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Quantum Mechanics of rf Spin Echo in Nuclear magnetic resonance
Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton
(Date: January 08, 2017) The quantum mechanics is used to understand the essential points for the spin echo method of nuclear magnetic resonance (NMR). A static magnetic field is applied along a nuclear spin with
angular momentum I and the magnetic moment I , where is the gyromagnetic ratio. The
static magnetic field is applied along the z axis. The radio frequency (r.f.) field magnetic field is applied along the plane perpendicular to the z axis. Here we consider the case of spin 1/2 system. 1. Gyromagnetic ratio of nuclear spin (a) Electron Orbital angular momentum and orbital magnetic moment
mc
eBB
LL 2
L
L
L
μ
Spin angular momentum and spin magnetic moment
mc
eg
gg
eBe
BeS
S 2
S
S
S
μ
where 2g .0023193043617(15).
(b) Proton (neutron); nuclear spin
The gyro
where I Table-1:
omagnetic ra
gnIN
I
μ
I is the angu
Gyrom
atio is define
g NnNn
I
I
ular moment
magnetic ra
d by
cm
eg
pn
N
2
tum.
atio
c
2. Spin recession (a) Classical treatment
The magnetic moment of nuclear spin with angular momentum zI -s given by
zI I
When a static magnetic field is applied along the z axis, the Hamiltonian has the form of the Zeeman energy, given by
BIμ BH I
We consider the equation of motion for the angular momentum
Bdt
d IBμτ
L
)( BII
dt
d
When the magnetic field is applied along the +z axis,
0zI , yx BII , xy BII
leading to the equation
)()( yxxyyx iIIBiBIiBIiIIdt
d
or
tiyx eIBtiIiII 0
00 )exp(
where
B 0
(clockwise for 0 and counter clockwise for 0 ).
Fig. P
r.
Fig. P
(b) S
The t
applied a
recession of
f. field RrB
recession of
)(tRFresB with
pin precess
time evoluti
along the z ax
f nuclear sp
)(tRFres with
f nuclear spi
h 0ref
ion using qu
on operator
xis (in the ab
pin. 0 B
0ref .
in. 0 B
uantum me
is defined b
bsence of the
0B for
0B for
chanics
by exp(T
e r.f. field), t
0 . (Count
0 . (Clockw
)ˆtHi
. Wh
the Hamilton
ter-clockwise
wise directio
hen a static
nian is given
e direction)
on). The r.f.
magnetic fie
n by
. The
foeld
eld is
zz IBIH ˆˆˆ0
where B 0 and zzI 2
1ˆ . The state vector at the time 2t is related to the
state vector at the time t1 through the time evolution operator as
)()ˆ2
exp(
)()ˆexp(
)()ˆexp(
)()ˆexp()(
10
10
10
12
ti
tIi
tIi
tHi
t
z
z
z
Here we define the rotation operator as
)2
exp(0
0)2
exp()ˆ
2exp()(ˆ
i
ii
R zz
This means that spin rotates around the z axis through the angle 0 .
We note that
yei
ee
exR ii
i
i
z
4/4/
4/
4/ 1
2
1
1
1
0
0
2
1)
2(ˆ
xeee
e
e
exR ii
i
i
i
i
z
2/2/
2/
2/
2/
2/
1
1
2
1
2
1
1
1
0
0
2
1)(ˆ
3. r.f. magnetic field
The r.f. coil generates a field RFB along the tilted axis in Fig. During an r.f. pulse on a single
spectrometer channel, the magnitude of this field oscillates at the spectrometer reference angular frequency ref . Between pulses, the r.f. field is equal to zeo. If the pulse is perfectly rectangular,
then the r.f. field has the form:
The r
B
r.f. field is gi
)( BtRFres
RFRF
B
B
iven by
)(
(cos
t RFresnon
zRF
B
e
)()(
co)sin
tt RFlong
xRF
B
e
)
)os( t pref )
during an r.f. pulse. The r.f. field consists of a longitudinal component proportional to RFcos
plus a transverse component proportional to RFsin . It turns out to be useful to imagine that the
transverse oscillating field is actually a sum of two rating components. Both components rotae in the xy-plane, at the same frequency, but in opposite directions. Note that
)]sin()cos([sin2
1)( prefyprefxRFRF
RFres ttBt eeB
)]sin()cos([sin2
1)( prefyprefxRFRF
RFresnon ttBt eeB
)cos(cos)( prefzRFRFRFlong tBt eB
where
B 0
For positive , the angular frequency ref is negative (clockwise), and for negative , the angular
frequency ref is positive (counter-clock wise). The transverse part of the spin Hamiltonian is
approximated as
)]sin(ˆ)cos(ˆ[sin2
1
)(ˆˆ
prefyprefxRFRF
RFresL
tItIB
tH
BI
Note that
RFRFnut B sin2
1
is the nutation angular frequency.
4. S
tate vector
Rx z )(ˆ'
in the rotat
x)
tion frame aand Laborattory frame
')(ˆ xRx z
Here we use the notations
Rxx ' Rotating frame
Lxx ' Laboratory frame
Then we have
LzRxRx )(ˆ'
This relationship may be generalized. Any spin state, viewed from the rotating frame, is related to the spin state, viewed from the fixed frame through
)(ˆ zRR ,
RzR )(ˆ
Now we consider the Schrödinger equation in the laboratory frame,
Ht
i ˆ
tiRR
ti
Rt
it
i
zz
zR
)(ˆ)](ˆ[
)(ˆ
Here we note that
)ˆexp()ˆexp()(ˆzzz IiJ
iR
where
zzz IJ 2
ˆˆ , zzI
2
1ˆ
The derivative of )(ˆ zR with respect to t;
)(ˆˆ
)ˆexp(ˆ
)ˆexp()(ˆ
zzref
zz
zz
RIi
Iidt
dIi
Iit
Rt
where
refdt
d , refref t
Thus we have
RR
RzLzzref
RzLzRzref
Lzzzref
zzzrefR
H
RHRI
RHRI
HRRI
tiRRI
ti
ˆ
)](ˆˆ)(ˆˆ[
)(ˆˆ)(ˆˆ
ˆ)(ˆ)(ˆˆ
)(ˆ)(ˆˆ
The Schrödinger equation in the rotating frame is given by
RRRH
ti ˆ
with the Hamiltonian in the rotating frame,
)(ˆˆ)(ˆˆˆ zLzzrefR RHRIH
5. Precession in the rotating frame
The spin Hamiltonian in the static field is
zL IH ˆˆ0 .
So the rotating-frame Hamiltonian is
z
zref
zzzzrefR
I
I
RIRIH
ˆ
ˆ)(
)(ˆˆ)(ˆˆˆ
0
0
00
where
)( 00 ref
Note that
zzzzzzz IRRIRIR ˆ)(ˆ)(ˆˆ)(ˆˆ)(ˆ
6. Rotating-frame Hamiltonian
In the presence of a static field along the z axis, the Hamiltonian is given by
zzIH ˆ2
ˆˆ0
00
with
zzI 2
1ˆ
B 0
The Hamiltonian due to the r.f. field in the x-y plane is
)]sin(ˆ)cos(ˆ[sin2
1
)(ˆˆ
prfyprfxRFRF
RFresRF
tItIB
tH
BI
where
P Prf t ,
Then the resulting Hamiltonian is
)]sin(ˆ)cos(ˆ[sin2
1ˆ
ˆˆˆ
0
0
pypxRFRFz
RFL
IIBI
HHH
The Hamiltonian in the rotating frame is
)(ˆˆ)(ˆsin2
1ˆ)(
)(ˆ)(ˆ]sinˆcosˆ)[(ˆ)(ˆsin2
1ˆ)(
)(ˆ]sinˆcosˆ)[(ˆsin2
1)(ˆˆ)(ˆˆ
)(ˆˆ)(ˆˆˆ
0
0
0
PzxPzRFRFzref
PzPzPyPxPzPzRFRFzref
zPyPxzRFRFzzzzref
zLzzrefR
RIRBI
RRIIRRBI
RIIRBRIRI
RHRIH
where
xpzpypxpz IRIIR ˆ)(ˆ]sinˆcosˆ)[(ˆ
)(ˆ)(ˆ)(ˆ zPzPz RRR
)(ˆ)(ˆ)(ˆ zPzPz RRR
((Formula))
ypxppzxpz IIRIR ˆ)sin(ˆ)cos()(ˆˆ)(ˆ
ypxppzypz IIRIR ˆ)cos(ˆ)sin()(ˆˆ)(ˆ
or
)(ˆ]ˆ)sin(ˆ))[cos((ˆˆpzypxppzx RIIRI
)(ˆ]ˆ)cos(ˆ)sin()[(ˆˆpzypxppzy RIIRI
Thus we have
)(ˆˆ)(ˆsin2
1ˆˆ0 PrefzxprefzRFRFzR RIRBIH
Note that
ref 00
refref t , prfp t
0 is the resonance offset. Note that the time dependence has vanished from this expression.
This is the point of the rotating frame. It transforms a time-dependent quantum-mechanical problem into a time-independent one. ((Note))
For 0 , 0ref (definition)
)ˆsinˆ(cossin2
1ˆ
)(ˆˆ)(ˆsin2
1ˆˆ
0
0
ypxpRFRFz
PzxpzRFRFzR
IIBI
RIRBIH
For 0 , ref (definition)
)ˆsinˆ(cossin2
1ˆ
)(ˆˆ)(ˆsin2
1ˆˆ
0
0
ypxpRFRFz
PzxpzRFRFzR
IIBI
RIRBIH
where
ypxpPzxpz IIRIR ˆsinˆcos)(ˆˆ)(ˆ
and
)ˆsinˆ(cos
ˆ)sin(ˆ)cos()(ˆˆ)(ˆ
ypxp
ypxpPzxpz
II
IIRIR
Combining these two equations, we get the final result
)ˆsinˆ(cosˆˆ 0ypxpnutzR IIIH
where
RFRFnut B sin2
1
_____________________________________________________________________________ For 0
((Note))
ref for 0
0ref for 0
7. x-pulse
We Consider a strong pulse of frequency ref , duration p and phase 0p (an ‘x-pulse’, in
the usual NMR jargon). The amplitude of the pulse is specified through the nutation frequency
nut .
The Ham
H
For p
2tp
H
with
and
R
or
R
where
miltonian in t
ˆˆ0 zR IH
0 and 0
1t ,
xnutR IH ˆˆ
xRRt ˆ)( 2
12 ttp .
exp)(ˆpxR
c)(ˆ
px
iR
the rotation-f
(cosnut
0 ref
Rp t )()( 1
p( pnut Ii
co2
sin
2cos
p
p
i
i
frame is
ˆsinˆpxp II
0 (resonanc
exp()ˆx iI
2os
2sin
p
pi
)ˆyI
ce), the rotat
e)ˆxpnut Ii
ting-frame H
)ˆexp( xpIi
Hamiltonian
n during the time
pnutp
Now we start with
)()]([ˆ)( 111 ttRt zR
)()]([ˆ)( 222 ttRt zR
12 ttp ,
where
tref ,
Thus we get
RpxRtRt )()(ˆ)( 12
or
)()]([ˆ)(ˆ)()]([ˆ1122 ttRRttR zpxz
or
)()]([ˆ)(ˆ)]([ˆ)( 1122 ttRRtRt zpxz
Note that
)ˆexp()ˆexp())((ˆzrefzz ItiIitR
Using the Mathematica, we get the expression
R
8. P
In ge
H
with p
the time
H
with
(ˆ)]([ˆ2 xz RtR
Pulse of gene
neral, the Ha
ˆˆ0 zR IH
0 For 0
12 ttp ,
(cˆnutRH
RRt
p
ˆ)( 2
([ˆ) 1zp tR
eral phase
amiltonian in
(cosnut
0 ref
is given by
sinˆcos xpI
p tp
)()( 1
)]2(
2
1
2
1
1i
i
ie
e
n the rotatin
ˆsinˆpxp II
0 (resonan
)ˆn yp I
R
sin
2cos
)2 1t
pi
refp
refp
ng frame is
)ˆyI
nce condition
22
1
2(2
1
ip
i
e
ie
n), the rotatin
2cos
sin)2 1
p
t
refp
refp
ng-frame Ha
2
p
amiltonian dduring
12 ttp .
and
2cos
2sin
2sin
2cos
)]ˆsinˆ(cosexp[)(ˆ
pip
ipp
ypxppp
p
p
p
ei
ei
IIiR
where
pnutp
Now we start with
)()]([ˆ)( 111 ttRt zR
)()]([ˆ)( 222 ttRt zR
tt ref )( , 12 ttp ,
_________________________________________________________________________
RpRtR
p)()(ˆ)( 12
)()]([ˆ)(ˆ)()]([ˆ1122 ttRRttR zpz p
or
)()]([ˆ)(ˆ)]([ˆ)( 1122 ttRRtRt zpz p
Note that
)ˆexp()ˆ)(exp())((ˆzrefzz ItiItitR
Thus we have
]2
sin2
cos)[cos()sin(2
cos
)sin(2
cos]2
sin2
cos)[cos(
)]([ˆ)(ˆ)]([ˆ
)(2
1)(
2
1
)(2
1)(
2
1
12
ppp
i
ppi
ppipp
p
i
zpz
ieie
ieie
tRRtR
refpprefpp
refpprefpp
p
9. Off-Resonance effect
For 000 ref (resonance condition), the rotating-frame Hamiltonian during the time
12 ttp , is given by
)ˆsinˆ(cosˆˆ0 ypxpnutzR IIIH
with 0p
RpRtRt
p)()(ˆ)( 12
with
12 ttp .
and
eff
effp
effpeffp
effp
ip
effp
effp
ip
eff
effp
effp
zpypxpppoff
iei
eii
IiIIiR
p
p
)2
sin()
2cos()
2sin(
)2
sin()
2sin(
)2
cos(
]ˆ)ˆsinˆ(cosexp[)(ˆ
0
0
0
where
202 nuteff
pnutp
Now we start with
)()]([ˆ)( 111 ttRt zR
)()]([ˆ)( 222 ttRt zR
12 ttp
_________________________________________________________________________
RpRtR
p)()(ˆ)( 12
)()]([ˆ)(ˆ)()]([ˆ1122 ttRRttR zpz p
or
)()]([ˆ)(ˆ)]([ˆ)( 1122 ttRRtRt zpz p
Note that
)ˆexp()(ˆzz IiR
where
tref
eff
peffeff
peffi
effp
ti
peffp
effp
ti
peffp
eff
peffeff
peffi
zpz
ieei
eiie
tRRtR
refprefpp
effprefpprefp
p
)]2
cos()2
sin([)2
sin(
)2
sin()]2
cos()2
sin([
)([ˆ)(ˆ)]([ˆ
02
])2(2[2
])2(2[2
02
12
1
1
10. Rabi formula Suppose that
01 t , and tt 2 , tp
z)0(
eff
ti
effp
eff
effeff
effti
eff
effeff
effti
eff
ti
effp
eff
ti
effp
eff
effeff
effti
zpz
t
et
i
ttie
ttie
t
et
i
t
et
itt
ie
RRtRt
refp
ref
refrefp
effrefpref
p
]2[2
02
02]2[
2
])(2[202
)2
sin(
)]2
cos()2
sin([
0
1
)]2
cos()2
sin([)2
sin(
)2
sin()]2
cos()2
sin([
)0()]0([ˆ)(ˆ)]([ˆ)(
with
tnutp
The probability of finding the system in the state
1
0z ,
effp
ipeff
p
effp
ipeff
p
eff
peffeff
peffi
Lp
prefp
prefp
pref
ei
ei
ie
z
]2[2
]2[2
02
)2
sin(
)2
sin(
)]2
cos()2
sin([
1 0)(
2sin
)(
)2
sin()(
2
202
2
2]2[
22
peff
nut
nut
eff
ipeff
p
Lp t
eiz
prefp
where
202 nuteff
11. Density operator:
2cos
2sin
2sin
2cos
)(
i
iRx
0
1z ,
1
0z
The density operator is defined by
2*
21*
2
2*
11*
1
2221
1211ˆCCCC
CCCC
ˆˆ 12*
21
1]ˆ[ Tr , 12211
11 , 22 ; Populations of states 1 and 2 .
11 , 22 ; Coherences of states 1 and 2 .
Note that
1ˆ111 , 2ˆ112
1ˆ221 , 2ˆ222
)(2
1
2
1
]10
01[
2
1
]ˆˆ[
2211
2221
1211
2221
1211
Tr
Tr
ITrI zz
cos
)(2
1
2
1
]01
10[
2
1
]ˆˆ[
21
1221
1211
2221
2221
1211
Tr
Tr
ITrI xx
sin
)(2
1
2
1
]0
0[
2
1
]ˆˆ[
21
1221
1211
2221
2221
1211
ii
ii
iiTr
i
iTr
ITrI yy
where
ie2121
12. The density operator in thermal equilibrium
)2
exp(0
0)2
exp(
)2
exp()2
exp(
1
0
0
)ˆexp(1
ˆ
0
0
00
22
11
HZeq
where
00 B
zzz IBIH ˆ2
ˆˆˆ 000
with
zzI 2
1ˆ
zI is the magnetic moment of nuclear spin.
zzIH ˆ2
ˆˆ 00
with
zzI 2
1ˆ
From the condition that 1]ˆ[ Tr
12211
((Approximation))
z
eq HZ
ˆ4
12
12
10
02
1
2
1
)ˆexp(1
ˆ
0
0
0
Here we note that
BB )(0
Thus we have
zzzeq IBIBB ˆ~
2
11
2
1ˆ2
12
1ˆ
41
2
1ˆ
where
BB ~.
13. The density operator in the rotating frame
The rotating frame density operator is defined by
RRR ˆ
Using the relations
LzRR )(ˆ , )(ˆ zLR R
The density operator in the rotating frame is related to that in the laboratory frame as
)(ˆˆ)(ˆ
)(ˆ)(ˆ
)(ˆ)(ˆˆ
zz
zz
zzR
RR
RR
RR
or
)(ˆˆ)(ˆˆ zRz RR
We have
1]ˆ[)](ˆˆ)(ˆ[]ˆ[ TrRRTrTr zzR
from the property of Tr. Note that
2221
1211ˆ
2
2
0
0)ˆ2
exp()ˆexp()(ˆi
i
zzz
e
eiI
iR
2
2
0
0)ˆ2
exp()ˆexp()(ˆi
i
zzz
e
eiI
iR
Thus we get
2221
1211
222
221
212
211
2
2
2
2
2221
1211
2
2
0
0
0
0
0
0ˆ
i
i
ii
ii
i
i
i
i
i
i
R
e
e
ee
ee
e
e
e
e
e
e
The thermal equilibrium density operator contains only populations, and is the same in both frames
eqReq ˆ)ˆ( ,
since
zeq IB ˆ~2
11
2
1ˆ
is diagonal. 14. Free particle without relaxation
Suppose that there is no r.f. field between t1 and t2. We note that
)()( ttR
We still assume that rf remains unchanged. For ptt 12
zIH ˆˆ0
Then we have
)()(ˆ
)()ˆexp(
)()ˆexp(
)()ˆexp()(
10
10
10
112
tR
tIi
tIi
tHi
tt
z
z
z
Density operator
)(ˆ)(ˆ)(ˆ
)(ˆ)()()(ˆ
)()()(ˆ
010
0110
222
zz
zz
RtR
RttR
ttt
where
0
0
2
2
00
0
0)ˆ2
exp()(ˆi
i
zz
e
eiR
0
0
2
2
00
0
0)ˆ2
exp()(ˆi
i
zz
e
eiR
Suppose that
zR IBtt ˆ~2
11
2
1)(ˆ)(ˆ 11
Thus we have
)(ˆ
ˆ~
2
11
2
1
)(ˆ)(ˆˆ~
2
11
2
1
)(ˆˆ)(ˆ~
2
11
2
1
)(ˆ)ˆ~
2
11
2
1)((ˆ
)(ˆ)(ˆ)(ˆ)(ˆ
1
00
00
00
0102
t
IB
RRIB
RIRB
RIBR
RtRt
z
zzz
zzz
zzz
zz
)(ˆ
ˆ~2
11
2
1
)(ˆ)(ˆˆ~2
11
2
1
)(ˆˆ)(ˆ~2
11
2
1
)(ˆ)ˆ~2
11
2
1)((ˆ
)(ˆ)(ˆ)(ˆ)(ˆ
1
00
00
00
0102
t
IB
RRIB
RIRB
RIBR
RtRt
R
z
zzz
zzz
zzz
zRzR
15. The magnetization vector using the density operator
]Re[)(2
]ˆˆ[ 122112
xx ITrM
]Im[)(2
]ˆˆ[ 122112
iITrM yy
)(2
]ˆˆ[ 2211
zz ITrM
where
1]ˆ[ Tr , ( )12211
ˆˆ ( *1221 )
Thus we have
)2
1(2
111 zM
, )
21(
2
122 zM
)(1
12 yx iMM
, )(1
21 yx iMM
leading to the density matrix as
σM ˆ1
12
1)ˆˆˆ(
11
2
1ˆ
zzzzxx MMM .
16. Magnetization vector (II) in general case
M
M
M
________17. S
(
[R
R
R
R)( 3
M
tM
x
x
)( 3
M
tM
x
y
(
[2
[2
)( 3
M
tM
z
z
__________trong radio
)cos()(
cos)(Re
(Re{[(Re
)(Re[
)](Re[
02
212
12
212
312
t
t
t
et
ti
)sin()(
c)([Im
Im{[(Re
)(Im[
)](Im[
02
212
12
212
312
t
t
et
t
)(
)(
)(
2
22311
22311
t
t
t
___________o-frequency
sin)()
Im)s(
(Im)
]
2
0
122
0
tM
it
y
co)(
Re)os(
Im)(
]
2
0
122
0
tM
it
y
i
)](
)](
3
3
t
t
__________pulse
)n(
sin()(
)][cos((
0
0212
02
t
t
)s(
sin()(e
)][cos((
0
212
212
t
t
___________
)]
sin()
0
0
i
)]
sin()
0
00
i
__________
)]
)]0
___________
_____________
RpRtRt
p)()(ˆ)( 22
)(ˆ)(ˆ)(ˆ)(ˆ 12 pRpR ppRtRt
where
)()(
)()(
0
0)()(
)()(
0
0
)]([ˆ)(ˆ)]([ˆ)(ˆ
2221)(
12)(
11
)(2
)(2
2221
1211
)(2
)(2
tte
tet
e
ett
tt
e
e
tRttRt
ti
ti
ti
ti
ti
ti
zzR
or
)()(
)()()(ˆ
2221)(
12)(
11
tte
tett
ti
ti
R
where
)(2
)(2
0
0]ˆ)(2
exp[]ˆ)(exp[)]([ˆt
i
ti
zzz
e
eti
Iti
tR
and
)()(
)()(
)]([ˆ)(ˆ)]([ˆ)(ˆ
2221)(
12)(
11
tte
tet
tRttRt
RRti
Rti
R
zRz
______________________________________________________________________________ ((Diagram))
zeq IB ˆ~2
11
2
1ˆˆ1
p
p
)(ˆˆ)(ˆ~
2
11
2
1ˆ2 pzp pp
RIRB
where
xpppppzppp
pppi
pp
pppppi
pzp
IiI
ie
ieRIR
p
p
pp
ˆsinsin1)cos1(cossinˆ)sincos(cos
)sincoscos(sinsin
sinsin)sincos(cos
2
1)(ˆˆ)(ˆ
22
18. Example: x
2
pulse with 0p
For the x)2/( with 0p . we have
)ˆexp()(ˆ)(ˆ0 xppxp IiRR
p
with
2
nutp
At 1tt ,
zeqR IBtt ˆ~2
11
2
1ˆ)(ˆ)(ˆ 11 (in thermal equilibrium)
)ˆ2
exp()2
(ˆxpx IiR
The density operator in the rotating frame is
y
xzx
xzx
xRxR
IB
RIRB
RIBR
RtRt
ˆ~
2
11
2
1
)2
(ˆˆ)2
(ˆ~
2
11
2
1
)2
(ˆ)ˆ~
2
11
2
1)(
2(ˆ
)2
(ˆ)(ˆ)2
(ˆ)(ˆ 12
or
yR IBt ˆ~
2
11
2
1)(ˆ 2 .
where
yxzx IRIR ˆ)2
(ˆˆ)2
(ˆ
(using Mathematica)
We note that
where
R
19. E
From the
We note
since the
12
1
12
1
[ˆ
[ˆ)(ˆ 2
z
z
R
Rt
zyz RtR ˆˆ)]([ˆ
Example:
e above discu
R t 12
1)(ˆ 2
that
R tt (ˆ)(ˆ 22
re is no r.f. f
[sin(~
2
1
)([ˆ~2
12
11
2
1)]((
ˆ)(ˆ)](
2
2
22
ref
z
R
tB
tRB
t
Rtt
zi
t)]([
x
2
pul
ussion,
yIB ˆ~2
1
B~
2
11
2
1)
field in the p
cos(ˆ)
([ˆˆ]
([ˆ)ˆ~
)]([
2
2
2
x
zy
zy
z
It
tRI
RIB
t
tiie
ieref 0
0
se, followed
yIB ˆ~
period betwe
]ˆ)
)]
)](
2
2
yref It
t
ti ref
sin0
d by a free p
een t2 and t3,
xref t cˆ)n(
precession in
yref t ˆ)cos(
nterval
0 tref .
)()()]([ˆ)( tttRt zR
Then we have
]ˆ)cos(ˆ)[sin(~
2
11
2
1
]ˆ)cos(ˆ)[sin(~
4
11
2
1
0
0~4
11
2
1
)(ˆˆ)(ˆ~2
11
2
1
)(ˆ)(ˆ)(ˆ)(ˆ
230230
230230
230230
23022303
230
230
yx
yx
i
i
zyz
zz
IIB
B
ie
ieB
RIRB
RtRt
or
]ˆ)cos(ˆ)[sin(~
2
11
2
1)(ˆ 2302303 yx IIBt
_____________________________________________________________________________ ((Diagram))
zeq IB ˆ~2
11
2
1ˆˆ1
x
2
yIB ˆ~
2
11
2
1ˆ2
)].sin(ˆ)cos(ˆ[~
2
1ˆ 003 xy IIB
where
yxzx IRIR ˆ)2
(ˆˆ)2
(ˆ
sinˆcosˆ)(ˆˆ)(ˆyxzxz IIRIR
cosˆsinˆ)(ˆˆ)(ˆyxzyz IIRIR
______________________________________________________________________________ In the case of the presence of transverse relaxation and longitudinal relaxation,
21 /230230
/ ]ˆ)cos(ˆ)[sin(~
2
1ˆ)1(~
2
11
2
1)(ˆ Tt
yxzTt eIIBIeBt
______________________________________________________________________________ ((Diagram))
zeq IB ˆ~
2
11
2
1ˆˆ1
x)2
(
yIB ˆ~
2
11
2
1ˆ2
21 ]ˆ)sin(ˆ)[cos(~
2
1ˆ)1(~
2
11
2
1ˆ 00
/3
Txyz
T eIIBIeB
where
yxzyz IIRIR ˆ)cos(ˆ)sin()(ˆˆ)(ˆ0000
______________________________________________________________________________
20. Example: 0p . x)( pulse
For the x)( with 0p . we have
)ˆexp()(ˆ)(ˆ0 xxp IiRR
p
The density operator in the rotation frame is
z
xzx
xzxR
IB
RIRB
RIBRt
ˆ~2
11
2
1
)(ˆˆ)(ˆ~2
11
2
1
)(ˆ)ˆ~2
11
2
1)((ˆ)(ˆ 2
where
zxzx IRIR ˆ)(ˆˆ)(ˆ (using Mathematica).
______________________________________________________________________________
21. Transverse relaxation time 2T
The phenomological equation for the polulation is given by
])1
exp[()()(2
0121221 T
itt
])1
exp[()()(2
0112212 T
itt
We consider the following case. At t = t1, eq ˆˆ . the x
2
pulse is applied between t1 and t2.
After that the r.f. pulse is turned off.
Just after
So we ge
or
eq
2
11
2
1
r the x
2
p
12
1)(ˆ 2t
et
( 2312 tt
( 2321 tt
( 2311 tt
IB z
0
42
1
ˆ~
pulse is turne
4
2
1
ˆ~2
1i
IB y
exp[~
4) B
i
exp~
4) B
i
2
1) ,
B
B
~4
1
2
1
0~
4
1
ed off,
2
1~4
~42
1
Bi
Bi
)1
[(2
0 T
i
)1
p[(2
0 T
i
( 322 t
B~
~4
] 2
eBi T
~4
])
eBi
2
1)2 t
)[cos( 0
)[cos( 02
T
)]sin( 0 i
)sin() 0 i
)]
or
________
22. L
The p
We cons
After tha
)(ˆ 23 tt
)(ˆ 23 tt
__________
Longitudina
phenomolog
t 11211 ([)(
t 22222 [)(
ider the foll
at the r.f. pul
[~
4
12
eB T
~2
11
2
1eB
___________
l relaxation
ical equation
eqt 111 ex])(
eqt 2212 e])(
lowing case.
se is turned
)cos([
2
1
0i
cos([2Te
__________
n time 1T
n for the pop
T1
)xp(
T1
)exp(
. At t = t1,
off.
)]sin( 0
sin(ˆ)0 yI
___________
pulation is gi
eq11
eq22
eq ˆˆ . the
c[~
4
12
ieB T
]ˆ)0 xI
__________
iven by
x pulse
2
1
si)cos( 0
___________
is applied b
)]in( 0
__________
between t1 an
_____
nd t2.
B
BIB z
eq
~4
1
2
10
0~
4
1
2
1
ˆ~2
11
2
1
Just after the x pulse is turned off,
B
BIBt z ~
4
1
2
10
0~
4
1
2
1
ˆ~2
11
2
1)(ˆ 2
Using the above equations
)21(~
4
1
2
1
~4
1
2
1}
~4
1
2
1~4
1
2
1{)(
1
12311
T
T
eB
BeBBtt
Similarly, we have )( 311 t as
)21(~
4
1
2
1
~4
1
2
1}
~4
1
2
1~4
1
2
1{)(
1
1322
T
T
eB
BeBBt
where
23 tt
In summary we get the expression
zT IeBt ˆ)21(
~2
11
2
1)(ˆ 1
3
23. rf spin echo method for determination of T1
where
R
eq 12
1ˆˆ1
xR ˆ)(ˆˆ 12
B~
2
11
2
1ˆ3
xR ˆ)2
(ˆˆ4
xzx RIR (ˆˆ)(ˆ
zIB ˆ~2
11
x
xR2
1)(ˆ
1
T IeB ˆ)21(~
1
x)2
(
xR )2
(ˆˆ3
zI)
zIB ˆ~2
11
zI
B 1(~
2
11
2
1 y
T Ie ˆ)2 1
R
24. r.
xzx RIR (ˆˆ)2
(ˆ
.f. spin echo
eq 12
1ˆˆ1
B~
2
11
2
1ˆ2
3ˆ[
~2
1ˆ yIB
4ˆ[
~2
1ˆ yIB
yI)2
o for determ
zIB ˆ~2
11
x
2
yIB ˆ~
2
1
0 )2
1cos(
y
0 )2
1cos(
mination of T
02
1sin(ˆ
xI
2
1sin(ˆ
xI
T2
220 )] Te
220 )] Te
where
R
R
R
________REFEREM.H. Le
W APPEND
xzx RIR (ˆˆ)2
(ˆ
yxy RIR (ˆˆ)(ˆ
yyy RIR (ˆˆ)(ˆ
__________ENCES evitt, Spin DWiley & Son
DIX
yI)2
xI)
yI)
___________
Dynamics, Bs).
__________
Basics of N
___________
Nuclear Mag
__________
gnetic Reson
___________
nance, secon
__________
nd edition (
____
(John
Mathematica
Clear "Global` " ;
x PauliMatrix 1 ;
y PauliMatrix 2 ;
z PauliMatrix 3 ;
Rx : MatrixExp2
x ;
Ry : MatrixExp2
y ;
Rz : MatrixExp2
z ;
Rz
2 , 0 , 0, 2
Rx
Cos2
, Sin2
, Sin2
, Cos2
Ry
Cos2
, Sin2
, Sin2
, Cos2
Rxy : MatrixExp pCos
2x
Sin2
y ;
k1 Rxy p ExpToTrig FullSimplify;
k1 MatrixForm
Cos p2
Sin p2
Cos p Sin p
Sin p2
Cos p Sin p Cos p2
k11 Rxy 2 ExpToTrig FullSimplify;
k11 MatrixForm
Cos p2
Sin p2
Sin p2
Cos p2
k12 Rxy 0 ExpToTrig FullSimplify;
k12 MatrixForm
Cos p2
Sin p2
Sin p2
Cos p2
k13 Rxy ExpToTrig FullSimplify;
k13 MatrixForm
Cos p2
Sin p2
Sin p2
Cos p2
((From Lecture Note on rf spin echo method))
Roff :
MatrixExp pCos
2x
Sin2
y
0 p2
z ;
Roff :
MatrixExp pCos
2x
Sin2
y
0 p2
z ;
s1
Roff p . p2 p2 02 p eff ,
1
p2 p2 02
1p eff
FullSimplify;
s1 MatrixForm
Cos p eff2
0 Sinp eff
2eff
p Cos p Sin p Sinp eff
2p eff
p Cos p Sin p Sinp eff
2p eff
Cos p eff2
0 Sinp eff
2eff
