May 3, 2008
Electron in a rotating magnetic field
Rotating frame
Suppose we have an electron at rest in a magnetic field
B = B0 (sin α cos ωtî + sin α sin ωtĵ + cos αk̂)
So the field vector points along the polar angle α and rotates about the zaxis with frequency ω. It is convenient to transform into the rotating frame.
~ = B0 cos αk̂ + B0 sin αî. At a later time
At t = 0, the magnetic field is B
the field has rotated about the z-axis by an angle θ = ωt. If we rotate the
spinor about the z-axis we can move to a frame in which the hamiltonian is
independent of time. A rotation about z is accomplished with
iθσz /2
R(θ) = e
And since
i(ωt)σz /2
=e
=
iωt/2
e
ω1 h̄
cos α
B0
H=
sin αe−iωt
2
0
0
e−iωt/2
sin αeiωt
− cos α
(1)
then
R−1 HR = H0
−iωt/2
e
=
0
ω1 h̄
=
2
=
ω1 h̄
2
0
eiωt/2
−iωt/2
e
0
ω1 h̄
2
0
eiωt/2
cos α
sin α
sin α − cos α
cos α
sin αe−iωt
cos αeiωt/2
sin αe−iωt/2
sin αeiωt
− cos α
iωt/2
e
e−iωt/2
0
sin αeiωt/2
− cos αe−iωt/2
and H0 is the time independent hamiltonian in the rotating frame.
Then
∂
Hχ = RH0 R−1 χ = ih̄ χ
∂t
1
0
and
∂
χ
∂t
we need to be careful in the next step.
H0 R−1 χ = ih̄R−1
∂
Since R−1 does not commute with ∂t
We find that
∂
∂ −1
∂R−1
(R χ) =
χ + R−1 χ
∂t
∂t
∂t
and also
∂R−1
iω e−iωt/2
ω
0
=−
= −i σz R−1
iωt/2
0
−e
∂t
2
2
With this in mind we can write
H0 R−1 χ = ih̄R−1
∂
∂
h̄ω
χ = ih̄ (R−1 χ) −
σz R−1 χ
∂t
∂t
2
and
∂
∂ 0
χ = ih̄ χ0
∂t
∂t
h̄ω
0
−1
0
where χ = R χ and H = H0 + 2 σz . Now we have a time independent
hamiltonian. To solve we can compute eigenvalues and eigenvectors to get
χ0 (t) and then transform back to the lab frame.
0
Another strategy is to construct the time translation operator e−iH t/h̄ .
First write H 0 /h̄ in the form n̂ · σλ/2
H 0 χ0 = ih̄
H0 =
h̄
2
ω1 cos α + ω
ω1 sin α
ω1 sin α
−ω1 cos α − ω
=
h̄
n̂ · σλ
2
where
nz =
nx =
ny
ω1 cos α + ω
(ω12
1
+ 2ω1 ω cos α + ω 2 ) 2
ω1 sin α
1
(ω12 + 2ω1 ω cos α + ω 2 ) 2
= 0
and
1
λ = (ω12 + 2ω1 ω cos α + ω 2 ) 2
Then
0
Q(t) = e−iH t/h̄ = e−in̂·σλt/2 = cos(λt/2) − in̂ · σ sin(λt/2)
2
=
cos(λt/2) − i(ω1 cos α + ω) sin(λt/2)/λ
−iω1 sin α sin(λt/2)/λ
iω1 sin α sin(λt/2)/λ
cos(λt/2) + i(ω1 cos α + ω) sin(λt/2)/λ
Finally
0
1
0
0
χ(t) = R(t)χ (t) = R(t)Q(t)χ (t = 0) = R(t)Q(t)
Adiabatic approximation
If we return to the original hamiltonian H, and define ψn such that
H(t)ψn (t) = En (t)ψn (t)
then
ψ1 (t) = χ+ (t) =
cos(α/2)
sin(α/2)e−iωt
and ψ2 (t) = χ− (t) =
sin(α/2)eiωt
− cos(α/2)
are eigenvectors of the hamiltonian(Equation1 ) with eigenvalues E± =
cos(α/2)
±h̄ω1 /2. If our initial state is χ+ =
, then
sin(α/2)
χ(t) = R(t)Q(t)χ+
cos(λt/2) − i(ω1 cos α + ω) sin(λt/2)/λ
−iω1 sin α sin(λt/2)/λ
= R(t)
−iω1 sin α sin(λt/2)/λ
cos(λt/2) + i(ω1 cos α + ω) sin(λt/2)/λ
cos(α/2)
×
sin(α/2)
 h
= 
h
i
(cos(λt/2) − i ω1λ+ω sin(λt/2)) cos(α/2) eiωt/2
i
(cos(λt/2) − i ω1λ−ω sin(λt/2)) sin(α/2) e−iωt/2
= (cos(λt/2) − i



ω1 + ω cos α
ω sin α
sin(λt/2))eiωt/2 χ+ (t) − i
sin(λt/2)e−iωt/2 χ− (t)
λ
λ
In the adiabatic limit, ω ω1
ω
cos α) sin(λt/2))eiωt/2 χ+ (t)
ω1
= (cos(λt/2) − i sin(λt/2))eiωt/2 χ+ (t)
χ(t) = (cos(λt/2) − i(1 −
= e−iλt/2 eiωt/2 χ+ (t)
3
and
χ(t) → e−iλt/2 eiωt/2 χ+ (t)
∼ e−i(ω1 +ω cos α)t/2 eiωt/2 χ+ (t)
Berry phase
The dynamic phase is θ+ = −ω1 t/2. The remaining phase is geometric
γ+ = (ω/2)(− cos α +1). Unfortunately I started out rotating in the negative
φ direction. If ω changes sign then γ+ = (ω/2)(cos α − 1) and Berry’s phase
(cos α − 1) = π(cos α − 1). 1/2 of the solid angle subtended by the
is (ω/2) 2π
ω
tip of magnetic field vector.
We could also get the geometric phase by
Z
γ = i hχ+ |
= i
=
Z
Z
∂χ+ 0
idt
∂t0
0
( cos(α/2) sin(α/2)eiωt )
0
0
−iω sin(α/2)e−iωt
dt0
ω sin2 (α/2)dt0
= −π(1 − cos α)
That last − comes from integrating backwards.
Q(t) →=
cos(λt/2) + i cos α sin(λt/2)
i sin α sin(λt/2)
i sin α sin(λt/2)
cos(λt/2) − i cos α sin(λt/2)
4