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Pulsed Nuclear Magnetic Resonance Experiment
Bin LI
Dept. of Physics & Astronomy, Univ. of Pittsburgh, Pittsburgh, PA 15260, USA March 18th, 2001 Introduction
1. Simple Resonance Theory Magnetic resonance is a phenomenon found in magnetic systems that possess both magnetic moments and angular momentum. The term resonance implies that we are in tune with a natural frequency of magnetic system, here corresponding to the frequency of gyroscopic precession of the magnetic moment in an external static magnetic field. Nuclear Magnetic Resonance (NMR) is a very important one among them, where the magnetic system is a nucleus. Since the nuclear magnetic resonance frequencies fall typically in the radio frequency region, we use the term radio frequency. Through studying the nuclear magnetic resonance of a nucleus, we will get enough information about processes at the atomic and nuclear level, so NMR is also a very significant tool to explore the tiny internal structure of macromolecule with high resolution in modern science. Generally, we consider a nucleus that consists of many particles coupled together
so that in any given state, the nucleus possesses a total magnetic momentum and a total
angular momentum . In fact, the two vectors may be taken as parallel, so that we can
write: , where
→
µ→
J→→
⋅= Jγµ γ is called the “gyromagnetic ratio”. For any given state of a nucleus, knowledge of the wave function would in principle enable us to compute both →
µ and . In quantum mechanics, both and are treated as operators, and we can
define , where
→
J→
µ→
J→→
⋅= IJ h I is a dimensionless angular momentum operator, which has
eigenvalues of either integer or half-integer. As we know, any component of →
I (for example ) commutes withzI 2I , so we can specify simultaneous eigenvalues of both 2I and . Let us call the eigenvalues zI I and respectively. Of course, m may be any of the
m12 +I values, III −− ,......,1, . So we can get any component of the operator µ
and I along an arbitrary ' direction, which is: z ><>=< Im'||ImIm'||Im '' zz Ihγµ
Now if we introduce an external magnetic field B, which will produce interaction
energy . Taking the field to be along the z-direction, we have a very simple Hamiltonian:
→→
⋅− Bµ
zIB ⋅⋅⋅−=Η hγ with the allowed Eigen energy mBE ⋅−= hγ , m=I, I-1, ……, -I. So the interaction between the magnetic field and the nucleus will result transitions between different energy levels as spectral absorption. Most commonly coupling used to produce nuclear magnetic resonance is an altering transverse magnetic field applied perpendicular to the static field at z-direction. If the amplitude of the altering field is , we get an additional perturbation term in the previous Hamiltonian of
zB'xB
)cos(' tIB xxper ⋅⋅−=Η ωγh . The operator has matrix elements between states and ' , as
will vanish unless xI m m
>< Im'||Im xI 1' ±= mm . Consequently, we will get that the allowed transitions are only between the adjacent levels and give the absorption frequency 0ω : zBE ⋅⋅=∆=⋅ hh γω 0 , or zB⋅= γω0 . 2. Motion of Single Spin System Now we start our discussion with the motion of a spin in an external magnetic
field →
B , and→
B will produce a torque on the magnetic momentum of amount . So
the angular momentum will satisfy the following equation:
→
µ→→
× Bµ→
J
→→
→
×= Bdt
Jd µ
Since , we have →→
⋅= Jγµ→→
→
×= Bdt
d γµµ .
This equation will hold no matter whether or not →
B is time dependent, it tells us
that at any instant, the changes in are perpendicular both and →
µ→
µ→
B . If →
B is time
independent, the angle between and→
µ→
B does not change, the vector therefore generates a cone (please refer to Figure 1).
→
µ
→
B
→
µ
Fig. 1
Consider a general vector time function , here are corresponding to the unit axes vectors. Originally, they should be constant, but now we think they can change with time. As their lengths are fixed, they can at most
rotate. We assume that they rotate with an instantaneous angular velocity , then we
have
)()()()( tFktFjtFitF zyx
→→→→
++=→→→
kji ,,
→
Ω→→
→
×Ω= idt
id , same with the unit vector and . →
j→
k
So the total time derivative of is therefore →
)(tF
dtkdF
dtdFk
dtjdF
dtdF
jdt
idFdt
dFi
dttFd
zz
yy
xx
→→
→→
→→
→
⋅++⋅++⋅+=)(
)( zyxzyx FkFjFi
dtdFk
dtdF
jdt
dFi
→→→→→→→
++×Ω+++=
→→
→
×Ω+= FtFδδ
Where the tFδδ
→
represents the time rate of change of→
F with respect to the
coordinate system . By using this formula, we can rewrite the equation of motion
for in terms of a coordinate system rotating with an arbitrary angular velocityΩ :
→→→
kji ,,→
µ→
→→→→
→
×=×Ω+ Bt
γµµδµδ , or further )()(
γγµγµ
δµδ
→→→→→→
→
Ω+×=Ω+×= BB
t.
From this equation, we can see that the motion of in the rotating coordinate obeys the same equation as in the laboratory frame if we just replace the actual magnetic
field
→
µ
→
B by the effective magnetic field γ
→→→ Ω+= BBe .
We can now readily solve for the motion of in a static field →
µ→
B = by choose a
specific , which makes . That is, . So in this rotating referent frame,
we have
0Bk→
→
Ω 0=→
eB→→
−=Ω kB0γ
,0=
→
tδµδ →
µ remains fixed with respect to , but in the laboratory frame,
rotates at angular frequency .
→→→
kji ,,→
µ
→→
−=Ω kB0γ 0B⋅γ is called “Larmor Frequency”. Now we
can reach a surprise conclusion that the classical precession frequency is identical in magnitude with the angular frequency needed for magnetic resonance absorption, as found by view of quantum mechanics.
→
Ω
Nerveless, instead of having just one single spin, if we have a group of spins with
moments , for the kth spin, their total magnetic moment can be written as:
. If the spins do not interact with each other, it is easily to prove that the
equation
k
→
µ→
µ
∑→→
=k
kµµ
)(→→
→
×= Bdt
d γµµ still holds. So the experimentally determined bulk magnetization
is simply the expectation value of the total magnetic moment. Therefore, the classical equation correctly describes the dynamics of the magnetization, provided the spins may be thought of as not interacting with one another. 3. Alternating Magnetic Fields Now let’s consider the effect of the altering magnetic field along the x-direction (generally, we create this altering field by using a AC-coil):
→→
⋅= itBtB x )cos(2)( 1 ω
And this magnetic field can be divided into two rotating components, each of amplitude , one rotating clockwise and the other rotating anti-clockwise. 1B
We denote the two rotating fields by and : RB→
LB→
)]sin()cos([1 tjtiBB R ⋅+⋅=→→→
ωω
)]sin()cos([1 tjtiBB L ⋅−⋅=→→→
ωω Near to the resonance region 0ωω ≈ , one of the components will contribute heavily to transition absorption, the other can be neglected since its frequency is different by ω2 . Here we assume we only have the field , and the component of frequency LB ω
along the z-axis, so . We may therefore write the transverse magnetic filed as: →→
⋅= kωω
)]sin()cos([)( 11 tjtiBtB −= ωω
Meanwhile we have a static filed along z-direction, zB 0BkB z =
So we get the equation: )]([ 1 tBBkdt
dz
→→→→
+×= γµµ
So in a rotating frame with angular velocity referent to the lab frame , we
have:
→→
−=Ω kω
])[()]()[( 101
→→→→→→→→
+−⋅×=+Ω+⋅×= iBkBtBBt γ
ωγµγµδµδ
Now effBt
γµδδµ
×= , where 110 )( BiBkBiBkBeff ⋅+∆⋅=⋅+−⋅=γω ,
γω
−≡∆ 0BB
The equation 1BiBkBeff ⋅+∆⋅= tells us in the rotating frame, the magnetic moment precesses in a cone of fixed angle about the direction of at angular frequency
effB
effBγω = . We notice that the motion of the moment is periodic. If it is initially oriented along the z-direction, it will returns to that direction periodically. As we increase its angle with the z-direction, its magnetic potential energy in the laboratory changes. All the energy it takes to tiltµ away from is returned in a complete cycle of 0B µ around the
cone. There is no net absorption of the energy from the altering field but rather alternately receiving and returning of energy.
Fig. 2
Z
Y
X
If the resonance condition is fulfilled exactly ( 0B⋅= γω ), the effective field is then simply , so the magnetic moment that is parallel to the static filed initially will then precess in the y-z plane. (please refer to Figure. 2). That’s to say, it will precess but remaining always perpendicular to . Periodically, it will be lined up opposed to . If we induce the field as a pulse with duration time instead of a continuous wave front, the magnetic moment would precess through an angle
1Bi ⋅
1B 0B
1B wt
wtB ⋅⋅= 1γθ . If we choose a which will satisfy wt πθ = , the pulse will simply invert the moment, such a pulse is
referred to in the literature as a “180 degree pulse”. If 2πθ = , the magnetic moment is
turned from the z-direction to the y-direction, and the pulse is called “90 degree pulse”. Without other effection, following the turn-off the 90-degree pulse , the moment would then remain at rest in the rotating frame, and hence precess in the laboratory perpendicular to the static field . But in fact that is not true, the magnetic moment will go back to the initial z-direction after enough long time due to relaxation, it will bring the spin system back into Boltzmann equilibrium.
1B
0B
4. Equilibrium Magnetization and Bloch Equation In practice, we can’t measure the magnetic momentum for a single spin; instead, we can only measure the macroscopic magnetization for the whole sample. Now we will try to understand the equilibrium distribution of spin system under the uniformly constant magnetic field along the z-axis . 0B
As we know, the energy of the magnetic dipole interaction is: 00 BJBJE z ⋅⋅−=⋅⋅−= γγ If the energy difference between two adjacent states is very small, we can consider the distribution is continuous (Classical Boltzmann Statistics). θγ cos0BJE ⋅⋅−= So the probability that any spin particle in the orientation ),( ϕθ in space is given
by: )cos()(),( 0 θγ
ϕθTk
JBExpCTk
EExpCPBB
⋅=−⋅= , where C is the normalized
constant.
Generally, in room temperature, we have 10 <<Tk
JB
B
γ, then from the equation
1sin)cos(),( 0
0
2
0=⋅⋅=⋅Ω ∫∫ ∫ θθ
γθϕϕθ
ππ
TkJB
ExpddCPdB
, we get π41
≈C .
Thus we get:
Tk
BJTk
JBJddJBB
z 3)cos1(
4cossin 0
20
0
2
0
⋅⋅=+
⋅⋅>=< ∫∫
γθ
γπθθθϕ
ππ
So in thermal equilibrium, only the z-component of the magnetization is non-zero, the x-component and y-component both should be zero due to the azimuthal symmetry.
→→→→
⋅⋅⋅
+⋅+⋅>=< kTk
BJjiJ
B300 0
2γ
For a system with N non-interaction spins, we have the total angular momentum:
Tk
BJNJNJ
B30
2
0⋅⋅
⋅>=<⋅=γ
So the equilibrium magnetization for the spin system should be:
Tk
BJNJM
B30
22
00⋅⋅⋅
=⋅=γ
γ
Now if we induce an altering magnetic field (as we discussed before ), the magnetization will deflect from the equilibrium value.
)(tBx
From the previous equation )(→→
→
⋅×= Bdt
d γµµ , we know there should be a similar
equation for magnetization M, here, we include the thermal relaxation. So the magnetization must wish to be parallel to , the x- and y- components must have a tendency to vanish.
0B
Thus
zzz BM
TMM
dtdM
)(1
0 ×⋅+−
= γ
xxx BM
TM
dtdM
)(2
×⋅+−= γ
yyy BM
TM
dtdM
)(2
×⋅+−= γ
This set of equations is called Bloch Equations. Here we introduce times, is called longitudinal relaxation time which is different from which is the transverse relaxation time.
1T
2T
Now, we want to figure out what’s the solution of the Bloch Equations when the altering transverse magnetic field is turned on. In order to make the calculation easier, we immediately transform to the coordinate frame rotating at angular velocity k⋅−=Ω ω ,
taking along the rotating x-axis and denoting 1Bγω
−0B by . 0b
So and ∧∧∧→
++= yMxMzMtM yxz)(∧∧→
+= xBzbB 10
Then in rotating coordinate
1
01 T
MMBM
dtdM z
yz −
+−= γ
2
0 TM
bMdt
dM xy
x −= γ
2
01 )(TM
bMBMdt
dM yxz
y −−= γ
In experiment, we choose value low enough to avoid the saturation that will kill the resonance absorption signal. As we know, and must vanish as ,
and is in a steady state, it differs from to the order of . Therefore, we replace by in the last two equations, and neglect the transient term, we get
1B
xM yM 01 →B
zM 0M 21B
zM 0M
122
20
20200 )(1
)()( B
TT
TM x ωωωω
ωχ−+−
=
122
20
200 )(11)( B
TTM y ωω
ωχ−+
=
Where 0χ is the magnetic susceptibility, which is equal to TkJN
B3
22γ .
So we can see in the rotating referent frame with angular velocityω , the magnetization is a constant. Therefore, in the lab frame, it must rotate at frequency ω . Typically, we choose the orientation of the coil along the fixed X-direction, we can calculate the time dependent component of magnetization along the coil direction. XM By referring to Figure.3, we get )sin()cos( tMtMM yxX ⋅+⋅= ωω
Y
My
Mx
X
Fig. 3 If we regard the magnetic field as a linear filed, we will get )cos()( 0 tBtB XX ⋅= ω 10 2 BBX ⋅= Then we see both of and are proportional to , and we can write xM yM 0XB
)cos()())sin(")cos('()( 0"'
0 tBjBtttM XXX ⋅⋅+=⋅+⋅= ωχχωχωχ
Where the quantities 'χ and "χ should be:
22
20
2020
0
)(1)(
)(2
'T
TT
ωωωω
ωχ
χ−+−
=
22
20
200
)(11)(
2"
TT
ωωω
χχ
−+=
It is convenient to regard both and as being the real parts of the complex functions and . Then if we define
)(tBX )(tM X
)(tB CX )(tM C
X "')( χχωχ −= I and write , we find , and ,
where
tIX
CX eBtB ω
0)( = )()()( tBtM CX
CX ⋅= ωχ )(Re)( 0
tIXX eBtM ωωχ ⋅=
1−=I . Ordinarily, if a coil of inductance is filled with a material of susceptibility0L 0χ , the inductance is increased to )1( 00 χ+L , since the diamagnetic material will increase the magnetic flux by factor )1( 0χ+ . Similarly, the complex susceptibility produce a flux change from the original value , 0L )](1[0 ωχ+= LL Denoting the coil resistance in the absence of a sample as R , the coil impedance when including sample becomes RLLIZ +++⋅= ")'1( 00 ωχχω . The real part of the susceptibility 'χ therefore changes the inductance, where as
the imaginary part "χ modifies the resistance. The fractional change in resistance RR∆ is:
R
LRR "0ωχ=
∆ .
If we assume that uniform magnetic field occupy a volume of , the peak stored magnetic energy produced by an altering current, whose peak value is , is
V0i
VBiL X ⋅=⋅ 20
200 2
121
While, we can get the average power dissipated in the nuclei is:
"21
21
02
02
0 χω ⋅⋅⋅⋅=∆⋅⋅=∆ LiRiP
So we find VBP X ⋅⋅⋅⋅=∆ "21 2
0 χω . This equation provides a simple connection
between the power absorbed, "χ , and the strength and the frequency of the altering field. So if we can use some detecting system to measure the energy dissipated in the sample during the resonance absorption, we can measure "χ . The following figure (Figure. 4) shows the theoretical relationship between 'χ ,
"χ with the oscillation frequency of the altering magnetic field. “ 'χ vs. )( 0 ωω − ” is called the dispersion curve, and the other one ------ “ "χ vs. )( 0 ωω − ” is called the absorption curve. We can see when the frequency is near to the resonance frequency, the absorption curve will get close to its peak value.
"χ
'χ
Fig By using sensitive electronic circuit, susceptibilities to electronic signals and meastune it well, we can get clear experimental sare corresponding to the “In-Phase Signal” an 5. Pulsed NMR Experiment and Free Induc Since the static magnetic field has a smcoil, so from the definition of the Larmor Fredistribution of the values of Larmor Frequendenoted by )( 0ωg , the number of particles dn
00 ωω d+ is: 00 )( ωω dgNdn ⋅⋅=
. 4
we can change both of these two magnetic ure them by using digital oscilloscope. If we ignals, where the absorption and dispersion d “Out-Phase Signal” respectively.
tion Decays
all inhomogeneity across the sample in the quency 00 B⋅= γω , we know there will be a cies as well. If the distribution function is with Larmor Frequencies between 0ω and
In order to solve the problem of the time dependence of the net magnetization
, we first estimate the average angular momentum of the spin system: →
)(tM
Instead of using , we have ∑→→
=k
kjJ
00 )()()()( 00 ωωωω dgtJNdntJtJ ⋅⋅=⋅= ∫∫→→→
Now we examine the response of our sample to a rotating magnetic field of
magnitude , angular speed and its lasting time . Let us view the behavior of the spin system from the reference frame rotating with , and this time due to convenience,
we choose along the direction. (please refer to Figure 5.)
1B−
0ω wt
1B
1B→
− j
Fig. 5
So now the effective field in the rotating frame becomes:
, where →→→
−∆= jBkBBeff 10 γω−
−=∆ 00 BB .
At the resonance case, 1
21
20 ])[( BB <<∆−
is a good approximation, we can set
. →→
−= jBBeff 1
Just before the application of the rotating pulsed field, we have ,
where is the so-called isochromat. After we induced the transverse field, each of
→→
= kjj )0()0(0ω
)(0
tjω→
i
k
effB→
j
the isochromats will precess about the axis in time by an amount →
j wt wtB ⋅⋅= 1γθ , so
that . )sin(cos)0()(0
→→→
⋅+⋅⋅= ikjtj w θθω
So the total angular momentum over the sample is:
)sin)(cos0()()( 000
→→→
⋅+⋅⋅⋅= ∫ ikjgdNtJ w θθωωω
)sin(cos)0(→→
⋅+⋅⋅= ikNj θθ )sin(cos)0(→→
⋅+⋅⋅= ikJ θθ where )0()0( NjJ = . So we find
for →→
= iJtJ w )0()( °= 90θ ( pulse) °90
for →→
−= kJtJ w )0()( °= 180θ ( pulse) °180
After the pulse turned off , we get and each isochromat
precesses about the axis at an angular speed of
)( wtt >→→
∆= kBBeff 0→
k 00 B∆⋅=∆ γω . Thus
]))(sin())([cos(sin)0(cos)0()( 000
→→→→
−∆−−∆+= jttittjkjtj ww ωωθθω
If we redefine , and do the integral over wttt −=1 )( 0ωdn , we find
])sin()()cos()([sin)0(cos)0()( 0100010010 ωωωωωωθθω ∆⋅∆∆−∆⋅∆∆⋅+=+ ∫∫→→→→
dtgjdtgiJkJttJ w
Generally, )( 0ω∆g is a symmetric function of 0ω∆ , so the second integral
vanishes, while it predicts that the component of the angular momentum is time
invariant and the or (if we choose the rotation field along another orientation) components decays to zero slowly.
→
k→
i→
j
For simplicity, we assume that the function of )( 0ωg is a constant a2
1 in the range
of ; other wise, it has value of zero: )( 000 aa +<<−−−
ωωω So we will get
=0100 )sin()( ωωω ∆⋅∆∆∫
+∞
∞−
dtg ∫+
−
=∆⋅∆a
a
dta
0)sin(21
010 ωω
=0100 )cos()( ωωω ∆⋅∆∆∫
+∞
∞−
dtg ∫+
−
=∆⋅∆a
a atatdt
a 1
1010
sin)cos(
21 ωω
So we get
→→→
⋅+⋅=+ iat
atJkJttJ w
1
11
sinsin)0(cos)0()(0 θθω .
Since 00 JM ⋅= γ , so the magnetization has the same time dependent function. Since the orientation of the coil is along the y-direction, the magnetization along that direction in the lab frame should be )sin()cos()( tMtMtM ijX ⋅+⋅= ωω , so the
average magnetization along that fixed direction is 1
10 sinsin
2||
atatM
M X θ= . So it will
decay after the pulse as time increases. Meanwhile, we also know that the amplitude of Fid signal will be proportional to the magnetization of the direction of the coil, so we know the fid signal will also decay to zero after the pulsed transverse magnetic field is turned off. Now let us consider the effection of a second pulse of the transverse field. We assume that the second pulse comes at time τ after the first pulse. This time in order to differentiate the two pulses, the rotation produced by the first pulse is labeled by 1θ , and that by the second is labeled by 2θ . This time, we still choose the second along the
direction. To simplify future discussion, we think the and components of have died to zero due to the transverse relaxation. So just before the 2
1B→
− j→
i→
j )(tJ→
nd pulse comes, we
have . Use the same analysis as we used for discussing the response of the system to the first pulse, the response to the second pulse should be:
→→
= jJtJ 1cos)0()( θ
∫ ∆∆∆+=++→→→
020021212 )cos()((sincos)0(coscos)0()'( ωωωθθθθτ dtgiJkJttJ w
))sin()( 0200 ωωω ∆∆∆− ∫→
dtgj
=→→
⋅+ iat
atJkJ
2
22121
sinsincos)0(coscos)0( θθθθ
It shows how the angular momentum, or magnetization of the sample functions as time after the 22t
nd pulse. From this equation, we can draw the conclusion that the 2nd
fid has the same shape as the fid curve, and its amplitude is proportional to 21 sincos θθ , instead of proportional to 1sinθ as the first fid. So if °= 901θ or °= 1802θ or both, the second fid is zero. However, this is true just for the case there is no longitudinal relaxation; that is to say, after the first pulse, the angle between the magnetization and the static field is unchanged, stays 1θ . But for the real experiment, the ideal case is not true. After the first pulse, the magnetization wishes to go back to the z-direction (the direction of magnetic field), so the angular 1θ will decreases with time, when the 2nd pulse comes, it will become )'(1 τθ . So we get the first fid and the second fid are the following respectively:
→→
⋅=+ iat
atJttF w
1
11111
sinsin)0()( θ
=++ iat
atJttF w
2
221222
sinsin'cos)0()( θθτ
As we know, when it is in resonance, the average resonance frequency range ------
should be the smallest value, the fid signal will decay to the zero slower, so the wiggle of the fid curve will become smaller (Figure 6 ----- Indication of Resonance). a
Indication of Resonance
Resonance
Fig. 6
After finding the Resonance condition, we focus our attention on two specific cases.
(1) The "9090" −−τ pulse sequence(Figure 7):
o90121 ≈== wtBγθθ
τ
Fig. 7
More ever, we can see if we add "9090" °−° pulses ( == 9021 θθ ) to the spin system, the amplitude of the first fid is the maximal of all the values it could be. The amplitude of the 2nd fid depends on the delay time from the 2nd pulse to the 1st pulse ------τ . When the delay time τ is a constant, we can tune our detecting circuit until we get the maximum of the second fid, when it shows the second pulse is also a 90 pulse. Using Bloch Equations, let 01 =B (since the pulse is turned off), and change
magnetization M to , so we have: J
1
0
TJJ
dtdJ zz −
=
2
0 TJ
BJdt
dJ xy
x −∆= γ
2
0 TJ
BJdt
dJ yx
y −∆−= γ
Solving this equation, we get:
1))0(()( 00Tt
zz eJJJtJ−
−+=
)cos()0()( 02 teJtJ Tt
xx ω∆=−
(where 00 B∆⋅=∆ γω ) So we can see that the longitudinal angular momentum will go back to the
equilibrium value as function of
)(tJ z
0J )1( 1Tt
e−
− after the pulsed transverse field is turned off;
the transverse angular momentum will decay as )(tJ x2Tt
e−
from its initial value after pulse ( has a similar performance as ). )0(xJ )(tJ y )(tJ x
The fid curve is: →−→
⋅−= itfeJtF TFid )()1(),( 1
0
τ
τ Where function should be )(tf
→
−−−−
itta
tta)(
)(sin
2
2
ττ .
So we can also measure the amplitudes of a series of second fid curves with
differentτ , there should satisfy a relation of )1( 1Teτ
−
− . By using this method, we can
measure the longitudinal relaxation time .1T
(2) The "18090" −−τ pulse sequence (Figure 8):
τ
o90111 ≈= wtBγθ o180222 ≈= wtBγθ
Fig. 8
Now let’s discuss about another special case "18090" °−° pulses and the
phenomenon of “echo”. Firstly, we follow a typical isochromat . At the time)(0
tjω→
τ , before the application of the second pulse, this isochromat will have accumulated an
azimuthal phase τωφ 0∆−= relative to the axis. The inducing of the 2→
i nd pulse will
make it nutate by about the axis, so that its phase immediately following the second pulse is now
°180→
− jτωπφ 0' ∆−= (please refer to Figure 9, I didn’t show it here). As the
accumulation of the transverse phase should be proportional to the time, so at the time 2tt += τ , the phase should be 200" tωτωπφ ∆+∆−= .
Thus we have the transverse component differential:
]))(sin())()][cos(('sin[)]('cos[)0()( 2020222120
→→→
−∆+−−∆++=+ jtitttJtj τωπτωπθτθτω
×=⋅⋅=⋅=+ ∫∫→→→
)'sin()'cos()0()()()()( 21002 00 θθωωτ ωω JdgtJNdntJtJ
]))(sin()())(cos()([ 02000200 ωτωωωτωω ∆⋅−∆∆+∆⋅−∆∆− ∫∫→→
dtgjdtgi As we know before, the integral
≈∆⋅−∆∆+∆⋅−∆∆− ∫∫→→
]))(sin()())(cos()([ 02000200 ωτωωωτωω dtgjdtgi→
−−
ita
ta)(
)(sin
2
2
ττ
So
→→
−−
×=+ ita
taJtJ
)()(sin
)'sin()'cos()0()(2
2212 τ
τθθτ
If is different with2t τ , since a is a significant number, this term will go to zero quickly. So generally, we can see nothing after turning off the 2nd fid signal. But
when τ=2t , the factor )(
)(sin
2
2
ττ
−−
tata will become quite large ---- 1, which is very
significant compared to the other cases: τ≠2t , it will increase the probability for us to detecting the fid signal. This phenomenon (when τ=2t ) is called spin refocusing, the signal shows up at time ττ 22 =+= tt is called echo.
Here, the transverse angular momentum will decay as )(tJ x2Tt
e−
from its initial value after pulse. )0(xJ Now, based on the previous discussion, we can estimate the fid curves and echo including the spin-relaxation. The echo should be like the following term:
→→
⋅−⋅−= itgTtJtF Echo )()exp(')(2
0
Where should be )(tg)2(
)2(sin)(
)(sin
2
2
ττ
ττ
−−
=−−
tata
tata .
So using echo signal, we can measure the echo peak values of different delay timeτ , then fit the experimental data, we can obtain the transverse relaxation time .
The height of echo should decay as function of
2T
)2exp(2Tτ
− .
After this point, we see the main experiment we can do for pulsed NMR:
1. For the "9090" −−τ pulse sequence:
“ )1( 1Teτ
−
− vs. τ ”, fid curve fitting to get . 1T
2. For the "18090" −−τ pulse sequence:
“ )2exp(2Tτ
− vs. τ ”, echo curve fitting to get . 2T
Experiment SET-UP This experiment is evolved a lot of equipments including a complicated structure of circuit, now we will introduce the essential parts briefly. The following block diagram (Figure 10.) shows the main contents. F A E H B C D G I N S
Fig. 10
A: Pulse Programmer B: Modulator C: Power Amplifier D: Protection and Coupling Circuit E: Receiver F: Oscillator G: Phase Detector H: Oscilloscope I: Sample in coil N (S): North Pole or South Pole of Permanent Magnetic Filed The Oscillator (F) outputs a continuous radio frequency signal. The frequency can be adjusted by different scales. When we are doing pulsed NMR experiment, we can change the frequency until we are close to the Larmor Frequency 0ω of the resonance. The Pulse Programmer (A) generates D.C pulses that are used to determine when the RF (radio frequency) pulse are applied to the sample to excite sample from equilibrium state. The programmer can determine the repetition rate of series of identical pulse sequences, the length of each of the D.C pulse ( and ), and the delay time between the two individual pulses.
1wt 2wt
The Modulator (B) modulates the signal from the Oscillator with the pulses from the programmer. It can be looked as a switch that opens and closes according to the pulse signals from the programmer. The main function of Power Amplifier (C) is to take the RF pulses from the modulator and amplifies them from about 0.1 volts to 100 volts, but keeping the same shape. The 100 volts out of power amplifier will damage the receiver and cause it to “block”, so that the signal following the RF pulse is unobservable. The Protection Circuit (D) connects the resonant circuit of the probe to the power amplifier during the excitation pulse, while disconnecting the receiver. Following the excitation pulse, it connects the resonant circuit of the probe to the receiver and disconnects it from the transmitter. So spurious noise from the transmitter doesn’t reach the receiver, and almost all of the signal from the resonant circuit goes the receiver. The Receiver (E) is a high gain, low noise, radio frequency amplifier. It takes the signal coming from the sample coil and amplifies it from 1 Voltµ ~ 100 Voltµ to about 0.1 volt, then send it to the phase detector. The Phase Detector (G) takes the signal from the receiver and changes it to D.C. It can be viewed as a switch that opens and closes in synchronism with the reference signal from the oscillator. In fact, the output of the phase detector seems as the vector sum of the signal from the receiver and the reference signal. There is one thing I have to mention left ------ the resonance circuit around the sample. When the RF frequency is tuned to the Larmor Frequency 0ω , it uses the power
from the power amplifier to create an oscillating magnetic field to excite the sample. It can also pick up a signal from the precessing magnetization to relay to the receiver. Experimental Data and Discussion In our experiment, we used three different concentrations of solution. The following are data table and plots.
2CuBr
Concentration 0.0375 M ( ) 2CuBr90 Deg ~ 90 Deg Sequence 90 Deg ~ 180 Deg Sequence Delay Time 1 (ms) Amplitude of 2nd
Pulse (mV) Delay Time 2 (ms) Amplitude of Echo
(mV) 5 15.6 10 136 10 17.2 15 132 50 18.8 20 128 100 21 25 120 200 26.6 30 96 300 32.4 35 80.8 400 34.8 40 61.6 45 43.2 50 27.2 60 13.4 70 5.8 Concentration 0.15 M ( ) 2CuBr90 Deg ~ 90 Deg Sequence 90 Deg ~ 180 Deg Sequence Delay Time 1 (ms) Amplitude of 2nd
Pulse (mV) Delay Time 2 (ms) Amplitude of Echo
(mV) 5 3.92 5 24.8 10 4.64 10 21.6 50 6.4 20 16 100 7.36 30 10.5 200 8.64 50 3.6 500 9.8
Concentration 0.75 M ( ) 2CuBr90 Deg ~ 90 Deg Sequence 90 Deg ~ 180 Deg Sequence Delay Time 1 (ms) Amplitude of 2nd
Pulse (mV) Delay Time 2 (ms) Amplitude of Echo
(mV) 5 8 5 76 10 20.8 10 36.8 50 44.4 15 17.2 100 48 20 8.6 200 49.6 25 3.8 300 50.4
Longitudinal Relaxation Time T1 Measurement for 0.0375 M CuBr2 (90~90
Pulse Sequence)
05
10152025303540
0 100 200 300 400 500
Relaxation Time (ms)
Am
plitu
de o
f 2nd
Pul
se (m
V)
0.0375 MFit
Where we get the fit curve is )1(3113)( 3902
τ
τ−
−⋅+= eVFid .
Longitudinal Relaxation Time T1 Measurement for 0.15 M CuBr2 (90~90 Pulse Sequence)
0
2
4
6
8
10
12
14
0 100 200 300 400 500 600
Relaxation Time (ms)
Am
plitu
de o
f 2nd
Pul
se (m
V)
0.15 MFit
Where we get the fit curve is )1(94)( 2752
τ
τ−
−⋅+= eVFid
Longitudinal Relaxation Time T1 Measurement for 0.75 M CuBr2 (90~90 Pulse
Sequence)
0
10
20
30
40
50
60
0 100 200 300 400Relaxation Time (ms)
Am
plitu
de o
f the
2nd
Pul
se (m
V)
0.75 MCuBr2
Where we get the fit curve is )1(50)( 302
τ
τ−
−⋅= eVFid
Transverse Relaxation Time T2 Measurement for 0.0375 M CuBr2 (90~180 Pulse Sequence)
y = 308.69e-0.049x
R2 = 0.8996
0
50
100
150
200
250
300
0 20 40 60 80Relaxation Time (ms)
Am
plitu
de o
f Ech
o (m
V)
0.0375 M
Where we get the fit curve is: 8.402
9.308)(τ
τ−
⋅= eVehco .
Transeverse Relaxation Time T2 Measurement for 0.15 M CuBr2 (90~180 Pulse
Sequence)
y = 33.995e-0.043x
R2 = 0.9816
0
5
10
15
20
25
30
0 10 20 30 40 50 60
Relaxation TIme (ms)
Am
plitu
de o
f Ech
o (m
V)
0.15 M
Where we get the fit curve is: 5.462
0.34)(τ
τ−
⋅= eVehco .
Transeverse Relaxation Time T2 Measurement for 0.75 M CuBr2 (90~180 Pulse
Sequence)
y = 161.93e-0.1489x
R2 = 0.9995
0102030405060708090
0 5 10 15 20 25 30
Relaxation TIme (ms)
Am
plitu
de o
f Ech
o (m
V)
0.75 M
Where we get the fit curve is: 43.132
93.161)(τ
τ−
⋅= eVehco . After fitting our experimental curves, we obtained the longitudinal relaxation time and transverse relaxation time for the three different samples:
Concentration of CuBr2 Longitudinal Relaxation Time T1 (ms)
Transverse Relaxation Time T2 (ms)
0.0375 M 390 40.8 0.150 M 275 46.5 0.750 M 30 13.43
Adjustment and tuning of the apparatuses of this experiment is really a challenge job, we try our best to improve the signal, but it turns out that we couldn’t finally make it as good as we desired. Tuning both of the fid curves and echoes need a lot of diligence and patience, Professor Donna Naples (Faculty Member in Dept. of Physics & Astronomy, Univ. of Pittsburgh) also afforded me lots of help which I really feel appreciated, however, unfortunately the whole system seems not working consistently and appropriately. So the best suggestion here is to update the whole experiment set up, and make it work well. Looking at our experiment data, we can see that as the concentration of the salt solvent increases, both of the longitudinal and transverse relaxation times decrease. Since our data are not good, so it is quite difficult for us use it to make a plot of “Relaxation Time vs. Solution Concentration”. The most valuable thing of doing this experiment is to get a chance to review all the physics details in pulsed NMR experiment and give a reasonable analysis and estimation.
