Understanding Spin
Relaxation using Conformal
Transformations
Manuel Berrondo
Dept. Physics & Astronomy
Brigham Young University
Provo UT 84602
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
In collaboration with:
Wes Krueger
“The purpose of models is not to fit the data
but to sharpen the questions”
S. Karlin
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Introduction
 Dynamics
of Heisenberg model with
dissipation
 Look first at spin precession and
relaxation w/ constant B field
 Dissipation term (Gilbert) can be
solved exactly w/stereographic
projection
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Spin precession and relaxation
dS
 S  B   S  (S  B)
dt
preserves S2, chosen to be 1.
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Conformal Transformations in
3
R
Inverse of a vector a ≠ 0
a
1
aˆ
a
a

 2  2,
a
a
a
a  a a
2
a-1
0
a
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2
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Mapping plane onto sphere
w
r
1
1
nˆ
1
nˆ 

w  w  nˆ 

  2
 
w  nˆ
w  nˆ 2
 w  nˆ 2 
2
r
 nˆ , with inverse :
w  nˆ
2
2 (r  nˆ )
w
 nˆ  BYU2 Physics
 nˆ
r  nˆ
r  1  2 nˆ  r
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Stereographic Projection
of S on the unit sphere, S2=1
 Projection onto plane has two
components: W  R2
 Tip
2Ω
S 
,
S   S nnˆ
2
1 W
S  S  S ,
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S
Ω
,
1  Sn
1  W2
Sn 
2
1 W
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Mixed Representation
S   (1  S n ) Ω
 From
equation of motion:
dS
 S  B   B (nˆ  S nS)
dt
we get:
dS n
2
  B (1  S n )
dt
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Equation of Motion for W (t)
  Ω  B   BΩ
Ω
with solution:
Ω(t )  e
  B t  t B
e Ω(0),
2 Ω(t )
2
S  (t ) 
,
S
(
t
)

1

S
(
t
)
n

2
1  W (t )
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