Lecture 17
Short Description
The adiabatic approximation...
Description
Lecture Lecture 17: The adiabatic adiabatic approxima approximation: tion: 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 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.
Additiona Additionall reading if you wish: Griffiths Griffiths 10.2, Sakurai Sakurai sup. 1 Earlier, Aharonov and Bohm explained that particles are able to detect a magnetic field even in regions regions away away from their trajectorie trajectories. s. We will look lo ok at insight insightss of Berry and of Aharonov Aharonov and Bohm – and their relationships.
Classical Classical analogue: analogue: pendulum pendulum Start Start at the North North pole with a pendulu pendulum m in your hand. Walk toward towardss the equator equator,, along along a great great circle. circle. Then Then turn left, left, walk around the equator by the angle ϑ. ϑ . Then turn left again and return to the North North pole. What What will happen with with the pendulu pendulum’s m’s swing? swing? It will clearl clearly y rotate rotate by ϑ. Note Note that this angle ϑ angle ϑ is equal to the solid angle Ω subtended by the path:
1 ϑ (4πR 2 ) = ϑR = ϑR 2 = ΩR Ω R2 A = (4πR 2 2π This statement is clearly true for any path. Draw a picture. A Foucaul oucaultt pendulu pendulum m is an examp example. le. It is located located at latitu latitude de ϑ. Earth’s Earth’s rotation rotation effectiv effectively ely causes causes the pendulum to move move with respect to a non-rotatin non-rotating g “ghost” within Earth. But the effect is identical. After one day, the pendulum subtends the solid angle 2π
1
Ω= dφ 0
cos ϑ
(cos(θ )) = 2π 2π(1 d(cos(θ
− cos ϑ)
which which is also also the angle by which which the axis of the pendulum pendulum rotates rotates after one day day. In the rotatin rotatingg 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.
Quantum Quantum mechanics mechanics:: closed closed path in the space of Hamilton Hamiltonians ians 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 i.e. i.e. adiabat adiabatic ically ally.. And the phase phase we’re e’re inter interes ested ted in is nothin nothingg else else than than the complex complex quan quantum tum-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 called Berry’s phase. In other words, the latitude and longitude are replaced by external parameters on which the Hamiltonian depends. Finally, we should try to calculate Berry’s phase. Recall that the standard phase associated with time-independent Hamiltonians is t/ih ¯ ψ(t) = ψn eE n t/i¯ 1
What about time-dependent Hamiltonians? The most straightforward answer is to replace E n t in t the exponent – which is not well-defined because E n changes – by the integral tif E n (t )dt . Note that it gives the same answer if the energy is constant. A natural guess – Ansatz – is therefore
t
ψ(t) = ψ (t)e 0
n
dt E n (t )/i¯ h iγ (t)
e
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 ∂ψ(t) = H (t)ψ(t) i¯h ∂t The left-hand side contains a time-derivative of a product. It is clear that all resulting terms will be proportional to θ n (t) exp(iγ (t)), so let us divide the resulting equation by this factor (both phases) immediately: i¯h
∂ψ n (t) + E n (t)ψn (t) ∂t
− h¯ dγ dt(t) ψ (t) = H (t)ψ (t) ≡ E (t)ψ (t) n
n
n
n
n
Differential equation for the phase You see that the terms containing E n (t) cancel and we have, after dividing by i¯h, ∂ψ n (t) = ∂t
−i dγ dt(t) ψ (t) n
n
We can obtain a formula for the time-derivative of γ itself if we multiply the equation by ψn :
|
dγ n ∂ψn = i ψn dt ∂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 it is a the size of various fields etc. – and we formally unify all these parameters into a vector R; 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 of time. What is the change of the eigenstate with time? Recall the a vector-valued function R(t) rules for derivatives of composite functions: ∂ψn (t) d R = Rψ ∂t dt
∇ ·
≡ ∂ψ ∂R
∂R 1 ∂ψn ∂R 2 + + . . . ∂R 2 ∂t 1 ∂t
n
Now we can simply integrate the equation for dγ n /dt to get
|∇ ψ (R) dR ∆γ = i ψ (R) n
n
R n
where d R came from dt dR/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 n must be real. Indeed, the whole factor in the integrand 2
|∇ ψ (R)
is pure imaginary, making the whole integral real (point by point). Why is it pure imaginary? Note that n
R ψn (R)
|ψ (R) = 1 ψ (R) n
0 = R ψn ψn
⇒
n
∇ |
Using the Leibniz rule,
∇ | |∇ |∇ + ψ |∇ ψ ψ , and because the real part The last sum is nothing else than twice the real part of ψ |∇ vanishes, the object itself must be pure imaginary. This proof was mathematically identical to the operator acted on a different proof that the momentum operator is Hermitean even though the ∇ ⇒
0 = R ψn ψn + ψn R ψn = ψn R ψn n
∗
R
n
R n
n
space.
Analogy with the magnetic flux We should notice that our formula for the phase
|∇ ψ (R) dR ∆γ = i ψ (R) n
n
R n
is reminiscent of the magnetic flux through a surface Σ:
· Φ = B · dArea = (∇ × A) · dArea = A dl Σ
Σ
∂ Σ
The 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
|∇ ψ (R) dR = − V · dS ∆γ = i ψ (R) n
n
R
n
S
n
n is our counterpart of the magnetic field B: where V n V
≡ Im ∇ × ψ |∇ ψ R
n
R n
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
n = Im R ψn V
∇ | × |∇ ψ R n
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: n = V
Im ∇ ψ |ψ × ψ |∇ ψ R
n
m
m
R
n
m =n
where we omitted the term m = n because it vanishes, being the imaginary part of a manifestly real number.
3
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,
ψ |∇ ψ = ψ E |∇− H E |ψ m
m
R n
R
n
n
m
Why is it true? It is because ˆ ψn = E n ψn H
⇒ (∇
|
ˆ ψn + H ˆ R ψn = ( R E n ) ψn + E n R ψn
R H )
|
|∇ ∇ | |∇ and because the last equation can be multiplied by ψ |: ˆ )|ψ + ψ |H ˆ |∇ ψ = ψ |(∇ E )|ψ + ψ |E |∇ ψ ψ |(∇ H m
m
R
n
m
R
n
m
R
n
n
m
n
R n
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 R ψn and the “new form” of the equation follows n as directly. Using this new form, we may also rewrite V
−
n = V
|∇
Im ψ |∇ H |ψ × ψ |∇ H |ψ n
R
m
(E n
m =n
−
m E m )2
R
n
Example: particle in rotating magnetic field magnetic field whose direction is slowly Consider an atom with an arbitrary spin in a constant B B is indeed three-dimensional. The Hamiltonian and the corresponding moving. In this case, R energy eigenvalues are
≡
ˆ (B) = H
h −gµS · B/¯
⇒
E m =
−gµmB
The gradient of the Hamiltonian in the parameter space (parameterized by the magnetic field) is
∇
ˆ
B H =
h −gµS/¯
We always measure m with respect to the axis given by the instantaneous value of the magnetic m , we will need the matrix elements of the spin S . Only m = m contribute field. To calculate V m . Recall that the matrix element s, m S s, m only has nonzero matrix elements for m = m to V (which we do not need) and for m = m 1, and these nonzero matrix elements are those from the raising and lowering operators:
s, m |S |s, m = s, m |S |s, m =
m = m + 1 : m = m
±
||
−1:
¯ h 2
¯ h 2
m, we have Substituting into our formula for V g 2 µ2 /2 ( [s(s + 1) V m =
−
s(s + 1) − m(m + 1)(e − ie ) s(s + 1) − m(m − 1)(e + ie ) x
y
x
y
− m(m + 1)] + [s(s + 1) − m(m − 1)]) e
z
g 2 µ2 B 2
=
m ez B2
Note that the parameters g and µ canceled and the result has a simple geometric form. The Berry’s phase is simply the flux 1 ˆ ∆γ m = m B dB = mΩ C B 2
−
·
4
−
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) rather than the ordinary threewere taken in the three-dimensional space of possible values of B 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 ) = −1 0; at this point • the formula for the phase breaks down if we probe the singular point at B = m
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 point
Aharonov-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 (r) and B( r), and the force acting classical physics, the electromagnetic field is fully described by E on a particle is the Lorentz force = e(E + v B). F
×
0 and B = 0 along the trajectory of your particle, the force simply vanishes. There is no If E = field. You can still express E, B in terms of the potentials φ, A = E
−∇ φ − ∂ ∂tA ,
B =
∇ × A
but we only use them for convenience, and it does not matter that φ, A may be nonzero. As long and B are as E 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 − ∂ ∂tΛ ,
A
→ A + ∇Λ
which will change φ, A but not E, B, and therefore physics is unchanged. The situation in quantum mechanics is analogous – there are also E, and gauge B,φ, A fields 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? 5
Recall that the quantum Hamiltonian is constructed from the classical Hamiltonian by adding hats: 2 ˆ = 1 pˆ q H A(ˆr ) + qφ(ˆr) 2m Recall that ( pˆ q A) = mvˆ is proportional to the actual velocity and its expectation value is invariant under the gauge transformations. These transformations also act on the wavefunction
−
−
ψ(r)
iqΛ(r)/¯ h
→ ψ(r)e
and we will study the combined transformation later. Once again, gauge transformations change ˆ such a way that the combination is not changed as we will see. they also change p in φ, A but What Aharonov and Bohm have shown is that the particle is affected by φ and especially A and B vanishes even though E 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 = A(r,θ,φ) eˆφ 2πr
Φ = πa2 B,
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 = 1 H 2m
−h¯ ∇ + 2i¯hq A · ∇ + q A 2
2
2
2
where we could be sloppy about the ordering of the mixed term because the commutator is propor 0 in our case. Because we only allow the particle to move in the φ direction, we tional to A = should now replace eˆφ d . b dφ With this substitution, the Schr¨odinger equation reads
∇·
∇→
1 2m
−
¯ 2 d2 ¯ q Φ d h h q Φ + i + 2πb b2 dφ2 πb2 dφ
2
ψ(φ) = Eψ(φ)
and it is a homogeneous (=without the absolute term) linear differential equation of the general form d2 d α 2 + β + γ ψ(φ) = 0. dφ dφ
Such equations have solutions of the type ψ = exp(λφ) where λ are the solutions of the characteristic quadratic equation 2
2
αλ + βλ + γ = 0 In our case α =
√ − β ± β − 4αγ λ =
−
¯2 h , b2
⇒
2α
¯ q Φ h β = i 2 , πb 6
2
q Φ − 2mE γ = 2πb
which leads to the solutions λ = i
q Φ
±
2π¯h
√
b 2mE . ¯ h
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 q Φ n = 2π¯h
±
√
b 2mE ¯ h
¯2 h E = E n = n 2mb2
or
−
q Φ 2π¯h
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 0. That could not happen in classical physics. If you go to the classical limit, the spacing of B = 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 0) Schr¨odinger equation – including the nonvanishing A-terms – in a field-free region (where A = may be obtained as r r ) dr Λ(r) ψ = e iqΛ/¯h ψ , A(
∇×
≡
·
r0
0 in the same region. where ψ is a solution of the simpler equation with A = odinger equation Proof: We start with the assumed validity of the simpler Schr¨ 2
pˆ
2m
+ V (r)
−
∂ i¯h ψ = 0 ∂t
The important step we want to prove is ˆ ( pˆ q A)ψ = ( pˆ q A)eiqΛ/¯h ψ = e iqΛ/¯h ( p)ψ
−
−
In the last step, the term q A exp(iq Λ/¯h)ψ cancelled the opposite term coming from the gradient of the phase exp(iq Λ/¯h) in the Leibniz rule. You can use the same step again, replacing ψ by ( pˆ q A)ψ, to see that ˆ = e iqΛ/¯h ( p) ˆ 2 ψ ( pˆ q A)2 ψ = ( pˆ q A)2 eiqΛ/¯h ψ = ( pˆ q A)eiqΛ/¯h ( p)ψ
−
−
−
−
−
In this derivation, all operators act on everything on the right. You see that everytime we move ˆ Therefore the simple, the phase exp(iq Λ/¯h) to the left, we transform ( pˆ q A) to the simple ( p). Schr¨odinger equation for ψ multiplied by the phase exp(iq Λ/¯h) is completely identical to the A-free “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 ¯ q (∂ Λ/∂t)ψ times the usual phase on the “time-derivative” side, and this term would cancel h against the “electrostatic potential energy” term q Φψ for Φ = (∂ Λ/∂t).
−
−
7
Aharonov-Bohm interference pattern The statement we have just proved means that by changing the phase of ψ appropriately, we can 0 in the field-free regions. However, it is not quite right because the “appropriate” essentially set A = 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 = 0. Note that the interference patterns disappear in the classical never visits the space with B 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 around the solenoid. Let us start from the final steps and the box – whose central position is R – the result of Berry’s derivation: the overall phase will be computed from our oldest integral formula for ∆γ n : q A( R) dR = q Φ ∆γ n = i ψn R ψn dR = ¯h ¯ h which is the Aharonov-Bohm result for the phase. The last step is clear but the previous step is not. The required identity is q ψn R ψn = i A( R). ¯ h the position of the box, to define our adiabatic parameters, How do we prove it? We consider R, and we define the corresponding eigenstates associated with the same eigenvalues using our previous trick r r ) dr Λ(r) ψn = eiqΛ/¯h ψn , A(
|∇ ·
·
|∇ −
≡
·
R
This implies that
∇
∇
= −i q A( R)e − R)] ¯ h
+ e − R) ∇ ψ (r − R) The desired identity is obtained by multiplying this ket-equation by ψ |. You can see that you get what you need from the left-hand side and the first term of the right-hand side. The second term on = −∇ because ψ only the right hand side will vanish after being contracted with ψ | because ∇ ∇ in a box vanishes. depends on r − R and because the expectation value of the momentum # · R ψn =
iqΛ/¯ h ψn (r R [e
iqΛ/¯ h
ψn (r
iqΛ/¯ h
R
n
n
n
R
r
r
8
n
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