NMR Spectroscopy: Principles and...
Transcript of NMR Spectroscopy: Principles and...
NMR Spectroscopy: Principles and Applications
Nagarajan Murali
Advanced Tools
Lecture 4
Advanced Tools – Quantum Approach
We know by now that NMR is a branch of Spectroscopy and the MNR spectrum is an outcome of nuclear spin interaction with external magnetic field and electromagnetic radiation. This is a subject best tackled by quantum mechanics.
We also know that NMR is a bulk effect and it is indeed very hard to describe NMR of a real sample by quantum theory as it becomes a complex many body problem. But, surprisingly, we can focus on a single spin or few coupled spins and then predict the course of NMR experiment to represent our real systems. The statistical ensemble of nuclear spins behave exactly like the isolated model spin system.
Interaction of Spin with External Magnetic Field
We learnt in Lecture 1, the interaction of nuclear magnetic moment m with external magnetic field B0 is known as Zeeman interaction and the interaction energy known as Zeeman energy is given as:
NMR is a branch of spectroscopy and so it describes the nature of the energy levels of the material system and transitions induced between them through absorption or emission of electromagnetic radiation.
IIIm
B
where
mmBE
BE
I
II
z
,...,1,
2 000
00
0
IBIBμ 00
Resonance
Resonance Method implies that the frequency of irradiation is same as that of the separation of energy levels in frequency unit.
Detection of Resonance is reduced to the measurement of frequency at which there is detectable change in the rate of transition of spin state.
Hamiltonian
In quantum mechanics the energy of a system is an important quantity and is described by an operator with a special name Hamiltonian. Operators are important in that that we observe their magnitude when a suitable experiment is performed to measure them. Such measurable quantities are called observables and the operator corresponding to such quantities are called observable operators. For a single spin the Hamiltonian is given as
The terms that are operators have a “hat” over them. We knew this interaction is called Zeeman energy and this term is called Zeeman term.
00 BIBμH
Hamiltonian
The static field is applied along the z-axis of the laboratory frame. Then the Hamiltonian is reduced to
Only the operator Iz is relevant as the scalar product is non-vanishing for this term. The possible eigen values mI of the Iz operator is –I,-I+1,…,I (i.e. 2I+1 values).
0BzIBIH 0
Zeeman Energy
Thus the interaction energy of a single spin in a static field applied along the z-axis of the laboratory frame is
ImBE 0
Magnetic Field
Energy
1HI=1/2
15NI=1/2
27AlI=5/2
Magnetic Field
Magnetic Field
mI takes (2I+1) values from –I to +I in steps of 1.
Resonance of a single spin I=1/2
For I=1/2, mI can be +1/2 or -1/2. Usually the 1/2 state is referred as a and the -1/2 state as b.
12
1
2
1
2
1
2
1
2
1
2
1
000
00
0
0
0
ba
ba
abba
b
a
m
hBE
BBEEE
BE
BE
mBE I
Resonance of a single spin I=1/2
Quantum mechanics says that a transition is allowed if the change of quantum number m is +1 or -1. So the transition between a to b state is allowed and is called a one quantum or single quantum transition.
12
1
2
1
000
ba
ba
m
hBE
Hamiltonian in Frequency Units
Since energy can be expressed in terms of frequency unit, we can also write the Hamiltonian in frequency unit. Since,
We can write the Hamiltonian as,
12
1
2
1
000
ba
ba
m
hBE
Hzin
or
s radin
0
1-0
z
z
IH
IH
Spin I=1/2 Hamiltonian Summary
m state energy units
Frequency rad / s
Frequency Hz
+1/2 a -(1/2)ħB0 (1/2)0 (1/2)0
-1/2 b (1/2)ħB0 -(1/2)0 -(1/2)0
Two Spins I=1/2 Hamiltonian
m1 m2 state Frequency Hz
+1/2 +1/2 aa (1/2)01+(1/2)02
+1/2 -1/2 ab (1/2)01-(1/2)02
-1/2 +1/2 ba -(1/2)01+(1/2)02
-1/2 -1/2 bb -(1/2)01-(1/2)02
Let us extend this idea of writing Hamiltonian to two spins each with spin I=1/2
Hzin 202101 zz IIH
Two Spins I=1/2 Energy Levels
The energy level diagram could be presented as below (a) Two proton spins and (b) 13C-1H pair.
Hzin 202101, 21mmE mm
HamiltonianTwo Spins I=1/2 and J Coupling
Let us extend this idea of writing Hamiltonian to two spins each with spin I=1/2 and J Coupling between them.
The J coupling is also known as scalar coupling as the operators involved are expressed as a scalar product. Note that there is no applied field dependent term in this interaction – means the coupling value is independent of field strength. The nuclear magnetic moments sense the presence of other spins trough the chemical bond between them.
Hzin 2112202101 IIIIH zz
J
Hzin 21212112202101 zzzz IIIIIIIIH
yyxxJ
1202012112202101 case for the Hzin JJ zzzz IIIIH
Energy LevelsTwo Spins I=1/2 and J Coupling
m1 m2 state Frequency Hz
+1/2 +1/2 aa (1/2)01+(1/2)02+(1/4)J12
+1/2 -1/2 ab (1/2)01-(1/2)02-(1/4)J12
-1/2 +1/2 ba -(1/2)01+(1/2)02-(1/4)J12
-1/2 -1/2 bb -(1/2)01-(1/2)02+(1/4)J12
The energy levels of the two spins each with spin I=1/2 and J Coupling between them is then written as
1202012112202101, case for the Hzin 21
JmmJmmE mm
Energy LevelsTwo Spins I=1/2 and J Coupling
The energy level diagram directly predicts the NMR spectrum.
Left: Energy level diagram; dark arrows show spin 1 flip and its transitions and light arrow show flip of the spin 2 and its transitions. Right: The NMR spectrum resulting from the four transitions.
Energy LevelsTwo Spins I=1/2 and J Coupling
The NMR spectrum has all the information of the energy levels or the Hamiltonian of the system.
Transition Spin states Frequency Hz
12 aaab -02-(1/2)J12
34 babb -02+(1/2)J12
13 aaba -01-(1/2)J12
24 abbb -01+(1/2)J12
A Bit More of Quantum Mechanics
So far we saw that the energy of interaction of nuclear spins with magnetic field and with each other can be expressed by a Hamiltonian that in turn expressed in terms of the various components of the spin angular momentum. With this description we could calculate the transitions. We did not describe how we excite the transition and what happens during the NMR experiment such as one pulse experiment or spin echo experiment.
To understand the time course of a NMR experiment we have to follow the system as it evolves during the experiment. We have to understand how the spin system states are defined in quantum mechanics and how they change in an experiment and how we can observe the spin system state.
Wave Function: The state of the SpinIn quantum theory, we don not know a priori whether a spin (say I=1/2) is in
the a state (+1/2) or in the b state (-1/2). So we write in general the state to be a superposition of the two possible state
The coefficients Ca(t) and Cb (t) are numbers (complex numbers in general) and their phase can vary in time and their magnitudes give rise to the value of the observable quantities in NMR. This function (t) is called a wave function and as it is a complex function we can also write its complex conjugate as
Then
ba ba )()()( tCtCt
)()()( ** tCtCtba
ba
number ajust is )()()()(
)()()()()()()()(
)()()()()()(
**
****
**
tCtCtCtC
tCtCtCtCtCtCtCtC
tCtCtCtCtt
ba
baba
ba
ba
bbaa
ba
bbabbaaa
baba
Properties of Wave FunctionWe can then compute a number as a bra|ket product
In this calculation we have used the fact that the spin states are orthogonal.
number ajust is )()()()(
)()()()()()()()(
)()()()()()(
**
****
**
tCtCtCtC
tCtCtCtCtCtCtCtC
tCtCtCtCtt
ba
baba
ba
ba
bbaa
ba
bbabbaaa
baba
1
0
0
1
bb
ab
ba
aa
Expectation ValuesOnce we define the state of the system by a wave function, we can
now ask what is the chance that we have a component of the spin angular momentum along z-axis and is given by the expectation value of the spin angular momentum operator along z-axis.
Since the denominator is a number we can set it to equal to 1 saying the wave function is normalized, all we have to do is compute
And we will use the fact that the state a and b are eigen states of Iz
)()(
)(ˆ)(
tt
tt zz
II
)(ˆ)( tt zz II
2
1
2
1bbaa zz II
Expectation ValuesThen the expectation value of Iz is
We have used the orthogonal property of the spin states a and b in evaluating the above value.
The above equation means that the average value of the component along z-axis is given by the difference in the probability of finding the spin in the +1/2 state or -1/2 state when many measurements are made.
)()(2
1)()(
2
1)( ** tCtCtCtCtz bbaa I
)()()()(2
1)( ** tCtCtCtCtz bbaa I
Expectation ValuesWe can now see what happens to the spin angular momentum
components in the transverse plane and using the fact that
Again the expectation value for these components also depend on the coefficients that are probability functions.
2
1
2
1abba xx II
2
1
2
1abba ii yy II
)()()()(2
1)( ** tCtCtCtCtx abba I
)()()()(2
1)( ** tCtCtCtCity abba I
Bulk MagnetizationWe are now ready to arrive at the bulk magnetization that is induced
when a collection such individual spins are exposed to a magnetic field. The bulk magnetization along z-axis then sum of the expectation value of the z-component of the individual spins.
The bar above the functions on the right hand side indicates sum over the ensemble of spins.
N
iiiii
N
iizz tCtCtCtCttM
1
**
1
)()()()(2
1)()( bbaa I
)()()()(2
1)( ** tCtCtCtCNtM z bbaa
N
izzzzN
NtM1
1 where)( III
PopulationsThe probability difference that gives component along z- axis now
can be interpreted as population difference in the two states that give rise to the z-magnetization
and
)()(* tCtCNn aaa
ba nntM z 2
1)(
)()(* tCtCNn bbb
Bulk Magnetization – Transverse PlaneWe can in the same way compute magnetization along x-axis and y-
axis.
At equilibrium there is only z-magnetization and no magnetization in the transverse plane. This means that the ensemble average on the right goes to zero.
This is called the random phase approximation or that there is no coherence between the spin states.
)()()()(2
1)( ** tCtCtCtCNNtM xx abba I
)()()()(2
1)()( ** tCtCtCtCNitNtM yy abba I
0)()()()( ** tCtCtCtC abba 0)()()()( ** tCtCtCtC abba
Time Evolution in Quantum MechanicsWe discussed in the vector model how magnetization rotates in the
presence of applied fields (both DC and RF). We wrote the equation of motion as the time derivative of the magnetic moment equal to the torque on the moment. In the same way, the motion of the spins can be expressed in terms of its state undergoing change effected by the interaction Hamiltonian.
Suppose if we consider a single spin I=1/2 in the rotating frame then the Hamiltonian is
And
)()(
tidt
td
H
zIH
ba ba )()()( tCtCt
Time Evolution in Quantum MechanicsThen,
By multiplying on either side by <a| and <b| and using <a| a > = <b| b > = 1, and <a| b > = <b| a > = 0, we can simply separate the above equation into two equations on the coefficients as
)()(
tidt
td
H
ba
baba
ba)()(
)()(tCtCi
dt
tCtCdz
I
baba baba
zz tCitCidt
tdC
dt
tdCII
)()()()(
)()(
and )()(
tCidt
tdCtCi
dt
tdCb
ba
a
aa2
1zI
bb
2
1zI
Time Evolution in Quantum MechanicsThe solution of these two equations is straightforward,
Since we know now the value of the coefficients C’s at any time t we can evaluate the expectation values of the spin angular momentum components in the x, y, and axis
)(2
1)( and )(
2
1)(tCi
dt
tdCtCi
dt
tdCb
ba
a
titieCtCeCtC
2
1
2
1
)0()( and )0()( bbaa
)()()()(2
1)( ** tCtCtCtCtx abba I
)()()()(2
1)( ** tCtCtCtCity abba I
)()()()(2
1)( ** tCtCtCtCtz bbaa I
Time Evolution in Quantum MechanicsLet us just evaluate one of these in detail
titix eCCeCCtCtCtCtCt )0()0(
2
1)0()0(
2
1)()()()(
2
1)( ****
abbaabbaI
titCCtitCC sincos)0()0(2
1sincos)0()0(
2
1 **abba
)0()0(
2
1)0()0(
2
1sin)0()0(
2
1)0()0(
2
1cos ****
baababba CCiCCitCCCCt
)0(sin)0(cos)( yxx ttt III
Time Evolution in Quantum MechanicsSimilar calculations can be done for the other components and in
summary we have
Free evolution does not affect the z-component. The x and y components rotate in the xy plane. These results are same as we got from the vector model.
)0(sin)0(cos)( yxx ttt III
)0(sin)0(cos)( xyy ttt III
)0()( zz t II
Time Evolution of Bulk Magnetization in Quantum Mechanics
We know that the value of the bulk magnetizations are given by their respective expectation values and thus we can compute the state of the magnetization components at any time t.
)0(sin)0(cos)(
)0(sin)0(cos)()(
yxx
yxxx
MtMttM
NtNttNtM
III
)0(sin)0(cos)(
)0(sin)0(cos)()(
xyy
xyyy
MtMttM
NtNttNtM
III
)0()(
)0()()(
zz
yzz
MtM
NtNtM
II
Time Evolution Due to RF Pulse in Quantum Mechanics
So far we worked with a rotating frame Hamiltonian corresponding to just the applied static magnetic field.
Now let us say we have an RF field also along x-axis and for simplicity let us also assume that we are on-resonance (=0). So the new Hamiltonian in the rotating frame in the presence of RF along x-axis is
Where 1 is the amplitude of the RF in units of radians/sec. We can repeat the calculations in the same line as before and the results for the magnetization components can be given by analogy.
zIH
xIH
1
Time Evolution Due to RF Pulse in Quantum Mechanics
The effect of RF along x axis then,
Under x-pulse the magnetization precess in the zy plane as we saw in the vector model. For example if we set 1t=/2 and at t=0 with magnetization only along +z axis, we will end up along –y axis.
)0()(
)0()()(
xx
xxx
MtM
NtNtM
II
)0(sin)0(cos)(
)0(sin)0(cos)()(
11
11
zyy
zyyy
MtMttM
NtNttNtM
III
)0(sin)0(cos)(
)0(sin)0(cos)()(
11
11
yzz
yzzz
MtMttM
NtNttNtM
III
Operator Formalism
We have collected so much knowledge in quantum frame work to describe the spins interacting with the magnetic field, RF, and among themselves and now we see that if we follow the evolution of the components of the spin angular momentum operator, we can predict the course of the magnetization of a spin in any NMR experiment. To facilitate that, we will describe the state of the spin system directly by these spin angular momentum operators. It is then said that we have described the system by a density operator (as opposed to describing the state by a wave function).
For a single isolated spin system, the arbitrary state can be given by the operator (t),
zzyyxx tatatat IIIρ
)()()()(
Operator Formalism
Now we can write the time evolution of this density operator (Liouville-von Neumann equation) to describe the spin system.
Note that in the above the order in which the terms are written is important since
zzyyxx tatatat IIIρ
)()()()(
HρρHρ
)()()( ttitdt
d
titi eet HHρρ
)0()(
HρρH
)()( tt
Operator Formalism – Rotations
Let us say at t=0 ax=1 ay=az=0, then
Also Let us say the Hamiltonian is
Then
The whole calculation can be written in a notation as below
xIρ
)0(
tix
tititi zz eeeetIIHH
Iρρ )0()(
zIH
ttt yx sincos)( IIρ
tt yxt
xz
sincos III
I
)()0( tt
ρρH
Operator Formalism – Rotations
Let us say the Hamiltonian is
Then
Using the notation we described before
titititi xx eeeetIIHH
ρρρ 11 )0()0()(
xIH
1
xt
xx II
I
1
tt zyt
yx
11 sincos 1
IIII
tt yzt
zx
11 sincos 1
IIII
Rotations -Summary
Let us now summarize the rotations under pulses about different axes and flip angles.
zy
xII
I
2
yz
xII
I
- 2
zx
yII
I
- 2
xz
yII
I
2
yyx II
I
-
zzx II
I
-
xxy
III
-
zzy
III
-
Rotations -Summary
Positive rotations are counter-clockwise (as shown) and negative rotation is clockwise.
Operators – Two Spins
We can extend the same representation of the state of the system by operators to two spins I1 and I2.
The two unit operators E1 and E2 do not represent any observable. We need also the products of the spin operators of the two spins to describe the system. Noting that the products with the E operator is the same as the 8 operators above, we have
zyxzyx 22221111 IIIEIIIE
zzyzxz
zyyyx
zxyxxx
y
212121
21212
212121
2 2 2
2 2 2
2 2 2
IIIIII
IIIIII
IIIIII
Hamiltonian for Two SpinsLet us recall the Hamiltonian for two spins I1 and I2
Let us convert the above in to the rotating frame Hamiltonian in angular frequency units.
The time evolution of the two spin system is under the influence of the above Hamiltonian and can be calculated in separate steps.
We have already seen the evolution of single spin operators under the Zeemann term and RF pulses. For two spins the respective spin operators evolve separately with these terms as if they are independent. The evolution under the coupling part of the Hamiltonain is slightly more involved.
1202012112202101 case for the Hzin JJ zzzz IIIIH
122121122211 2 case for the rad/secin 2 JJ zzzz IIIIH
)( )0( 21122211 2t
tJttρρ zzzz IIII
Evolution under Coupling-Two Spins
Let us now illustrate the evolution under the coupling part of the Hamiltonian
This kind of evolution is called a bi-linear rotation as the rotation is about the product of two operators both along the z-direction. Note the second term on the right, it is also a product term in which the state of spin 1 is represented by its y-component which also senses the state of the z-component of the coupled second spin.
2 2112 zz IIH
JJ
tJtJ zyxtJ
x 12211212
1 sin2cos2112
IIII zzII
Evolution under Coupling-Two Spins
We can now illustrate all of such evolutions:
ztJ
z
zxytJ
y
zyxtJ
x
tJtJ
tJtJ
12
1
12211212
1
12211212
1
2112
2112
2112
sin2cos
sin2cos
II
IIII
IIII
zz
zz
zz
II
II
II
ztJ
z
zxytJ
y
zyxtJ
x
tJtJ
tJtJ
22
2
12121222
2
12121222
2
2112
2112
2112
sin2cos
sin2cos
II
IIII
IIII
zz
zz
zz
II
II
II
Evolution -Two Spins
Let us now see the physical meaning of the se terms that we have collected. Let us say at t=0 we have I1x and I2x and see the evolution. As one can see if we follow one spin through the evolution under various parts of the Hamiltonian we can write for the other spin by induction
However, We can detect only the Ix or Iy operators and not the product operators
tJtJtt
tJtJtt
tt
zxytJ
y
zyxtJ
x
yxt
x
122112112
11
122112112
11
11111
sin2cossinsin
sin2coscoscos
sincos
2112
2112
11
IIII
IIII
III
zz
zz
z
II
II
I
Evolution -Two Spins
The observed signal of the spin 1 is thus come from the term
In the observed signal S(t) the x-component is the real part and the y-component is the imaginary part.
tJttJt yx 12111211 cossincoscos II
)exp(expexp2
1
)exp(expexpexp2
1
)exp(expcos
)exp(cossincoscos)(
121121
11212
112
121121
RttJitJi
RtttJitJi
RttitJ
RttJtitJttS
Evolution -Two Spins
Fourier transform S(t) will then give real and imaginary part of the spectrum with lines at the two frequencies in the exponent.
Note the horizontal axis in units of . The multiplets are in same phase with respect to each other in both the real and imaginary parts.
)exp(expexp2
1)( 121121 RttJitJitS
Evolution -Two Spins
Thus the evolution of I1x under a two-spin Hamiltonian gives a in phase doublet spectrum
Thus the operator I1x is called the in-phase x operator.
zzz III
I 211211 21
tJtx
Evolution -Two SpinsSuppose at t=0 if we have the operator 2I1xI2z, then we can calculate
the resulting spectrum in a similar way.
Noting again that we can only detect Ix or Iy components, the detected signal is
tJtJtt
tJtJtt
tt
xzytJ
zy
yzxtJ
zx
zyzxt
zx
121122112
211
121122112
211
12112121
sincos2sin2sin
sincos2cos2cos
sin2cos22
2112
2112
11
IIIII
IIIII
IIIIII
zz
zz
z
II
II
I
)exp(expexp2
1
)exp(expexpexp2
1
)exp(expsin
)exp(sincossinsin)(
121121
11212
112
121121
RttJitJi
RtttJitJii
i
RttitJi
RttJtitJttS
Evolution -Two SpinsUp on Fourier transform the spectrum would be as below,
The mulitiplets are in opposite phase in both the real and imaginary part and so the operator 2I1xI2z is called anti-phase x component of the spin 1.
zzz III
II 211211 2212
tJtzx
Summary of Physical Nature of Operators - Two Spins
x- is absorption y- is absorption
Summary of Physical Nature of Operators - Two Spins
yx
yx
22
2
, 2spin on ion magnetizat -y and - xphase-in
2spin on ion magnetizat-z
, 1spin on ion magnetizat -y and - xphase-in
1spin on ion magnetizat-z
II
I
II
I
z
11
1z
zz
xyyxyyxx
yx
yx
21
21212121
2222
2121
2 population equlibrium-non
,2,22,2 coherences quantum multiple
2,2 2spin on ion magnetizat -y and - xphase-anti
2,2 1spin on ion magnetizat -y and - xphase - anti
II
IIIIIIII
IIII
IIII
zz
zz
Multiple Quantum Coherences
xyyx
yyxx
xyyx
yyxx
2121y
2121x
2121y
2121x
22 )(ZQy - Quantum Zero
22 )(ZQ x - Quantum Zero
22 )(DQy - Quantum Double
22 )(DQ x - Quantum Double
IIII
IIII
IIII
IIII
(1/2)01+(1/2)02+(1/4)J12
(1/2)01+(1/2)02+(1/4)J12
(1/2)01-(1/2)02-(1/4)J12-(1/2)01+(1/2)02-(1/4)J12
0201
0201
(ZQ)
(DQ)
Summary of Physical Nature of Operators - Three Spins
(a) J12=10Hz, J13=2Hz, (b) J12=7Hz, J13=10Hz, (c) J12=5Hz, J13=5Hz