The second part of the Hamiltonian can be written as
H 1 =
−i~eBz
4 m(
−y∂
∂x+x∂
∂y)
+
e^2 B^2 z
8 m(
x^2 +y^2)
+gμBSzBzThe first part ofH 1 can be identified as the angular momentum operator:
Lz=−i~(
x∂
∂y−y∂
∂x)To get the energy associated withH 1 , it is necessary to calculate the matrix elements〈Ψ|H 1 |Ψ〉. It
turns out, that in most cases spherical symmetric waves are of specific interest. For those waves one
can write the expectation value ofr^2 as
〈
r^2〉
=〈
x^2〉
+〈
y^2〉
+〈
z^2〉
= 3〈
x^2〉
=^32(〈
x^2〉
+〈
y^2〉)So when we use waves of the described form and calculate the matrix elements, we get for the energy
(using the Bohr magnetonμB= 2 em~e):
E 1 = (〈Lz〉+g〈Sz〉)μBBz+
e^2 B^2 z
12 m〈
r^2〉The right-side term depends on the average position of the electrons. It also doesn’t matter from
which side one applies the magnetic field, there is always an increase in energy. So this part stands
for thediamagneticresponse. This is because the energy in a magnetic problem is calculated by
E=−μ·B,so only when the magnetic moment and the magnetic field are antiparallel the energy increases.
The left side-term depends on the orbital angular momentum and the spin angular momentum, which
can either be positive or negative. So it is on the magnetic quantum number (for angular momentum
l= 1→ml=− 1 , 0 , 1 ) whether the energy increases or decreases. A good example is the Zeeman
effect, where the two states for spin +1/2 and -1/2 split up and one goes up in energy and one goes
down. Now the state that goes down in energy has a lower energy and therefore is more likely to be
occupied. This leads to a magnetization of the system because there are still both states, but the lower
one is more likely to be occupied than the upper one and there the magnetic moments are aligned
with the magnetic field which is responsible for theparamagneticeffect.
9.3.1 Diamagnetic Response
At the beginning, we need the formula for the diamagnetic moment:
μ=−∂E
∂B
=−
e^2 Bz
6 m〈
r^2〉This description is often used in ionic crystals, where the two atoms have an electric closed shell
configuration. For a filled shell with L=0 and S=0 the paramagnetic part vanishes and one is left with
the diamagnetic part. Then the magnetic moment can be calculated like described above.
The minus sign indicates, that the magnetic moment is opposite to the applied field. The energy can
be calculated asE=−μ·Band always increases with an applied field.
It is not easy to synthesize magnetic molecules, because stable molecules have a closed shell configu-
ration. However there are molecules that do not have a closed shell configuration (e.g. OH), which