Lectur
Lecture
e 17:
17: The adiabat
The adiabatic
ic appro
approxima
ximation:
tion:
Berry’s phase & Aharonov-Bohm effect (11/17/2005)
Berry’s phase & Aharonov-Bohm effect (11/17/2005)
In classical physics, the axis of the motion of a pendulum rotates a bit if we walk around America In classical physics, the axis of the motion of a pendulum rotates a bit if we walk around America with the pendulum. In 1984, M.V. Berry noticed a similar effect – a geometric phase – in quantum with the pendulum. In 1984, M.V. Berry noticed a similar effect – a geometric phase – in quantum mechanics that no one had observed for 59 years.
mechanics that no one had observed for 59 years. Addit
Additionaional reading if you wish: l reading if you wish: GriffiGriffiths 10.2, Sakuths 10.2, Sakurai sup. rai sup. 11
Earlier, Aharonov and Bohm explained that particles are able to detect a magnetic field even in Earlier, Aharonov and Bohm explained that particles are able to detect a magnetic field even in regio
regions awans away from y from their trajecttheir trajectorieories. s. WWe will loe will look at ok at insiginsighthts s of Berry and of Berry and of Aharonoof Aharonov and v and BohmBohm –
– and and their their relationships.relationships.
Clas
Classical analogu
sical analogue:
e: pendul
pendulum
um
Sta
Start at rt at the Nortthe North h pole with a pole with a pendpenduluulum m in in yyour hand. our hand. WWalk towalk towardards s the equthe equatorator, , aloalong a ng a gregreatat circ
circle. le. TheThen turn n turn leftleft, walk around the equator by the angle, walk around the equator by the angle ϑ ϑ. Then turn left again and return to. Then turn left again and return to the Nor
the North pole. th pole. WhaWhat will happen witt will happen with h the pendthe penduluulum’s swinm’s swing? g? It will cleIt will clearlarly y rotrotate byate by ϑϑ. . NoNotete that this angle
that this angle ϑ ϑ is equal to the solid angle Ω subtended by the path: is equal to the solid angle Ω subtended by the path: A A = = (4(4πRπR22))11 22
ϑϑ 22ππ
= = ϑR ϑR22 = Ω= ΩRR22This statement is clearly true for any path. Draw a picture. This statement is clearly true for any path. Draw a picture.
A
A FFoucoucaulault t pendpenduluulum m is an is an exexampample. le. It is It is locatlocated at ed at latlatituitudede ϑϑ. . EarthEarth’s rotatio’s rotation effecn effectivtivelyely cause
causes s the pendulum to the pendulum to movmove with respect to e with respect to a a non-rnon-rotatinotating “ghost” within Earth. g “ghost” within Earth. But the effectBut the effect is identical. After one day, the pendulum subtends the solid angle
is identical. After one day, the pendulum subtends the solid angle Ω = Ω =
22ππ 0 0 dφdφ
11 coscos ϑϑdd(cos((cos(θθ
)) = 2)) = 2ππ(1(1
−
−
coscos ϑϑ)) whiwhich is ch is alsalso o the angle by whicthe angle by which h the axis of the axis of the penduthe pendulum rotatlum rotates after one es after one dadayy. . In In the rotathe rotatintingg frame, we typically attribute the very same angle to the Coriolis’ force, but you can see that its frame, we typically attribute the very same angle to the Coriolis’ force, but you can see that its origin may also be argued to be geometric.
origin may also be argued to be geometric.
Quan
Quantum mech
tum mechanics
anics:
: close
closed
d path in
path in the space
the space of
of Hami
Hamilton
ltonians
ians
In quantum mechanics, we replace the pendulum by any quantum mechanical system we like, and In quantum mechanics, we replace the pendulum by any quantum mechanical system we like, and the closed path is replaced by a path in the space of Hamiltonians. We will follow this path slowly the closed path is replaced by a path in the space of Hamiltonians. We will follow this path slowly i.e
i.e. . adiadiabatabaticicallyally. . And the And the phaphase se wwe’re’re e ininterteresested ted in in is is notnothinhing g elselse e thathan n the complthe complex ex ququanantum tum--mechanical phase of the physical system we study. You can see that this is a special case of Bailes’ mechanical phase of the physical system we study. You can see that this is a special case of Bailes’ phase we discussed last time but because the path is closed, the resulting change of the phase is phase we discussed last time but because the path is closed, the resulting change of the phase is called Berry’s phase.
called Berry’s phase.
In other words, the latitude and longitude are replaced by external parameters on which the In other words, the latitude and longitude are replaced by external parameters on which the Hamiltonian depends.
Hamiltonian depends.
Finally, we should try to calculate Berry’s phase. Recall that the standard phase associated with Finally, we should try to calculate Berry’s phase. Recall that the standard phase associated with time-independent Hamiltonians is
time-independent Hamiltonians is
ψ
What about time-dependent Hamiltonians? The most straightforward answer is to replace E nt in
the exponent – which is not well-defined because E n changes – by the integral
ttif E n(t)dt. Note
that it gives the same answer if the energy is constant. A natural guess – Ansatz – is therefore ψ(t) = ψn(t)e
t 0dt
E
n(t)/i¯heiγ (t)
where we had to use ψn(t) – the eigenstate of the Hamiltonian H (t) whose character depends on
time; the usual “dynamical” phase θn(t), expressed using an integral; and Berry’s phase eiγ (t) to
account for the phase errors we may have overlooked. What should we do with this Ansatz? Much like with other Ans¨atze, we should plug it into an equation. In this case it is the time-dependent Schr¨odinger equation
i¯h∂ψ(t)
∂t = H (t)ψ(t)
The left-hand side contains a time-derivative of a product. It is clear that all resulting terms will be proportional to θn(t) exp(iγ (t)), so let us divide the resulting equation by this factor (both phases)
immediately:
i¯h∂ψn(t)
∂t + E n(t)ψn(t)
−
¯hdγ n(t)
dt ψn(t) = H (t)ψn(t)
≡
E n(t)ψn(t)Differential equation for the phase
You see that the terms containing E n(t) cancel and we have, after dividing by i¯h,
∂ψn(t)
∂t =
−
idγ n(t)
dt ψn(t)
We can obtain a formula for the time-derivative of γ itself if we multiply the equation by
ψn|
:dγ n
dt = i
ψn|
∂ψn∂t
Don’t forget our goal: to compute the total phase increment ∆γ resulting from a closed path in the space of Hamiltonians. Imagine that the Hamiltonian depends on several parameters – such as the size of various fields etc. – and we formally unify all these parameters into a vector R; it is a vector in a d-dimensional space where d can be any integer and the space has nothing to do with the space where we live. The closed path in the space of possible Hamiltonians is determined by a vector-valued function R(t) of time. What is the change of the eigenstate with time? Recall the rules for derivatives of composite functions:
∂ψn(t) ∂t =
∇
Rψ·
d R dt≡
∂ψn ∂R1 ∂R1 ∂t + ∂ψn ∂R2 ∂R2 ∂t + . . . Now we can simply integrate the equation for dγ n/dt to get∆γ n = i
ψn( R)|
∇
Rψn( R)
d Rwhere d R came from dt d R/dt. If the space of vectors R were one-dimensional, you can see that ∆γ n
will be zero because the contributions from the two parts of the path (back and forth) will cancel; they only differ by the sign.
As we mentioned last time, the total probability – the squared norm of the wavefunction – is conserved. This means, in our case, that ∆γ must be real. Indeed, the whole factor in the integrand
ψn( R)|
∇
Rψn( R)
is pure imaginary, making the whole integral real (point by point). Why is it pureimaginary? Note that
ψn( R)|
ψn( R)
= 1⇒
0 = ∇
R
ψn|
ψn
Using the Leibniz rule,
⇒
0 =
∇
Rψn|
ψn
+
ψn|
∇
Rψn
=
ψn|
∇
Rψn
∗ +
ψn|
∇
Rψn
The last sum is nothing else than twice the real part of
ψn|
∇
Rψn
, and because the real partvanishes, the object itself must be pure imaginary. This proof was mathematically identical to the proof that the momentum operator is Hermitean even though the
∇
operator acted on a different space.Analogy with the magnetic flux
We should notice that our formula for the phase
∆γ n = i
ψn( R)|
∇
Rψn( R)
d Ris reminiscent of the magnetic flux through a surface Σ: Φ =
Σ B·
dArea =
Σ(∇ ×
A)·
dArea =
∂ Σ A·
dlThe last step used Stokes’ theorem. Applying Stokes’ theorem in the opposite direction to our situation and assuming that R lives in a three-dimensional space for a moment, we see that
∆γ n = i
ψn( R)|
∇
Rψn( R)
d R =−
S
V n
·
dS where V n is our counterpart of the magnetic field B:
V n
≡
Im ∇
R×
ψn|
∇
Rψn
We can bring the last expression to a more symmetric form if we realize that
∇ ×
(f ∇
g) = ∇
f×
∇
g + f (∇ ×
∇
g) = ∇
f×
∇
g and therefore
V n = Im
∇
Rψn| × |
∇
Rψn
Note that the cross-product with itself would vanish if the gradient were real, but because it is complex, it is a cross product with someone else (its complex conjugate) and it does not vanish. Let us also insert a complete set of states:
V n =
m=nIm
∇
Rψn|
ψm ×
ψm|
∇
Rψn
where we omitted the term m = n because it vanishes, being the imaginary part of a manifestly real number.
Instead of calculating gradients of the eigenvectors ψm, it may be more convenient to find an
expression that involves the gradient of the Hamiltonian itself (and its matrix elements). To achieve this new form of the formula, we first prove that for m
= n,
ψm|
∇
Rψn
=
ψm|
∇
RH|
ψn
E n
−
E mWhy is it true? It is because ˆ
H
|
ψn
= E nψn⇒
(∇
R ˆH )|
ψn
+ ˆH|
∇
Rψn
= ( ∇
RE n)|
ψn
+ E n|
∇
Rψn
and because the last equation can be multiplied by
ψm|
:
ψm|
(∇
RH ) ˆ|
ψn
+
ψm|
H ˆ|
∇
Rψn
=
ψm|
(∇
RE n)|
ψn
+
ψm|
E n|
∇
Rψn
The first term on the right-hand side vanishes because it is proportional to a vanishing inner product. The second terms on both sides give (E n
−
E m)
ψm|
∇
Rψn
and the “new form” of the equation followsdirectly. Using this new form, we may also rewrite V n as
V n =
m=nIm
ψn|
∇
RH|
ψm ×
ψm|
∇
RH|
ψn
(E n−
E m)2Example: particle in rotating magnetic field
Consider an atom with an arbitrary spin in a constant B magnetic field whose direction is slowly moving. In this case, R
≡
B is indeed three-dimensional. The Hamiltonian and the corresponding energy eigenvalues areˆ
H ( B) =
−
gµ S·
B/¯h⇒
E m =−
gµmBThe gradient of the Hamiltonian in the parameter space (parameterized by the magnetic field) is
∇
B ˆH =−
gµ S/¯hWe always measure m with respect to the axis given by the instantaneous value of the magnetic field. To calculate V m, we will need the matrix elements of the spin S . Only m
= m contributeto V m. Recall that the matrix element
s, m|
S |
s, m
only has nonzero matrix elements for m = m(which we do not need) and for m = m
±
1, and these nonzero matrix elements are those from the raising and lowering operators:m = m + 1 :
s, m|
S |
s, m
= 2h¯
s(s + 1)−
m(m + 1)(ex−
iey)m = m
−
1 :
s, m|
S |
s, m
= ¯h2
s(s + 1)−
m(m−
1)(ex + iey)Substituting into our formula for V m, we have
V m = g2µ2/2 (
−
[s(s + 1)−
m(m + 1)] + [s(s + 1)−
m(m−
1)]) ez g2µ2B2 = mez B2Note that the parameters g and µ canceled and the result has a simple geometric form. The Berry’s phase is simply the flux
∆γ m =
−
m
1
and it is simply given by (
−
m) times the solid angle surrounded by the closed path. The result is completely analogous to the classical result for the pendulum.The problem was mathematically equivalent to the flux of the magnetic field of a point-like magnetic monopole except that the contours (and other geometric objects such as the solid angle) were taken in the three-dimensional space of possible values of B rather than the ordinary three-dimensional space. Finally, you should notice that
•
the total Berry’s phase was independent of the magnitude of B. Instead, it was determined purely by the solid angle, a geometric quantity. The cancellation of all the dimensionful parameters – and the independence on the speed of the changes – is what we mean when we say that the effect is purely geometrical•
in the special case mΩ = π – which can occur both for fermions as well as bosons – we have exp(iγ m) =−
1•
the formula for the phase breaks down if we probe the singular point at B = 0; at this point the states are degenerate in energy (many states have the same energy).•
even if we keep B
= 0 and avoid the degenerate point, you can see that the actual phase is measured by the solid angle defined with respect to the degenerate pointAharonov-Bohm effect
Much earlier, in 1959, a special example of the Berry’s phase was figured out by David Bohm and his Israeli student Yakir Aharonov. When Aharonov was defending his PhD thesis, the committee agreed that the content was nonsensical but Aharonov had made such a good job in defending it that he had to pass. Nevertheless, the effect was observed in 1960 and today virtually everyone believes that the effect obviously exists. It has become a part of introductory lectures on quantum mechanics like this one.
What’s the problem? The effect involves a charged particle moving outside a solenoid. In classical physics, the electromagnetic field is fully described by E (r) and B(r), and the force acting on a particle is the Lorentz force
F = e( E + v
×
B).If E = 0 and B = 0 along the trajectory of your particle, the force simply vanishes. There is no field. You can still express E, B in terms of the potentials φ, A
E =
−
∇
φ−
∂ A∂t , B =
∇ ×
Abut we only use them for convenience, and it does not matter that φ, A may be nonzero. As long as E and B are zero, we say that there is no field and no physical effect caused by such a field. For example, you can always change φ, A by the so-called gauge transformation
φ
→
φ−
∂ Λ∂t , A
→
A + ∇
Λwhich will change φ, A but not E, B, and therefore physics is unchanged.
The situation in quantum mechanics is analogous – there are also E, B,φ, A fields and gauge transformations. But the quantum mechanical case differs in certain subtle features that actually allow us to “see” some “parts” of φ, A even though they do not affect E, B. What do we mean?
Recall that the quantum Hamiltonian is constructed from the classical Hamiltonian by adding hats: ˆ H = 1 2m
pˆ−
q A(ˆr)
2 + qφ(ˆr)Recall that (ˆ p
−
q A) = mˆv is proportional to the actual velocity and its expectation value is invariant under the gauge transformations. These transformations also act on the wavefunctionψ(r)
→
ψ(r)eiqΛ(r)/¯hand we will study the combined transformation later. Once again, gauge transformations change φ, A but they also change ˆ p in such a way that the combination is not changed as we will see.
What Aharonov and Bohm have shown is that the particle is affected by φ and especially A even though E and B vanishes everywhere along the trajectory of the particle. Consider a very long solenoid – draw a long cylinder whose cross section is a circle of radius a. It can be produced by a wire (with a current) densely wrapped on the cylinder. Outside the solenoid, the magnetic field is zero. Inside the solenoid, the magnetic field is parallel to the axis of the solenoid. It is constant and its value is B. The total flux and A outside the solenoid can be calculated:
Φ = πa2B, A(r,θ,φ) = Φ 2πrˆeφ
Check that the contour integral of A around the solenoid gives you the flux Φ as required by Stokes’ theorem.
Now we want to study a charged particle. Imagine a bead on a wire of radius b surrounding the solenoid; b > a. Add the new wire to your picture. The Hamiltonian for the charged particle is
ˆ
H = 1
2m
−
¯h2
∇
2+ 2i¯hq A·
∇
+ q 2A2
where we could be sloppy about the ordering of the mixed term because the commutator is propor-tional to
∇ ·
A = 0 in our case. Because we only allow the particle to move in the φ direction, we should now replace
∇ →
ˆebφdφd . With this substitution, the Schr¨odinger equation reads1 2m
−
¯h 2 b2 d2 dφ2 + i ¯hq Φ πb2 d dφ +
q Φ 2πb
2
ψ(φ) = Eψ(φ)and it is a homogeneous (=without the absolute term) linear differential equation of the general form
α d 2 dφ2 + β d dφ + γ
ψ(φ) = 0.Such equations have solutions of the type ψ = exp(λφ) where λ are the solutions of the characteristic quadratic equation αλ2 + βλ + γ = 0
⇒
λ =−
β±
√
β 2−
4αγ 2α In our case α =−
¯h 2 b2 , β = i ¯hq Φ πb2 , γ =
q Φ 2πb
2−
2mEwhich leads to the solutions λ = i
q Φ 2π¯h±
b ¯h√
2mE
.Because the wavefunction must be periodic, ψ(φ + 2π) = ψ(φ), we see that λ must be i times an integer, i.e. λ = in. We see that
n = q Φ 2π¯h
±
b ¯h
√
2mE or E = En = ¯h2 2mb2
n−
2π¯hq Φ
2 for n = 0,±
1,±
2, . . . When the flux was zero, n and−
n corresponded to a particle going in opposite directions with the same speed, and these two states had the same energy. A nonzero flux Φ breaks the degeneracy. You see that the spectrum depends on the flux Φ even though the particle is confined to a wire in which
B = 0. That could not happen in classical physics. If you go to the classical limit, the spacing of eigenstates becomes infinitely dense and you won’t be able to determine n accurately, i.e. you won’t be able to see any effect of Φ as ¯h
→
0.Gauge transformation
You see that the flux modified the spectrum in a rather simple way. In fact, the solutions of the full Schr¨odinger equation – including the nonvanishing A-terms – in a field-free region (where
∇×
A = 0) may be obtained asψ = eiqΛ/¯hψ, Λ(r)
≡
rr0
A(r)
·
drwhere ψ is a solution of the simpler equation with A = 0 in the same region.
Proof: We start with the assumed validity of the simpler Schr¨odinger equation
pˆ2 2m + V (r)−
i¯h ∂ ∂t
ψ = 0 The important step we want to prove is(ˆ p
−
q A)ψ = (ˆ p−
q A)eiqΛ/¯hψ = eiqΛ/¯h(ˆ p)ψIn the last step, the term
−
q A exp(iq Λ/¯h)ψ cancelled the opposite term coming from the gradient ofthe phase exp(iq Λ/¯h) in the Leibniz rule. You can use the same step again, replacing ψ by (ˆ p
−
q A)ψ, to see that(ˆ p
−
q A)2ψ = (ˆ p−
q A)2eiqΛ/¯hψ = (ˆ p−
q A)eiqΛ/¯h(ˆ p)ψ = eiqΛ/¯h(ˆ p)2ψIn this derivation, all operators act on everything on the right. You see that everytime we move the phase exp(iq Λ/¯h) to the left, we transform (ˆ p
−
q A) to the simple (ˆ p). Therefore the simple,
A-free Schr¨odinger equation for ψ multiplied by the phase exp(iq Λ/¯h) is completely identical to the
“complicated” equation for ψ. Of course, the 1/2m factor can be added to the equation above and the V ψ and E ψ terms in the equation are unaffected by the phase redefinition. QED.
Incidentally, we could have also derived ψ from ψ times the phase whose Λ was time-dependent.
In that case, the Schr¨odinger equation for ψ (without the potential Φ) would include an extra term
−
¯hq (∂ Λ/∂t)ψ times the usual phase on the “time-derivative” side, and this term would cancelAharonov-Bohm interference pattern
The statement we have just proved means that by changing the phase of ψ appropriately, we can essentially set A = 0 in the field-free regions. However, it is not quite right because the “appropriate” phase for ψ(r) depends on the contour that ends at r. The two contours that avoid the solenoid in opposite directions lead to phases that differ by q Φ/¯h where Φ is the magnetic flux. Aharonov and Bohm realized that this phase shift can be measured by an interference (double-slit) experiment where a solenoid is inserted in between the two slits. Draw a figure of the experiment.
The position of the interference maxima depends on the magnetic flux Φ even though the particle never visits the space with B
= 0. Note that the interference patterns disappear in the classical limit, together with all these extra phases. In other words, if you send ¯h→
0, all the phases are essentially random – being proportional to a large number 1/¯h.Aharonov-Bohm effect as an example of Berry’s phase
The Aharonov-Bohm phase γ = q Φ/¯h can also be calculated as a special example of Berry’s phase we started with today. Imagine that we have a particle in a box of medium size and we slowly move the box – whose central position is R – around the solenoid. Let us start from the final steps and the result of Berry’s derivation: the overall phase will be computed from our oldest integral formula for ∆γ n: ∆γ n = i
ψn|
∇
Rψn ·
d R = q ¯h
A( R)·
d R = q Φ ¯hwhich is the Aharonov-Bohm result for the phase. The last step is clear but the previous step is not. The required identity is
ψn|
∇
Rψn
=−
iq
¯hA( R).
How do we prove it? We consider R, the position of the box, to define our adiabatic parameters, and we define the corresponding eigenstates associated with the same eigenvalues using our previous trick ψn = eiqΛ/¯hψn , Λ(r)
≡
r R A(r)·
dr This implies that
∇
Rψn = ∇
R[eiqΛ/¯hψn (r−
R)] =−
i q ¯hA( R)e iqΛ/¯hψ n(r−
R) + e iqΛ/¯h∇
Rψn (r−
R)The desired identity is obtained by multiplying this ket-equation by
ψn|
. You can see that you getwhat you need from the left-hand side and the first term of the right-hand side. The second term on the right hand side will vanish after being contracted with
