Download Landau Levels

Landau Levels
Yen Lee Loh
Started 2012-2-8, touched 2012-7-24, 2013-2-1
1
Setup
2
Prelude: 3D hydrogen atom wavefunctions
3
2D Harmonic Oscillator using Cartesian Coordinates
Consider a 2D harmonic oscillator with M = k = — = W = 1. The Hamiltonian is
` 1 ` 2 ` 2 `2 `2
H = 2 Ipx + py + x + y M.
This is the sum of two 1D oscillator Hamiltonians for the x and y directions. We can solve this by defining ladder
operators
`
`
`
1
ax =
Ix + ipx M
`
ay =
2
1
2
`
`
Iy + ipy M.
Then the Hamiltonian can be factored as
` `
`
H = nx + ny + 1
` `†`
` `
`
` `†
`†
where ni = ai ai . It can easily be shown that Ani , ai E = -ai and Ani , ai E = -ai . One can then derive the raising and
`†
lowering relations (recall that for the 1D HO, a †n\ = n + 1 †n + 1\). Following the usual arguments, one finds that
a complete set of energy eigenstates can be chosen to be °nx ny ] where nx , ny = 0, 1, 2, … The eigenenergies are
Enx ny = nx + ny + 1.
` †
` †
The ground state is †00\. All excited states can be generated by successive application of ax and ay , i.e.,
` †n ` †n
1
°nx ny ] =
Hax L x Hay L y †00\.
nx ! ny !
See Fig. 1(a). The wavefunctions can be found using the usual argument for the 1D harmonic oscillator. In the
`
`
position representation, px ª -i ¶∂x . The ground state wavefunction satisfies ax †0\ = 0, so Hx + ¶∂x L y0 HxL = 0. Solving
2
this differential equation and normalizing the wavefunction gives y0 HxL = p-1ê4 e-x ë2 . Thus, we may formally write
yn HxL =
=
1
n!
B
1
2
p-1ê4 2-nê2
n!
n
2
Hx - ¶∂x LF Ap-1ê4 e-x ë2 E
2
Hx - ¶∂x Ln e-x ë2
and so ynx ny Hx, yL = ynx HxL yny HyL. In terms of special functions, one can write yn HxL =
2n n!
Hn HxL are Hermite polynomials.
x2
1
Hn HxL e- 2 where
p
n!
=
N3-LandauLevels.nb
and so ynx ny Hx,
n!
2
Hx - ¶∂x Ln e-x ë2
yL = ynx HxL yny HyL. In terms of special functions, one can write yn HxL =
Hn HxL are Hermite polynomials.
5
Energy E=n+1
4
3
` †
ay
ny
2
` †
ax
nx
1
†00\
0
5
Energy E=n+1
4
3
`†
b
`†
a
2
1
0
-4
-3
-2
-1
0
1
2
3
4
3
4
Angular quantum number m
5
4
3
2
1
0
-4
-3
-2
-1
0
1
2
x2
1
2n n!
Energy E=n+1
2
2
p-1ê4 2-nê2
Hn HxL e- 2 where
p
1
0
N3-LandauLevels.nb
-4
-3
-2
-1
0
1
2
3
4
3
4
3
Angular quantum number m
5
Energy E=n+1
4
3
2
1
0
-4
-3
-2
-1
0
1
2
Angular quantum number m
(a)
(c)
(b)
Fig. 1:
` †
` †
(a) For the 2D oscillator, starting from the ground state †00\, one can use the ladder operators ax and ay to
generate excited states °nx ny ] by increasing the vibrational energy in the x and y directions.
`†
`†
(b) Alternatively, use the ladder operators a and b to generate excited states †nm\ by adding energy while
changing angular momentum by ±1.
(c) Landau level eigenstates are identical to 2D oscillator eigenstates, but with energy shifted by -m. Thus,
panel (c) is a sheared version of panel (b).
ü Code for Fig. 1
ü Code for Fig. 2
4
2D Harmonic Oscillator using Polar Coordinates
It will be instructive to solve the 2D HO in a different way. Define ladder operators
`
`
`
`†
` † ` †
`
` `
1
1
1 `
a=
Iax + iay M
a =
Iax - iay M = 2 Ix + ipx - i y + py M
`
b=
2
1
2
`†
b =
`
`
Iax - iay M
2
1
2
` † ` †
Iax + iay M
` `
` `
`
and let n = nx + ny , so that H = n + 1. Calculate the following commutators:
` `†
` ` †
` ` †
`†
1
An, a E =
IAnx , ax E - iAny , ay EM = +a ,
2
` `†
``
`` ` †
`` `
`
` `
`
`
` `
1 ` `
1 ` `
1 ` `
ALz , a E = Ax py - y px , ax E = 2 Ax py - y px , x + ipx - i y + py E = 2 Ax py , ipx - i yE - 2 Ay px , x + py E
`
`
`
`
`
`
`
`
`†
1
1
1
= 2 I-py - xM - 2 I-i y + i px M = 2 I-py - x + i y - i px M = -a .
`†
This means that the ladder operator a increases the total energy quantum number n by 1 and decreases the magnetic
`
quantum number m (the eigenvalue of Lz ) by 1. Similarly, it can be shown that
`†
` `†
Bn, b F = +b ,
` `†
`†
BLz , b F = +b .
`†
`†
So the ladder operator b increases both n and m by 1. We may therefore start from the ground state †00\ and apply a
`†
and b to generate excited states, i.e.,
` † Hn-mLê2 ` † Hn+mLê2
1
†nm\ =
Ha L
Hb L
†00\.
J
n+m
2
N! J
n-m
2
N!
See Fig. 1(b). The wavefunctions can be found by starting with
-1ê4 -x2 ë2
-1ê4 -y2 ë2
-1ê2 -r2 ë2
4
quantum number m (the eigenvalue of Lz ) by 1. Similarly, it can be shown that
`†
` `†
Bn, b F = +b ,
N3-LandauLevels.nb
` `†
`†
BLz , b F = +b .
`†
`†
So the ladder operator b increases both n and m by 1. We may therefore start from the ground state †00\ and apply a
`†
and b to generate excited states, i.e.,
` † Hn-mLê2 ` † Hn+mLê2
1
†nm\ =
Ha L
Hb L
†00\.
J
n+m
2
N! J
n-m
2
N!
See Fig. 1(b). The wavefunctions can be found by starting with
2
2
2
y00 Hr, fL = Ip-1ê4 e-x ë2 M Ip-1ê4 e-y ë2 M = p-1ê2 e-r ë2
and applying the raising operators in the position representation.
We have
`† 1
`† 1 `
`
` `
1
a = 2 Ix + ipx - i y + py M ª 2 Ix + ¶∂x -iy - i ¶∂y M and b = 2 Ix + ¶∂x +iy + i ¶∂y M. Thus, formally,
1
ynm Hr, fL =
J
n+m
2
J
N! J
n-m
2
J
2
N! J
J
x+¶∂x +iy+i ¶∂y Hn+mLê2
Ix + ¶∂x -iy - i ¶∂y M
n-m
2
N
2
Hn-mLê2
2
n+m
N
2
2
Ip-1ê2 e-r ë2 M
N!
-n
=
x+¶∂x -iy-i ¶∂y Hn-mLê2
Hn+mLê2 -r2 ë2
Ix + ¶∂x +iy + i ¶∂y M
e
.
N! p
If desired, one can rewrite the ladder operators in terms of ¶∂r and ¶∂f instead and work entirely in terms of polars. In
1
terms of special functions, it turns out that ynm Hr, fL =
J
n-m
2
N! J
rm UI
n+m
2
m-n
,
2
r2
1 + m, r2 M e- 2 eimf where U is the
N! p
confluent hypergeometric function as defined in Mathematica (which is a polynomial in r). The eigenfunctions
ynm Hr, fL are illustrated in Fig. 3.
4
r4
Radial
quantum 3
number
n
2
r4 - 3 r2
r3
r4 - 4 r2 + 2
r3 - 2 r
r2
such that
energy 1
En =n+1
0
r4 - 3 r2
r4
r3 - 2 r
r3
r2 - 1
r
r2
r
-4
-3
-2
-1
1
0
I
1
1
1
µ
n-m
2
2
M! I
3
2
e- 2 r eimf
n+m
2
M! p
4
Angular quantum number m
Fig. 3: 2D oscillator eigenfunctions ynm Hr, fL.
To summarize: We have considered a 2D harmonic oscillator with M = k = — = W = 1. The Hamiltonian is
` 1 ` 2 ` 2 `2 `2
`
`
H = 2 Ipx + py + x + y M. One can find simultaneous eigenfunctions of H and Lz , labelled by radial quantum
number n and angular quantum number m, where n = 0, 1, 2, … and m = -n, -n + 2, -n + 4, …, n:
1
ynm Hr, fL =
J
n-m
2
N! J
rm UI
n+m
2
N! p
m-n
,
2
r2
r2
1
1 + m, r2 M e- 2 eimf =
J
n-m
2
N! J
Pnm HrL e- 2 eimf ,
n+m
2
N! p
En = Hn + 1L —W .
where U is the confluent hypergeometric function as defined in Mathematica. The functions Pnm HrL are polynomials in
r. The energy depends only on n. The eigenfunctions are illustrated above. Each eigenfunction has n radial nodes and
m angular nodes.
ü Code for Fig. 3
N3-LandauLevels.nb
5
ü Code: Verify that the wavefunctions defined above are indeed eigenfunctions of H
5
Landau Levels Wavefunctions in Symmetric Gauge (using Polar
Coordinates)
Consider a particle of mass m and charge q in an electric field E = -! V - ¶∂t A and a magnetic field B = ! µ A,
where VHrL is the electric scalar potential and AHrL is the magnetic vector potential.
`
`
` 2 `
`
1 `2
1
The quantum mechanical Hamiltonian is H = 2 M P + V = 2 M Ip - qAM + V .
`
Here p is the ordinary momentum operator, which is not gauge-invariant (in the position representation it is the partial
derivative operator -i !),
` `
`
whereas P = p - qA is the canonical momentum, which is gauge-invariant (in the position representation it is the
covariant derivative -i D).
`
Consider a 2DEG in a uniform perpendicular magnetic field B = B ez .
`
`
`
1
1
1
Use the “symmetric gauge” A = 2 B µ r = 2 BI-y ex + x ey M = 2 Br ef .
`
It is straightforward to verify that B = ! µ A = B ez .
This gauge choice is good when the potential is radially symmetric, VHrL = VHr, fL = VHrL.
qB
Let M = — = 1 and qB = 2, so that wc = M = 2. The Hamiltonian is
`
`
`2 1 `
`2 `
1 `
H 2DEG = 2 Ipx + yM + 2 Ipy - xM = H HO - Lz .
`
`
`
Therefore, a simultaneous eigenfunction of H HO and Lz is also an eigenfunction of H 2DEG , albeit with a different
energy, E2DEG = EHO - m. Ultimately, one finds that the 2DEG eigenfunctions, labelled by radial quantum number n
and angular quantum number m, where n = 0, 1, 2, … and m = -n, -n + 1, -n + 2, …, are
ynm Hr, fL =
Enm = In +
r2
1
n! Hn+mL! p
1
M —wc .
2
rm UI-n, 1 + m, r2 M e- 2 eimf ,
(Exercise: Repeat the above derivation keeping physical units.)
Fig. 4: Landau level wavefunctions ynm Hr, fL in symmetric gauge.
3
r3
Radial
quantum
number
n
r4 - 3 r2
2
r2
such that
Landau
level
energy
r5 - 6 r3 + 6 r6 - 9 r4 + 18 r72 - 12
6 r5 + 36 r38 - 24
15 r 6 + 60 r94 - 18
60 r72 + 90 r5 - 120 r3
r3 - 2 r
r4 - 4 r2 + 2
r5 - 6 r3 + 6 r r6 - 8 r4 + 12 r2r7 - 10 r5 + 20 r3
r
r2 - 1
r3 - 2 r
r4 - 3 r2
r5 - 4 r3
1
r
r2
r3
1
1
En = In + 2 M —wc
0
1
n! Hn+mL!
1
2
e- 2 r eimf µ
6
N3-LandauLevels.nb
3
r3
Radial
quantum
number
n
r4 - 3 r2
2
r2
such that
Landau
level
energy
r5 - 6 r3 + 6 r6 - 9 r4 + 18 r72 - 12
6 r5 + 36 r38 - 24
15 r 6 + 60 r94 - 18
60 r72 + 90 r5 - 120 r3
r3 - 2 r
r4 - 4 r2 + 2
r5 - 6 r3 + 6 r r6 - 8 r4 + 12 r2r7 - 10 r5 + 20 r3
r
r2 - 1
r3 - 2 r
r4 - 3 r2
r5 - 4 r3
1
r
r2
r3
0
1
2
3
1
1
En = In + 2 M —wc
0
1
n! Hn+mL!
-3
1
2
e- 2 r eimf µ
-2
-1
Angular quantum number m
Fig. 5: Landau level wavefunctions ynm Hr, fL in symmetric gauge.
These eigenfunctions form Landau levels. First examine the lowest Landau level wavefunctions (n = 0 according to
this scheme). y00 has some zero-point kinetic energy (the spread of the Gaussian); y01 represents a cyclotron orbit
that closes on itself after one cycle; y02 represents a cyclotron orbit closing after two cycles; and so on. Now go to the
higher Landau levels. Every time n increases by 1, the number of radial nodes increases by 1. It is quite similar to the
hydrogenic orbitals. Note that cyclotron orbits in the “wrong direction” are not allowed at the lowest energy. Each
unit of angular momentum in the “wrong direction” costs the same as one unit of radial “kinetic energy”.
ü Code for Fig. 4
ü Code for Fig. 5
6
Landau Levels Wavefunctions in Landau Gauge (using Cartesian
Coordinates) – unfinished
7
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