Advanced Solid State Physics

N+=

1

2

∫EF

−μB

D(E+μ 0 μBB)dE≈

1

2

∫EF

0

D(E)dE+

1

2

μ 0 μBBD(EF)

N−=

1

2

∫EF

μB

D(E−μ 0 μBB)dE≈

1

2

∫EF

0

D(E)dE−

1

2

μ 0 μBBD(EF)

HereN+is the number of spins parallel to the magnetic field andN−is the number of spins opposite
to the magnetic field. One can see, as mentioned above, that the only difference occurs at the fermi-
surface and that there is the same number of electrons for both spin orientations in the fermi sphere.
Now the magnetization can be calculated.

M=μB(N+−N−) (71)

M=μ 0 μ^2 BD(EF)B (72)

So in this case the magnetization can be calculated as a function of the magnetic field. The suscepti-
bility can be obtained by taking the derivative and it turns out to be temperature independent.

χ

μ 0

=

dM

dB

χPauli=μ^2 BD(EF)

So, Pauli paramagnetism is much smaller than atomic paramagnetism and it is temperature indepen-
dent. However, the Pauli paramagnetism is^23 times bigger than the diamagnetic part.

Landau Levels, Landau diamagnetism
The first thing to do is to find the microscopic states for free electrons in a magnetic field. So we take
the Schroedinger equation with zero electric field (V=0).

1

2 m

(−i~∇−qA)^2 Ψ(r) =EΨ(r)

Using the Landau gaugeA=Bzxyˆone gets

1

2 m

(

−~^2 ∇^2 +i 2 ~qBzx

d

dy

+q^2 Bz^2 x^2

)

Ψ =EΨ.

So y and z don’t appear in the Hamiltonian and so the solution can be written as a product of functions
of x,y and z (separating variables) and has the form

Ψ =eikyyeikzzg(x).

The equation for g(x) is

1

2 m

(

−~^2

∂^2 g(x)

∂x^2

+~^2 k^2 y+~^2 k^2 z− 2 ~qBzkyx+q^2 Bz^2 x^2

)

g(x) =Eg(x).

Let’s complete the square

1

2 m

(

−~^2

∂^2 g(x)

∂x^2

+q^2 B^2 z

(

x−

~ky

qBz

) 2 )

g(x) =

(

E−

~^2 k^2 z

2 m

)

g(x).