Transcript The Quantum Mechanics of MRI

The Quantum Mechanics of MRI
Part 1: Basic concepts
David Milstead
Stockholm University
Background reading
• Fundamentals of Physics, Halliday,
Resnick and Walker (Wiley)
• The Basics of NMR, J. Hornak
(http://www.cis.rit.edu/htbooks/nmr/inside.
htm )
Outline
• What is quantum mechanics
– Wave-particle duality
– Schrödinger’s equation
– Bizarreness
• uncertainty principle
• energy and momentum quantisation
• precession of angular momentum
What is quantum mechanics ?
Basic concepts
•
•
•
•
What is a wave ?
What is a particle ?
What is electromagnetic radiation ?
What happens to a magnet in a magnetic
field ?
What is a wave ?
Double slit diffraction
Properties of waves:
Superposition, no localisation (where is a wave??)
What is electromagnetic radiation ?
EM radiation is made up of electromagnetic waves
of various wavelengths and frequencies
Radio frequency is of most interest to us
Circularly polarised light
Rotating magnetic field
Radiation can be produced and filtered to produce a rotating magnetic
field.
Light is a wave!
Photon
Photons
Quanta of light:
Energy E  hf
h=Planck's constant=6.626 10
f =frequency
v
=
f
34
 =wavelength, v  speed
Js
Wave-particle Duality
Particles also show wave-like properties
Wave-length  =
h
p
Energy=hf
f =frequency, p=momentum, h=6.6 1034 Js Planck's constant.
We’ve just learned a basic result of quantum mechanics.
Now we move onto some maths.
Fundamental equation of quantum
mechanics
  x   wave function
E =particle energy
U =potential energy of particle
Free particle
A particle not confined or subject to forces.
Re(  x )
U  0 ; Fixed E
 2
 E
2m x 2
   A  cos kx  i sin kx 
2
x
  constant - particle can be anywhere!
2
2
p2
p
;k 
; E


2m
 fixed wavelength and momentum
h
Well defined wavelength and momentum but no preferential
position.
"Uncertainty on position": x  
"Uncertainty on momentum": p  0
Wave function of a confined particle
x
Eg particle trapped in a tiny region of space. How do we model the
wave function ?
p
p
Solution to Schrödinger’s equation would be
sum lots of sine waves with different wavelengths/momenta.
x
x
Similarly: py y  h pz z  h
Question
A 12-g bullet leaves a rifle at a speed of 180m/s. a) What is the wavelength of this
bullet? b) If the position of the bullet is known to an accuracy of 0.60 cm (radius of
barrel), what is the minimum uncertainty in its momentum? c) If the accuracy of
the bullet were determined only by the uncertainty principle (an unreasonable
assumption), by how much might the bullet miss a pinpoint target 200m away?.
h 6.6 1034
(a) Wavelength   
 3.06 1034 m
p 0.012 180
h 6.6 1034
(b) p 

 1031 kgms 1
x
0.006
p
1031
(c) Displacement
200 
 200 
 9 1031m Tiny!
p
0.12 180
Because h is so small, quantum effects are never seen for macroscopic
objects.
Hydrogen atom
Solve Schrödinger’s equation for an electron around a
proton in a hydrogen atom.
The electron is confined due to a Coulomb potential.
e2
U
4 0 r
e  proton charge
Component along one direction:
Lz  ml
ml  -l ,-  l  1 ,...., l  1, l
(2l  1) different states.
where
h

2
Where is the electron ?
Wave functions
Current loops and magnetic dipoles
+
-
Current loop
N
S
Bar magnet
Orbiting electron as a current loop
Tiny current loop.
dq charge e
current I 


dt period T
Magnetic moment   IAnˆ
e
A  area of loop
1
v
ev
ev
evR

   IA 
A
 R2 
T 2 R
2 R
2 R
2
Angular momentum: L  me vr
 l  
R
e
L (parallel to angular momentum)
2me
Gyrom agnetic ratio:  
l
L
=-
e
2me
nˆ into the page.
eml
eLz the electron
Recall
that
U 
Bz 
Bz has fixed values of LZ
2m
2m
An atom in a magnetic field
• l=1 and therefore 2l+1 states
E
(1)
2 z Bz
(2)
(3)
U  -l • B  -lz Bz
(1) ml  1 lz  
e
2me
lz  
eml
2me
; U
e
Bz (l antiparallel with B)
2me
(2) ml  0 lz  0 ; U  0
(3) ml  1 lz 
e
e
Bz ; U  
Bz (l parallel with B)
2me
2me
Two states with shifted energy  one state with no energy shift.
l=1
l=0
What happens when a circulating electron is placed
in a magnetic field.
We know that LZ , L2 are fixed and known.
L2x  L2y  L2  L2Z   l  l  1  ml2 
2
 constant
z
By symmetry, same spread of results of
Lx or Ly when measurements of made.
i.e. just as likely to find angular momentum X and Y .
y
Magnetic field introduces a torque
  l  B  l B sin 
 
x
dL
dt
LZ
B
d
1 dL
1


l B sin   l
dt L sin  dt L sin 
L
2me
eB
L
l   
e
2me
=
X
i.e. the angular momentum vector precesses around the magnetic
field (z -axis) with frequency . This is Larmor precession.
Gyromagnetic ratio 
:  

B

e
2me
U
Angular momentum precession
LZ constant and L precesses around z-axis
Question
Is it possible for an electron in a hydrogen atom to have L  LZ ?
Show that this value can be approached for large values of l.
Would L  LZ be possible using classical physics ?
No - it can never happen otherwise we would "know" values of Lx , Ly
i.e. Lx  Ly  0
L2  l  l  1
LZ  ml
L
2
For large l L2  l 2
2
Ll
Max LZ  l
LZ (comes close)
In classical physics there is no restriction on what we can and can't know.
The mathematics of spin angular momentum is identical to orbital
angular momentum.
FN
Magnetic field
FS
FN
Ignore force not parallel to North-South axis.
FS
Magnetic field
Magnetic field
Magnetic field
Magnetic field
Zeeman effect with orbital and spin
angular momentum
In the presence of
a magnetic field,
multiplicities of
”spectral lines”
appear
2B B
1
2
number of lines: n  2s  1  2
Eg for l  0, s 
Larmor precession for spin
Identical to orbital case.

s
;


S
B
 s  spin magnetic moment
Electron:  e  
e
ge
2me
z
g e  electron g-factor  2, me  electron mass.
Nucleus:
e
N 
gN
2m p
(next lecture)
y
x
g  nuclear g-factor, m p  proton mass.
SZ
S
S sin
Summary
• Established basic quantum mechanics
theory needed for NMR
– wave-particle duality
• Light is either photons or electromagnetic waves
– Schrödinger’s equation and the wave function
at the heart of QM predictions
– Energy and angular momentum are quantised
• Larmor precession
– Angular momentum comes in two varieties
(orbital and spin)
The Quantum Mechanics of MRI
Part 2: Understanding MRI
David Milstead
Stockholm University
Outline
• Spin - reminder
• Fermions and bosons
• Nuclear energy levels
Identical in form to orbital case.
S 2  s ( s  1)


2
Spin
S z  ms

B
s
S
 s  spin magnetic moment
Electron:  e  
e
ge
2me
z
g e  electron g-factor  2, me  electron mass.
y
Nucleus:
e
N 
gN
2m p
(next lecture)
x
g  nuclear g-factor, m p  proton mass.
S
SZ
S sin
Gyromagnetic ratio
Why do they have different values ?
Fermions and bosons
• Fermions
– Spin 1/2, 3/2, 5/2 objects
– Electrons, protons and neutrons have spin 1/2
– Tricky bit comes when combining their spins
to form the spin of, eg, an atom or a nucleus
• Bosons
– Integer spin objects
Similar shell structure for nuclear
physics as for atomic physics
• Need to fill up, shell by shell
• Pairs of protons and neutrons cancel
each other’s spins.
• Pauli’s exclusion principle ensures
that many shells are filled.
• Nuclei with uneven (even) atomic
number have half-integer (integer)
spin
• Nuclei with even atomic and mass
numbers have zero spin.
• Unpaired neutrons/protons provide
the spin for MRI.
Question
Nature would prefer all electrons to be in
the lowest shell and all nucleons
protons/neutrons) in the lowest shell ?
Why doesn’t this happen ?
Usefulness for MRI
• Need isotopes with unpaired protons (to
produce signal for MRI)
• Most elements have isotopes with nonzero nuclear spin
• Natural abundance must be high enough
for MRI to be performed.
Spins of various nuclei
Now we can understand MRI
We know the basics:
A uniform magnetic field B 0
A short pulse of a rf field B1
A system out of equilibrium
 measurement of return to equilibrium.
Why is
(a) B1  B0 ?
(b) The rf pulse shorter than recovery time ?
No magnetic field
Apply an external magnetic field
B0
Spins precess at Larmor
frequency .
Gyromagnetic ratio:  

B
Precessions incoherent:
Total spin S  S Z , S x  S y  0
SZ  
1
2
In fact, there are continual transitions
and interactions from thermal energy.
Another look at the system
Split into ”spin-zones”.
For uniform system we can regard the macroscopic system as giving
a single magnetisation.
Conventional to talk about magnetisation M :
N
M  s
; N  no. dipoles, V  volume
V
Putting together what we’ve learned
Need to know its energy splitting.

s
S

e
gN
2m p
S =spin angular momentum g  nuclear g-factor , m p  proton mass.
E   sZ B0 ;  sZ   S z 
 E  2 sZ B0 
g pe
2m p
e
1
e
gN 

gN
2m p
2
4m p
B0  0
 Frequency of light needed for excitation 0 
Larmor frequency:    B0 
g pe
2m p
B0
eB0
g N  0
2m p
How many nuclei can be excited ?
0
0
Energy level population
Nature has a preference for the lowest energy states.
In thermal equilibrium the lowest states are a bit more
populated than the higher energy states.
2
x
Taylor expansion: e x  1  x   ...
2!
 E 
N
E
 e
  1
N
k BT
 k BT 
k B  Boltzmann's constant = 1.38  10 23 J/K
T =room temperature
E 
g pe
2m p
310 K
B0   B0  0
B0  1.5 T   42.576 106  2  E  4.2  10 26 J
N

 1.000009
N
 N  N g pe
N  N  N   N 1 
B0

 N  2 2m p k BT
Changing the spin populations
Tiny differences in, eg, 0.02ml of water
expect 6×1015 more in parallel state.
Question
Chemists can excite certain samples using UV light. Would you expect
N
to be substantially different in this case ? What does this imply for
N
the relative size of the sample of a MRI scan and a test made by
chemists with infrared light ?
3 108
UV-light  10 m  f  
 3  1016 s 1
8
 10
 E 6.6 1034  3  1016  2 10 17 J
8
c
N
E
2 1017
 1
 1
 1  5000
N
k BT
1.4 1023  300
In fact, approximation no longer valid since it is no longer a
small difference!!
A chemist can use far smaller samples since the energy gap
is larger and there are far more in the lower energy state.
Exciting the nuclei - Rf pulse
Rotating magnetic field B1
Typical pulse duration ~1ms.
Two ways to think about the pulse. Both are needed to understand MRI.
Excitation
B0
B1
Pulse of rf-
Rotating magnetic field B1
The rf pulse acts in two ways:
(1) The photons are absorbed, saturating the system and reducing M Z  0.
(2) The rotating magnetic field B1 acts on the magnetisation vector
to rotate into the complex plane.
Individual dipole:    s  B1 ;
Use magnetisation:
 
dS 1 d  s

  s  B1
dt  dt
1 dM
 M  B1 - simple form of the Bloch equations.
 dt
Rotating frame of reference
B0
B1
X’
Easier to understand if a rotating frame of reference
(with Larmor frequency) is used.
Rotate x, y co-ordinates  x ', y '
Magnetic field constant on x ' axis.
y’
Different types of pulses
90o
180o
Duration of pulse and size of B1 determine angle of rotation of magnetisation:
Spins precess around B1

t
Consider a full rotation of   2 in 1ms
   B1 
  2  42.58 MHz/T
1
1

103 42.58 106
1T
 B1  B0 .
 B1 
B0
105 T
Question
When applying B1 a new magnetic field is obtained:
Bn  B1  B0 .
Why don't the protons just align with respect to that field ?
B0
Bn
What happens next ?
Consider a 90o rotation.
The magnetisation vector is in the x - y plane.
The rf -pulse is turned off.
The system must return to equilibrium.
We have two components: M Z and M xy .
Longitudinal relaxation
Spin-lattice effect:
higher energy state interacts with lower energy state due and lose
energy through rotation and vibration .
Relaxation times for different materials
Transverse relaxation
In the laboratory frame
Dipole moments are initially in phase.
Mxy precesses and decays.
As it precesses phase decoherence occurs.
Complicated process: contributing factor non-uniform magnetic
field over sample  different precession rates for different regions.
Free induction decay
As the transverse magnetisation decays, a changing magnetic
field is produced:

 =an emf is produced in a receiver.
t
The signal is a decaying sinusoidal wave with lifetime T2* .
Measurement of gives information on composition of sample.
Free induction decay
Easier to interpret a single line on a "frequency" spectrum.
Use a Fourier transform to move from damped exponential to
signal.
Summary
• Basic quantum mechanics at the heart of
nuclear magnetic resonance
– Angular momentum quantisation
– Energy quantisation
• Features of a MRI experiment
investigated.