Introduction to Magnetic Resonance David J. Keeble

Introduction to Magnetic Resonance

David J. Keeble

Magnetic Resonance

Magnetic moments?

Magnetic

What matters is matter with moments

Matter: 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

2

Q

L M

2

Q

M

μ L

The ratio of the magnetic moment, m, to the angular momentum L is called the gyromagnetic ratio, g (or magnetomechanical ratio).

μ 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 ResonanceMagnetic

The electron

Classical Physics:2

Q

L M

For an electron Quantum Mechanics tells us there is an intrinsic angular momentum of . 1

2 24 24.

2638 10 A

1 1

2 2 22 m

2 Be e

e e

m m

Q

M

μ L

Dirac’s Relativistic Quantum Mechanics 24 21

9.28 10 A m2

2 B B μ

2eg define a ‘g-factor’

e e Bg μ S 1

2S 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


The electron

24 29.2740097(2) 10 A m2B

e

e

m

Dirac’s Relativistic Quantum Mechanics: 2eg

e e Bg μ S 1

2S where here we let

341.05457173(5) 10 J s

2.00231930436153(53)eg

24 19.2740097(2) 10 J TB

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


The proton

The proton is composed of three quarks (uud)

24 19.2740097(2) 10 J TB

Quark Charge Spin

Up, u +2/3 1/2

Down, d -1/2 1/2

2 161.41060674(3) 10 J Tp

The neutron

The neutron is composed of three quarks (udd)

The intrinsic angular momentum of the proton2

I

The intrinsic angular momentum of the neutron2

I

26 10.9662365(2) 10 J Tn

Magnetic moment


Magnetic moment


Magnetic moments?


The proton26 11.41060674(3) 10 J Tp

26 10.9662365(2) 10 J Tn

The neutron

24 19.2740097(2) 10 J T2B

e

e

m

27 15.0507835(1) 10 J T2N

p

e

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.


e e Bg μ SMagnetic moment


24 19.2740097(2) 10 J TB

2.00231930436153(53)eg

μ L

e e Bg

11 1 11.76085971(4) 10 rad s Te

e Be eg

μ

S remember above we define S as a dimensionless number above

The electronMagnetic moments?

e eμ S

The proton26 11.41060674(3) 10 J Tp Magnetic moment


27 15.0507835(1) 10 J TN

p p Ng μ I Define a proton g –factor p

pN

gI

5.58569471(5)pg


p Np pg

I

8 1 12.67522201(6) 10 rad s Tp

p pμ I

The proton26 11.41060674(3) 10 J Tp Magnetic moment


27 15.0507835(1) 10 J TN

p p Ng μ I Define a proton g –factor p

pN

gI

5.58569471(5)pg

μ L

Nuclear Isotopes

We 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.


Nuclear 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


moment (μ/μN)

63Cu 62.9295989 (17) 69.17 (3) 3/2 2.223365Cu 64.9277929 (20) 30.83 (3) 3/2 2.3817


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


The proton26 11.41060674(3) 10 J Tp

26 1928.47643(2) 10 J Te The electron

658.210685(5)e p

658e p

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 μ J Here you can choose to pull the h-bar into the angular momentum operator definition.

ˆˆ e e Bg μ Jˆˆ n n Ng μ I

Here you can’t since h-bar is included in the magneton.


Magnetic 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


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 μ J

What effect does this have on the energy, E, of our particles carrying magnetic moments?

Magnetic Resonance

Magnetic moment


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 μ 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 B

Here we’ve moved the dipole in from infinite and rotated it. Then as long as B is zero at infinityU 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


Resonance?

0ˆB kB

ˆˆ μ 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 J B ˆ

e BH g J B

ˆn NH g I B

Quantum Mechanics

Magnetic Resonance

Magnetic moment


Resonance?

0ˆB kB

ˆˆ μ J

ˆH μ BˆH J B

ê BH g J B

ˆn NH g 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


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 Resonance

Resonance?

0ˆB kB

ˆ ˆH 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 B I

3 0 0 0

0 1 0 01ˆ0 0 1 02

0 0 0 3

z

I

Magnetic moment


1

03

02

0

0

11

02

0

0

01

12

0

0

03

02

1

3 0 0 0

0 1 0 0ˆ0 0 1 02

0 0 0 3

z

I

0ˆ ˆ

zH B I

Magnetic Resonance

Resonance?

0ˆB kB


0

3 0 0 0 0

0 1 0 0 03ˆ0 0 1 0 02 2

0 0 0 3 1

H B

Magnetic moment


0 0 0

0 0

0 03 3 3

0 02 2 2 2

3 1

B B B

0ˆ ˆ

zH B I

3 0 0 0

0 1 0 0ˆ0 0 1 02

0 0 0 3

z

I

ˆn n nH E

1

03

02

0

0

11

02

0

0

01

12

0

0

03

02

1

Magnetic Resonance

Resonance?

0ˆB kB


Magnetic moment


0ˆ ˆ

zH B I

3

2

3

2

1

2

1

2

32

0

3

2E B

12

0

1

2E B

32

0

3

2E B

12

0

1

2E B 0 IE B m

Magnetic Resonance

Resonance? 0 IE B m

3

2I

3

2

3

2

1

2

1

2

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 const

kT

0

0

exp

expi

e B i

i Je B i

m J

g B m

kTP

g B m

kT

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 Resonance

Resonance?

0

0

exp

expi

e B i

i Je B i

m J

g B m

kTP

g B m

kT

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’:

2

2 21

ˆmm xw B m I m E E

2

2 21

ˆmm xw B m I m g

0 IE B m

3

2I

3

2

3

2

1

2

1

2

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(w)

B1 is the magnitude of a magnetic field oscillating at frequency w perpendicular to B0

Magnetic Resonance

Resonance?

0

0

exp

expi

e B i

i Je B i

m J

g B m

kTP

g B m

kT

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’:

2

2 21

ˆmm xw B m I m E E

0 IE B m

3

2I

3

2

3

2

1

2

1

2

The other important consequence of this expression is that the term in the square brackets defines the ‘selection rules ‘ for these transitions.

1m

Magnetic Resonance

Resonance?

Magnetic moment


0ˆB kB

ê BH g S B

1

2S

0ˆ

e B zH g B S

1

2

1

2

Two eigenstates

1 01 1ˆ0 12 2z z

S

1

0

0

1

ˆ2z S

ˆ2z S

0

ê B zH g B S

0

1

2e BE g B

0

1

2e BE g B

How about a single electron, S = 1/2, placed in a magnetic field B.

Magnetic Resonance

Resonance?

1

2

1

2

Magnetic moment


0ˆB kB

ê BH g S B

1

2S

0ˆ

e B zH g B S0

1

2e BE g B

0

1

2e BE g B GHzmT 71.448

fB

g

0e BE hf g B

Electron Paramagnetic Resonance (EPR)

BH B g S

Zeeman

9.5 GHz

0.34

34 GHz

1.22

94 GHz

3.36

B (T)

S = 1/2g = 2

1

2SM

1

2SM

Quantitative, Sensitivity ~ 1010 spins


BH B g S

Zeeman

ê BH g 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

BH B g S

Zeeman

Symmetry

Hyperfine & Nuclear Zeeman

,i i n n i ii

gI 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)


BH B g S

Zeeman Hyperfine & Nuclear Zeeman

,i i n n i ii

gI 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

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Introduction to Magnetic Resonance David J. Keeble