Thus in the rotating frame the magnetic field becomes time-independentwhile the z-magnetic field component is reduced by the frequency of rotation

1

RoF

xRoFyRoF

zRoF

On-resonance, when , there is only an x-components to the field.In such a case the magnetization performs a precession around the x-directionwith a rotation frequency .

0)( 0

rotating frame

1

How to generate this B1 RF irradiation field in the laboratory frame:

x

y

xt cos2 1

)sin(cos)sin(cos 11 ytxtytxt

xy

z

Ignore because it is off-resonance!

Top view

B0

tItI cos)( 1

tBtB cos)( 11

18

u-of-o-nmr-facility.blogspot.com/2008/03/prob...

Doty Scientific

National High Magnetic Field Laboratory

Bird cage

Thus the magnetic field in the laboratory frame :

Becomes in the rotating frame:

yxz

ytxtz

RoF

LAB

sincos

)sin()cos(

11

110

ttttt yRoFxRoFLABx sin)(cos)()(

Although the signal detection in the laboratory frame is along the direction of the coil:

In NMR we measure the magnetization in the rotating frame: )(

)(

,

,

t

t

yRoF

xRoF

in:

out:

A sample with an overall )(tM the S/N voltage at the coil is:

0

2/1

21/ M

kTVQ

fNS s

f =noise of apparatus =filling factor =frequency =band width Q =quality factor Vs =sample volume 19

2d-i complex numbersA large part of our discussion will deal with precessions of vectors, and with expressionsthat need complex numbers. Here we introduce these numbers and give some necessary rules.

Let us suppose that there are “numbers” that in fact are composed of two numbers:

xxc ibax

2222

)()()).(()).((

)()(/

)()()).((.

)()()()(

yy

xyyx

yy

yxyx

yyyy

yyxx

yy

xxcc

xyyxyxyxyyxxcc

yxyxyyxxcc

bababa

ibabbaa

ibaibaibaiba

ibaibayx

babaibbaaibaibayx

bbiaaibaibayx

12 iThese complex numbers have their own mathematics based on :

A common practice is to present the values of the x- and y-components of the magnetization as if they are one complex number:

yxcxy i

)sin(cos titxycxy

A precessing magnetization around the z-direction has as xy-components the value:

This can be written asti

xycxy e

with the definition

sincos iei

a

b real axis

imaginary axis

1

Re

Im

Re

Im

y

x

20

2d. Necessary concepts for QM description

There is no way of explaining NMR without some Quantum Mechanics:

Here we give rules that can help us to “understand” what we are measuring in NMR.

Because the spin behave according to the theory of QM, the necessary “rules” we need

to proceed are: (We are all used to presenting spectroscopy by , but why? )

1. A stationary quantum system experiencing a time independent environment (Hamiltonian)

will preside in one of its discrete (constant energy) eigenstates.

A spin-1/2 system, like the proton or an electron, in a magnetic field

has only two eigenstates with energies according to the z-components of their magnetization

. The allowed components of the angular momentum are with .

The way to present these energies is by an energy level diagram and the eigen states by :

02/1 21 E

ozz BIBE 0

02/1 21 E

zB //0

. The actual length of the angular momentum is

but we can only determine its z-component.

m 2/1m

)1( III

)1( II

)1( II

2/1

2/1

In fact we do not know what the phase of its xy- component is.

2/1m

2/1m

mI ,

m is called the magnetic quantum number

When the spin is in one of these states it is stationary 21

2d-ii Wavefuntions, eigenstates and observables

5

2. A dynamic quantum system can find itself in a linear superposition of its eigenstates.

for a spin ½:

3. The result of a measurement of an observable O can be obtained by a calculation of the form

knowing the value of the elements with

In analogy, a spin-I can occupy eigenstates with energies

and

)12( I 0)( mmE

023

021

021

023

2/3;2/3

2/1;2/3

2/1;2/3

2/3;2/3

2/15spin=3/2:

IIIIm ,1,...,1,

21

21)( 2/12/1 )/(

2/1)/(

2/1

tEitEi ecect

mIect tEim

I

Im

m ;)( )/1(

In general:

mOmecc

mecOmec

tOttO

tEEimm

I

Imm

I

Im

tEim

I

Im

tEim

mm

mm

'

'

)()(()(

)(/*'

',

//*'

'

'

'

mOm'

mm

mm

OmOm

mmmm

'

'

'

''

22

k

nkkn

mllkknlk

D

DAD

1

1

,

;

the eigenvalues of the Hamiltonian are the energies of the system

nmmn aA

nmnnmn aADD 1

When the representation of an operator is diagonal, then the basis set is the eigenbasis set and the matrix elements the eigenvalues.

When the representation of an operator is not diagonal then after diagonalizationwe obtain the eigenfunctions and eigenvalues

Nnnn

nmmnNnn

c,1

,1;

define a basis set spanning Hilbert space that is orthonormal. This enables a matrix representation of any operator and a vector representation of any wave function.

nc

cc

1

2d-iv Matrix representation

nnn

n

aa

aa

A

1

111

23

d-v The Schrodinger Equation

)()()( ttHitdtd

a time dependent Hamiltonian doesn’t have eigenvalues/energiesThe solution of the Schrodinger equation defines an evolution operator

)0()0,()( tUt

If the Hamiltonian is time independent then

222

)/(

211)0,( tHHtietU tHi

If the Hamiltonian is time dependent then, with the Dyson operator T:

tHiNn

Nt

dtHin

t

o eTetU

)/(

,10

)/(lim)0,(

))2/1(( tnHH n :)(tHt tn

)()()( tctHitcdtd

)0()()( ctUtc

Vector-matrix representation

24

d-vi. The angular momentum operators, definitions

zyxpI p ,,

zyx IiII ],[ an cyclic permutation

sincos zyIi

yIi IIeIe xx

in the eigenfunction representation:

IIImmImmII z ,1,....,,

1,)1()1(2

1,)1()1(2

,

1,)1()1(211,)1()1(

21,

mImmIIimImmIIimII

mImmIImImmIImII

y

x

1,)1()1(,

1,)1()1(,

mImmIImII

mImmIImII

yx iIII

25

:1IIn the basis set of Iz:

02202

20

21;

02202

20

21;

10

1

iii

iIII yxz

The linear angular momentum operators do not span the whole 3x3 Hilbert space

0

021;

0110

21;

1001

21

ii

III yxz

:2/1I In the basis set of Iz:

Pauli matrices span the whole 2x2 Hilbert space

001000100

21

100020001

)(21

02202

20

21

02202

20

21

22222yxyxz

yzzyxzzx

IIIII

iii

iIIIIIIII

The missing operators are:

26

2/cos2/sin2/sin2/cos

;2/cos2/sin2/sin2/cos

00

],[;],[;],[

2/1002/1

;02/

2/0;

02/12/10

)2/(

)2/(

yx

z

IiIi

i

iIi

yxzxzyzyx

zyx

ei

ie

ee

e

iIIIiIIIiIII

Ii

iII

The spin Hamiltonian 2d-vii A nucleus in a magnetic field

IBH LAB 00. with the magnetic field in the z-direction of the lab. frame

and an RF irradiation:

and the rotating frame:xzLAB ItIH )cos(2 10

yxzRoF IIIH sincos 11 For spin=1/2 :

sincos yxIi

xIi IIeIe zz

|

|

cyclic permutations:

Pauli matrices

27

Remember that

1)/()/()0,( DDeetU titHi

1 DDHDHDwhen the Hamiltonian is time independent and

NN

ji

ij

HH

HH

H

11

is Hemitian *jiij HH

NN

ji

ij

UU

UU

tU

11

)( is unitary *1

1 1)()(

jiij UU

tUtU

28

2d-viii The spin-density operator

An arbitrary function can be expended in a basis set

This defines a set of coefficients with

Let us define an operator with matrix elements

nNnn tct

,1

)()(

)()(1)()( tctHtcdtdtcn

)()()( * tctct mnnm

)()()()()()(

)()()()()()()()(

**

****

tHtctcitctctHi

tctHtcitctctHitctcdtd

kmknk

mknkk

kmknk

mknkk

mn

)(),()( ttHitdtd

NN

ji

ij

t

11

)( Is Hermitian

29

2d-ix Ensemble Average and thermodynamicsConsider an ensemble average of the matrix elements of the spin-density

In the representation of the eigenbasis of the Hamiltonian the solution for the density matrix elements

)()(

)0()(*/*

/

tcetc

cetc

mti

m

nti

n

mm

nn

then

randommmnn inm

tinm eet )0()( )(/

with the random phase approximation

0)(0)( eqeq nnnm

)()()( * eqceqceq mnnm

the populations

11

22

N1,N1

NN

/)( jjiiij

coherences 30

Thus the equilibrium density operator can be defined by the populations satisfying the Boltzman statistics

kTeZ

/1)0(

....}11{11 / HkTZ

eZ

kTHeq

In NMR we solve the Liouville-von Neuman equation:The signal is proportional to an “observable”

OtTrtOt

tcOtctOttO

nmmnmn

mnmmnn

)()()(

)()()()()(

,

,

*

9999924.011014.4)300()1038.1(

106.71015.3)103()1005.1(

021123

602618340

0

kTeJKJKkT

kTJsJs

kT

IIItttt

IIIttttt

tttt

TrI

z

yxz

*121222112211

121222112211

122221

1211

)()()()(

)Im(Re2)()()()(21)(

1)()()()(

31

This requires a definition of the populations of the eigenstates.

At thermal equilibrium these populations follow the Boltzmann distribution:

2/1N

2/1NkTkTE

kTE

eee

NN /

/

/

2/1

2/1 0

2/1

2/1

)(21

2/12/1 NNIensemblez

and the populations become

)2

1(21)

21(

21 0

2/10

2/1 kTNN

kTNN

The thermal-equilibrium ensemble magnetization becomes

kTBN

kTNNNM

ensemblez 422)(

21 0

220

2/12/1

32

…………..

This require a definition of the populations of the eigenstates.

2/1N

2/1N

The thermal-equilibrium ensemble magnetization becomes

kTBN

kTNNNM

ensemblez 422)(

21 0

220

2/12/1

The magnitude of at 50-300K is very small. kT

0

9999924.01106.7

1014.4)300()1038.1(1015.3)103()1005.1(

060

21123

2618340

0

kTe

kT

JKJKkTJsJs

kT

and the populations become )2

1(21)

21(

21 0

2/10

2/1 kTNN

kTNN

Despite these very small values the ensemble nuclear magnetization can be detected.

A similar derivation can be presented for a spin higher than ½ in a magnetic field.

The Boltzman distribution at high temperature results in

)1(12

1 0

kTmN

INm

and the bulk magnetization becomes

I

Im

I

Imm

I

Immz

I

Im kTm

IN

kTmm

INmNIM 0

20

0 12)1(

12

)1(3

022

0 IIkTBNM

I

Im

SSSm )12()1(312

33

2d-x NMR is a journey along the matrix elements of the reduced spin density matrix

Coherences that are off diagonal element are detected. Populations that are the diagonal element are not detected

iiiE

ZkT

tt )()(

)(),()( ttHitdtd

)()0()()( 1 tUtUt ')'(

0)(dttHi

t

TetU

))()(()( tOtTrtO

))(()( ,, yxRRyx ItTrtI NMR signals

)sincos(2cossinsincos

2/cos2/sin2/sin2/cos

11

2/cos2/sin2/sin2/cos

1111

11

11

11

11

11

tItIttitit

ttitit

ttitit

yz

Example of pulse on I=1/2

34

011 HkTZeq

)(,)( 0 tHitdtd

tiHtiH eet 00 )0()(

))(( OtTrO

2d-xi (high temperature) NMR on S=1/2

zIH 0

zzyyxx IsIsIs )0()0()0()0( ttIitIi zz eet 00 )0()(

tstsI

tstsI

yxy

yxx

00

00

cos)0(sin)0(

sin)0(cos)0(

)0(zz sI

The Larmor precession for spin I=1/2

zeq IkTZ

01

35

)(,)()( ,10 tHIitdtd

RoFRoFzRoF

))(( OtTrO RR

y

zzyyxx

IH

IsIsIs

11

)0()0()0()0(

tstsI

sI

tstsI

xzRoFz

yRoFy

zxRoFx

11

11

sin)0(cos)0(

)0(

sin)0(cos)0(

NMR signals in the Rotating frame

“quadrature detection”

RoFyRoFx IiItSignal ,,)(

36x y

z

I

A symbolic summary without explicitly calculating the reduced density matrix

zAI)0(Define the spin system by its equilibrium density operator:

)cos(2 10 tIIH xz

xz IIH 10 )(

)sincos()(:)0( tItIAtAIIH yxxz

)sincos()(:)0( tItIAtAIIH yzzx

AMAI zz

AMAI xx

The environment of the spin system is defined by the Hamiltonian:

and in the rotating frame:

/ xx BIH

/ zz BIH

Here the Hamiltonians H correspondto magnetic fields Bp in the p-direction

The density operator can be “measured” by calculating “observables”resulting here in the magnetization {Mp}

kTBA /20

According to the rotation properties of the angular moment operator The Hamiltonian “rotates” the spin density operator

This result is like an indication that the classical motion of the magnetization for

a spin-1/2 behaves like the QM result. For spin-1/2 we can use the vector picture

derived before for the magnetization precession around any external magnetic field.

For spins higher than ½ the number of elements of the angular momentum components

becomes larger and we will have to show that these components also result in a

precession motion of its classical magnetization.

For further reading about QM, take any book on an “Introduction to Quantum Mechanics”

38

The transverse components of the bulk magnetization can only become zero when all individual

spins behave the same and there is no phase-scrambling. Thus, when we manipulate the

spins simultaneously in an equal manner, the bulk magnetization behaves like a single spin.

Just as we described the motion of a single spin in the laboratory and rotating frame,

we can present the bulk magnetization in the lab and rotating frame.

1

RoFM 0

xRoFyRoF

zRoF

rotating frame

0

0

1

0M

xy

z

Laboratory frame

The bulk magnetization during an RF irradiation field.

The return of the bulk magnetization to thermal equilibrium is governed by thermal motions.

We distinguish between two mechanisms, th0e dephasing of the transverse component

with a typical time constant T2 and the buildup of the longitudinal component to thermal

equilibrium with a time constant T1.

RoFM

xRoFyRoF

zRoF

1T

2T

2e T1 and T2 Relaxation

A simplistic way of understanding the action of the T1 and T2 relaxation mechanisms is to consider

thermally fluctuating magnetic fields in the laboratory frame. These fields will rotate the individual

components of the magnetization and a dephasing process will decrease the magnitude of the

transverse magnetization and Increase the longitudinal magnetization towards equilibrium.

The interaction of the spins their thermal bath will results in a Boltzmann population of their spin levels.

That there is a difference between the two relaxation times can be understood by realizing that the

individual magnetizations are precessing around the external magnetic field. To influence the direction

of the z-magnetizations the small fluctuating fields must be constant for a sufficient time. To influence

the phases of the xy-magnetizations the fluctuating field must be have components that vary at the order

of the Larmor frequency. For a randomly fluctuating field in the lab. frame .

RoFM

xRoFyRoF

zRoF

)(tz

ttxyttyx yxyx sin))((cos))((

Slow fluctuating components, contribute to T1

Fast fluctuating components at the order of contribute to T1 and T20/2 39

23

During an NMR experiment the nuclei are irradiated by a linear oscillating small magnetic field:

)cos(2)( ,10 ttH xz

40