Download The (Integer) Quantum Hall Effect

The (Integer) Quantum Hall Effect
AQM Lectures 12-13
(Dated: 18 April 2007)
I.
THE HALL EFFECT
Consider a slab of conductor carrying a current I along the y-axis, and let B be a uniform magnetic field in the
z-axis. A voltage VH , known as the Hall voltage, will arise along the x-extent of the slab. Essentially, this voltage is
due to the deflection of charges by the magnetic field.
In an equilibrium situation, the net force on a charge carrier with drift velocity vd will be zero, meaning that the
electric and magnetic components of the Lorentz force will cancel. Thus,
Ex = −
q
vd B z ,
|q|
(1)
where Ex is in the positive x-direction for negative charge carriers, but in the negative x-direction for positive charge
carriers. The drift velocity vd is related to the current density by
Jy = nqvd ,
(2)
where n is the number density of charge carriers and q is the charge of each carrier. Eliminating the drift velocity
leads to
Ex = −
Bz
Jy .
nq
(3)
Thus, the electric field across the x-extent of the slab is proportional to the current density. The coefficient of
proportionality
ρxy = −
Bz
,
nq
(4)
has the form of a resistivity. Note that this model predicts that a plot of the resistivity vs. magnetic field should yield
a constant slope 1/(nq), which is confirmed by experiments at high temperatures. In the next section, however, we will
observe that this linear relationship between ρxy and Bz breaks down at high magnetic fields and low temperatures.
We can express the current density in terms of the total current I = Jy (dh) through the slab, which has crosssectional area given by the product of the thickness (z-extent) d and height (x-extent) h. The voltage that arises
across the x-extent of the slab is given by Ex h. We then have the relation
µ
¶
Bz
VH =
I.
(5)
nqd
That is, the Hall voltage is proportional to the current, and the coefficient of proportionality is itself proportional
to the strength of the magnetic field. Note that the sign of the Hall voltage is dependent on the sign of the charge
carriers.
II.
A QUANTUM TREATMENT OF THE HALL EFFECT
A.
Landau quantization
We first consider the quantum mechanics of a single (spinless) charge carrier, with charge q and mass m. This
charge is confined to move in the x − y plane.
We must express the magnetic field B = (0, 0, B), constant and uniform in the z-direction, in terms of a vector
potential. Specifically, we must choose a gauge. For this problem, it is simplest to choose the Landau gauge: it is
straight forward to show that the vector potential
A = (0, Bx, 0) ,
(6)
2
results in B = ∇ × A = (0, 0, B).
The time-independent Schrödinger equation for the charged particle, in the position representation in two dimensions
(i.e., square-integrable functions of x and y) is then
Ã
µ
¶2 !
~2 ∂ 2
1
∂
−
+
−i~
− qBx
ψ(x, y) = Eψ(x, y) .
(7)
2m ∂x2
2m
∂y
We note that if ψ(x, y) is a solution to this equation, then so is ψ(x, y + d). In other words, the solutions are invariant
under translations in the y-direction. This observation suggests that we should look for solutions of the form
ψ(x, y) = eiky φ(x) ,
(8)
i.e., eigenstates of momentum in the y-direction. If we use this expression in the Schrödinger equation above, we
obtain
µ
¶
~2 ∂ 2
1
2
−
+
(~k
−
qBx)
φ(x) = Eφ(x) .
(9)
2m ∂x2
2m
Defining ω = qB/m and x0 = ~k/(eB), this equation takes the form
µ
¶
~2 ∂ 2
1
2
2
−
+ mω (x − x0 ) φ(x) = Eφ(x) .
2m ∂x2
2
(10)
This equation is precisely that of a one-dimensional quantum harmonic oscillator with angular frequency ω centred
about the equilibrium position x0 . Note that ω = qB/m is precisely the cyclotron frequency of a particle with charge
q and mass m in a magnetic field B.
We have already investigated the solutions to the 1-D quantum harmonic oscillator. (The addition of a nonzero equilibrium position x0 is not tricky to incorporate, and it does not effect the energy level structure.) The
energy eigenstates φn,k (x) are labeled by a non-negative integer n, and their energy eigenvalues are ~ω(n + 1/2); the
dependence on k arises only through the equilibrium position x0 = ~k/(eB).
The energy eigenstates for the full two-dimensional problem are
ψn,k (x, y) = eiky φn,k (x) ,
(11)
where we note again that the energy eigenvalue is independent of k (it depends only on n). Thus, the spectrum is
highly degenerate – for each n, there are many energy eigenstates (with different y-momenta k) with the same energy.
These energy levels are called Landau levels, and the higher Landau levels can be loosely viewed as “copies” of the
lower ones.
To calculate this degeneracy, we must determine the number of orthogonal wavefunctions that have the same
energy ~ω(n + 1/2). Consider the lowest level, with n = 0; the wavefunctions φ0,k (x) are Gaussians centred about
x0 = ~k/(qB). Define Lx and Ly to be the dimensions of our 2-dimensional conducting sheet. The fact that Ly is
finite ensures that only quantized y-momentum values are allowed:
k=
2πm
,
Ly
m an integer.
(12)
However, the range of integers m cannot be arbitrarily large, because the equilibrium position of the oscillator
x0 = ~k/(qB) must lie within the x-extent of the conductor. As k is quantized, there is a fixed separation between
the different values of x0 given by
x0 (km ) − x0 (km−1 ) =
h
2π~
=
.
qBLy
qBLy
The number M of integers m must be limited to those which “fit” within Lx , i.e.,
³ h ´ qBL L
qBA
Φ
x y
=
=
=
,
M = Lx /
qBLy
h
h
Φ0
(13)
(14)
where Φ0 = h/q is the magnetic flux quantum for charges q. Thus, the number of orthogonal wavefunctions with
the same energy E0 = ~ω/2 is equal to the number of magnetic flux quanta that permeate the conduction sheet. A
similar calculation holds for the higher energy levels n > 0: they all have degeneracy Φ/Φ0 .
3
B.
Landau levels
So far, we have considered only a single charged particle. What happens if there are many charged particles in the
sheet? To answer this, we make use of some standard techniques from quantum statistical mechanics and condensed
matter physics.
One would expect, naively, that placing many charges (say, electrons) in a metal would cause them to interact very
strongly through the Coulomb force, and that the resulting energy eigenstates would look very different from the
single-particle energy eigenstates. Landau showed the remarkable result that, in many systems (this works especially
well in metals), electrons behave essentially as free electrons even though there are many of them in the system –
the main noticeable change is that the electron mass m must be changed to a effective mass m∗ , but otherwise the
single-particle wavefunctions are still basically valid.
In this situation, we are able to treat the system as a degenerate electron gas – the electrons are “free” and do not
interact with each other, but we must take into account the Pauli exclusion principle, and cannot give two electrons
the same wavefunction because they are fermions. At zero temperature (which is a very good approximation for the
quantum Hall systems we are investigating) the state of the whole system is obtained by placing electrons, one at a
time, in the lowest available energy eigenstates until we have filled N states.
For a normal metal, where there is a fairly continuous spectrum of wavefunctions, this process results in filled states
up to the Fermi energy. Our two-dimensional quantum Hall system, though, is very different. The first M = Φ/Φ0
electrons go into the lowest Landau level n = 0; the next M electrons go into the n = 1 Landau level, etc.
C.
The integer quantum Hall effect
Consider the situation where exactly p Landau levels are filled, where p is an integer. The number of charges in a
sheet of area A is then Nq = pΦ/Φ0 = pqBA/h. The number density per unit area is
n=
pqB
.
h
(15)
In our discussion so far, there hasn’t been a net current in any direction; we considered only the magnetic field (in
the z-direction). However, we have still described the physics of the Hall effect – it’s just that we’ve worked in the
rest frame of the charged particles. To obtain predictions for a non-zero current in the y-direction, we simply have to
Lorentz-boost our results by the drift velocity vd in the −y-direction. If vd is non-relativistic (so γ ' 1), the Lorentz
transformation is:
E = (0, 0, 0) → E0 = (vd B, 0, 0)
B = (0, 0, B) → B0 = (0, 0, B) .
(16)
(17)
In this frame, there is a non-zero electric field Ex = vd B and a current density Jy = nqvd . By eliminating vd we
obtain
Ex = −
Bz
Jy ,
nq
(18)
just as in the classical case. (Note, however, that our current density and number density are now per unit area, not
volume.) Using our expression for the number density if p Landau levels are filled, we find that
Ex = −
h
Jy ,
pq 2
for p Landau levels filled.
(19)
The coefficient of proportionality
ρxy = −
h
,
pq 2
(20)
has the form of a resistivity. However, it is independent of the magnetic field. If the magnetic field is varied, this
resistivity will remain constant.