M:/FYS-KJM4740/2008/Solution to exercises-2009.doc
FYS-KJM4740
MR Spectroscopy and Tomography – Part I
Solutions Manual to Exercises presented in the Lecture
Eddy W. Hansen
The new solid-state NMR spectrometer (located at SINTEF Oslo) which will be available for
researchers at KI/UiO/Sintef from spring 2009.
Department of Chemistry
University of Oslo
January 2009
1
Preface
This booklet is designed to serve as a Solutions Manual to the exercises presented in the NMR
course FYS-KJM4740, Part I in “MR Spectroscopy and Tomography”. The major reason for
preparing such a booklet was a sincere request from the students. Also, such an extensive and
detailed collection of solved problems is believed to be of help to students who meet the
challenging world of MR for the first time.
Relative to the first edition which was prepared in 2007 some few corrections have been made in
this second edition. I have no doubt however, that in spite of strenuous efforts, there remain errors
of one sort or another. I will be pleased and grateful to hear from any reader who discovers errors
(eddÿwh@kjemi.uio.no)
Eddy W. Hansen
UiO January 2009
2
Exercise 1.1
A charge q moves in a circular loop with frequency ν. According to classical electromagnetic
theory, a magnetic dipole moment μ is generated, given by;
μ = i⋅ A
(1)
where i represents the current and A the area enclosed by the circular loop (of radius R). Hence:
i = qυ =
qυ⋅ 2π qω
=
2π
2π
and A=πr2
⇒
1
2
μ = qωr2
(2)
From the definition of the angular momentum (L); L = R x mv, we obtain;
L = r.mv.sinθ = rmvsin(π/2) = rmv = rm.ω r = mωr2 (see Figure 1)
(3)
Combining Eqs 1 and 3;
μ=
q
L = γL
2m
(4)
3
Exercise 1.2
We start by differentiating the angular momentum L with respect to time;
r
r
r
r s r
r dr
r r d (mv) r
r r r
dL d r
= [r × mv] =
× mv + r ×
= v × mv + r × F = 0 + r × F = τ
dt dt
dt
dt
(1)
The symbol F represents the force and τ represents the torque (dreiemoment).
From classical physics we know that a magnetic dipole moment (μ) within a magnetic field B
experiences a torque, given by;
τ=μxB
(2)
From the previous exercise;
μ = γL
(3)
By combining Eqs 1 – 3, we obtain;
dμ
= γμ × B
dt
(4)
Solution to Eq 4
Let us choose B = B0k. Upon inserting this into Eq. 4, we obtain;
i
dμ y
dμ x
dμ z
k = γ μx
i+
j+
dt
dt
dt
0
j
μy
0
k
μ z = γμ y B0 i − γμ x B0 j + 0k
B0
Resulting in the following 3 equations:
dμ x
= γμ y B0
dt
dμ y
= −γμ x B0
dt
dμ z
=0
dt
(5a )
(5b)
(5c)
If multiplying Eq. 5b by i ( = − 1 ) and adding to Eq. 5a, the following result appears;
4
(5)
d ( μ x + iμ y )
dt
= −iγ ( μ x + iμ y ) B0
(6)
Since we can always write the complex number μx + iμy on the form;
μ x + iμ y = μ⊥eiωt
( = μ⊥ cos(ωt ) + i μ⊥ sin(ωt ))
(7)
We obtain, by inserting Eq.7 into Eq. 6;
iωμ⊥ = −iγμ⊥ B0
⇔
ω = −γB0
(8)
Equation 8 (right) represents the basic NMR equation, or the Larmor equation and shows that the
magnetic moment rotates clockwise around the static magnetic field with a frequency ω. The
component of the magnetic moment along the z-axis is constant and independent on time (Eq. 5c).
5
Exercise 1.3
We will tentatively assume the spin population ratio N-/N+ between the two energy levels shown
in the Figure follows a Boltzmann distribution, i.e.;
N−
hγB
⎡ ΔE ⎤
⎡ hγB ⎤
= exp ⎢−
= exp ⎢−
≈1−
⎥
⎥
N+
kT
⎣ kT ⎦
⎣ kT ⎦
The last term is obtained by a Taylor expansion (we have previously shown that
(1)
hγB
<<1)
kT
Energy diagram
NΔE= ħω = ħγΒ (where we have used ω = γΒ in the last term)
N+
We also have:
N+ + N-= N0
Where N0 is the total number of spins (or NMR active nuclei) in the sample.
(2)
Combining Eqs 1 and 2 we obtain;
N
N ⎡ hγB ⎤
1
= 0 ⎢1 +
N+ = 0 ⋅
2 1 − hγB / 2kT
2 ⎣ 2kT ⎥⎦
The last term in Eq 3a is obtained by noting that:
1
=1+ x
for
x << 1
1− x
Substituting Eq 3a into Eq 2 we obtain:
N 0 ⎡ hγB ⎤
(3b)
The difference in the number of spins (Δn) between the two energy levels is therefore:
γhBN 0
(3c)
N− =
1−
2 ⎢⎣ 2kT ⎥⎦
Δn = N + − N − =
2kT
The observable, macroscopic magnetization M for a spin-1/2 particle is thus;
M = μΔn =
6
(γh ) 2
BN 0
4kT
(3a)
Exercise 1.4
We simply replace B0 in Eq. 8 (see Exercise 1.2) by Beff = ωeff/γ = B0 + ω/γ and obtain the
following solution for the magnetic moment (μ) in the xy-plane;
μ = μ⊥e
iω eff t
= μ ⊥ cos(ω eff t ) + i μ ⊥ sin(ω eff t )
If on resonance, i.e.; ωeff = 0 (ω/γ = − B0 = ω0/γ), we obtain the solution;
μ = μ⊥e
iω eff t
= μ⊥ + i 0
This means that the magnetic dipole is located along the x’-axis in the rotating frame of reference.
In this frame, the magnetic dipole remains constant and independent of time.
If not on resonance, the magnetic dipole (in the rotating frame of reference;
ω0 = γB0) will have the following components;
μ = μ⊥e
iω eff t
= μ ⊥ cos(ω − ω 0 )t + i μ ⊥ sin(ω − ω 0 )t
or;
μ x' = μ ⊥ cos(ω − ω 0 )t
μ y ' = μ ⊥ sin(ω − ω 0 )t
1,5
"on resonance"
magnetization
1
0,5
0
-0,5
-1
"off-resonance"
-1,5
0
2
4
6
8
10
tim e
Normally, the experiment is set up to observe the signal along x’ (rotating frame)..
7
Exercise 1.5
We consider the motion in the rotating frame of reference, on resonance. This means that
i'
dμ y '
dμ x '
dμ z '
j+
k = γ μx
i+
dt
dt
dt
B1
dμ x'
=0
dt
dμ y '
= γμ z ' B1 = ω1μ z '
dt
dμ z
= −γμ y ' B1 = ω1μ y '
dt
j'
k
μy
μ z = 0i'+γμ z ' B1 j '−γB1μ y ' k
0
0
(5a )
(5b)
(5c)
We can easily see that the following solution satisfies the differential equations (by insertion);
μ y ' = μ 0 sin(ω1t )
μ z ' = μ 0 cos(ω1t )
The magnetic moment is rotating around the x’-axis with frequency ω1 = γB1
8
Exercise 2.1
dM'/dt = γM' x Beff – u/T2 – v/T2 + (Mo-Mz)/T1
Beff = B1i' + (B – ωref/γ)k = (ω1/γ)i'+((ω-ωref)/γ)k
dM z
du
dv
i′ +
j′ +
k =γ
dt
dt
dt
i'
j'
u
ω1 / γ
v
0
dt
(2)
k
M −Mz
u
v
Mz
i′ −
j′ + 0
k
−
T2
T2
T1
(ω 0 − ω ref ) / γ
du
u
= (ω − ω ref ) v −
dt
T2
dM 'y
(1)
(3)
(3a )
= −(ω − ω ref )u + ω1M z −
v
T2
(3b)
M −Mz
dM z
= −ω1 v + 0
dt
T1
(3c)
Case 1
On resonance (ω = ωref)
du
u
=−
dt
T2
(3a )
dv
v
= ω1M z −
dt
T2
M −Mz
dM z
= −ω1M z + 0
dt
T1
(3b)
(3c)
9
Exercise 2.2
Using the last set of equations in Exercise 2.1 with ω1 = 0, we obtain;
du
u
=−
dt
T2
(3a )
dv
v
=−
dt
T2
(3b)
M −Mz
dM z
=+ 0
dt
T1
(3c)
Solution with initial constraints v(0) = M0, u(0) = 0 and Mz(0) = 0
u du
t dt
du
u
du
dt
=−
⇔
=−
⇔ ∫
= −∫
⇔ u = u (0) ⋅ exp[− t / T2 ] = 0
dt
T2
u
T2
u
T2
u (0)
0
v dv
t dt
dv
v
dv
dt
=−
⇔
=−
⇔ ∫
= −∫
⇔ v = v(0) ⋅ exp[− t / T2 ] = M 0 ⋅ exp[− t / T2 ]
dt
T2
v
T2
v
T
v(0)
0 2
M z'
Mz
t dt
dM z M 0 − M z
dM z
dM x'
dt
=
⇔
=
⇔ ∫
=∫
⇔ − ln[M 0 − M z ]
= [t / T1 ]
dt
T1
T
0
M 0 − M z T1
M z (0) M 0 − M z 0 1
M z = M 0 ⋅ [1 − exp(−t / T1 )]
Schematics outline of the experiment.
10
Exercise 2.3
Initial conditions. You first apply an rf-pulse (B1) such that Mz(0) = - M0.
According to Exercise 2.2, we may write;
M z'
Mz
t dt
dM x'
dM z M 0 − M z
dM z
dt
=
⇔
=
⇔
=∫
⇔ − ln[M 0 − M z ]
= [t / T1 ]
∫
dt
T1
T
−M 0
M 0 − M z T1
− M z (0) M 0 − M z 0 1
[
M z = M 0 ⋅ 1 − 2 exp(−t / T1)
]
11
Exercise 2.4
2τ c
1
∝ J (ω ) =
T1
1 + ω 2τ c2
(1)
We differentiate Eq. 1 with respect to τc and obtain;
2 − 2ω 2τ c2
d (1 / T1 )
=
2
dτ c
⎡1 + ω 2τ 2 ⎤
⎢⎣
⎥⎦
(2)
We set Eq 2 equalt to 0 and obtain;
τc =
1
ω
We calculate the second derivative of 1/T1 and obtain:
d 2 (1 / T1 )
f (ω ; τ c ) =
=−
dτ c2
4ω 2τ c ⎡3 − ω 2τ c2 ⎤
⎥⎦
⎢⎣
3
⎡1 + ω 2τ 2 ⎤
⎢⎣
⎥⎦
(3)
(4)
f (ω ; τ c = 1 / ω ) = −ω < 0
From Eq. 4 we conclude that 1/T1 has a maximum at τ c =
τc =
1
ω
1
ω
, i.e., T1 has a minimum at at
(5)
Note; When increasing the magnetic field strength (increasing ω), the minimum in T1 shifts to
smaller correlation time (Eq 5), i.e., to higher temperature. From Eq 1 we see that the minimum in
T1 increases with increasing temperature.
12
Exercise 2.5
13
Exercise 2.6
dM'/dt = γM' x Beff – u/T2 – v/T2 + (Mo-Mz)/T1
Beff = B1i'+(B + ωref/γ+gz)k = (ω1/γ)i'+((ω + ωref + gz)/γ)k
(1)
(2)
Setting B1 = 0 and ω = -ωref (on resonance)
i' j ' k
dM z
du
dv
u
v M0 −M z
i+
j+
−
+
k = γ u' v M z −
dt
dt
dt
T2 T 2
T1
0 0 gz
du
u
= γgzv −
dt
T2
dv
v
= −γgzu −
dt
T2
(3a )
(3b)
M −Mz
dM z
=γ 0
dt
T1
(3c)
Concerning the transversal magnetization, we multiply Eq. 3b with i (= − 1) and add this to Eq. 3a
to obtain M ⊥ = u + iv :
dM ⊥
= (γgzi − 1 / T2 ) = α ( z )i − 1 / T2
dt
(4)
α(z)t = γgzt represents the phase angle and increases with increasing z. The solution to Eq 4 can
be easily found:
M ⊥ = M 0 eα ( z )ti ⋅ e −t / T2
14
Exercise 2.7
Energy
Energy
⇓
N↓
W↓↑
⇑
N↑
W↑↓
Ni: Number of spins in level i
Wij: Transition probability from level I to j
From general rate-process analysis we may write:
dN ↓
dt
dN ↑
dt
= −W↓↑ N ↓ + W↑↓ N ↑
(1a)
= −W↑↓ N ↑ + W↓↑ N ↓
(1b)
Subtracting Eq 1a from Eq 1b gives:
d (N↑ − N↓ )
= −2W↑↓ N ↑ + 2W↓↑ N ↓
dt
(2)
We introduce the following parameters:
ΔN = N ↑ − N ↓
N = N↑ + N↓
(3a)
(3b)
Inserting Eqs 3a and 3b into Eq 2 gives:
d ( ΔN )
= −W↑↓ (ΔN + N ) + W↓↑ ( N − ΔN )
dt
(4)
15
At equilibrium, the left hand side of Eq 4 must be equal to 0 (
N = ΔN 0
W↓↑ + W↑↓
W↓↑ − W↑↓
d ( ΔN )
= 0 ), i.e.;
dt
(5)
Inserting Eq 5 into Eq 4 results in:_
d (ΔN )
= −(W↑↓ + W↓↑ )(ΔN − ΔN 0 )
dt
(6)
The spin-lattice relaxation rate 1/T1 is defined as the sum of the transition probabilities, i.e.;
1
= W↑↓ + W↓↑
T1
(7)
Likewise, we recall that the magnetization is proportional to the difference in spin population, i.e.;
M 0 = kΔN 0
M = kΔN
(8)
Inserting Eqs 7 and 8 into Eq 6 gives:
M − M0
dM z
=− z
dt
T1
Eq 9 is identical to the corresponding Equation presented by the Bloch Equation !!!
16
(9)
Exercise 3.0
17
Exercise 3.1 – Some introductory remarks and comments
From basic quantum mechanics one may show that for a nucleus of spin I, a number of (2I+1)
different spin functions exist. These functions are simply denoted:
|I,m> for m = -I, -I + 1, .., 0, … I – 1, I
In particular, for I = ½ (1H, 13C, 31P, 19F,…) we have only two spin functions denoted,
respectively:
|α> = |1/2,1/2>
and |β> = |1/2, -1/2>
Again, from basic quantum mechanics the following operator properties may be defined
Īz|α> = 1/2|α>
Īz|β> = –1/2|β>
(3.1a)
It is frequently useful to apply the so called shift operators, or the “raising” ( Iˆ + ) and “lowering”
( Iˆ − ) operators, respectively:
Iˆ ± I , m = [I ( I + 1) − m( m ± 1)]1 / 2 I , m ± 1
Hence:
Iˆ + α = Iˆ + 1 / 2,1 / 2 = 0
Iˆ − α = Iˆ − 1 / 2,1 / 2 = 1 ⋅ [1 / 2,−1 / 2 ) = (β )
Iˆ + β = Iˆ + 1 / 2,−1 / 2 = 1 ⋅ 1 / 2,1 / 2 = α
Iˆ − β = 0
We define:
Iˆ x = 1 / 2 ⎡ I + + I − ⎤
⎢⎣
⎥⎦
Iˆ y = −i / 2 ⎡ I + − I − ⎤
⎢⎣
⎥⎦
Hence;
Īx|α> = 1/2|β>
Īy|α> = i/2|β>
Īx|β> = 1/2|α>
Īx|β> = -i/2|α>
(3.1b)
(3.1c)
What is the classical energy (E) of two interacting magnetic dipoles (μA and μB) in a magnetic
field B0?
r
r
r
r
r
v
E = − μ A ⋅ B0 − μ B ⋅ B0 − μ A ⋅ μ B
18
The corresponding quantum mechanical energy operator H is:
ˆ = −γhIˆ B − γhIˆ B − γhIˆ ⋅ γhIˆ
Η
zA 0
zB 0
zA
zB
= − ω 0 hIˆ zA − ω 0 hIˆ zB − γ 2 h 2 Iˆ zA ⋅ Iˆ zB
(ergs)
= −ν A Iˆ zA −ν B Iˆ zB − J AB Iˆ zA ⋅ Iˆ zB
( Hz )
The “constant” JAB is denoted the coupling constant.
Uttrykker produktet av spinnoperatorene IA og IB ved operatorene I+ og I-:
[
][
Iˆ A ⋅ IˆB = Iˆ XA i + IˆYA j + IˆZA k ⋅ IˆXB i + IˆYB j + IˆZB k
]
Iˆ A ⋅ IˆB = IˆXA Iˆ XB + IˆYA IˆYB + IˆZA IˆZB
[
= 1 / 4[Iˆ
= 1 / 2[Iˆ
][
]
[
][
]
Iˆ A ⋅ IˆB = 1 / 4 Iˆ A+ + Iˆ A− ⋅ IˆB+ + IˆB− − 1 / 4 Iˆ A+ − Iˆ A− ⋅ IˆB+ − IˆB− + IˆZA IˆZB
Iˆ A ⋅ IˆB
Iˆ A ⋅ IˆB
]
+
A
⋅ IˆB+ + Iˆ A+ ⋅ IˆB− + Iˆ A− ⋅ IˆB+ + Iˆ A− ⋅ IˆB− − Iˆ A+ ⋅ IˆB+ + Iˆ A+ ⋅ IˆB− + Iˆ A− ⋅ IˆB+ − Iˆ A− ⋅ IˆB− + IˆZA IˆZB
+
A
⋅ IˆB− + Iˆ A− ⋅ IˆB+ + IˆZA IˆZB
]
In short, we may write the Hamiltonian for a two-spin system as:
[
]
ˆ = −υ Iˆ − ν ⋅ Iˆ + J / 2 Iˆ + ⋅ Iˆ − + Iˆ − ⋅ Iˆ + + J Iˆ Iˆ
Η
A ZA
B
ZB
AB
A
B
A
B
AB ZA ZB
19
Exercise 3.2
The spin-functions |α> and |β> are defined as orthonormal eigenfunctions of the z component of
the spin operator Iz (see Eq 3.1) and satisfy the following equations:
∫ α ( X ) α ( X ) dτ = ∫ ααdτ = ∫ β ( X ) β ( X ) dτ = ∫ ββ dτ = 1
∫ α ( X ) β ( X ) dτ = ∫ αβ dτ = ∫ β ( X ) α ( X ) dτ = ∫ βα dτ = 0
(3.1a)
(3.1b)
X refers to the actual nucleus in question. For a two-spin system (X = A, B) we may define 4
product functions Θ i (i = 1 − 4) ;
Θ1 = α ( A)α ( B) = αα
Θ 2 = α ( A) β ( B) = αβ
Θ 3 = β ( A)α ( B ) = βα
Θ 4 = β ( A) β ( B ) = ββ
(3.2a)
(3.2b)
(3.2c)
(3.2d)
Since these spin-functions (Θ i ) are orthonormal (see Eq 3.1), we may construct a set of
orthonormal eigenfunctions ( Ψi ), defined as a linear combination of Θ i , i.e.;
Ψ1 = C11 Θ1 + C12 Θ 2 + C13 Θ 3 + C14 Θ 4
Ψ2 = C 21 Θ1 + C 22 Θ 2 + C 23 Θ 3 + C 24 Θ 4
Ψ3 = C 31 Θ1 + C 32 Θ 2 + C 33 Θ 3 + C 34 Θ 4
Ψ4 = C 41 Θ1 + C 42 Θ 2 + C 43 Θ 3 + C 44 Θ 4
(3.3a)
(3.3b)
(3.3c)
(3.3d)
Since these equations are eigen-functions to the Hamiltonian, the following equations must be
valid for each i (=1 – 4):
ˆ Ψ =E Ψ
Η
i
i
ˆ
ˆ Θ +C Η
ˆ
ˆ
= C i1Η Θ1 + C i 2 Η
i3 Θ 3 + C i 4 Η Θ 4
2
= EC i1 Θ1 + EC i 2 Θ 2 + EC i3 Θ 3 + EC i 4 Θ 4
We multiply by Θ j and integrate over the entire spin space:
20
(3.4)
∫ Θ j Ηˆ Ψi dτ = E ∫ Θ j Ψi
c
∫ C i1 Θ j Ηˆ Θ1 dτ + ∫ C i 2 Θ j Ηˆ Θ 2 dτ + C i3 ∫
= EC i1 ∫
ˆ Θ dτ + C
Θj Η
3
i4 ∫
ˆ Θ dτ
Θj Η
4
Θ j Θ1 dτ + E ∫ C i 2 Θ j Θ 2 dτ + EC i3 ∫ Θ j Θ 3 dτ + EC i 4 ∫ Θ j Θ 4 dτ
c
C i1 H j1 + C i 2 H j 2 + C i3 H j 3 + C i 4 H j 4
= EC i1δ j1 + EC i 2δ j 2 + EC i3δ j 3 + EC i 4δ j 4
For j = 1 to 4, the following set of linear equations arise:
C i1 H 11 + C i 2 H 12 + C i3 H 13 + C i 4 H 14 = EC i1
C i1 H 21 + C i 2 H 22 + C i3 H 23 + C i 4 H 24 = EC i 2
C i1 H 31 + C i 2 H 32 + C i3 H 33 + C i 4 H 34 = EC i3
C i1 H 41 + C i 2 H 42 + C i3 H 43 + C i 4 H 44 = EC i 4
(3.4)
Eq 3.4 can be formulated in matrix notation, i.e.;
⎛ H 11 − E
⎜
r
⎜ H 21
ˆ
H ⋅C = 0 ⇔ ⎜
H 31
⎜
⎜ H
41
⎝
H12
H 22 − E
H 32
H 13
H 23
H 33 − E
H 42
H 43
⎞ ⎛ C i1 ⎞
⎟
⎟ ⎜
⎟ ⎜ Ci 2 ⎟
⎟⋅⎜C ⎟ = 0
⎟ ⎜ i3 ⎟
H 44 − E ⎟⎠ ⎜⎝ C i 4 ⎟⎠
H 14
H 24
H 34
(3.5)
r
A non-trivial solution (C ≠ 0) to Eq 3.5 exists only and only if the determinant of Ĥ is
identical to 0, i.e.:
⎛ H 11 − E
⎜
⎜ H 21
⎜ H
31
⎜
⎜ H
41
⎝
H 12
H 22 − E
H 13
H 23
H 32
H 42
H 33 − E
H 43
⎞
⎟
⎟
=0
H 34 ⎟
⎟
H 44 − E ⎟⎠
H 14
H 24
(3.6)
Which is equivalent to Eq 27.
We will not calculate all the terms Hpq (we leave this to the student!) but we illustrate how
this can be performed (see Exersice 3.0):
⎡ ˆ+ ˆ− ˆ− ˆ+ ⎤
ˆ
ˆ ˆ
ˆ Θ = −υ Iˆ
Η
1
A ZA Θ1 −ν B ⋅ I ZB Θ1 + J AB / 2 ⎢⎣ I A ⋅ I B + I A ⋅ I B ⎥⎦ Θ1 + J AB I ZA I ZB Θ1
ˆ α ( A)α ( B ) = −υ Iˆ
Η
A
ZA
α ( A)α ( B) −ν B ⋅ IˆZB α ( A)α ( B) +
− α ( A)α ( B ) + J
ˆ− ˆ+
ˆ ˆ
J AB / 2 Iˆ + ⋅ Iˆ B
AB / 2 I A ⋅ I B α ( A)α ( B ) + J AB I ZA I ZB α ( A)α ( B )
A
21
ˆ α ( A)α ( B ) = − 1 υ α ( B ) α ( A) − 1 ν ⋅ α ( A) α ( B ) +
Η
A
B
2
2
J AB
J
1
1
⋅ 0 ⋅ β ( B) + AB β ( A) ⋅ 0 + J AB ⋅ α ( A) ⋅ α ( B )
2
2
2
2
ˆ Θ = (−υ / 2 −ν / 2 + J
Η
1
A
B
AB / 4) φ1 = E1 Θ1 1
(3.7 a )
Likewise:
ˆ Θ = (−υ / 2 + ν / 2 − J
Η
2
A
B
AB / 4) Θ 2 + J AB / 2) Θ 3
(3.7b)
ˆ Θ = (υ / 2 −ν / 2 − J
Η
3
A
B
AB / 4) Θ 3 + J AB / 2) Θ 2
(3.7c)
ˆ Θ = (υ / 2 + ν / 2 + J
Η
4
A
B
AB / 4) Θ 4 = E 4 Θ 4
4
(3.7 d )
We easily see that H12 = H21 =0, H13 = H31 = 0, H14 = H41 = 0, H42 = H24 = 0, H43 = H34 = 0
because ∫ Θ1 Θ 2 dτ = ∫ Θ1 Θ 3 dτ = ∫ Θ1 Θ 4 dτ = ∫ Θ 2 Θ 4 dτ = ∫ Θ 3 Θ 4 dτ = 0
22
Exercise 3.3
Because two of the spin-functions (Θ1 and Θ4) are eigenfunctions while Θ2 og Θ3 are not.
23
Exercise 3.4
If we consider the total z-component of our spin-operator (two-spin system AB), as defined by:
Fˆ z = Iˆ zA + Iˆ zB
(3.7)
we notice that:
Fˆ z Θ1 = Iˆ zA Θ1 + Iˆ zB Θ1
= Iˆ zA α ( A)α ( B ) + Iˆ zB α ( A)α ( B )
= 1 / 2 α ( A)α ( B) + 1 / 2 α ( A)α ( B)
= 1 ⋅ α ( A)α ( B)
= 1 ⋅ Θ1
Fˆ z Θ 2 = Iˆ zA Θ 2 + Iˆ zB Θ 2
= Iˆ zA α ( A) β ( B ) + Iˆ zB α ( A) β ( B)
= 1 / 2 α ( A) β ( B) − 1 / 2 α ( A) β ( B)
= 0 ⋅ α ( A) β ( B) .
= 0⋅ Θ2
Fˆ z Θ 3 = Iˆ zA Θ 3 + Iˆ zB Θ 3
= Iˆ zA β ( A)α ( B ) + Iˆ zB β ( A)α ( B)
= − 1 / 2 β ( A) β ( B) + 1 / 2 β ( A)α ( B )
= 0 ⋅ β ( A)α ( B)
= 0 ⋅ Θ3
Fˆ z Θ 4 = Iˆ zA Θ 4 + Iˆ zB Θ 4
= Iˆ zA β ( A) β ( B) + Iˆ zB β ( A) β ( B)
= − 1 / 2 β ( A) β ( B) − 1 / 2 β ( A) β ( B )
= −1 ⋅ β ( A) β ( B )
= −1 ⋅ Θ 4
This means that the eigenvalues Fz of the operator F̂z are the same for the two spin-functions
Θ 2 and Θ 3 , implying that a linear combination of these two functions will define the actual
eigenfunction
24
Exercise 3.5
Two find the two remaining energies, we must solve the matrix equation:
⎛ H 22 − E
⎜⎜
⎝ H 32
H 23 ⎞ ⎛ C i 2 ⎞
⎟⋅⎜
⎟=0
H 33 − E ⎟⎠ ⎜⎝ C i3 ⎟⎠
A non-trivial solution exists if and only if the secular determinant is 0, i.e.:
H 22 − E
H 32
H 23
=0
H 33 − E
(3.8)
Using Eqs. 7a-d, we can calculate Hij, i.e.:
1
1
1
1
H 23 = Θ 2 Hˆ Θ 3 = Θ 2 Hˆ ( ν A − ν B − J AB ) Θ 3 + J AB Θ 2
2
2
2
4
=
1
J AB
2
(3.9a)
Likewise, we derive the following results:
1
1
(ν A −ν B ) − J AB
2
4
1
1
H 33 = (ν A −ν B ) − J AB
2
4
H 22 = −
(3.9b)
(3.9c)
Inserting Eqs. 3.9a–c into Eq 3.8 gives:
J
1
E 2 = − AB −
J 2 + (ν A −ν B ) 2
AB
4
2
J
1
E 3 = − AB +
J 2 + (ν A −ν B ) 2
AB
4
2
(3.10)
25
Exercise 3.6
If introducing the following short hand notations:
V = ν A +ν B
C = J 2 + (υ A − υ B ) 2
AB
we obtain from Exercises 3.1 and 3.4 the following energy level diagrams:
Table 1. Energy levels and corresponding “allowed” transitions
Level
Energy
Fz Transition
Wave function
1
V/2 + J/4
-1
* *
|Ψ1> = a11|Θ1>
* * *
2
C/2-J/4
0
|Ψ2> = a21|Θ2> + a22|Θ3>
*
*
*
3
-C/2-J/4
0
|Ψ3> = a31|Θ2> + a32|Θ3>
*
*
4
-V/2+J/4
1
|Ψ4> = a44|Θ4>
Hence, the following transition may be easily derived:
ΔE1→2 = E1 – E2 = (V – C + J)/2
ΔE1→3 = E1 – E3 = (V + C + J)/2
ΔE2→4 = E2 – E4 = (V – C + J)/2
ΔE3→4 = E3 – E4 = (V – C - J)/2
26
Exercise 3.7
In order to determine the eigenfunctions (Table 1) we must determine the constants aij (Table 1)
This can be easily performed by noting that these functions are orthonormal, i.e.:
∫ Ψi Ψ j dτ = δ ij
(= 1
if
i = j,
=0
if
i ≠ j)
One may easily show that this results in the following equations (show this!)
|Ψ1> = |Θ1>
|Ψ2> = cosθ|Θ2> + sinθ|Θ3>
|Ψ3> = -sinθ|Θ2> + cosθ|Θ3>
|Ψ4> = |Θ4>
Vi skal bestemme intensitetet (overgangssannsynligheten) Im=-1 ⇒ m= 0 mellom nivå 3 og 4 og
benytter resultatet fra kvantemekanikken;
[
]
I m = −1→m =0 = ( ∫ψ m = −1 Iˆ−A + Iˆ−B ψ m =0 dτ ) 2
Vi beregner først;
Iˆ−Aψ m =0 = Iˆ−A (cosθ ⋅ α ( A) β ( B ) + Iˆ−A sin θ ⋅ β ( A)α ( B))
= cosθ ⋅ β ( B) Iˆ Aα ( A) + sin θ ⋅ α ( B ) Iˆ A β ( A)
−
−
= cosθ ⋅ β ( B) β ( A) − sin θ ⋅ α ( B ) ⋅ 0
= cosθ ⋅ β ( B) β ( A)
Tilsvarende finner vi for;
Iˆ−Bψ m =0 = Iˆ−B (cosθ ⋅ α ( A) β ( B ) + Iˆ−B sin θ ⋅ β ( A)α ( B))
= cosθ ⋅ α ( A) Iˆ B β ( A) + sin θ ⋅ β ( A) Iˆ Bα ( B)
−
−
= cosθ ⋅ α ( A) ⋅ 0 − sin θ ⋅ β ( A) ⋅ β ( B )
= − sin β ( A) β ( B)
Innsatt i første likning;
27
[
]
I m = −1→m =0 = ( ∫ψ m = −1 Iˆ−A + Iˆ−B ψ m =0 dτ ) 2
= ( ∫ β ( A) β ( B ) ⋅ [cosθ ⋅ β ( B) β ( A) − sin θ ( A) β ( B)]dτ ) 2
= (cosθ ∫ β ( A) β ( A)dτ A ⋅ ∫ β ( B ) β ( B )dτ B − sin θ ∫ β ( A) β ( A)dτ A ⋅ ∫ β ( B) β ( B )dτ B ) 2
= (cosθ − sin θ ) 2 = cos 2 θ − 2 sin θ cosθ + sin 2 θ
= 1 − sin(2θ )
28
Exercise 4.1
r
v
If introducing a gradient field g along any direction r in space we obtain:
v ∂B r ∂B y r ∂B z r
g= xi+
j+
k
∂y
∂z
∂x
r rr r r
r = x i + yj + zk
(1)
(2)
Without loss of generality we consider the component of the gradient field only in the direction of
the external field, i.e., along the z-direction. Hence,
r r r
∂B
g ⋅ r = g ⋅ k = z z = z ⋅ g 0 (t )
∂z
(3)
Eq 3 implicitly assumes that the field gradient ∂B z / ∂z is constant (= g0), and hence independent
on the space-coordinates). This implies that the total magnetic field B along the z-axis is:
B z = B0 + z ⋅ g 0 (t )
(4a)
Hence on resonance,
the following magnetic field exists within the rotating frame of reference:
r
r
(4b)
Beff = zg 0 k
Again, within the rotating frame of reference:
i'
dM z
dv
du
j′ +
k =γ u
i′ +
dt
dt
dt
0
j'
v
0
k
M −Mz
v
u
2
j′ + 0
k + D∇ M
Mz −
i′ −
T2
T1
T2
zg 0
∂u
u
= γg 0 z ⋅ v −
+ D∇ ⋅ ∇u
∂t
T2
(5a)
∂v
v
= γg 0 z ⋅ u −
+ D∇ ⋅ ∇v
∂t
T2
(5b)
M − M0
∂M z
=− z
+ D∇ ⋅ ∇M z
∂t
T1
(5c)
Since we are interested only in the transversal magnetization (uv-plane), we will apply
a “complex-number-approach”, by introducing the complex magnetization M̂ defined by:
)
(6)
M = M x + iM y
Multiplying Eq 5b by the complex number i and adding Eq 5a, we obtain:
)
)
) M
)
∂M
= −iγg 0 zM −
+ D∇ 2 M
T2
∂t
(7)
29
Exercise 4.2
From Eq 7:
)
)
) M
)
∂M
= −iγg 0 zM −
+ D∇ 2 M
∂t
T2
(7)
Noting that Bz may be a function of both z and t (Eq 4) we will look for a solution in which also
M̂ is a function of z and t and independent of x and y. This implies that:
)
) ∂2M
∇ ⋅ ∇M =
∂z 2
(8)
By inserting Eq 9 into Eq 8, we easily derive Eq 10:
)
)
M ( z, t ) = e −t / T2 m( z, t )
−
)
∂m
1 − t / T2 )
)
e
m + e − t / T2
= −iγzg 0 e − t / T2 m − e − t / T2
T2
∂t
(9)
)
)
m
∂ 2m
+ e − t / T2 D
T2
∂z 2
(10)
Which simplifies to:
)
)
∂m
∂ 2m
)
= −iγzg 0 m + D
∂t
∂z 2
30
(11)
Exercise 4.3
We will try to find a solution to Eq 11 of the form:
)
m = Ψ (t ) ⋅ Ω( z , t )
(12)
t
⎡
⎤
Ω( z, t ) = exp ⎢− iγz ∫ g o (t ' )dt '⎥
0
⎣
⎦
(13)
where:
Since Ω( z , t ) is a complex function of modulus 1, it simply represents a phase shift of the
)
observable (and complex) magnetization m . Hence, the modulus (absolute value) of the
)
complex magnetization m will be identical to Ψ (t ) (see Eq 14) and thus represents the
observable signal intensity.
)
abs (m) = abs(Ψ (t ) ⋅ Ω( z , t )) = Ψ (t )abs(Ω( z , t )) = Ψ (t ) ⋅ Ω( z , t ) ⋅ Ω* ( z , t ) = Ψ (t ) ⋅ 1 = ψ (t )
*; complex conjugate
(14)
Noting that:
)
1 ∂Ψ ⎤
∂m
∂Ω
∂Ψ
Ψ ∂Ψ ) ⎡
=ψ
+Ω
= ΨΩ ⋅ [− iγzg 0 (t )] + Ω
= m ⋅ ⎢− iγzg 0 (t ) +
∂t
∂t
∂t
Ψ ∂t
Ψ ∂t ⎥⎦
⎣
)
t
⎤
⎡
∂m
∂Ω
=ψ ⋅
= Ψ ⋅ Ω ⋅ ⎢− iγ ∫ g 0 (t ' )dt '⎥
∂z
∂z
0
⎦
⎣
)
∂ 2m
∂ 2Ω
2
t
t
⎡
⎤
⎤
)⎡
=Ψ
= ΨΩ ⋅ ⎢− iγ ∫ g 0 (t ' )dt '⎥ = m ⎢− iγ ∫ g 0 (t ' )dt '⎥
2
2
∂z
∂z
0
0
⎣
⎦
⎣
⎦
(15a)
(15b)
2
(15c)
We obtain the following simple Eq for ψ when substituting Eqs 15a) –c) into Eq 11:
31
2
t
⎡
⎤
⎡t
⎤
1 dΨ
= D ⎢− iγ ∫ g 0 (t ' )dt '⎥ = − Dγ 2 ⎢ ∫ g 0 (t ' )dt '⎥
Ψ dt
0
⎣
⎦
⎦
⎣0
2
⇔
2
⎡t
⎤
dΨ
= − Dγ 2 ⎢ ∫ g 0 (t ' )dt '⎥ dt
Ψ
⎣0
⎦
⇔
2 ⎤
⎡
t t'
⎤
2 ⎡
⎢
Ψ (t ) = exp − Dγ ∫ ⎢ ∫ g 0 (t ' ' )dt ' '⎥ dt '⎥
⎢
⎥
0 ⎣0
⎦
⎣
⎦
32
(16)
Additional 1
Let us consider the phase term of the magnetization (Eq. 1);
t
⎤
⎡
Ω( z , t ) = exp ⎢− iγz ∫ g o (t ' ) dt '⎥
⎥
⎢
0
⎦
⎣
(1)
After the gradient pulse has been on for a time τ, the phase angle ϕ at position z can be
written:
τ
θ = −γz ∫ g 0 (t ' )dt '
(2)
0
)
This means that the magnetization m can be written;
)
m = Ψ (t ) ⋅ exp[−iθ ]
(3)
What happens to the phase when applying a π-pulse (rf-pulse) along the x-axis?
Figure 1A
A constant gradient field G0 (=δBz/δz) is applied along the z-axis (along B0). At time τ, the phase
angle θ1 will be:
τ
θτ = γG0 z ∫ dt = γG0 zτ
(1)
0
The magnetization M(τ) will then be:
M (t ) = M 0 ⋅ exp[−t / T2 ]⋅ exp[iϑt ] ⇒ M 0 ⋅ exp[−τ / T2 ]⋅ [iγG0τ ]
(2)
What will be the phase angle θ2 after the π-pulse?
The angle will be θ2 = - θτ so the magnetization just after the pulse will be:
33
M t , after = M 0 ⋅ exp[− τ / T2 ]⋅ [− iγG 0τ ]
Hence, the effect of a π-pulse (regarding the change in phase angle) is the same as changing the
sign of the gradient pulse, i-e-, changing G = G0 to G = – G0
In short, we may consider the following analogous situation:
Figure 1B
How can we express the phase angle θ as a function of time when considering the pulse-gradient
scheme in Figure 1B?
Figure 1C
t
Figure a) shows how the phase angle θ = γ ∫ g 0 (t ' )dt ' changes with time t and reveals a discontinuity in θ
0
at t = τ because of the π-pulse. However, since we are only interested in the time variation of the square of
34
2
⎡t
⎤
the phase angle, as given by the integral ⋅ ⎢ ∫ g 0 (t ' )dt '⎥ (see Eq 16) we may alternatively consider the
⎣0
⎦
integral illustrated on Figure b).
35
Additional 2
36