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3. Magnetism
3.1 Coupling of matter to a magnetic field: Diamagnetism and
paramagnetism
An external magnetic field can couple to matter and electrons in two different ways (we consider the non-relativistic case):
(1) through the minimal coupling, expressed as
e~
~p → ~p − A(~
r),
c
where V(~r) is the lattice potential. This Hamiltonian can be derived out
of the Dirac equation.
Let us consider a static magnetic field applied along the c direction:
(3.1)
~ is the vector potential
where ~p is the momentum of the electron and A
of the electromagnetic field,
(2) and through the spin of the electron, as
~,
~ = − eh ~S · B
gµB~σ · B
mc
where g ≈ 2 is the gyromagnetic factor for the free electron,
The minimal coupling (3.1) is responsible for the diamagnetism in the
system. The coupling to the spin, eq. (3.2), is the Zeeman coupling and
is responsible for the paramagnatism in the system. From equations (3.1)
and (3.2), we can write the Hamiltonian of an electron in a magnetic field
~ as
B
2
~
~p − ec A
2
eh
~ + eh 1 dV ~l · ~σ + V(~r),
H=
(3.3)
−
~σ · B
2 2
2m
2mc
}
|4m c r dr{z
spin-orbit coupling
(3.2)
|e|h
>0
2mc
is the Bohr magneton and ~σ are the Pauli matrices.
µB =
~ = (0, 0, B),
B
~ × ~r) = 1 (−By, Bx, 0) (symmetric gauge)
~ = 1 (B
A
2
2
and
e2
e ~ 2
e ~ ~
~2
~p − A
= p2 −
~p · A + A · ~p + 2 A
c
c
c
e2 B2 2
eB
(px y + ypx − py x − xpy ) +
(x + y2 ).
= ~p 2 +
2c
4c 2
Then, the Hamiltonian (3.3) becomes
H=
⋆ We will be using throughout the course atomic units. Since the distance between atoms in solids is of a few angstrom, the natural lengthscale
is the Bohr radius a0 :
h2
= 0.5 angstrom = 0.5 × 10−8 cm.
a0 =
me2
The energy scale is given in
E0 =
e2
me4
=
= 0.43 × 10−10 erg = 2 Ry = 27.2 eV.
h2
a0
57
eh2 1 dV ~
e2 B2 2
p2
(x + y2 ) +
l · ~σ + V(~r) .
+ µB (lz + σz )B +
2
2m
8mc
4m2 c2 r dr
(3.4)
We note that
∂
∂
h
x
−y
= hlz ,
Lz = xpy − ypx =
i
∂y
∂x
1
∂
∂
lz =
x
,
−y
i
∂y
∂x
h 1 0
h
.
Sz = σz =
2
2 0 −1
58
Here, ~L = h~l is the orbital angular momentum and ~
S = h2 ~σ is the spin.
From eq. (3.4), we have that the external magnetic field couples linearly
to the orbital angular momentum and the spin.
~ = (0, 0, B), then
If B
∂
1X
χzz =
Mz =
∂B
Z n
How does the system react to the application of external magnetic field?
It is to be expected that the system will magnetize, therefore we should
analyze the thermal expectation value of the magnetization:
~ = h~µi = Tr(ρ~µ) = − ∂ FS ,
M
~
∂B
where FS is the free energy density of the system that can get magnetized.
Reminder :
−
with Z =
∂Mα
∂ ∂
=−
FS ,
∂Bβ
∂Bβ ∂Bα
n
Here, En are the eigenvalues of the electron system in the presence of the
external magnetic field. Then,
X ∂En
~ = − ∂FS = P −1
M
e−βEn ,
−βEn
~
~
∂B
∂
B
ne
n
which corresponds to
~ = h~µi = −µB h~l + ~σi = eh h~L + 2~Si .
M
2mc
The magnetization is obtained out of the thermal average of the magnetic
moment ~
µ. Since the electrons have negative charge (e < 0), the total magnetic moment has the opposite sign to orbital (~L) and spin (~
S) moments.
59
−βEn
∂En
∂B
X ∂En
n
∂B
2
∂2 En
−
∂B2
!2
e−βEn
!
e−βEn
.
being the partition function in the canonical ensemble.
En = E0n + µB hn|lz + σz |niB
e 2 B2
+
hn|(x2 + y2 )|ni
8mc2
X |hn|lz + σz |mi|2 B2
,
+ µ2B
E0n − E0m
The static (isothermal) magnetic susceptibility is then given by
where χαβ is a tensor with components α and β. If we consider the equilibrium state, to which the system relaxes after application of a magnetic
field, we can handle the system within equilibrium thermodynamics:
X
FS = −kB T ln
e−βEn .
ne
BT
In order to obtain the eigenvalues of the Hamiltonian (3.4), we apply perturbation theory up to the 2nd order in the magnetic field (see QM II
course). The result reads as
~ B
~.
dFS = −SdT − Md
χαβ =
P
1
Z2 k
1
kB T
m6=n
where E0 are the eigenvalues of the unperturbed Hamiltonian
In the limit B → 0,
∂En
= µB hn|lz + σz |ni,
∂B
2
X |hn|lz + σz |mi|2
e2
∂ En
2
2
2
hn|y
+
x
|ni
+
2
µ
.
=
B
∂B2n
4mc2
E0n − E0m
(3.5)
m6=n
In the absence of collective magnetism due to interaction among the electrons, we have for B → 0 that
1 X ∂En
1X
Mz = −
=−
µB hn|lz + σz |ni = hµz i = 0,
Z n ∂B
Z n
i. e., the magnetization disappears, since the magnetic moments thermally
compensate.
From eq. (3.5) we can divide the static susceptibility into three terms:
χzz = χC + χvV + χdia ,
with
χC =
µ2B
kB T
P
hµ2 i
σz |ni)2 e−βEn
= z >0
−βE
n
kB T
ne
z+
n (hn|l
P
60
(3.6)
being the paramagnetic contribution,
χvV = −2
1 XX
|hn|lz + σz |mi|2 −βEn
e
>0
µB
Z n
E0n − E0m
(3.7)
In this section, we considered one electron under the influence of a magnetic
field. We will generalize these results to a system of N non-interacting
electrons.
m6=n
3.2 Paramagnetism of localized magnetic moments
the van Vleck susceptibility and
e2
e2
2
2
hx
+
y
i
=
−
hr2 i < 0
8mc2
6mc2
the diamagnetic contribution.
χdia = −2
(3.8)
The paramagnetic contribution, eq. (3.6), is positive and has a 1/T -temperature
dependence:
χC =
C
,
T
with
C=
µ2B 1 X
hn|(lz + σz )|ni2 e−βEn
kB Z n
being the Curie constant. Please note that even when the magnetization is
zero,
X
hn|(lz + σz )|ni = 0,
n
χC is different from zero and the system is in the paramagnetic state.
The paramagnetic contribution, eq. (3.7), is constant and positive since
the eigenvalues of the excited states |mi are larger than the ground state
energy: E0m > E0n . This term will be significant when
kB T ≪ E0m − E0n
i.e., at low temperature. For kB T ≫ E0m − E0n , the van Vleck term has a
1/T -behavior like the Curie term.
The diamagnetic contribution, eq. (3.8), is negative, which implies that
the magnetization of the system for small fields has opposite sign to the
magnetic field. This is a purely quantum effect. It comes from the term
~ . Usually, the paramagnetic contribution is larger than the
~p → ~p − ec A
diamagnetic contribution, but if the total angular momentum disappears,
there is no paramagnetic contribution and we are left with only the diamagnetic contribution (see below). This is called the Larmor diamagnetism.
61
We consider a system of N atoms or ions with partially filled electron shells.
The total magnetic moment per atom/ion is given by
~J = ~L + ~S.
In this case, we are dealing with localized magnetic moments, which correspond to the electrons in the partially filled shells and are localized on
the atoms/ions, i.e. we are dealing with insulators. Let us consider the
following two cases.
1) The electron shell has J = 0, which would be so in, for instance, shells
with S = 2 and L = 2:
~J = ~L + ~S,
|J| = |L − S|, |L − S| + 1, . . . , L + S,
J = |L − S| = 0.
In this case, the linear term hn|~L+~
S|ni disappears, and only the higher
order terms (van Vleck paramagnetism and Larmor diamagnetism)
contribute.
2) If the shell does not have J = 0, the linear term does not disappear
and will dominate the magnetic behavior of the system. Please note
that the ground state is (2S + 1)-fold degenerate in zero field. We can
define the magnetic moment as
~µ = −gµB~J,
where g is the Landé factor
g=1+
J(J + 1) + S(S + 1) − L(L + 1)
.
2J(J + 1)
62
We consider now the simplest Hamiltonian where we neglect the diamagnetic contribution:
X
X
~ =
H=−
~µB
gµB Jz B,
i
i
~ = (0, 0, B), and calculate the susceptibility for the N atoms:
with B
Z = Tr e−βH = Tr e−β
=
=
N
Y
i=1
N
Y
i=1
P
i
µB BJz
=
N X
+J
Y
e−βµB BmJ
i=1 mJ =−J
1 − e−gµB B(2J+1)β
1 − e−gµB Bβ
Abbildung 3.1: The Brillouin function.
1
1
egµB B(J+ 2 )β − e−gµB B(J+ 2 )β
1
1
egµB Bβ 2 − e−gµB Bβ 2
.
so that
BJ (x) ≈
Then, the free energy is
1
F = −kB T ln Z = −NkB T ln
1
egµB B(J+ 2 )β − e−gµB B(J+ 2 )β
1
egµB Bβ 2
−
M≈N
1
e−gµB Bβ 2
and
and the magnetization
1
J
x
,
3
gµB JBβ
(gµB )2 J(J + 1)
J+1
gµB J
=N
B
J
3
3k B T
⇒
χ =
C
T
with C = N
J(J + 1)
(gµB )2 .
3k B
For spin- 12 systems,
where
2J + 1
BJ (x) =
coth
2J
2J + 1
x
2J
1
− coth
2J
x
2J
is the Brillouin function. This function is shown in Fig. 3.1. For x ≪ 1
it is linear and for x ≫ 1 BJ (x) → 1 and M → NgµB J, the saturation
magnetization.
1
J=S= ,
2
Then,
g = 2,
and
µB B
M = NµB tanh
kB T
1 x
+ + ...
x 3
63
C=N
B1/2 (x) = tanh(x)
For small magnetic fields,
coth(x) ≈
1+
∂M (gµB )2 J(J + 1)
χ =
=
N
∂B B→0
3k B T
∂
M = − F,
∂B
1
1
M = NgµB (J + ) coth βgµB B(J + )
2
2
= NgµB JBJ (gJµB Bβ),
.
64
µ2B
.
kB
3.3 Pauli paramagnetism of conduction electrons
In the previous section, we considered localized moments. In the present
section, we shall deal with conduction electrons in a metal. The electrons
have a spin and the application of a magnetic field induces their magnetization. Therefore, we also expect a paramagnetic contribution from the
conduction electrons.
Let us consider a minimal model where the spin of the conduction electrons
is coupled to an external magnetic field:
X
X †
H=
ε~k c~†kσ c~kσ + µB B
(c~k↑ c~k↑ − c~†k↓ c~k↓ ).
(3.9)
~k,σ
Abbildung 3.2: Energy spectrum of the conduction electrons in an external magnetic field.
~k
The z-component of all electron spins is
1X †
1X †
Sz =
(c~R↑ c~R↑ − c~†R↓ c~R↓ ) =
(c~k↑ c~k↑ − c~†k↓ c~k↓ )
2
2
~R
~k
and
µz = −gµB Sz = −µB
X
(c~†k↑ c~k↑ − c~†k↓ c~k↓ ).
where ρ0 (ε) is the density of states of the free electrons. For small B, we
perform a Taylor expansion:
Z
Z
df
df
df
2
2
= 2µB B dε ρ0 (ε) −
,
−
= δ(ε−εF ).
M = −2µB B dε ρ0 (ε)
dε
dε
dε
Then,
~k
We calculate now the magnetization due to the application of an external
~ = (0, 0, B):
magnetic field B
X †
M = hµz i = −µB
(hc~k↑ c~k↑ i − hc~†k↓ c~k↓ i).
(3.10)
~k
M = 2µ2B Bρ0 (εF )
and
χPauli = 2µ2B ρ0 (εF )
The Hamiltonian (3.9) can be rewritten as
X
X
H=
(ε~k + µB B)c~†k↑ c~k↑ +
(ε~k − µB B)c~†k↓ c~k↓
is the Pauli susceptibility (only valid for small B). Therefore, one expects
in metals at low temperature a constant contribution to the susceptibility.
so that we have free electrons with slightly displaced one-particle energies:
ε ± µB B (see Fig. 3.2).
Therefore, the expectation value in eq. (3.10) can be calculated through
the Fermi function:
Z
M = −µB dε [f(ε + µB B)ρ0 (ε + µB B) − f(ε − µB B)ρ0 (ε − µB B)] ,
In the previous section, we dealt with the coupling of the spins of conduction electrons to an external magnetic field. The conduction electrons
have also a well defined ~p, and therefore we will also have a diamagnetic
contribution due to the diamagnetic coupling to the magnetic field:
65
66
~k
~k
3.4 Landau diamagnetism
e~
~p → ~p − A
.
c
In order to analyze only this coupling, we consider spinless fermions in a
magnetic field. In this case, the Hamilton operator is
2
~ ri )
N
~pi − ec A(~
X
H =
2m
i=1
X 1 e2 ~ 2
e ~
e~
~pi − ~pi Ai (~ri ) − Ai (~ri )~pi + 2 A (~ri ) .
=
2m
c
c
c
i
~ = (0, 0, B) and the Landau gauge,
For the magnetic field given as B
~ = (0, Bx, 0),
A
~ =0
divA
~ = A~
~ p,
~pA
⇒
We are left with a displaced harmonic oscillator for the x coordinate:
x − x0
ϕ(x) = ϕn
,
λ
where
hcky
hky
=
,
mω
eB
r
r 0
h
hc
=
.
λ =
mω0
eB
x0 =
ϕ(x) are the Hermite polynomials. The eigenvalues of H are
the Hamiltonian reads
H=
X
i
!
p2iy
p2ix
p2iz
eB
e2 B2 2
x .
+
+
−
piy xi +
2m 2m 2m mc
2mc2 i
Since we are considering non-interacting electrons, we concentrate on the
H for one electron:
Hi =
p2x
2m
+
py
m 2
ω x−
2 o
mωo
2
+
p2z
2m
,
where
eB
mc
is the cyclotron frequency, which corresponds to the classical frequency of
particles in a magnetic field due to the Lorentz force in the xy plane. We
consider the following one-particle wavefunction:
ωo =
En,ky ,kz = hω0
= ECϕ(x)eiky y eikz z .
67
1
n+
2
+
h2 k2z
,
2m
with n being the Landau quantum number.
Instead of free electrons with dispersion
ε~k =
h2 k2
,
2m
we now have free electrons with one-particle energies En,ky ,kz , which are
determined through three quantum numbers: the Landau quantum number
n, ky and kz . The energy, however, has no explicit dependence on ky .
Therefore, we have degeneracy in the Landau level. This degeneracy is
determined from the following condition for x0 :
x0 =
Ψ(~r) = Cϕ(x)eiky y eikz z .
Here, C is a normalization constant, and the y- and z-dependences are
given in plane waves since the Hamiltonian does not depend explicitly on
either y or z. Then,
!
2
hky
m 2
p2x
h2 k2z
HΨ(~r) =
+ ω0 x −
Cϕ(x)eiky y eikz z
+
2m
2
mω0
2m
hky
6 Lx .
mω0
On the other hand, we have periodic boundary conditions in y:
ky =
2π
ly
Ly
with ly ∈ N.
Then,
2 πh
ly 6 Lx
mω0 Ly
⇒
ly 6
mω0 Lx Ly
,
2 πh
68
which implies that the number of allowed ly , defining the degeneracy of a
Landau level, is
mω0 Ly Lx
|e|B Lx Ly
=
.
2 πh
c 2πh
with y = µ − hω0 x. The grand canonical potential then becomes



Zµ

(hω0 )2 d
kB T m 

Φ = 2 3V 
dy g(y) −
g(y)  .
2π h  −∞
24 dy
µ
{z
}
|
B−independent
Let us calculate the contribution of the Landau diamagnetism to the susceptibility by considering the thermodynamic relations. We can work either in
the canonical or in the grand canonical ensemble:
X Φ = −2kB T
ln 1 + e−β(εα −µ)
α
Z
∞
Lz
|e|B Lx Ly X = −2kB T
dkz
ln 1 + e−β(εα −µ)
2 πh
c 2πh n=0
Z
∞
h 2 k2
kB T V |e|B X
−β(hω0 (n+ 21 )+ 2mz −µ)
,
= − 2 2
dkz ln 1 + e
2π h c
n=0
which can be written as
Φ=
−|e|kB T V B
1
g(µ − hω0 (n + )),
2
2
2π h c
2
n=0
with
kB T m
Φ0 (T , µ) = 2 3 V
2π h
h2 e2 B2 ∂2
Φ0 (T , µ),
24m2 c2 ∂µ2
Zµ
kB T m
dy g(y) = 2 3 V
2π h
−∞
Zµ
−∞
Z
dy dkz ln 1 + e
β(y−
h 2 k2
z
2m )
Then,
M=−
∂2 Φ0
e2 h2
∂Φ
B
=
2
2
∂B
12m c
∂µ2
χ=
∂M
e2 h2 ∂2 Φ0
.
=
∂B
12m2 c2 ∂µ2
Therefore, the Landau susceptibility is
h 2 k2
β(µ−x− 2mz )
g(µ − x) = dkz ln 1 + e
.
Z
In order to perform the sum over the Landau quantum number, we can
use the Euler-Maclaurin formula:
Z∞
∞
X
1
1
F(n + ) =
F(x)dx + F ′ (0).
2
24
0
n=0
Then,
n=0
Φ = Φ0 (T , µ) −
and
∞
X
where
∞
X
It can be written as
1 d
g(x)
g(µ − hω0 x)dx +
24 dx
0
x=0
Zµ
hω0 d
1
dy g(y) −
,
=
g(y)
hω0 −∞
24 dy
y=µ
1
g(µ − hω0 (n + )) =
2
Z∞
69
χLandau =
1 2 ∂2 Φ0
.
µ
3 B ∂µ2
For Φ0 , we directly take the grand canonical potential for free electrons
without magnetic field:
X Φ0 = −2kB T
ln 1 + e−β(ε~k −µ) ,
~k
X
X e−β(ε~k −µ)
∂Φ0
1
= −2
= −2
−β(ε
−µ)
β(ε~k −µ) + 1
~
k
∂µ
1
+
e
e
~k
~k
X
= −2
f(ε~k ),
~k
∂2 Φ0
∂µ2
X df
= 2
dε~k
~k
T →0
−→
−2ρ0 (εF ).
70
.
Then,
with
1
χLandau = − χPauli .
3
p=
The total contribution of the conduction electrons to the susceptibility is
2
χtotal = χPauli ,
3
|e|BLx Ly
.
2πhc
If we have a system of N electrons, pBn0 electrons fill n0 Landau levels,
and the remaining N − pBn0 electrons fill the (n0 + 1)th level. The total
energy is then given by
Etot =
and it is paramagnetic.
n
0 −1
X
n=0
3.5 De Haas van Alphen effect
The de Haas van Alphen effect is the periodic variation of the magnetic
susceptibility as a function of the inverse of the magnetic field. With this
effect one can measure the Fermi surface of metals as well the effective
mass of conduction electrons. The De Haas van Alphen effect is based on
the energy quantization of the electrons due to the external magnetic field.
1
pBhω0 (n + ) + hω0 (N − pBn0 ).
2
For larger B, the filling of the (n0 + 1)th level decreases linearly. The
situation is graphically expressed in Fig. 3.3.
Let us consider the kx − ky plane perpendicular to the magnetic field.
Without application of a magnatic field, both kx and ky are good quantum
numbers, i. e., the electronic states are characterized through lattice points
in the kx − ky plane. In a magnetic field, the states are degenerate and
characterized by the Landau quantum number n. There are
|e|BLx Ly
2πhc
degenerate states with energy hω0 (n + 12 ).
In order to understand quantum oscillations in the presence of a magnetic
field, let us consider the two-dimensional free electron gas in the x−y plane
and a magnetic field applied along the z-direction. The Landau states are
quantized and, since there is no third dimension, there is no continuum
contribution. The energy eigenvalues are
1
En = hω0 (n + ),
2
and the degeneracy for every level is
Abbildung 3.3: Filling of the Landau levels in different magnetic fields.
If
B=
N
pn0
⇒
1
pn0
=
,
B
N
the (n0 + 1)th level won’t be filled anymore, and the Fermi energy is then
at the n0 th level. Therefore, for the total energy and magnetization we
expect a B1 -periodic behavior.
3.6 Quantum Hall effect
Since in two dimensions only discrete highly degenerate Landau levels are
present, one can in such cases analyze the Landau quantization in detail.
|e|BLx Ly
mω0 Lx Ly
=
= pB,
2 πh
2πhc
71
72
The density of states of a two-dimensional spinless non-interacting electron
gas in a strong magnetic field is given by
mω0 Lx Ly X
1
δ E − hω0 (n + ) .
ρ2d (E) =
2 πh
2
n
This density of states consists of delta peaks at the Landau energies hω0 (n+
1
2 ) weighted by the degeneracy.
With the help of semiconductor physics, it is possible to create a purely
two-dimensional electron gas. This is done on heterostructures of p-doped
GaAs and n-doped Ga1−x Alx As. Using the molecular beam epitaxy (MBE)
technique, one can grow alternately the two semiconductors in order to
form a heterostructure. Each layer has a width of about several nanometers.
Ga1−x Alx As is n-type doped, which generates extra mobile electrons in its
conduction band. These electrons migrate to fill the holes at the top of
the GaAs valence band and partially end up as states near the bottom
of the GaAs conduction band. There is of course a positive charge left on
the donor impurities which attracts these electrons to the interface and
”bends”the bands. This is the source of electric field in the system.
Abbildung 3.4: The scheme of the energy bands of the Ga1−x Alx As/GaAs
heterostructure. The charge density of the system is ∼ 1011 cm−2 .
The charges bend in the y-direction until the created Ey -field compensates the Lorentz force:
Ey =
Uy
vx
jx B
1
I
B=
=
B=
.
c
nec nec Ly Lx
Ly
The transfer of electrons from Ga1−x Alx As to GaAs continues until the dipole layer formed from the positive donors and the negative inversion layer
is sufficiently strong. This dipole layer gives rise to a potential discontinuity, which makes the Fermi level of GaAs equal to that of Ga1−x Alx As.
The electronic states perpendicular to the separation layer are localized so
that in the layer between GaAs and Ga1−x Alx As there is a two-dimensional
electron gas.
A possible way to investigate this two-dimensional electron gas is by considering the Hall effect. For that we shall introduce this effect briefly.
Reminder : Hall effect.
Under application of an electric field along the x-direction and a homogeneous magnetic field along the z-direction, the electrons in a
conductor of length Lx and width Ly , moving at velocity vx in the
x-direction, feel a Lorentz force in the y-direction:
e
FL = vx B.
c
Here, Uy = UH is the Hall voltage, which is measured. It is determined
by
IB
.
Lx
The proportionality coefficient rH ,
UH = rH
rH =
73
1
,
nec
74
is called the Hall coefficient and can be either positive or negative
depending on whether the carriers are electrons or holes.
In the classical Hall effect, we have that the Hall resistance UIH is proportional to the magnetic field. In the two-dimensional electron system that
we introduced, for large enough magnetic fields and low temperatures, this
proportionality is not anymore fulfilled. In fact, what is observed is that
the Hall resistivity ρxy shows steps and plateaux at quantized values of ieh2 ,
where i is an integer. ρxy is not linear in B, but remains constant over a
large interval of B and jumps at a critical B to the next quantized value
h
(i−1)e2 (see Fig. 3.5).
As a function of the Fermi energy, the Hall conductivity of the two-dimensional
electron gas in a magnetic field follows a step-like behavior. The quantized
values for the Hall conductivity are the values, where the Fermi energy
falls in the gap between two discrete Landau levels.
There must be a mechanism that explains why the Fermi energy either
remains in a gap between two Landau levels or in a region of occupied
states and the current-carrying state does not change as a function of
magnetic field.
Abbildung 3.5: Schematic representation of the measured Hall resistance.
While the Hall resistivity shows quantized values, the longitudinal resistivity ρxx = 0.
In the basis of the Landau states, the matrix elements of the current operator are given by
r
√
e
mhω0 √
hnky |jx |lky′ i =
n + 1δn,l−1 − nδn,l+1 δky ky′ ,
m
r2
√
√
h
hnky |jy |lky′ i = −eω0
n + 1δn,l−1 + nδn,l+1 δky ky′ .
2mω0
Out of these matrix elements, one can derive the Hall conductivity:
σxy =
e2
e2
(n + 1) = (n + 1).
2 πh
h
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Two-dimensional systems are not completely perfect, but have impurities
and impurity scattering is possible. Through this impurity scatering and,
in general, a disorder potential, the degeneracy of Landau levels is lifted
and the Landau levels extend to Landau bands.
If the disorder is not very strong, there are still gaps between the Landau
bands for various Landau values n. In every Landau band, there are delocalized states in the middle of the band, i.e., where the original eigenvalue
hω0 (n + 12 ) resided, and localized states at the boundaries of the Landau bands. The localized states do not contribute to the current transport,
which means that when the Fermi energy is in the region of localized states
the diagonal conductivity σxx disappears and the non-diagonal Hall conductivity σxy keeps the value that it takes in the gap between two Landau
levels. Without disorder we cannot account for the plateau nature of the
quantum Hall effect.
The observation that in the quantum Hall effect the quantization of the
conductivity is an integer multiple of a universal constant means that this
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behavior can only be dependent on a very robust property of the twodimensional systems: the geometry. The quantization of the conductivity
is obtained in a one-dimensional channel. In the quantum Hall effect current
arises from the states at the edge: EDGE STATES.
Up to now, we have talked about the integer quantum Hall effect. When
the quality of samples is high enough, at low T and high B one observes
non-integer steps:
ρxy =
h
,
fe2
with f = qp (p and q are integers, q is odd). In order to understand the
fractional quantum Hall effect, one has to include electronic correlations.
The excitations in such a system can be described through quasiparticles
with non-integer charge. Laughlin won the Nobel Prize in 1998 for suggesting an ansatz wavefunction to describe the fractional quantum Hall
effect.
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