Fermi surface and de haas van alphen effect ppt

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Fermi surface and De Haas-van Alphen Effect
Tewodros Adaro
July 23, 2021
Fermi surface and De Haas-van Alphen Effect
Tewodros Adaro Fermi surface and De Haas-van Alphen Effect

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Contents
Contents
Onsager theory: Semiclassical quantization of orbits in a
magnetic field
Physical meaning for the quantization of the magnetic flux
Definition of the magnetic length
Physical meaning of the Onsager relation
Extremal orbit contributing to the dHvA signal
Landau tubes with the quantum number n
Density of states using the magnetic length
Derivation of the dHvA frequency
Energy dispersion of the Landau level
Simple model to understand the dHvA effect
Derivation of the oscillatory behavior in a 2D model

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magnetic field
magnetic field
The Onsager-Lifshitz idea was based on a simple semi-classical
treatment of how electrons move in a magnetic field, using the
Bohr-Sommerfeld condition to quantized the motion. The
Lagrangian of the electron in the presence of electric and magnetic
field is given by
L =
1
2
mv2
− q(ϕ −
1
c
v.A) (1)
where m and q are the mass and charge of the particle.
Canonical momentum:
p =
∂L
∂v
= mv +
q
c
A (2)

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magnetic field
Mechanical momentum:
π = mv = p −
q
c
A (3)
The Hamiltonian:
H = p.v−L = (mv+
q
c
A).v−L =
1
2
mv2
+qϕ =
1
2m
(p−
q
c
A)2
+qϕ
(4)

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magnetic field
The Hamiltonian formalism uses A and ϕ and not E and B,
directly. The result is that the description of the particle depends
on the gauge chosen. We assume that the orbits in a magnetic
field are quantized by the Bohr-Sommerfeld relation
π = mv = ~k = p −
q
c
A = p +
e
c
A (5)
and I
p.dl = (n + γ)2π~ (6)
where q = −e(e > 0) is the charge of electron, n is an integer, and
γ is the phase correction: γ = 1/2 for free electron.

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magnetic field
I
p.dl =
I
~k.dl −
e
c
I
A.dl = (n + γ)2π~ (7)
Note we assume ro = 0. Then we get
I
~k.dl = −
e
c
I
rXB.dl =
e
c
B.
I
rXdl =
e
c
B.2An =
2e
c
Φ (8)
where I
(rXdl) = 2(areaenclosedwithintheorbit)n (9)
and Φ is the magnetic flux contained within the orbit in real space,
Φ = B.An.

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magnetic field
Figure: In this figure, we mean that dr = dl . The area of triangle OAB is
equal to |rXdl|/2.
and
rXdl = nrdrsinθ = nOC.AB = [2(areaof∆OAB)]n
where OC = r, AB = drsin(π − θ) = drsinθ

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magnetic field
Then we have
H
(rXdl) = 2(areaenclosedwithintheorbit)n
On the other hand,
−
e
c
I
(A.dl) = −
e
c
I
(▽XA).da = −
e
c
I
(B.da) = −
e
c
Φ (10)
by the Stokes theorem. Then we have
I
(p.dl) =
2e
c
Φ −
e
c
Φ =
e
c
Φ = (n + γ)2π~ (11)

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magnetic field
It follows that the orbit of an electron is quantized in such a way
that the flux through it is
Φn(r) = (n + γ)
2π~c
e
= 2Φo(r + γ), (12)
where Φois a quantum fluxoid and is given by
Φo =
2π~c
e
=
hc
2e
= 2.0678X10−7
Gausscm2
(13)
Note that -2e corresponds to the charge of the Cooper pair

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According to the Onsager theory, the magnetic flux inside the orbit
of electron is quantized as
Φn(r) = BSn(r) = (n + γ)
2π~c
e
= 2Φo(r + γ) (14)
where Sn(r) is the area of the electron orbit, and n is an integer.
Using the relation k =
eB
c~
r =
1
l2
r
it is found that the area Sn(k)of the orbit of electron in the k-space
(normal to B) is related to the area Sn(r)through
Sn(k) =
1
Bl4
BSn(r) =
e2B
c2~2
Φn(r) =
2πeB
~c
(n + γ) (15)
implying that Sn(k) is also quantized for the fixed magnetic field B.

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Figure: Quantization of the magnetic flux inside the orbit of electron (the
real space). The number of the magnetic flux inside the orbit is only an
integer.

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When the magnetic field B is applied to the z axis (normal to the
plane), what is the area A where the quantum flux 2Φo penetrates?
2Φo =
2π~c
e
= BSn(r) (16)
Then we get
Sn(r) =
π~c
eB
= πl2
(17)
where
l2
=
c~
eB
(18)
In other words, l is the radius of the circle inside which the
quantum fluxoid Φo passes through.

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We consider the Landau tubes whose cross sections by plane
perpendicular to the direction of the magnetic field has the same
area for the n-th Landau tube,
Sn(k) =
2πeB
~c
(n +
1
2
) (19)
and are bounded by the curves of constant energy (
~2k2
z
2m
=
~2k2
//
2m
) at the height kz in the k-space. The energy at kzof
the Landau tube is given by

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.
ε(n, kz) = (n +
1
2
)~ωc +
~2k2
z
2m
(20)
whereωc is the cyclotron angular frequency,
ωc =
eB
mc
(21)

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Figure: Landau tube in the k-space. It is represented by the value of
kn(k⊥). The bottom and top parts of the Landau tube intersects with
the Fermi surface. The magnetic field is directed along the z axis.

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Here we show how to draw the Landau tubes:
We choose the quantum number n (the integer) such
~2k2
⊥
2m
= ~ωc(n +
1
2
) (22)
where k⊥(kn)n is the in-plane wave number.
The energy of the cylinder (so-called Landau tube) at the
point denoted by n and kz is given by
ε(n, kz) = (n +
1
2
)~ωc +
~2k2
z
2m
(23)

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The intersection of the constant-kz plane with the Fermi surface
(ε = εf) forms a cross section (circle with the radius kn). The
Landau tube can be constructed such that the top and the bottom
are parts of the cylinder as shown in Fig-4 The radius of the
Landau tube denoted by the quantum number increases with
increasing the magnetic field.

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Figure: Landau tubes with the fixed n. The magnetic field is increased
from (a) to (g)

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The level density will have a sharp peak whenever ε is equal to the
energy of an extremal orbit, satisfying the quantization condition.
The reason for this is as follows. The above figures depict the
Landau tube with the same n for different B. As the magnetic field
B increases, the area of the Landau tube increases. The number of
the states D(ε)dε is proportional to the area of the portion of
Landau tube contained between the constantenergy energy
surfaces ε and ε + dε. The area of such a portion is not extremal
in (a) to (f). The area of the portion as shown in (g) becomes
extremal when there is an extremal orbit of energy ε on the
Landau tube. Evidently, the area of the portion of the tube is
enhanced as a result of the very slow energy variation of the level
along the tube near the given orbit.

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Figure: The number of the states D(ε)dε which is proportional to the
area of the portion of Landau tube contained between the constant
energy surfaces ε and ε + dε. The region shaded by green.

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From the above discussion, we notice that the extremal value of
the area of the crosssection of the Fermi surface, A(εf, kz), at the
actual Fermi level εf, at the slice kz, where kz is the component of
the wave vector along the direction of magnetic field B. The
extremal cross section is defined by
∂A(εf, kz)
∂kz
= 0 (24)
at kz = ko

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Figure: Extremal orbits 1 and 2. A(εf, kz) has a local maximum for the
orbit 1 and has a local minimum for the orbit 2.

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•Landau tubes with the quantum number n
Here we draw a series of Landau tubes with different quantum
number n for each fixed magnetic field B. We start with the
relation given by
E =
~2k2
z
2m
+ ~ωc(n +
1
2
) (25)
where
ωc =
eB
mc
(26)
For simplicity, we assume that~ = 1, m = 1. Then we get
E =
1
2
k2
z + ~ωc(n +
1
2
) (27)

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For a fixed n (0 or positive integer) and a fixed E,
kz = ±
p
2E − ωc(2n + 1) (28)
The Landau tube consists of the cylinder with a radius
k⊥ = kn =
p
(2n + 1)ωc (29)
and the length of cylinder
|kz| ≤
p
2E − ωc(2n + 1) (30)
We choose; E = 12.ωc is changed as a parameter, where
~ = 1, m = 1. In the quantum limit, there is only one state with n
= 0 inside the Fermi surface.

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Figure: The last figure shows the quantum limit where only the Landau
tube with n = 0 exists inside the Fermi surface.

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Fundamentals of the Landau levels
We now consider the case when kz = 0.
Figure: This regular periodic motion introduces a new quantization of the
energy levels (Landau levels) in the (kx, ky) plane, corresponding to those
of a harmonic oscillator with frequency ωc and energy

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This regular periodic motion introduces a new quantization of the
energy levels (Landau levels) in the (kx, ky) plane, corresponding to
those of a harmonic oscillator with frequency ωc and energy
ε(n) = ~ωc(n +
1
2
) =
~2k2
⊥
2m
(31)
where k⊥ is the magnitude of the in-plane wave vector and the
quantum number n takes integer values 0, 1, 2, 3, ...ωcis the
cyclotron angular frequency and is defined by
ωc =
eB
mc
(32)

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Each Landau ring is associated with an area of k space. The area
Sn is the area of the orbit n with the radius k⊥ = kn
Sn(k) = πk2
n =
2πeB
~c
(n +
1
2
) =
2π
l2
(n +
1
2
) (33)
Thus in a magnetic field the area of the orbit in k space is
quantized. Note that
l2 =
~c
eB
. l is given by
l =
256.556
p
B[T]
[µm] (34)
in the SI units.

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Figure: When B = 0, there isone state per area (
2π
L
)2
.L2
is the area of
the system.

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Figure: Quantization scheme for free electrons. Electron states are
denoted by points in the k space in the absence and presence of external
magnetic field B. The states on each circle are degenerate. (a) When
B = 0, there is one state per area (
2π
L
)2
.L2
is the area of the system. (b)
When B ̸= 0, the electron energy is quantized into Landau levels. Each
circle represents a Landau level with energy En = ~ωc(n +
1
2
)

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We consider the density of states for the 2D system
Figure: In a magnetic field the points in the (kx, ky) plane may be viewed
as restricted circles

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.
Do(ε)dε =
L2
(2π)2
2πkdk (35)
where the electron spin is neglected. The energy dispersion relation
is given by
ε⊥ =
~2
2m
k2
⊥ (36)
with
dε⊥ =
~2
m
k⊥dk⊥ (37)

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Then we get
Do(ε)dε⊥ =
L2
(2π)2
2π
m
~2
dε⊥ (38)
or
Do(ε) =
L2
(2π)2
2π
m
~2
=
mL2
2π~2
(39)
which is constant. Note that we neglect the electron spin, since
only the orbital motion is concerned.

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Now we calculate the number of states between the adjacent
energy levels εn and εn+1. Since△ε⊥ = εn+1 − εn = ~ω(which is
independent of the quantum number n. Then the number of states
is calculated as
D = D(ε⊥)△ε⊥ =
mL2
2π~2
△ε⊥ =
mL2
2π~2
~ωc
mL2
2π~2
~eB
mc
=
eBL2
2π~c
= ρB
(40)
Where ρ =
eL2
2π~c
and l2 =
c~
eB

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The use of the Onsager’s relation
In the dHvA we need the area of the orbit in the k-space. We
define Sn(r) as an area enclosed by the orbit in the real space (r)
and Sn(k) as an are enclosed by the orbit in the k-space. Then we
have a relation
Sn(r) = (
c~
eB
)2
Sn(k) = l4
Sn(k) (41)
and
Sn(k) = (n + γ)
2πe
~c
B (42)

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In the Fermi surface experiments we may be interested in the
increment △Bfor which two successive orbits, n and n+1, have the
same area in the k-space on the Fermi surface. Suppose that S(k)
is the Fermi surface in the momentum space. Then we have
Sn(k) = Sn+1(k) = S(k) (43)
or
(n + γ)
2πe
~c
Bn = (n + 1 + γ)
2πe
~c
Bn+1 = S(k) (44)
or
S(k)
Bn
= (n + γ)
2πe
~c
,
S(k)
Bn+1
= (n + 1 + γ)
2πe
~c
(45)
or
S(k)[
1
Bn
−
1
Bn+1
] =
2πe
~c
(46)
or
△(
1
B
) =
2πe
~cS(k)
=
1
F
(47)

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The dHVA frequency F is related to the area of the Fermi surface.
F =
~c
2πe
S(k) (48)
What is a typical value of△B which is experimentally observed?
△(
1
B
) =
| △ B|
B2
=
2πe
~cS(k)
(49)
For Au, S(k) = 4.8x1016cm−2 for the belly orbit of Au.
| △ B| =
2πe
~cS(k)
B2 = 0.198872[Bo(T)]2 (in units of G) where Bo is
in the units of T. When B = 5 T (which means Bo = 5), we have
| △ B| = 4.97G(= Oe).

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The use of Landau level
Suppose that n-th Landau level exists just below the Fermi energy.
The total number of states below the Fermi energy is
nρB =
L2
(2π)2
S(k) (50)
where S(k) is the 2D extremal cross section of the Fermi surface,
normal to B, and ρ is
ρ =
eL2
2πc~
(51)

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Then we get
n
e
c~
B =
1
2π
S(k) (52)
or
1
Bn
= n
2πe
~cS(k)
(53)
Similarly we get
1
Bn+1
= (n + 1)
2πe
~cS(k)
(54)

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Then we have
△(
1
B
) =
1
Bn+1
−
1
Bn
=
2πe
~cS(k)
=
1
F
(55)
This result is the same as that derived from the Onsager relation.

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Another method for the derivation: cyclotron frequency
Note that the formula for can also be derived from the
correspondence principle. The frequency for motion along a closed
orbit is
ωc =
eB
mcc
(56)
where mc is defined as
mc =
~2
2π
∂S
∂ε
(57)

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In the semiclassical limit, one should obtain equidistant levels with
a separation △ε equal to ~ωc . Hence
△ε = ~ωc =
~eB
mcc
=
~eB
c[
~2
2π
∂S
∂ε
]
=
2πeB
c~(∂S/∂ε)
(58)
or
∂S
∂ε
=
2πeB
c~
1
~ωc
(59)

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or
∂Sn(k)
∂ε
=
2πeB
c~
1
~ωc
(60)
or
Sn(k) =
2πeB
c~
1
~ωc
~ωc(n + γ) =
2πeB
c~
(n + γ) (61)

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We consider the energy dispersion of the Landau level with the
quantum number n as a function of kz.
ε(n, kz) = (n +
1
2
)~ωc +
~2k2
z
2m
(62)
Figure: Energy dispersion of the Landau levels with n and Kz for a 3D
electron gas in the presence of a magnetic field along the z axis.

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Consider the figure showing Landau levels associated with
successive values of n = 0, 1, 2, , s The upper green line represents
the Fermi level εF. The levels below εF are filled, those above are
empty. Since εFis much larger than the level-separation ~ωc , the
number n = s of occupied levels is very large. Let us assume that
the magnetic field is increased slightly. The level separation will
increase, and one of the lower levels will eventually cross the Fermi
level. The resulting distribution of levels is similar to the original
one except that the number of filled levels below εF is now
n = s − 1, instead ofn = s. Since n is large, this difference is
essentially negligible, so that one expects the new state to be
equivalent to the original one. This implies a periodic dependence
of the magnetization.

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Figure: Schematic energy diagram of a 2D free electron gas in the
absence and presence of B. At B = 0, the states below εF are occupied.
The energy levels are split into the Landau levels with (a) n = 0, 1, 2,…,
and s for a specified field and (b) n = 0, 1, 2,…, and s-1 for another
specified field. The total energy of the electrons is the same as in the
absence of a magnetic field.

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The energy level of each Landau level is given by (n +
1
2
)~ωc,
where n = 0, 1, 2, ….. Each on of the Landau level is degenerate
and contains ρB states. We now consider several cases.
Then = 0, 1, 2, …, s-1 states are occupied. n = s state is empty.
εF = ~ωcs.N = ρBs (63)

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Figure: Then = 0, 1, 2, …, s-1 states are occupied. n = s state is empty.

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The total energy is constant,
U = Uo = Σs−1
n=0ρB(n +
1
2
)~ωc = ~ωcρB[
1
2
s(s − 1) +
1
2
s] = ~ωcρB
1
2
s2
=
1
2
εFN
(64)
The case where the n = s state is not filled. We now consider the
case when ~ωcdecreases. This corresponds to the decrease of B.

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ε < ~ωc/2, where ε is the energy difference defined by the figure
below.
Figure: ε < ~ωc/2
~ωc/2 < ε < ~ωc

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εf = s~ωc + ε, with0 < ε < ~ωc (65)
Figure: 0 < ε < ~ωc
The n = 0, 1, 2, …, (s-1) levels are occupied and the n = s level is
not filled. The total number of electrons is N. The energy due to
the partially occupied n = s state is (N − ρBs)~ωc(s +
1
2
). Then
the total energy is

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U−Uo = Σs−1
n=0ρB(n+
1
2
)~ωc −
1
2
εFN+(N−ρBs)~ωc(s+
1
2
) (66)
U − Uo = ~ωcρB
1
2
s2
−
1
2
εFN + (N − ρBs)~ωc(s +
1
2
) (67)
where ρBs < N < ρB(s + 1) and s~ωc < εf = (s + 1)~ωc

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Here we introduceλ as λ = N − ρBs. The parameter λ satisfies the
inequality
0 < λ < ρB for
ρs
N
<
1
B
<
ρ(s + 1)
N
The parameter λ denotes the
number of electrons partially occupied in the n = s state.
The parameter ρBs is the total number of electrons occupied in
the n = 0, 1, 2,…, s-1 states for
ρs
N
<
1
B
<
ρ(s + 1)
N
. The n = 0,
1, 2, …, s states are occupied. n = s+1 state is empty.
Figure: εf = ~ωc(s + 1)

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In this case we have εf = ~ωc(s + 1), and N = ρB(s + 1) and
U = Uo
U = Uo = Σs−1
n=0ρB(n+
1
2
)~ωc = ~ωcρB[
1
2
s(s+1)+
1
2
(s+1)] = ~ωcρB
1
2
(s+1)2
=
1
2
εFN
(68)

Fermi surface and de haas van alphen effect ppt

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