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5.7. Berry phase
The adiabatic theorem tells us that if we are slow and if we start from an quantum eigenstate of H(t = 0), we will always stay on the same
quantum state. But it doesn’t mean that the wavefunction will remain the same. There could be a phase factor.
If we have a time-dependent Hamiltonian H(t). At each time, we can compute its eigenenergies and eigenwavefunctions. Now we focus on a
specific eigenstates (say the nth excited state), whose eigenenergy is E n and the eigenwavefunction is
χn (t)⟩.
The adiabatic theorem told us that if we change the Hamiltonian really slowly, then when we start from the state
χn (t)⟩, we will always stay
on the same quantum state. i.e., if we have initial condition
ψ(t = 0)⟩ =
χn (t = 0)⟩
(5.213)
then at later time
ψ(t)⟩ = ⅇⅈ ϕ(t) χn (t)
(5.214)
5.7.1. Example (same example discussed above)
If we rotate the B field really slowly, the system will always stay on the same quantum state. Say, we start from spin S parallel to B at t = 0. If
we rotate B field extremely slowly, then at a later time t > 0, the spin will still be parallel to B and it will never become anti parallel.
Q: At time t, the B field comes back to the initial direction (same as at t = 0). What is the wavefunction?
A: It must be the same wavefunction up to a phase factor
ψ(t = 2 π / ω) = ψ(t = 0) ⅇⅈ φ
(5.215)
Q: What is this φ?
A: it contains two parts
φ = φ 1 + φ2
(5.216)
First part, φ1 is a dynamic phase factor. For a static problem (H independent of t), we know that if we start from an eigenstate of H, the time
evolution is
ψn (t) = ψn ⅇ-ⅈ En t/ℏ
(5.217)
The phase factor here -En t / ℏ is our φ1 .
φ1 = -En t / ℏ
(5.218)
Comment #1: φ1 always arises in a quantum system (no matter H depends on time t or not).
Comment #2: φ1 is not a constant number. It depends on t.
The second part φ2 is what we called the Berry phase. It has a geometric original and thus is also known as the geometric phase.
For the example discussed above, we can show that
ψ(t) = A(t) χ+ (t) + B(t) χ- (t)
(5.219)
where
ψ(t) =
cos  λ2t  - ⅈ
cos  λ2t  - ⅈ
cos
χ+ =
ⅇⅈ ω t sin
sin
χ- =
α
2
α
2
and the coefficients
ω1 +ω
λ
sin λ2t  cos α2  ⅇ-ⅈ ω t/2
sin λ2t  sin α2  ⅇⅈ ω t/2
(5.220)
(5.221)
α
2
-ⅇⅈ ω t cos
ω1 -ω
λ
α
2
(5.222)
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λt
A = cos
B=ⅈ
ω
+ⅈ
2
sin
λ
ω cos α - ω1
sin
λt
2
λ
λt
2
 ⅇ-ⅈ ω t/2
(5.223)
ⅇ-ⅈ ω t/2 sin α
(5.224)
When we are doing everything really slowly, ω → 0, B = 0, and thus
ψ(t) = A(t) χ+ (t)
(5.225)
For ω → 0,
λ=
ω2 + ω1 2 - 2 ω ω1 cos α ≈ ω1 - cos α ω
(5.226)
As a result,
A = cos
λt
+ⅈ
2
ω cos α - ω1
ⅇ-ⅈ ω t/2 = cos
λ
λt
2
sin
- ⅈ sin
λt
2
λt
2
 ⅇ-ⅈ ω t/2 ≈ cos
λt
2
+ⅈ
ω cos α - ω1
ω1 - cos α ω
 ⅇ-ⅈ ω t/2 = ⅇ-ⅈ λ t/2 ⅇ-ⅈ ω t/2 = ⅇ-ⅈ
λ+ω
2
sin
λt
2

(5.227)
t
we have
ψ(t) = ⅇ-ⅈ
λ+ω
2
t
χ+ (t) = ⅇ-ⅈ 
ω+ω1 -cos α ω
2
t
ω1
χ+ (t) = ⅇⅈ - 2
The phase factor contains two parts. The first
expect the dynamic phase factor to be -E+ t / ℏ
t-
part - ω21
= - ω21 t
1-cos α
2
ω t
χ+ (t)
(5.228)
t is the dynamic phase factor. Because the energy of the χ+ state is E+ = +ℏω1 / 2, we
The second part is our Berry’s phase
φ2 = -
1 - cos α
2
ωt
(5.229)
At t = 2 π / ω (i.e. when the B field goes back to the initial direction),
φ2 = -
1 - cos α
2
ω
2π
= π (cos α - 1)
(5.230)
ω
When the B field rotate back to the initial direction, the Berry’s phase is a constant, independent of ω (of course we need to assume ω is very
small, but the value of ω doesn’t matter).
Comment #1: the Berry phase only arise when H changes.
Comment #2: the Berry phase is independent of details, e.g. how fast we rotate the B field, as long as it is slow enough; or how strong the B
field is; or what is the value of the g-factor.
5.7.2. Geometric meaning
We can define a unit vector
→
→
n→B = B  B
(5.231)
→
→
This unit vector is one point on the unit sphere. When we rotate the direction of B, this unit vector draws a line on the unit sphere. When B
finally comes back to the initial direction, the unit vector draws a close loop on the unit sphere.
This close loop enclose an area on the unit sphere, whose area is ΩB . The Berry’s phase is nothing but
φB = -s ΩB
(5.232)
where s is the spin quantum number. For electrons s = 1 / 2. For general quantum particles, s is integer (bosons) or half integer (fermions).
For the case we discussed above, when we rotate B around the z axis, the n→B draws a circle on the unit sphere. The (spherical) area enclosed by
this circle is
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α
ΩB =  2 π sinθ ⅆ θ = -2 π cosθ
0
θ=0
θ=α
= 2 π (1 - cos α)
(5.233)
Thus,
φB = -s ΩB =
1
2
× 2 π (1 - cos α) = π(cos α - 1)
(5.234)
Comment #1: in the example, we assume that n→B draws a perfect circle on the unit sphere. If the path is not a circle, all the conclusions
remains the same
Comment #2: We need to put a +/- sign in front of ΩB . The sign is determined by the direction (orientation) of the path. This is
because, if we reverse the path (reverse everything), the phase need to flip sign. The convention that we take here is, we use choose one
point inside the circle as our reference point. When n→B draw the circle, the ending point of n→B moves around my reference point. This
motion defines an angular momentum (relative to my reference point). If the angular momentum is positive (pointing from the center
of the unit sphere to outside), ΩB > 0. Otherwise, it is negative.
5.7.3. Quantization of spins
One thing the Berry phase tells us is that spins must quantize to integer or half-integer values. To see that, we look at the circle discussed above.
If we sit at the north pole, we find that n→B draws a circle around and the angular momentum is positive (pointing up). The area of the circle is
ΩB = +2 π (1 - cos α). The + sign here means that the angular motion has angular momentum along the positive direction (up).
If we sit at the south pole, we will also find that n→B draws a circle around, but the angular momentum is negative (pointing down, relative to the
south pole).
NOTE: the angular momentum is always along +z direction. For north pole, +z is up (from the center of the sphere to out side). But for south
pole, +z is down, pointing from outside the sphere towards the center.
Here, if we sit a the south pole, we find that the area enclosed by the circle is negative
π
ΩB ' = - 2 π sinθ ⅆ θ = 2 π cosθ
θ=α
θ=π
= 2 π (-1 - cos α) = -2 π(1 + cos α)
(5.235)
α
It is easy to notice that
ΩB - ΩB ' = ΩB + ΩB ' = area of the whole unit sphere = 4 π
(5.236)
Now, we have two different Berry phases
φB = -s ΩB
(5.237)
φB ' = -s ΩB ' = -s(4 π + ΩB ) = -4 π s - s ΩB
(5.238)
and
This two phases differ by 4 πs. But we know that they cannot be different (there is only one unique Berry phase). The only possible option here
is that they different by 2 πn
4 πs = 2 π n
(5.239)
s = n/2
(5.240)
So
If n is even, s is an integer.
If n is odd, s is a half integer.
5.7.4. a generic way to compute Berry phase φB
If we have a time-dependent Hamiltonian H(t). At each time, we can compute its eigenenergies and eigenwavefunctions. Now we focus on a
specific eigenstates (say the nth excited state), whose eigenenergy is E n and the eigenwavefunction is
χn (t)⟩.
The adiabatic theorem told us that if we change the Hamiltonian really slowly, then when we start from the state
χn (t)⟩, we will always stay