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TRANSCRIPT
The appearance of the Berry phase for the precession of nuclear spin with spin 1/2
Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton
(Date: February 08, 2017)
Here we use the spin-echo method which is used for the r.f. spin echo of nuclear magnetic resonance. We learned this method from a book of M.H. Levitt, Spin Dynamics Basic of Nuclear Magnetic Resonance). We present an exact solution the time dependence of the spin state of the nuclear spin, and discuss the adiabatic change of the nuclear spin. We solve the example 10.1 and problem 10.2 of the textbook of D.J. Griffiths (Introduction to Quantum Mechanics). We use the Mathematica for solving the problem. 1. Introduction
In order to understand the adiabatic approximation, we consider the time dependent behavior
of a nuclear spin (spin 1/2) in the presence of a magnetic field whose magnitude ( 0B ) is constant,
but whose direction sweeps out a cone, of opening angle of the magnetic field at constant
angular frequency . Note that the angle is fixed. The magnetic field is expressed by
)cos),sin(sin),cos((sin0 pp ttB B ,
where B0 is the magnitude of the magnetic field and p is the phase.
Fig. Magnetic field sweeps around in a cone, at the constant angular frequency . The case
with 0 (rotation of nuclear spin around the z axis in counter clockwise.
((Gyromagnetic ratio of nuclear spin))
Proton (neutron); nuclear spin
Fig. T
The gyro
where I
μ̂
where σ̂
H
where
The definition
omagnetic ra
gnI
II
μ
I is the angu
Iμ ˆˆ I ,
is the Pauli
ˆ2
1
ˆ
[si
ˆ
ˆˆ
1
1
0
z
z
I
I
B
H
BI
Bμ
cos11 B
n of the sign
atio is define
g NnN
I
I
ular moment
σI ˆ2
1ˆ
i operator. Th
[cos2
1
[cos(
cos(in
2
2 t
t
s , 2
n for the gyro
d by
cm
eg
pn 2
tum. So the m
he Hamilton
ˆ)s(
sinˆ)
sinˆ)
xp
xp
xp
t
I
I
sin0B
omagnetic ra
magnetic mo
nian is given
sin(
]ˆ)n(
sin(
yp
p
t
It
t
atio . (Lev
oment of the
n by the Zeem
]ˆ)
]
ˆcosˆ)
yp
zy II
vitt, 2008)
e nuclear spin
man energy a
]z
n is given by
as
y
2. Spin recession (Classical theory)
The magnetic moment of nuclear spin with angular momentum zI -s given by
zI I
When a static magnetic field is applied along the z axis, the Hamiltonian has the form of the Zeeman energy, given by
BIμ BH I ,
We consrotation)
d
d
When the
I
leading to
d
d
or
I
where
sider the eq
dt
d μτ
L
( BII
dt
d
e magnetic f
0zI ,
o the equatio
)( yx iIIdt
d
yx IiII 0 e
cos1 B
quation of m
B IB
) .
field is applie
yx BII ,
on
y BIiBI
Bti )exp(
motion for t
B ,
ed along the
, yI
( xx IBiI
tieI 10
the angular
e +z axis,
xBI
)yiI
(counter clo
momentum
ockwise for
m (Newton’s
0 ).
s second laww for
Fig. Precession of nuclear spin with the angular frequency( 0 B ) for 0 . (Clockwise
direction). (Levitt, 2008). 3. Spin precession using quantum mechanics
The time evolution operator is defined by )ˆexp(ˆ tHi
T
. When a static magnetic field is
applied along the z axis (in the absence of the r.f. field), the Hamiltonian is given by
zz IBIH ˆsinˆˆ100
where
sin01 B and zzI ̂2
1ˆ .
The state vector at the time t is related to the state vector at the time 0t through the time evolution operator as
)()ˆ2
exp(
)()ˆexp(
)0()ˆexp(
)0()ˆexp()(
11
11
1
0
tt
i
tIti
ttIi
ttHi
t
z
z
z
Here we define the rotation operator as
)2
exp(0
0)2
exp()ˆ
2exp()(ˆ
i
ii
R zz .
This means that spin rotates around the z axis through the angle t1 . The r.f. field is given by
)sin()cos([sin)( 0 pypxRF ttBt eeB ,
(counter-clockwise for 0 )
)]sin(ˆ)cos(ˆ[
)(ˆˆ
2 pypx
RFL
tItI
tH
BI
Note that
sin02 B ,
is the nutation angular frequency. 4. State vector in the rotation frame and Laboratory frame
xRx z )(ˆ' ,
')(ˆ xRx z .
Here we use the notations
Rxx ' Rotating frame
Lxx ' Laboratory frame
Then we have
LzRxRx )(ˆ'
This relationship may be generalized. Any spin state, viewed from the rotating frame, is related to the spin state, viewed from the fixed frame through
)(ˆ zRR ,
RzR )(ˆ
Note that for simplicity we use the notation of L
.
Now we consider the Schrödinger equation in the laboratory frame,
Ht
i ˆ
or
tiRR
ti
Rt
it
i
zz
zR
)(ˆ)](ˆ[
)(ˆ
Here we note that
)ˆexp()ˆexp()(ˆzzz IiJ
iR
where
zzz IJ ̂2
ˆˆ ,
The derivative of )(ˆ zR with respect to t;
)(ˆˆ
)ˆexp(ˆ
)ˆexp()(ˆ
zz
zz
zz
RIi
Iidt
dIi
Iit
Rt
where
dt
d, t
Thus we have the Schrödinger equation in the rotating frame
RR
RzLzz
RzLzRz
Lzzz
zzzR
H
RHRI
RHRI
HRRI
tiRRI
ti
ˆ
)](ˆˆ)(ˆˆ[
)(ˆˆ)(ˆˆ
ˆ)(ˆ)(ˆˆ
)(ˆ)(ˆˆ
The Schrödinger equation in the rotating frame is given by
RRRH
ti ˆ
with the Hamiltonian in the rotating frame,
)(ˆˆ)(ˆˆˆ zLzzR RHRIH
5. Rotating-frame Hamiltonian
In the presence of a static field along the z axis, the Hamiltonian is given by
zzIH ˆ2
ˆˆ 110
with
zzI ̂2
1ˆ
cos1 B
The Hamiltonian due to the AC (r.f.) magnetic field in the x-y plane is
)sinˆcosˆ(
)]sin(ˆ)cos(ˆ[ˆ
2
2
pypx
pypxRF
II
tItIH
where
P Pt ,
Then the resulting Hamiltonian is
)]sin(ˆ)cos(ˆ[ˆ
ˆˆˆ
21
0
pypxz
RFL
III
HHH
The Hamiltonian in the rotating frame is
)(ˆˆ)(ˆˆ)(
)(ˆ)(ˆ]sinˆcosˆ)[(ˆ)(ˆˆ)(
)(ˆ]sinˆcosˆ)[(ˆ)(ˆˆ)(ˆˆ
)(ˆˆ)(ˆˆˆ
21
21
21
PzxPzz
PzPzPyPxPzPzz
zPyPxzzzzz
zLzzR
RIRI
RRIIRRI
RIIRRIRI
RHRIH
where
xpzpypxpz IRIIR ˆ)(ˆ]sinˆcosˆ)[(ˆ
)(ˆ)(ˆ)(ˆ zPzPz RRR
)(ˆ)(ˆ)(ˆ zPzPz RRR
((Formula))
ypxppzxpz IIRIR ˆ)sin(ˆ)cos()(ˆˆ)(ˆ
ypxppzypz IIRIR ˆ)cos(ˆ)sin()(ˆˆ)(ˆ
or
)(ˆ]ˆ)sin(ˆ))[cos((ˆˆpzypxppzx RIIRI
)(ˆ]ˆ)cos(ˆ)sin()[(ˆˆpzypxppzy RIIRI
Note that these relations can be checked using the Mathematica.
Thus we have
)(ˆˆ)(ˆˆˆ20 PzxpzzR RIRIH
Note that
10 ,
t , pp t ,
pp .
0 is the resonance offset. Note that the time dependence has vanished from this expression.
This is the point of the rotating frame. It transforms a time-dependent quantum-mechanical problem into a time-independent one. The Hamiltonian in the rotation-frame is
)ˆsinˆ(cosˆˆ20 ypxpzR IIIH .
6. Off-Resonance effect
For 010 (resonance condition), the rotating-frame Hamiltonian is given by
)ˆsinˆ(cosˆˆ20 ypxpzR IIIH
with 0p . The state vector in the rotating-frame is
ReffoffRtRt )0()(ˆ)(
where
eff
eff
effeff
eff
i
eff
eff
i
eff
eff
eff
zypxpeffoff
t
ittei
teit
it
ItiIItitR
p
p
)2
sin()
2cos()
2sin(
)2
sin()
2sin(
)2
cos(
]ˆ)ˆsinˆ(cosexp[)(ˆ
02
20
02
where
212
22
02
2 eff
Now we start with
ReffRtRt
p)0()(ˆ)( , and )())((ˆ)( ttRt zR
Thus we get
)0()]0([ˆ)(ˆ)()]([ˆ zeffz RtRttRp
The state vector in the laboratory frame is given by
)0()]0([ˆ)(ˆ)]([ˆ)( zeffz RtRtRtp
Note that
)ˆexp()(ˆzz IiR
where
t
ti
eff
eff
effti
eff
eff
ti
eff
eff
ti
eff
eff
eff
zeffz
et
it
et
i
et
iet
it
RRtR
p
p
p
20)2(
22
]2(2220
)]2
sin()2
[cos()2
sin(
)2
sin()]2
sin()2
[cos(
)]0([ˆ)(ˆ)]([ˆ
When 1 (on resonance), 2 eff and 00
ti
ti
ti
ti
zeffz
et
et
i
et
iet
RRtR
p
p
p
22
11
22)2(
22
)2(2222
)2
cos()2
sin(
)2
sin()2
cos(
)]0([ˆ)(ˆ)]([ˆ
((Mathematica))
Clear "Global` " ;
x PauliMatrix 1 ;
y PauliMatrix 2 ;
z PauliMatrix 3 ;
Rx : MatrixExp2
x ;
Ry : MatrixExp2
y ;
Rz : MatrixExp2
z ;
Roff : MatrixExp 2 tCos
2x
Sin
2y
0 t
2z ;
s1 Roff p . 22 02eff ,
1
22 02
1
eff
FullSimplify;
s1 MatrixFormClear "Global` " ;
x PauliMatrix 1 ;
y PauliMatrix 2 ;
z PauliMatrix 3 ;
Rx : MatrixExp2
x ;
Ry : MatrixExp2
y ;
Rz : MatrixExp2
z ;
Roff : MatrixExp 1 tCos
2x
Sin
2y
0 t
2z ;
s1 Roff p . 12 02 eff,1
12 02
1
eff
FullSimplify;
s1 MatrixForm
7. The form of )(t with initial condition
2sin
2cos
)0(
Suppose that the initial condition is given by
2sin
2cos
)0(
Then we have
2
1
20)2(
22
]2(2220
20)2(
22
)2(2220
2sin)]
2sin()
2[cos(
2cos)
2sin(
2sin)
2sin(
2cos)]
2sin()
2[cos(
2sin
2cos
)]2
sin()2
[cos()2
sin(
)2
sin()]2
sin()2
[cos(
)0()]0([ˆ)(ˆ)]([ˆ)(
ti
eff
eff
effti
eff
eff
ti
eff
eff
ti
eff
eff
eff
ti
eff
eff
effti
eff
eff
ti
eff
eff
ti
eff
eff
eff
zeffz
et
it
et
i
et
iet
it
et
it
et
i
et
iet
it
RtRtRt
p
p
p
p
p
where
122
02
02
2 2eff ,
cos010
coscos 001 B , sinsin 002 B
00 B
Suppose that 0P . The matrix elements can be simplified as
ti
eff
eff
eff
ti
eff
eff
eff
ti
eff
eff
eff
eff
eff
ti
eff
eff
ti
eff
eff
eff
et
it
et
it
et
it
it
et
iet
it
20
200
200
22201
2cos)]
2sin()()
2[cos(
2cos)]
2sin(
)cos1()
2[cos(
}2
sinsin)2
sin(2
cos)]2
sin()2
{[cos(
2sin)
2sin(
2cos)]
2sin()
2[cos(
ti
eff
eff
eff
ti
eff
eff
eff
ti
eff
eff
effeff
eff
ti
eff
eff
effti
eff
eff
ti
eff
eff
effti
eff
eff
et
it
et
it
et
itt
i
et
it
et
i
et
it
et
i
20
200
2020
2020
20212
2sin)]
2sin()()
2[cos(
2sin)]
2sin(
)cos1()
2[cos(
]2
sin)2
sin(2
sin)2
cos(2
cos2
sin)2
sin(2
[
]2
sin)2
sin(2
sin)2
[cos(2
cossin)2
sin(
2sin)]
2sin()
2[cos(
2cos)
2sin(
Then we have
ti
effeff
eff
ti
eff
eff
eff
ett
i
et
it
t
20
20
2
1
2sin)]
2cos()
2sin()([
2cos)]
2sin()()
2[cos(
)(
Note that
1*22
*11
((Proof))
]2
sin)cos(
2[cos
2cos 2
2
200022*
11
tt eff
eff
eff
]2
sin)cos(
2[cos
2sin 2
2
200022*
22
tt eff
eff
eff
2sin]
2sin
)cos(
2cos
)cos([
2cos
]2
sin)cos(
2[cos
2sin
]2
sin)cos(
2[cos
2cos
222
2000
22
20002
22
200022
22
200022*
22*
11
t
t
tt
tt
eff
eff
eff
eff
eff
eff
eff
eff
eff
eff
where
10 , 21
20
2 2 eff , cos01
Her we get
20
2000 )()cos( , 2
02
000 )()cos(
Thus we have
2
022
0
220
220
220
220
22000
22000
cos2
)2
sin2
(cos2
2sin)(
2cos)(
2sin)cos(
2cos)cos(
eff
This leads to the relation
12
sin2
cos 22*22
*11
tt effeff
8 Calculation of 1 and 2 We note that
ti
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
et
it
it
et
it
it
it
et
it
it
20
20
201
)]2
sin(2
sinsin)2
sin(2
cos)cos(
2cos)
2[cos(
)]2
sin(2
sinsin)2
sin(2
coscos)2
sin(2
cos2
cos)2
[cos(
)]2
sin()2
sinsin2
cos(cos)2
sin(2
cos2
cos)2
[cos(
ti
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
et
it
it
et
it
it
it
et
it
it
20
20
202
)]2
sin(2
cossin)2
sin(2
sin)cos(
2sin)
2[cos(
)]2
sin(2
sincos)2
sin(2
cossin)2
sin(2
sin2
sin)2
[cos(
)]2
sin()2
sincos2
cos(sin)2
sin(2
sin2
sin)2
[cos(
where we use the trigonometry law
2sin
2sin
2coscos)
2cos(
2cos
and
2sin
2cos
2cossin)
2sin(
2sin
Then we have
)()2
sin(sin
)()]2
sin()cos(
)2
[cos()(
2
20
tet
i
tet
it
t
tieff
eff
tieff
eff
eff
where
2sin
2cos
)(
tie
t ,
2cos
2sin
)(
tie
t
The equation is exactly the same as Eq.(10.33) of Griffiths book. The probabilities of finding the
system in the states )(t and )(t are evaluated as
)cos(])cos(
1[2
1]
)cos(1[
2
1
)2
(sin)cos(
)2
(cos
)]2
sin()cos(
)2
[cos(
)()(
2
20
2
20
22
202
2
0
2
t
tt
ti
t
ttP
eff
effeff
eff
eff
eff
eff
eff
eff
)2
(sin)sin
()()( 222 tttP eff
eff
respectively. In the adiabatic region 0eff , P reduced to zero. Then we have
)()]2
sin()cos(
)2
[cos()( 20 tet
it
tt
ieff
eff
eff
In the adiabatic process ( 0 ),
0 eff ,
we have
})]cos1([2
exp{
]2
)(exp[
)()]2
sin()2
[cos()(
0
2
ti
ti
tet
it
t
eff
tieffeff
where
cos
)cos1(
cos21
cos21
0
00
00
02
0
2
0
eff
The first factor )2
exp( 0ti is the dynamical phase. The remaining factor is
])cos1(2
exp[)exp( ti
i ,
with the Berry’s phase,
)cos1(2
1 .
The solid angle is the surface area of the unit sphere, swept by the magnetic field is
)cos1(2sin2
00
dd
The phase of change during the period 2
T corresponds to the Berry phase and is given by
2)cos1()cos1(
2
1 T
The factor 1/2 is related to the value of spin (1/2).
9. Average
Now we calculate the expectation value
)]sin()sin()cos()2
1(sin)cos()cos(
)cos()2
1(cos)cos([sin
2
1
)(ˆ)(2
1ˆ
002
000
00022
2
tttt
tt
ttI
effeffeff
effeff
eff
x
)sin()cos()cos()2
1(sin)sin()cos(
)cos()2
1(cos)sin([sin
2
1
)(ˆ)(2
1ˆ
002
000
00022
2
tttt
tt
ttI
effeffeff
effeff
eff
yy
)]}cos()44()(4[cos
)2
1(cos)]3cos()2cos(44[2{
16
1
)(ˆ)(2
1ˆ
220
20
220
20
200002
t
t
ttI
effeffeff
eff
eff
zz
Note that
sinˆ0
txI
0ˆ0
tyI
cosˆ0
tzI
10. The form of )(t with initial condition
2cos
2sin
)0(
Suppose that the initial condition is given by
2cos
2sin
)0(
We note that this state can be rewritten as
)2
sin(
)2
cos(
2cos
2sin
)0(
Then we have
)2
sin(
)2
cos(
)]2
sin()2
[cos()2
sin(
)2
sin()]2
sin()2
[cos(
)0()]0([ˆ)(ˆ)]([ˆ)(
20)2(
22
)2(2220
ti
eff
eff
effti
eff
eff
ti
eff
eff
ti
eff
eff
eff
zeffz
et
it
et
i
et
iet
it
RtRtRt
p
p
p
The form of )(t with initial condition
2cos
2sin
)0(
can be obtained from the form of
)(t with initial condition
2sin
2cos
)0(
by replacing by . Thus we have
ti
effeff
eff
ti
eff
eff
eff
ett
i
et
it
t
20
20
2
1
2cos)]
2cos()
2sin()([
2sin)]
2sin()()
2[cos(
)(
Here we note that
ti
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
et
it
it
et
it
it
it
et
it
it
20
20
201
)]2
sin(2
cossin)2
sin(2
sin)cos(
2sin)
2[cos(
)]2
sin(2
sincos)2
sin(2
cossin)2
sin(2
sin2
sin)2
[cos(
)]2
sin()2
sincos2
cos(sin)2
sin(2
sin2
sin)2
[cos(
ti
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
eff
eff
ti
eff
eff
eff
eff
eff
et
it
it
et
it
it
it
et
it
it
20
20
202
)]2
sin(2
sinsin)2
sin(2
cos)cos(
2cos)
2cos([
)]2
sin(2
sinsin)2
sin(2
coscos)2
sin(2
cos2
cos)2
cos([
)]2
sin()2
sinsin2
cos(cos)2
sin(2
cos2
cos)2
[cos(
Using the basis given by
2sin
2cos
)(
tie
t ,
2cos
2sin
)(
tie
t
)(t can be rewritten as
)()2
sin(sin
)()]2
sin()cos(
)2
[cos()(
2
20
tet
i
tet
it
t
tieff
eff
tieff
eff
eff
In the adiabatic process ( 0 )
00 cos eff ,
})]cos1([2
exp{
]2
)(exp[
)()]2
sin()2
[cos()(
0
2
ti
ti
tet
it
t
eff
tieffeff
where cos0 eff and is the angle of the direction of the magnetic field from the z axis.
The first factor )2
exp( 0ti is the dynamical phase. The remaining factor is
])]cos1(2
exp[)exp( ti
i
with the Berry’s phase,
)cos1(2
1
The solid angle is the surface area of the unit sphere, swept by the magnetic field is
)cos1(2sin2
00
dd
The phase of change during the period 2
T corresponds to the Berry phase and is
2)cos1()cos1(
2
1 T
The factor 1/2 is related to the value of spin (1/2).
11. Summary
The Berry phase for a closed loop increases by the same amount no matter how one goes around it, provided only that one never go so fast as to invalidate the adiabatic approximation. For this reason, it is called a geometric phase. REFERENCES D.J. Griffiths, Introduction to Quantum Mechanics, second edition (Cambridge, 2017). E.S. Abers, Quantum Mechanics (Pearson, 2004). M.H. Levitt, Spin Dynamics Basic of Nuclear Magnetic Resonance, second edition (John Wiley & Sons, 2008).