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Introduction to Magnetic Resonance David J. Keeble

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Introduction to Magnetic Resonance. David J. Keeble. Magnetic Resonance. Magnetic. Magnetic moments?. What matters is matter with moments. Matter: Leptons and quarks . The simplest fundamental particle is the lepton the electron. Classical Physics :. - PowerPoint PPT Presentation

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Page 1: David  J.  Keeble

Introduction to Magnetic Resonance

David J. Keeble

Page 2: 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

Page 3: David  J.  Keeble

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

Page 4: David  J.  Keeble

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

Page 5: David  J.  Keeble

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

Page 6: David  J.  Keeble

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?

Page 7: David  J.  Keeble

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.

Page 8: David  J.  Keeble

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

Page 9: David  J.  Keeble

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.

Page 10: David  J.  Keeble

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

Page 11: David  J.  Keeble

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.

Page 12: David  J.  Keeble

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

Page 13: David  J.  Keeble

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?

Page 14: David  J.  Keeble

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.

Page 15: David  J.  Keeble

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

Page 16: David  J.  Keeble

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

Page 17: David  J.  Keeble

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

Page 18: David  J.  Keeble

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’.

Page 19: David  J.  Keeble

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

Page 20: David  J.  Keeble

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

Page 21: David  J.  Keeble

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

Page 22: David  J.  Keeble

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’

Page 23: David  J.  Keeble

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

mm xw B m I m E Eg

22 2

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

Page 24: David  J.  Keeble

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

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

Page 25: David  J.  Keeble

Magnetic ResonanceResonance?

Magnetic moment

Spin angular momentum

0ˆB kB

ˆe BH g m S B

12

S

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.

Page 26: David  J.  Keeble

Magnetic ResonanceResonance?

12

12

Magnetic moment

Spin angular momentum

0ˆB kB

ˆe BH g m S B

12

S

e B zH g B Sm0

12e BE g Bm

012e BE g Bm

GHzmT 71.448

fB

g

0e BE hf g Bm

Page 27: David  J.  Keeble

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

Page 28: David  J.  Keeble

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

Page 29: David  J.  Keeble

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)

Page 30: David  J.  Keeble

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