mc2

[I3 (ri · riK )− ri rTiK

r 3iK

]

MK ML hdsoKL =

(1

mc4

)[I3 (riK · riL)− riK rTiL

r 3iK r 3

iL

](diamagnetic spin-orbit)


Shielding tensorfor closed-shell molecule

σKµν =d2E

dmK ;µdB0;ν

∣∣∣∣B0={mL}=0

=⟨

0∣∣∣hdia

BK :µν

∣∣∣ 0⟩

+⟨⟨

horbB;µ; hpso

K ;ν

⟩⟩0

the diamagnetic contribution dominates in many cases, but this is lesssystematic than for magnetizabilities

the paramagnetic part vanishes for atoms

additional contributions appear in the open-shell case

basis sets needs to be flexible in outer core-inner valence region; inpractice flexible basis sets of triple zeta quality


Shielding: electron correlation effects can besizeable

HF CAS MP2 CCSD CCSD(T) Exp.

HF F 413.6 419.6 424.2 418.1 418.6 410±6

H 28.4 28.5 28.9 29.1 29.2 28.5±0.2

H2O O 328.1 335.3 346.1 336.9 337.9 344±17

H 30.7 30.2 30.7 30.9 30.9 30.05±0.02

NH3 N 262.3 269.6 276.5 269.7 270.7 264.5

H 31.7 31.0 31.4 31.6 31.6 31.2±1.0

CH4 C 194.8 200.4 201.0 198.7 198.9 198.7

H 31.7 31.2 31.4 31.5 31.6 30.61

F2 F -167.9 -136.6 170.0 -171.1 -186.5 -192.8

N2 N -112.4 -53.0 -41.6 -63.9 -58.1 -61.6±0.2

CO C -25.5 8.2 10.6 0.8 5.6 3.0±0.9

O -87.7 -38.9 -46.5 -56.0 -52.9 -42.3±17


DFT calculations of shieldings

Many commonly used functionals perform poorly for shieldingconstants

HF LDA BLYP B3LYP KT2 CCSD(T) Exp.

HF F 413.6 416.2 401.0 408.1 411.4 418.6 410±6

H2O O 328.1 334.8 318.2 325.0 329.5 337.9 344±17

NH3 N 262.3 266.3 254.6 259.2 264.6 270.7 264.5

CH4 C 194.8 193.1 184.2 188.1 195.1 198.9 198.7

F2 F -167.9 -284.2 -336.7 -208.3 -211.0 -186.5 -192.8

N2 N -112.4 -91.4 -89.8 -86.4 -59.7 -58.1 -61.6±0.2

CO C -25.5 -30.2 -19.3 -17,5 7.4 5.6 3.0±0.9

O -87.7 -87.5 -85.4 -78.1 -57.1 -52.9 -42.3±17

Functionals designed to work for shieldings (such as the KT2 andKT3 functionals of Keal and Tozer) do a good job


Reduced indirect spin-spin coupling tensorfor closed-shell molecule

KKL:µν =d2E

dmK ;µdmL;µ

∣∣∣∣B0={mL}=0

=⟨

0∣∣∣hdso

KL;µν

∣∣∣ 0⟩

+⟨⟨

hpsoK ;µ ; hpso

L;ν

⟩⟩0

+⟨⟨

hfcK :µ + hsd

K ;µ; hfcL:µ + hsd

L;µ

⟩⟩0

KDSOKL is in general very small and cancels the PSO contribution

KPSOKL is in general small, but may become substantial, for instance for

couplings between nuclei with multiple bonds

KFCKL dominates in most cases the isotropic coupling constants

KFC/SDKL vanishes for isotropic couplings, but often dominate

anisotropic couplings


Indirect spin-spin couplings in a large moleculeM. A. Watson, P. Sa lek, P. Macak, M. Jaszunski, and T. Helgaker,Chem. Eur. J. 10, 4627–4639 (2004)

JFCKL dominates at short distances, but decays exponentially

At large RKL distances positive JPSO and negative JDSO dominate,both decaying as R−2

KL ; the combined contribution as R−3KL .


JKL: computational requirements

Since JFCKL dominates, and it involves the triplet-excited states, only

methods t hat are stable toward triplet perturbations can be used:UHF, DFT, CCSD, MCSCF

Orbital relaxation may be required even for highly correlated wavefunctions

JKL in ethene (C2H4):

1JCC1JCH

2JCH2JHH

3JcisHH

3JtransHH

exp 68 156 -2 2 12 19RHF 1270 755 -572 -344 360 400CASSCF 76 156 -1 3 14 21


JKL: basis set requirements

The dominating contribution to JKL is JFCKL

Flexibility in the s orbitals required, as well as functions well aimed atdescribing the δ-function ⇒ Uncontracted s functions with additionalsteep functions needed

T. Helgaker, M. Jaszunski, K. Ruud and A. Gorska, Theor. Chem. Acc. 99 (1998) 175Trond Saue (LCPQ, Toulouse) Nuclear magnetic resonance Virginia Tech 2015 36 / 51

JKL: correlation effects

Due to the triplet operators, DFT often the only viable solution

RHF LDA BLYP B3LYP RAS Exp.

HF 1JHF 59.2 35.0 34.5 36.8 48.1 47.6

CO 1JCO 13.4 -65.4 -55.7 -44.9 -39.3 -38.3

N21JNN 175.0 32.9 -46.6 -32.9 -9.1 -19.3

H2O 1JOH 63.7 40.3 44.6 46.6 47.1 52.8

2JHH -1.9 -0.3 -0.9 -0.6 -0.6 -0.7

NH31JNH 61.4 41.0 49.6 52.6 50.2 50.8

2JHH -1.9 -0.4 -0.7 -0.8 -0.9 -0.9

C2H41JCC 1672.0 66.6 90.3 98.3 90.5 87.8

1JCH 249.7 42.5 55.3 54.7 50.2 50.0

2JCH -189.3 0.4 0.0 -0.4 -0.5 -0.4

1JHH -28.7 0.4 0.4 0.2 0.1 0.2

3JcisHH 30.0 0.8 1.1 1.1 1.0 0.9

3JtransHH 33.3 1.2 1.7 1.7 1.5 1.4

|∆| abs. 180.3 11.2 5.9 4.2 1.6

% 5709 72 48 20 14


Relativistic case: first-order Hamiltonians

hrelB

1

2(rO × ji )

hrelK

1

c2r 3iK

(riK × ji )

Relativistic current density operator: j = −ecα

σKµν =d2E

dmK ;µdB0;ν

∣∣∣∣B0={mL}=0

=⟨⟨

hrelK ;µ; hrel

B;ν

⟩⟩0

KKL:µν =d2E

dmK ;µdmL;µ

∣∣∣∣B0={mL}=0

=⟨⟨

hrelK ;µ; hrel

L;µ

⟩⟩0


Relativistic effects

The NMR parameters probes the electron density around nuclei and istherefore quite sensitive to relativistic effects.

The effect on absolute shieldings is quite more significant than for chemicalshifts.

Spin-orbit effects can be quitesignificant on shieldings and mayinduce triplet instabilities.


Indiret spin-coupling K(I,H) in iodoethane


NMR shielding σ(Hβ) in iodoethane


NMR shielding σ(Hβ) in iodoethaneSO-contribution only


Qualititative analogyM. Kaupp, O. L. Malkina, V. Malkin and P. Pyykko, Chem. Eur. J. 4(1998) 118

Fermi-contact mechanism of indirect spin-spin coupling

Interaction mechanics for spin-orbit shifts


K(H,I) vs. σSO(Hβ)in iodethane: 0◦


Inhomogeneous broadening

Magnetic resonance imaging (MRI) employs magnetic field gradients whichgives a positional dependency on the Larmor frequency.


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