Introduction to Magnetic Resonance David J. Keeble
Feb 24, 2016
Introduction to Magnetic Resonance
David J. Keeble
Magnetic ResonanceMagnetic moments?
MagneticWhat matters is matter with momentsMatter: Leptons and quarks
The simplest fundamental particle is the lepton the electron
What is the ratio of the magnetic moment, m, of a spinning sphere of mass M carrying charge Q, where the charge and mass are identically distributed, to the angular momentum L?
Classical Physics:
I d μ a m L r p r v
2Q
L Mm
2QM
μ L
The ratio of the magnetic moment, m, to the angular momentum L is called the gyromagnetic ratio, g (or magnetomechanical ratio).
gμ L
A thin uniform donut, carrying charge Q and mass M, rotates about its axis as shown below:
(a) Find the ratio of its magnetic dipole moment to its angular momentum. This is called the gyromagnetic ratio (or magnetomechanical ratio).(b) What is the gyromagnetic ratio for a uniform spinning sphere? [This requires no new calculation; simply decompose the sphere into infinitesimal rings, and apply the result of part (a).](c) According to quantum mechanics, the angular momentum of a spinning electron is . What then is the electron’s magnetic dipole moment in Am2?
12
z
Magnetic ResonanceMagneticThe electron
Classical Physics:2Q
L Mm
For an electron Quantum Mechanics tells us there is an intrinsic angular momentum of . 1
2 24 24.
2638 10 A1 1
2 2 22 m
2 Be e
e em m
QM
m
μ L
Dirac’s Relativistic Quantum Mechanics 24 21 9.28 10 A m2
2 B Bm m μ
2eg define a ‘g-factor’
e e Bg mμ S 12
S where now let
i.e. we’re here calling S the intrinsic angular momentum of the electron but the units, , are now assigned to the quantity we call the Bohr magneton
Magnetic moment
Spin angular momentum
Magnetic moments?
The Bohr magneton
Magnetic ResonanceMagneticThe electron
24 29.2740097(2) 10 A m2B
e
em
m
Dirac’s Relativistic Quantum Mechanics: 2eg
e e Bg mμ S 12
S where here we let
341.05457173(5) 10 J s
2.00231930436153(53)eg
24 19.2740097(2) 10 J TBm
Feynman, Schwinger, and Tomonaga applied quantum electrodynamics :
Magnetic moments?
The most precisely known quantity
NB: change of units – try using dimensional analysis to check this
Magnetic ResonanceMagnetic
The proton
The proton is composed of three quarks (uud)
24 19.2740097(2) 10 J TBm
Quark Charge Spin
Up, u +2/3 1/2
Down, d -1/2 1/2
2 161.41060674(3) 10 J Tpm
The neutronThe neutron is composed of three quarks (udd)
The intrinsic angular momentum of the proton 2I
The intrinsic angular momentum of the neutron 2I
26 10.9662365(2) 10 J Tnm
Magnetic moment
Spin angular momentum
Magnetic moment
Spin angular momentum
Magnetic moments?
Magnetic ResonanceMagneticThe proton
26 11.41060674(3) 10 J Tpm
26 10.9662365(2) 10 J Tnm
The neutron
24 19.2740097(2) 10 J T2B
e
em
m
27 15.0507835(1) 10 J T2N
p
em
m
The Bohr magneton
The nuclear magneton
Magnetic moments?
We define a similar quantity, the nuclear magneton where we substitute the mass of the proton, rather than the electron.
Experimental values
Comparing the measured magnetic moment values for the proton and neutron with the nuclear magnetron we see they are roughly of the same order.
Magnetic ResonanceMagnetic
e e Bg mμ SMagnetic moment
Spin angular momentum
24 19.2740097(2) 10 J TBm
2.00231930436153(53)eg
gμ L
e e Bgg m
11 1 11.76085971(4) 10 rad s Teg
e Be eg mg
μS
remember above we define S as a dimensionless number above
The electronMagnetic moments?
e egμ S
The proton26 11.41060674(3) 10 J Tpm Magnetic moment
Spin angular momentum
27 15.0507835(1) 10 J TNm
p p Ng mμ I Define a proton g –factor p
pN
gI
mm
5.58569471(5)pg
Magnetic ResonanceMagnetic
p Np pg
Im mg
8 1 12.67522201(6) 10 rad s Tpg
p pgμ I
The proton26 11.41060674(3) 10 J Tpm Magnetic moment
Spin angular momentum
27 15.0507835(1) 10 J TNm
p p Ng mμ I Define a proton g –factor p
pN
gI
mm
5.58569471(5)pg
gμ L
Nuclear IsotopesWe will be potentially interested in, normally stable, nuclear isotopes that possess a nuclear moment. Most isotope tables list nuclear spin and moment values, the nuclear g-value, defined in the same way as above may be given, or the simple ratio of the moment with the nuclear magneton, and/or the gyromagnetic ratio.
Magnetic ResonanceMagneticNuclear moments
Magnetic moments?
Isotope Atomic mass (ma/u) Natural abundance (atom %) Nuclear spin (I) Magnetic
moment (μ/μN)46Ti 45.9526294 (14) 8.25 (3) 047Ti 46.9517640 (11) 7.44 (2) 5/2 -0.7884848Ti 47.9479473 (11) 73.72 (3) 049Ti 48.9478711 (11) 5.41 (2) 7/2 -1.1041750Ti 49.9447921 (12) 5.18 (2) 0
Isotope Atomic mass (ma/u) Natural abundance (atom %) Nuclear spin (I) Magnetic
moment (μ/μN)
63Cu 62.9295989 (17) 69.17 (3) 3/2 2.223365Cu 64.9277929 (20) 30.83 (3) 3/2 2.3817
Isotope Atomic mass (ma/u) Natural abundance (atom %) Nuclear spin (I) Magnetic
moment (μ/μN)14N 14.003 074 005 2(9) 99.632 (7) 1 0.403760715N 15.000 108 898 4(9) 0.368 (7) 1/2 -0.2831892
Magnetic ResonanceMagnetic
The proton26 11.41060674(3) 10 J Tpm
26 1928.47643(2) 10 J Tem The electron
658.210685(5)e pm m
658e pm m
Magnetic moments?
Quantum Mechanics?
‘Observe’ magnetic moments
magnetic moment OPERATOR
μ̂
If we assume the non-interacting ‘particles’ each have a total angular momentum J
(A special case of the Wigner – Eckhart theorem)
ˆˆ constμ J
ˆˆ gμ J Here you can choose to pull the h-bar into the angular momentum operator definition.
ˆˆ e e Bg mμ Jˆˆ n n Ng mμ I
Here you can’t since h-bar is included in the magneton.
Magnetic ResonanceMagneticMagnetic moments in a bulk sample?
What we measure is the resulting macroscopic moment per unit volume V, due to the assemble of N magnetic moments in that volume – the Magnetization. 1 N
iiV
M μ 1 3J T m
Magnetic Resonance
Magnetic moment
Spin angular momentum
Resonance? So with an assemble of electron spins, or protons……..
Let’s put our magnetic moments into an external magnetic field , B
0ˆB kB
ˆˆ gμ J
What effect does this have on the energy, E, of our particles carrying magnetic moments?
Magnetic Resonance
Magnetic moment
Spin angular momentum
Resonance? So with an assemble of electron spins, or protons……..
Let’s put our magnetic moments into an external magnetic field , B
0ˆB kB
ˆˆ gμ J
Energy – to determine the quantum mechanical operator that allows us to predict the results of energy measurements we can start with the classical expression a substitute the appropriate observable operators.
Magnetic Resonance
Magnetic moment
Classical perfect magnetic dipole
0ˆB kB
Let’s first go back to the classical case and consider the forces acting on a loop area ab carrying current I, it’s not too difficult to establish that a torque must act and that it’s given by the expression :
I d μ a
N μ B
U d d
r r
F l μ B l μ B r μ BHere we’ve moved the dipole in from infinite and rotated it. Then as long as B is zero at infinity
U μ Bthe energy associated with the torque is :
The force on an infinitesimal loop, with dipole moment m, in a field B is: F μ B
Magnetic Resonance
Magnetic moment
Spin angular momentum
Resonance?
0ˆB kB
ˆˆ gμ J
N μ B U μ BClassical E&M
ˆH μ BU μ B
For a ‘static’ (it can rotate, but let’s not deal with translation) dipole moment m, in a field B we now have:
ˆH g J B ˆ
e BH g m J B
ˆn NH g m I B
Quantum Mechanics
Magnetic Resonance
Magnetic moment
Spin angular momentum
Resonance?
0ˆB kB
ˆˆ gμ J
ˆH μ BˆH g J B
ˆe BH g m J B
ˆn NH g m I B
So let’s remember the fundamental issues regarding J, L, S, and I in quantum mechanics: The algebraic theory of spin is identical to the theory of orbital angular momentum; we call it spin angular momentum. However, physically these are very different :
The eigenfunctions of orbital angular momentum are spherical harmonics we get from solving the differential equations that we get from the Schrödinger time-independent equation The eigenfunctions of spin angular momentum are expressed as column matrices. This physics emerges from Dirac equation, but we use it with the Schrödinger time-independent equation
Magnetic Resonance
Magnetic moment
Spin angular momentum
Resonance?
0ˆB kB
No spin stands lone – If they did the simple story we’ve developed would be it, and as we’ll learn we would measure ‘text book’ magnetic resonance spectra.
unfortunately?
Simple, elegant, understandable – but we’d be out of a job!
Spins couple – to eachother, to the orbital motion of the particles, to vibrations, to……………
But before we break out into the ‘real world’ let’s stick with our ideal isolated magnetic moments for a bit longer and look at the basic principles of ‘resonance’.
Magnetic ResonanceResonance?
0ˆB kB
ˆ ˆH g I B
Let’s consider an I = 3/2 nucleus placed in a magnetic field B.
ˆn n nH E
0ˆB kB 0
ˆ ˆzH Bg I
3 0 0 00 1 0 01ˆ0 0 1 020 0 0 3
z
I
Magnetic moment
Spin angular momentum
103020
011020
00112
0
003021
3 0 0 00 1 0 0ˆ0 0 1 020 0 0 3
z
I 0
ˆ ˆzH Bg I
Magnetic ResonanceResonance?
0ˆB kB
Let’s consider an I = 3/2 nucleus placed in a magnetic field B.
0
3 0 0 0 00 1 0 0 03ˆ0 0 1 0 02 20 0 0 3 1
H Bg
Magnetic moment
Spin angular momentum
0 0 0
0 00 03 3 30 02 2 2 23 1
B B Bg g g
0ˆ ˆ
zH Bg I
3 0 0 00 1 0 0ˆ0 0 1 020 0 0 3
z
I
ˆn n nH E
103020
011020
00112
0
003021
Magnetic ResonanceResonance?
0ˆB kB
Let’s consider an I = 3/2 nucleus placed in a magnetic field B.
Magnetic moment
Spin angular momentum
0ˆ ˆ
zH Bg I
32
32
12
12
32
032
E Bg
12
012
E Bg
32
032
E Bg
12
012
E Bg 0 IE B mg
Magnetic ResonanceResonance? 0 IE B mg
32
I
32
32
12
12
Consider and assembly of particles, each having total angular momentum ILet’s assume they are noninteracting – the greatest possible simplification
The probability that a dipole within the assembly at temperature T has potential energy Ei is, according to Boltzmann:
exp ii
EP constkT
0
0
exp
expi
e B i
i Je B i
m J
g B mkTP
g B mkT
m
m
Here:
Why Boltzmann statistics? It is the fact they are non-interacting, and hence distinguishable that’s key
The differences in population of the levels means that energy can be absorbed, there can be a net moving of spins ‘up’
Magnetic ResonanceResonance?
0
0
exp
expi
e B i
i Je B i
m J
g B mkTP
g B mkT
m
m
The probability that a dipole within the assembly at temperature T has potential energy Ei is, according to Boltzmann statistics. So at a finite temperature multiple levels can be populated
To get a transition from one level to another - we need to apply an oscillating magnetic field with the correct orientation with respect to the external magnetic field.
We can tackle this using time-dependent perturbation theory which can give us Fermi’s golden rule, which for our purposes can take the form for the probability per unit time that a paramagnet initially in state m will be found it state m’:
22 2
1ˆ
mm xw B m I m E Eg
22 2
1ˆ
mm xw B m I m gg
0 IE B mg
32
I
32
32
12
12
In this last expression, we’ve let the ‘real world’ butt in again and are assuming the there is a distribution of effective magnetic fields across our assembly giving a lineshape g()
B1 is the magnitude of a magnetic field oscillating at frequency perpendicular to B0
Magnetic ResonanceResonance?
0
0
exp
expi
e B i
i Je B i
m J
g B mkTP
g B mkT
m
m
We can tackle this using time-dependent perturbation theory which can give us Fermi’s golden rule, which for our purposes can take the form for the probability per unit time that a paramagnet initially in state m will be found it state m’:
22 2
1ˆ
mm xw B m I m E Eg
0 IE B mg
32
I
32
32
12
12
The other important consequence of this expression is that the term in the square brackets defines the ‘selection rules ‘ for these transitions.
1m
Magnetic ResonanceResonance?
Magnetic moment
Spin angular momentum
0ˆB kB
ˆe BH g m S B
12
S
0ˆ
e B zH g B Sm
12
12
Two eigenstates
1 01 1ˆ0 12 2z z
S
10
01
ˆ2z S
ˆ2z S
0
ˆe B zH g B Sm
012e BE g Bm
012e BE g Bm
How about a single electron, S = 1/2, placed in a magnetic field B.
Magnetic ResonanceResonance?
12
12
Magnetic moment
Spin angular momentum
0ˆB kB
ˆe BH g m S B
12
S
0ˆ
e B zH g B Sm0
12e BE g Bm
012e BE g Bm
GHzmT 71.448
fB
g
0e BE hf g Bm
Electron Paramagnetic Resonance (EPR)
BmH B g S
Zeeman
9.5 GHz
0.34
34 GHz
1.22
94 GHz
3.36
B (T)
S = 1/2g = 2
12SM
12SM
Quantitative, Sensitivity ~ 1010 spins
Electron Paramagnetic Resonance (EPR)
BmH B g SZeeman
ˆe BH g m S B ?
No spin stands lone …….
The expression on the right is the first, normally dominant, term in a general ‘spin’ Hamiltonian expression for EPR.
The left hand expression is exact for a mythical assembly of non-interacting ‘free’ electrons.
In a real sample those normally ‘special’ electrons that are not spin-paired and so are detectable by EPR will be occupying an orbital, an electronic state, that may also have some orbital angular momentum ‘character’ due to say to a spin-orbit interaction. In consequence, the true eigenstates of that electron involve angular momentum that is not purely spin.
This is messy so magnetic resonance experimentalists rapidly adopted the spin-Hamiltonian concept. The point of the spin-Hamiltonian is that you keep assuming that you are working with pure spin functions , you fold the nasty complications into the parameters – in this case you define a g-matrix that departs from ge in a way that allows you to still use those spin functions that we can express as simple column vectors.
The departure from ‘free’ is now characterized by the values in the g-matrix, the bonding character of the electronic state may now manifests itself as a g-value different from 2.0023
Electron Magnetic Resonance Spectroscopy
BmH B g S
Zeeman
Symmetry
Hyperfine & Nuclear Zeeman
,i i n n i ii
gmI A S B I
So armed with this spin-Hamiltonian concept we can develop terms which describe other important interactions between spins, for example the hyperfine interaction between magnetic nuclei and our electron spin(s)
Electron Paramagnetic Resonance (EPR)
BmH B g S
Zeeman Hyperfine & Nuclear Zeeman
,i i n n i ii
gmI A S B I
63Cu 69.2 % I = 3/2 m/mn = 2.2265Cu 30.8 % I = 3/2 m/mn = 2.38
PbTiO3 Cu (d9): S = 1/2Here is an example of a real EPR spectrum from a very low concentration of Cu2+ impurity ions substituting for Ti in the perovskite oxide PbTiO3.
At this orientation of the magnetic field with the crystal axes the g-value is ~ 2.34. It’s determining what the center field of the spectrum is. The hyperfine interaction with the magnetic Cu nuclei is defining the number of lines and the separation.
2I+1 lines