Download E3570: A particle on a disc with a homogeneous magnetic... levels

E3570: A particle on a disc with a homogeneous magnetic field - Landau
levels
Submitted by: Roee Enbar & Gil Sharon
The problem:
A particle is bound to move on the XY plane in the presence of a homogeneous magnetic field
perpendicular to the plane.
(1) Write the Hamiltonian in Cartesian coordinates.
(2) Show that the Hamiltonian is of a two dimensional harmonic oscillator + a Zeeman term.
(3) Write down the Hamiltonian in polar coordinates and identify the radial equation.
(4) What are the energy levels determined by the radial equation without the Zeeman term? What
is the degeneracy?
(5) What are the energy levels determined by the radial equation with the Zeeman term? What is
the degeneracy?
Notes:
• In question 4 the answer can be deduced from the known solution in Cartesian coordinates.
• In question 5 assume that the particle is bound by a disc of radius R.
• Show that the results are in agreement with the known results for Landau levels.
The solution:
(1)
~ = B ẑ = ∇
~ ×A
~
B
It is convenient to use the symmetrical gauge:
By
Bx
1~
~= −
A
,
,0 = B
× ~r
2
2
2
Therefore the Hamiltonian in Cartesian coordinates:
2
~
P~ − A
H=
2M
Using the triple multiplication associativity,
~=A
~ · P~ = 1 B
~ × ~r · P~ = 1 B
~ · ~r × P~ = 1 B
~ ·L
~ = 1 BLz
P~ · A
2
2
2
2
1~
1~
1 2 2 ~ x2 + y 2
A2 =
B × ~r ·
B × ~r =
r B − ~r · B
=
2
2
4
4
Finally,
H=
BLz
Px2 + Py2
P 2 + A2 − BLz
B2
=
+
x2 + y 2 −
2M
2M
8M
2M
1
The second term is the diamagnetic term and the last is the Zeeman term.
(2) For a two dimensional harmonic oscillator we have
Hosc =
taking ω =
H=
P2
1
+ M ω2 r2
2M
2
B
2M
our Hamiltonian is
P2
1
+ M ω 2 r2 − ωLz
2M
2
H = Hosc + Zeeman
(3)
−i
∂
∂2
= Lz ⇒ − 2 = L2z
∂φ
∂φ
P = −i∇ ⇒ P 2 = −∇2
The Laplacian in polar coordinates is
∂
1 ∂2
1 ∂
2
r
+ 2
∇ =
r ∂r
∂r
r ∂φ2
|{z}
|
{z
}
−L2z
−Pr2
Hence, the Hamiltonian in polar coordinates is
H=
Pr2
L2z
1
+
+ M ω 2 r2 − ωLz
2
2M
2M r
2
Working in a basis where Lz is diagonal, i.e
Lz |mi = m|mi
m = 0, ±1, ±2, . . .
For each m,
H(m) =
Pr2
m2
1
+
+ M ω 2 r2 − ωm
2
2M
2M r
2
And the radial equation is:
H(m) f (r) = Ef (r)
1 1 ∂
∂
m2
1
2 2
r
+
+ M ω r − ωm f (r) = Ef (r)
2M r ∂r
∂r
2M r2 2
(4) The energy levels of a 2-Dimensional harmonic oscillator are
E = ω (nx + ny + 1)
And the degeneracy
g(E) =
E
ω
2
Since without the Zeeman term the Hamiltonians are identical, we require the same energy levels and
degeneracies. In addition, we note that Hamiltonian contains only even functions of the quantum
number m. Therefore we deduce that the energies must be
E = ω(2ν + |m| + 1)
ν = 0, 1, 2, . . .
m = 0, ±1, ±2, . . .
(5) With the Zeeman term, we get
E = ω(2ν + |m| + 1) − ωm
= ω(2ν + |m| − m + 1)
Without being bounded by R, we would get an infinite degeneracy for positive m’s.
For ν = 0 only positive m’s will result in the ground state energy.
Being bounded by a disc of radius R, m will be limited and so the degeneracy will be finite.
The effective radial potential is a function of m
(m)
Vef f (r) =
m2
1
+ M ω2 r2
2
2M r
2
This potential has a minimum, rmin . The particle’s wave function is centered around rmin , and is
of a quantum (~) scale. The disc has a macroscopic scale. Therefore we can neglect the edge effects
and simply require that rmin be smaller than the disc’s radius R.
m2
d (m)
Vef f (r) = −
+ M ω2r = 0
dr
M r3
2
m = M ωrmin
rmin < R
m < M ωR2
Remembering that ω =
m<
B
2M
we get
BR2
=g
2
Where g is the degeneracy. which is exactly the known Landau degeneracy
g=
Lx Ly
B
2π
Since the area of disc A = Lx Ly = πR2 . Note that the above calculation is valid for ν = 0. For
ν > 0 we get additional degeneracy due to the negative values of m. However, for a strong magnetic
field or large R (g 1) this is negligible.
3