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Thus in the rotating frame the magnetic field becomes time-independent
while the z-magnetic field component is reduced by the frequency of rotation
zRoF
 RoF
1
xRoF
yRoF
rotating frame
(0   ) ,there
0
On-resonance, when
is only an x-components to the field.
In such a case the magnetization performs a precession around the x-direction
with a rotation frequency .1
How to generate this B1 RF irradiation field in the laboratory frame:
B0
B1 (t )  B1 cos t
I (t )  I1 cos t
21 cos t x
y
z
Ignore because it is off-resonance!
1 (cos t x  sin t y)  1 (cos t x  sin t y)
x
Top view
y
x
18
Bird cage
National High Magnetic
Field Laboratory
Doty Scientific
u-of-o-nmr-facility.blogspot.com/2008/03/prob...
in:
Thus the magnetic field in the laboratory frame :
LAB  0 z  1 cos(t   ) x  1 sin(t   ) y
Becomes in the rotating frame:
RoF   z  1 cos  x  1 sin  y
out:
In NMR we measure
the magnetization
in the rotating frame:
 RoF , x (t )
 RoF , y (t )
Although the signal detection in the laboratory frame is along the direction of the coil:
LABx (t )   xRoF (t ) cos t   yRoF (t ) sin t
A sample with an overall
M (t )
the S/N voltage at the coil is:

S/N 
f
1/ 2
Vs 
1
    Q  M 0
kT 
2
f =noise of apparatus
 =filling factor
 =frequency
=band width
Q =quality factor
Vs =sample volume
19
2d. Necessary concepts for QM description
2d-i complex numbers
A large part of our discussion will deal with precessions of vectors, and with expressions
that 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:
x c  a x  ibx
These complex numbers have their own mathematics based on
i 2 :  1
imaginary axis
x c  y c  (a x  ibx )  (a y  ib y )  (a x  a y )  i (bx  by )
b
real axis
a
x c . y c  (a x  ibx ).( a y  ib y )  (a x a y  bx by )  i (a x by  a y bx )
xc / yc 
(a x  ibx ) (a x  ibx ).( a y  ib y ) (a x a y  bx by ) (a x by  a y bx )


i
(a y  ib y ) (a y  ib y ).( a y  ib y )
a y2  by2
a y2  by2
Im
A common practice is to present the values of the
x- and y-components of the magnetization
as if they are one complex number:
y
 xyc   x  i y
x
Re
A precessing magnetization around the z-direction has as xy-components the value:
 xyc   xy (cos t  i sin t )
This can be written as
   xy e
c
xy
it
with the definition
Im
1
e i  cos   i sin 

Re
20
2d-ii Wavefuntions, eigenstates and observables
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
B0 // z
has only two eigenstates with energies according to the z-components of their magnetization
E   z B0  I z Bo
. The allowed components of the angular momentum are
m
with
m  1/ 2 .
The way to present these energies is by an energy level diagram and the eigen states by I, m
:
1
E1/ 2  0
2
E1/ 2
m  1 / 2
1
  0
2
m is called the
magnetic quantum number
m  1/ 2
. The actual length of the angular momentum is
I  I ( I  1)
but we can only determine its z-component.
1/ 2
I ( I  1)
In fact we do not know what
the phase of its xy- component is.
I ( I  1)
1/ 2
When the spin is in one of these states it is stationary
21
( 2 I eigenstates
 1)
with energies
In analogy, a spin-I can occupy
and
m   I ,  I  1,..., I  1, I
3
 0
2
1
 0
2
1
 0
2
3
 0
2
spin=3/2 :
E (m)  m0
3 / 2;3 / 2
3 / 2;1 / 2
15 / 2
3 / 2;1 / 2
3 / 2;3 / 2
2. A dynamic quantum system can find itself in a linear superposition of its eigenstates.
for a spin ½:
 (t )  c1 / 2e( i /  ) E
1/ 2 t
1
1
 c1 / 2e( i /  ) E1 / 2 t 
2
2
I
In general:
 (t )   cm ei (1/  ) E t I ; m
m
m I
3. The result of a measurement of an observable O can be obtained by a calculation of the form
O (t )  ( (t ) O  (t )
 I * i / Em ' t
  I

   cm ' e
m' O   cm ei / Em t m 

m '   I
 m   I

I

m , m '  I
cm* 'cm ei /  ( Em  Em ' ) t m' O m
m' m  m' m   m 'm
knowing the value of the elements
m' O m
with
m' O m  Om 'm
5
22
2d-iv Matrix representation
 
n
n 1, N
 
;  n  m   nm
c
n 1, N
n
 c1 
 
c  
c 
 n
n
n A m  anm
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 diagonalization
we obtain the eigenfunctions and eigenvalues

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.
 a11


A 

a
 n1
a1n 




ann 
 n D1 AD  m  ann nm
 n D 1  k  k A l l D  m ;
k ,l
 n    k  k D 1  n
k
the eigenvalues of the Hamiltonian
are the energies of the system
23
d-v The Schrodinger Equation
Vector-matrix representation
d
i
 (t )   H (t )  (t )
dt

d
i
c (t )   H (t ) c (t )
dt

a time dependent Hamiltonian doesn’t have eigenvalues/energies
The solution of the Schrodinger equation defines an evolution operator
 (t )  U (t ,0)  (0)
c (t )  U (t ) c(0)
If the Hamiltonian is time independent then
U (t ,0)  e
i ( H /  ) t
i
1
 1  Ht  2 H 2 t 2    

2
If the Hamiltonian is time dependent then, with the Dyson operator T:
t
U (t ,0)  Te o
i
( H /  ) dt
 lim  n1, N e i ( H n /  ) t
t 0
N 
H n  H (( n  1 / 2)t )
H (t ) :
t
nt
24
d-vi. The angular momentum operators, definitions
I  p  x, y, z
p
e
iI x
I ye
iI x
[ I x , I y ]  iI z
 I y cos   I z sin 
an cyclic permutation
in the eigenfunction representation:
I z I, m  m I, m
m  I ,....I  1, I
1
1
I x I , m  I ( I  1)  m(m  1) I , m  1  I ( I  1)  m(m  1) I , m  1
2
2
i
i
I y I , m   I ( I  1)  m(m  1) I , m  1  I ( I  1)  m(m  1) I , m  1
2
2
I   I x  iI y
I  I , m  I ( I  1)  m(m  1) I , m  1
I  I , m  I ( I  1)  m(m  1) I , m  1
25
I  1/ 2 :
In the basis set of Iz:
1 1 0 
1 0 1
1 0  i
 ; I x  
 ; I y  

I z  
2  0  1
2 1 0
2i 0 
Pauli matrices span the whole 2x2 Hilbert space
I  1:
In the basis set of Iz:
0
1



1
Iz   0
 ; Ix   2
2



 1


2
0
2

 0 i 2




1
2  ; I y  i 2 0  i 2 
2



0
i 2
0 

The linear angular momentum operators do not span
the whole 3x3 Hilbert space
The missing operators are:
 0

2


1
IxIz  IzIx   2
0
2
2



2
0


 1 0 0


1
I z2  ( I x2  I y2 )   0  2 0 
2
 0 0 1


 0
1
I yIz  IzI y  i 2
2


0 0
1
I x2  I y2   0 0
2
1 0

i 2

0
i 2

i 2
0 
1

0
0 
26
2d-vii A nucleus in a magnetic field
The spin Hamiltonian
H LAB   B0 .  0  I
with the magnetic field in the z-direction of the lab. frame
and an RF irradiation:
H LAB  0 I z  21 cos(t   ) I x
and the rotating frame:
H RoF  I z  1 cos I x  1 sin  I y
For spin=1/2 :
Pauli matrices
0 
 0 1/ 2
 0  i / 2
1 / 2
 ; I y  
 ; I z  

I x  
0 
1 / 2 0 
i / 2
 0 1/ 2
cyclic permutations:
| 
[ I x , I y ]  iI z ; [ I y , I z ]  iI x ; [ I z , I x ]  iI y
| 
 e ( i / 2 )
0 

e  
(  i / 2 ) 
e
 0

 cos  / 2 i sin  / 2 
 cos  / 2 sin  / 2 
iI y
 ; e  

 
 i sin  / 2 cos  / 2 
  sin  / 2 cos  / 2 
eiI x
iI z
e
iI z
I x e
iI z
 I x cos   I y sin 
27
Remember that
U (t ,0)  e
i ( H /  )t
 De
i (  /  )t
D
1
when the Hamiltonian is time independent and
HD  D
 H 11


H 
H
 ji


H ij
U 11


U (t )  
U
 ji


U ij
H  DD
H NN







is Hemitian



 is unitary

U NN 
1
H ij  H *ji
1
U (t )  U (t )  1
 
U ij  U
1 *
ji
28
2d-viii The spin-density operator
An arbitrary function can be expended in a basis set
 (t ) 
 c (t ) 
n 1, N
This defines a set of coefficients
n
cn (t )
with
n
d
1
c (t )   H (t )c (t )
dt

Let us define an operator with matrix elements
 nm (t )  cn (t )c (t )
*
m
 11


 (t )  

 ji



 ij
 NN



 Is Hermitian




d
i
i
*
cn (t )cm* (t )    H nk (t )c k (t )c m* (t )   cn (t ) H mk
(t )ck* (t )
dt
 k
 k
i
i
   H nk (t )c k (t )c m* (t )   cn (t )c k* (t ) H km (t )
 k
 k
d
i
 (t )   H (t ),  (t )
dt

29
2d-ix
Ensemble Average and thermodynamics
Consider 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
NN
N1,N1
22
11
cn (t )  e
 i /  nnt
c (t )  e
*
m
cn (0)
i /  mmt *
m
c (t )
then
 nm (t )  e
 i /  (  nn   mm ) t
 nm (0)  e
i random
 ij  ( ii   jj ) / 
with the random phase approximation
 nm (eq)  c n (eq)c (eq)
*
m
 nm (eq)  0
coherences
 nn (eq)  0
the populations
30
Thus the equilibrium density operator can be defined by
the populations satisfying the Boltzman statistics
1  / kT
 (0)  e
Z

0  (1.05  1034 Js )  (3  108 s 1 )  3.15  1026 J
kT  (1.38  10
 23
1
JK )  (300 K )  4.14  10
 21
J
0
kT
e

0
kT
 7.6  106
 1
0
 0.9999924
kT
1  H / kT 1
1
 eq  e
 {1  H  ....}
Z
Z
kT
In NMR we solve the Liouville-von Neuman equation:
The signal is proportional to an “observable”
O (t )   (t ) O  (t )   cn* (t )  n O  m cm (t )
n ,m
   mn (t )Onm (t )  Tr (t )O
n ,m
   (t ) 12 (t ) 


   12
 I   Tr   11

(
t
)

(
t
)
1

22

  21
1
 (t )  11 (t )   22 (t )  11 (t )   22 (t )I z  2(Re 12 I x  Im 12 I y )
2
* 
 11 (t )   22 (t )  11 (t )   22 (t )I z  12 I   12
I
31
This requires a definition of the populations
of the eigenstates.
At thermal equilibrium these populations follow
the Boltzmann distribution:
N 1/ 2 e  E1 / 2 / kT 0 / kT
  E1 / 2 / kT  e
N1/ 2 e
Iz

ensemble
N 1/ 2
N1 / 2
1
( N1/ 2  N 1/ 2 )
2
and the populations become


1
1
N1/ 2  N (1  0 ) N 1/ 2  N (1  0 )
2
2kT
2
2kT
The thermal-equilibrium ensemble magnetization becomes
0 N 2 2 B0
1
N
M z ensemble  ( N1/ 2  N 1/ 2 )   

2
2
2kT
4kT
…………..
This require a definition of the populations
of the eigenstates.
N 1/ 2
N1 / 2
32
The magnitude of
 0
at 50-300K is very small.
kT
0  (1.05 10 34 Js)  (3 108 s 1 )  3.15 10 26 J
kT  (1.38 10 23 JK 1 )  (300 K )  4.14 10 21 J


0

6
 7.6 10
e kT  1  0  0.9999924
kT
kT
0
N1/ 2 
and the populations become

1
N (1  0 )
2
2kT
N 1/ 2 

1
N (1  0 )
2
2kT
Despite these very small values the ensemble nuclear magnetization can be detected.
The thermal-equilibrium ensemble magnetization becomes
0 N 2 2 B0
1
N
M z ensemble  ( N1/ 2  N 1/ 2 )   

2
2
2kT
4kT
A similar derivation can be presented for a spin higher than ½ in a magnetic field.
The Boltzman distribution at high temperature results in
Nm 
m0
1
N (1 
)
2I  1
kT
and the bulk magnetization becomes
M0 
I

m I
Iz
m

I

m I
I
I
m0
m20
N
N
N m m  
 m(1  kT )   2I  1 m
2 I  1 m I
kT
 I
M0 
N  B0
I ( I  1)
3kT
2
2
I
m
m I
2
1
 S ( S  1)  (2S  1)
3
33
2d-x NMR is a journey along the matrix elements
of the reduced spin density matrix
 

 (t )    (t ) 


ZkT


d
i
 (t )   H (t ),  (t )
dt

 (t )  U (t ) (0)U 1 (t )
Ei  ii 
t
H ( t ') dt '

0
U (t )  Te
O (t )  Tr ( (t )O(t ))
i
Example of pulse on I=1/2
NMR signals
I x, y
R
(t )  Tr( R (t ) I x , y )
Coherences that are off diagonal element are
detected. Populations that are the diagonal element
are not detected
 cos 1 / 2t i sin 1 / 2t  1
 cos 1 / 2t  i sin 1 / 2t 




i
sin

/
2
t
cos

/
2
t

i
sin

/
2
t
cos

/
2
t

1
1
1

1
1



 cos 1t  i sin 1t 
  2( I z cos 1t  I y sin 1t )
 
i
sin

t
cos

t
1
1 

34
2d-xi (high temperature) NMR on S=1/2
d
i
1 1


H0
 (t )   H 0 ,  (t ) eq
Z kT
dt

 (t )  e
 iH0t
 (0)e
iH0t
O  Tr ( (t )O)
The Larmor precession for spin I=1/2
H   0 I z
 eq
1 0

Iz
Z kT
 (0)  sx (0) I x  s y (0) I y  sz (0) I z
 (t )  e
 i0 I z t
 (0)e
i0tI z t
I x  sx (0) cos 0t  s y (0) sin 0t
I y   sx (0) sin 0t  s y (0) cos 0t
I z  s z (0)
35
NMR signals in the Rotating frame
d
i
 RoF (t )    (0   ) I z  H1,RoF ,  RoF (t )
dt

O
R
 Tr ( R (t )O)
Signal (t )  I x ,RoF  i I y ,RoF
 (0)  sx (0) I x  s y (0) I y  sz (0) I z
H1  1I y
Ix
z
x
I
RoF
 sx (0) cos 1t  sz (0) sin 1t
Iy
RoF
 s y (0)
Iz
RoF
 sz (0) cos 1t  sx (0) sin 1t
y
“quadrature detection” 36
A symbolic summary
without explicitly calculating the reduced density matrix
Define the spin system by its equilibrium density operator:
 (0)  AI z
A  B02 / kT
The environment of the spin system is defined by the Hamiltonian:
H  0 I z  21 I x cos(t   )
and in the rotating frame:
H  (0   ) I z  1 I x
Here the Hamiltonians H correspond
to magnetic fields Bp in the p-direction
H  I x  Bx   / 
H  I z  Bz   / 
According to the rotation properties of the angular moment operator
The Hamiltonian “rotates” the spin density operator
H  I z   (0)  AI x :  (t )  A( I x cos t  I y sin t )
H  I x   (0)  AI z :  (t )  A( I z cos t  I y sin t )
The density operator can be
“measured” by calculating “observables”
resulting here in the magnetization {Mp}
  AI z  M z  A
  AI x  M x  A
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”
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.
z
0
zRoF
M0
x
1
Laboratory frame
y
0  
1
x
M 0RoF
yRoF
RoF
rotating frame
The bulk magnetization during an RF irradiation field.
38
2e T1 and T2 Relaxation
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.
zRoF
T1
M RoF
T2

yRoF
xRoF
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
zRoF
.
  z (t )
M RoF
yRoF
Slow fluctuating
components,
contribute to T1
xRoF
(x x  y y )(t ) cos t  (x y  y x )(t ) sin t
Fast fluctuating components at the order of   2 / 0
contribute to T1 and T2
39
During an NMR experiment the nuclei are irradiated
by a linear oscillating small magnetic field:
H (t )  0 z  21, x cos(t   )
23
40