nextnano^{3}  Tutorial
next generation 3D nano device simulator
2D Tutorial
Hole wave functions in a quantum wire subjected to a magnetic field
Note: This tutorial's copyright is owned by Stefan Birner,
www.nextnano.com.
Author:
Stefan Birner
If you want to obtain the input files that are used within this tutorial, please contact stefan.birner@nextnano.de.
> 2Dwire.in
(singleband approximation)
> 2Dwire_magnetic.in (singleband approximation including
magnetic field)
> 2Dwire_6x6kp.in
(6band k.p
approximation)
> 2Dwire_6x6kp_Burt.in
(6band k.p
approximation)
Quantum wire
Similar to the 1D confinement in a quantum well, it is possible to
confine electrons or holes in two dimensions, i.e. in a quantum wire.
The quantum wire consists of InAs (blue area) and is confined by GaAs barriers
(red area).
Its size is 10 nm x 10 nm whereas the whole simulation dimension is 30 nm x
30 nm.


The blue area is the quantum wire which consists of InAs. The quantum wire
rectangle has the size 10 nm x 10 nm.

This picture shows a possible configuration of rectangular grid lines.
Here, the grid spacing is 0.5 nm, thus the quantum wire (blue area) consists
of 21 x 21 = 400 grid points.

Quantum cluster
Assumptions:
1) We apply Dirichlet boundary conditions to our quantum cluster (psi = 0
at the boundary).
2) The quantum cluster is defined only in the area of the quantum wire,
i.e. from 10 nm to 20 nm.
These two conditions lead to an infinite GaAs barrier and thus the
wave function is forced to be zero at the InAs quantum wire boundaries. Of
course, this is not a realistic assumption but we simplify the sample to make
the tutorial easier.
The energy levels and the wave functions of a rectangular quantum wire of length 10 nm
with infinite barriers can be calculated analytically.
We assume that the barriers at the QD boundaries are infinite. This way we
can compare our numerical calculations to analytical results.
The potential inside the quantum wire is assumed to be 0 eV.
As effective mass we take the isotropic heavy hole effective mass of InAs, i.e. m_{hh}
= 0.41 m_{0}.
valencebandmasses = 0.41d0 0.41d0 0.41d0 !
isotropic heavy hole
effective
...
A discussion of the analytical solution of the 2D Schrödinger equation of a
particle in a rectangle (i.e. quantum wire) with infinite barriers can be found in e.g.
Quantum Heterostructures (Microelectronics and Optoelectronics) by V.V.
Mitin, V.A. Kochelap and M.A. Stroscio.
The solution of the Schrödinger equation leads to the following eigenvalues
(where m_{hh} is assumed to be negative):
E_{n1},_{n2} = h_{bar}^{2} pi^{2}
/ 2m_{hh }
( n_{1}^{2 }/ L_{x}^{2} + n_{2}^{2
}/ L_{y}^{2}) =
=
 9.1714667 * 10^{19} eVm^{2} ( n_{1}^{2
}/ L_{x}^{2} + n_{2}^{2 }/ L_{y}^{2}) =
=
 0.0091714667 eV
( n_{1}^{2
} + n_{2}^{2 } ) (if L_{x}
= L_{y} = 10 nm)
 E_{n1},_{n2} is the heavy hole energy in the two
transverse directions, or the total heavy hole energy for k_{z} = 0.
Note that the total heavy hole energy is given by E_{hh} = E_{n1},_{n2}
+ h_{bar}^{2 }k_{z}^{2 }/ 2m_{hh}
where k_{z} is the wavevector along the z direction leading to a
onedimensional E(k_{z}) dispersion.
 n_{1}, n_{2}^{2}
are two discrete quantum numbers (because we have two
directions of quantization).
 L_{x} and L_{y} are the lengths along the x and y
directions. In our case, L_{x} = L_{y} =
10 nm.
Generally, the energy levels are not degenerate, i.e. all energies are
different.
However, some energy levels with different quantum numbers coincide, if
the lengths along two directions are identical (E_{n1},_{n2} = E_{n2},_{n1}) or
if their ratios are integers. In our quadratic quantum wire, the two lengths are
identical.
Consequently, we expect the following degeneracies:
 E_{11} =
 0.018343 eV (ground state)
 E_{12} = E_{21} =
 0.045857 eV
 E_{22} =
 0.073372 eV
 E_{13} = E_{31} =
 0.091715 eV
 E_{23} = E_{32} =
 0.119229 eV
 E_{14} = E_{41} =
 0.155915 eV
 ...
 E_{18} = E_{81} = E_{47} = E_{74} =
 0.596145 eV (Here, the degeneracy is a coincidence.)
nextnano³ numerical results for a 10 nm quadratic quantum wire with 0.10 nm
grid spacing:
Output file name:
Schroedinger_1band/ev2D_vb001_qc001_sg001_deg001_dir_Kx001_Ky001_Kz001.dat
num_ev: eigenvalue [eV]:
(0.10 nm grid)
1 0.018341
= E_{11}
2 0.045845
(twofold degenerate) E_{12}/E_{21}
3 0.045845
(twofold degenerate) E_{12}/E_{21
} 4 0.073348
= E_{22
} 5 0.091653
(twofold degenerate) E_{13}/E_{31
} 6 0.091653
(twofold degenerate) E_{13}/E_{31
} 7 0.119156
(twofold degenerate) E_{23}/E_{32}
8 0.119156 (twofold degenerate) E_{23}/E_{32
} 9 0.155721
(twofold degenerate) E_{14}/E_{41
} 10 0.155721
(twofold degenerate) E_{14}/E_{41
} ... _{
} 42 0.593061
(fourfold degenerate) E_{18}/E_{81} or E_{47}/E_{74
} 43 0.593061
(fourfold degenerate) E_{18}/E_{81} or E_{47}/E_{74}
44 0.594144
(fourfold degenerate) E_{18}/E_{81} or E_{47}/E_{74
} 45 0.594144
(fourfold degenerate) E_{18}/E_{81} or E_{47}/E_{74}
Hole wave functions  Singleband effectivemass approximation (infinite GaAs
barriers)
> 2Dwire.in
The following figures show the charge densities (Psi²; Psi = hole
wave function) of the four lowest energy confined hole eigenstates in an
infinitely deep 10 nm x 10 nm InAs quantum wire. Due to the symmetry of the quantum wire, the
second and the third eigenstate are degenerate.


1^{st} eigenstate: 0.0183 eV 
2^{nd} eigenstate: 0.0458 eV 


3^{rd} eigenstate: 0.0458 eV 
4^{th} eigenstate: 0.0733 eV 
The heavy hole valence band edge energy of InAs is set to be located at 0 eV. The hole eigenvalues are:
schroedingermassesanisotropic =
yes 
= no 
= box 
1st eigenvalue =>
confinement energy: 0.018341 eV 
0.018341 eV 
0.018341 eV 
2nd eigenvalue =>
confinement energy: 0.045845 eV 
0.045845 eV 
0.045845 eV 
3rd eigenvalue =>
confinement energy: 0.045845 eV 
0.045845 eV 
0.045845 eV 
4th eigenvalue =>
confinement energy: 0.073348 eV 
0.073348 eV 
0.073348 eV 
Note that these wave functions were obtained by using a singleband
effective mass approximation for the holes. A more accurate and more
realistic treatmeant would have been to use 6band k.p. Note that the wire
has been assumed to be unstrained (which is a rather unphysical situation) for
the purpose to make this tutorial easier to understand.
Magnetic field  singleband approximation (infinite GaAs barriers)
> 2Dwire_magnetic.in
Here we use: schroedingermassesanisotropic =
yes ! 'yes', 'no',
'box'
We apply a magnetic field perpendicular to the (x,y) plane, i.e. along the
[001] direction.
$magneticfield
magneticfieldon =
yes
magneticfieldstrength = 1.0d0 ! 1
Tesla
magneticfielddirection = 0 0 1 ! [001]
direction
$end_magneticfield
This leads to an additional confinement in addition to the wire potential.
However, for the first and forth eigenstate, the confinement does not play an
important role whereas for the second and third it does. The effect is more
dominant onto the wave functions but not so pronounced onto the values of the
eigenenergies. Now the degeneracy of the 2^{nd} and 3^{rd}
eigenstate is slightly lifted in comparison to the case where no magnetic field
is applied.


1^{st} eigenstate: 0.0151 eV 
2^{nd} eigenstate: 0.0375
eV 


3^{rd} eigenstate: 0.0378
eV 
4^{th} eigenstate: 0.0602 eV 
Here, the probabilty density of the second eigenstate is plotted viewed from a
different perspective.
Note: Currently magnetic field only works for option schroedingermassesanisotropic =
yes .
Hole wave functions  6band k.p approximation (infinite GaAs barriers)
> 2Dwire_6x6kp.in
> 2Dwire_6x6kp_Burt.in
The following figures show the charge densities (Psi²; Psi = hole
wave function) of the four lowest energy confined hole eigenstates in an
infinitely deep 10 nm x 10 nm InAs quantum wire. This time we used 6band k.p theory to
describe the hole states. Here, the
second and the third eigenstate are no longer degenerate. Only the ground state
looks similar to the singleband effectivemass approximation results.
We used the following Luttinger parameters for InAs:
gamma_{1} = 20.0, gamma_{2} = 8.5, gamma_{3}
= 9.2
This corresponds to the Dresselhaus parameters L, M, N:
L =  55, M =  4, N =  55.2
6x6kpparameters = 55.000d0 4.000d0
55.200d0 ! L, M, N [hbar^2/2m](> divide by hbar^2/2m)
0.39d0
! delta_splitoff [eV]


1^{st} /2^{nd} eigenstate:
0.0342 eV 
3^{rd}/4^{th }eigenstate:
0.0367 eV 


5^{th}/6^{th} eigenstate:
0.0557 eV 
7^{th}/8^{th} eigenstate:
0.0581 eV 
The energies are contained in this file:
ev_hl_6x6kp_qc001_num_kpar001_2D.dat
num_ev: eigenvalue[eV]: cb:
hh: lh:
s/o:
1 0.34226968E01
0.000E+00 0.894E01 0.905E+00 0.519E02
2
0.34226968E01 0.000E+00 0.894E01 0.905E+00 0.519E02
3 0.36675499E01
0.000E+00 0.285E+00 0.710E+00 0.528E02
4
0.36675499E01 0.000E+00 0.285E+00 0.710E+00 0.528E02
5 0.55698284E01
0.000E+00 0.263E+00 0.718E+00 0.192E01
6
0.55698284E01 0.000E+00 0.263E+00 0.718E+00 0.192E01
7 0.58120024E01
0.000E+00 0.384E+00 0.604E+00 0.113E01
8
0.58120024E01 0.000E+00 0.384E+00 0.604E+00 0.113E01
We used:
schroedingerkpevsolv =
ARPACK
!
schroedingerkpdiscretization = boxintegration
!
kpvvtermsymmetrization =
yes
! 'yes' (= incorrect physics), 'no'
(= correct physics)
Comment: schroedingerkpevsolv = chearn
leads to very similar (almost identical) results for 6band k.p
for both options of kpvvtermsymmetrization =
yes/no .
However, using instead Burt's nonsymmetrized discretization
kpvvtermsymmetrization =
no
! 'yes', 'no'
leads to the following results:


1^{st} /2^{nd} eigenstate:
0.0383 eV 
3^{rd}/4^{th }eigenstate:
0.0419 eV 


5^{th}/6^{th} eigenstate:
0.0679 eV 
7^{th}/8^{th} eigenstate:
0.0768 eV 
num_ev: eigenvalue[eV]: cb:
hh: lh:
s/o:
1 0.38319736E01
0.000E+00 0.531E01 0.941E+00 0.606E02
2
0.38319736E01 0.000E+00 0.531E01 0.941E+00 0.606E02
3 0.41904230E01
0.000E+00 0.265E+00 0.735E+00 0.860E03
4
0.41904230E01 0.000E+00 0.265E+00 0.735E+00 0.860E03
5 0.67878674E01
0.000E+00 0.414E+00 0.584E+00 0.180E02
6
0.67878675E01 0.000E+00 0.414E+00 0.584E+00 0.180E02
7 0.76825362E01
0.000E+00 0.111E+00 0.874E+00 0.145E01
8
0.76825362E01 0.000E+00 0.111E+00 0.874E+00 0.145E01
The following is obsolete because the feature
valencebandmassesfromkp does not really make much sense.
Hole wave functions  Singleband effectivemass approximation (infinite GaAs
barriers)
> 2Dwire_vbmassesfrom6x6kp.in
Here we use a singleband effectivemass approximation but we use an
effectivemass tensor that is obtained from the Luttinger parameters.
valencebandmassesfromkp =
yes ! 'yes', 'no'
schroedingermassesanisotropic = yes ! 'yes',
'no', 'box'
We used the following Luttinger parameters for InAs:
gamma_{1} = 20.0, gamma_{2} = 8.5, gamma_{3}
= 9.2
This corresponds to the Dresselhaus parameters L, M, N:
L =  55, M =  4, N =  55.2
6x6kpparameters = 55.000d0 4.000d0
55.200d0 ! L, M, N [hbar^2/2m](> divide by hbar^2/2m)
0.39d0
! delta_splitoff [eV]
This leads to the following effectivemass tensor components for the heavy
hole:
m_{xx} = 0.333
m_{yy} = 0.333
m_{zz} = 0.333
m_{xy} =  0.943
m_{xz} =  0.943
m_{yz} =  0.943
The following figures show the charge densities (Psi²; Psi = hole
wave function) of the five lowest energy confined hole eigenstates in an
infinitely deep 10 nm x 10 nm InAs quantum wire. The degeneracy of the
second and the third eigenstate is now lifted.



1^{st} eigenstate: 0.0182 eV
(0.0182) 
2^{nd} eigenstate: 0.0407 eV
(0.0406) 




3^{rd} eigenstate: 0.0500 eV
(0.0500) 
4^{th} eigenstate: 0.0665 eV
(0.0662) 
5^{th} eigenstate: 0.0896 eV
(0.0895) 
In the brackets, the eigenvalues for schroedingermassesanisotropic =
box are given.
> 2Dwire_vbmassesfrom6x6kp_iso.in
If we had chosen
schroedingermassesanisotropic = no !
'yes', 'no', 'box'
then the offdiagonal elements of the effectivemass tensor are not taken
into account.
m_{xy} =  0.943 := 0 < assumed to be zero!!!
m_{xz} =  0.943 := 0 < assumed to be zero!!!
m_{yz} =  0.943 := 0 < assumed to be zero!!!
Due to the symmetry of the quantum wire and the effective mass tensor,
the second and the third eigenstate are degenerate.
The relevant wave functions in this case are:


1^{st} eigenstate: 0.0186 eV 
2^{nd} eigenstate: 0.0463 eV 


3^{rd} eigenstate: 0.0463 eV 
4^{th} eigenstate: 0.0741 eV 
 Please help us to improve our tutorial. Send comments to
support
[at] nextnano.com .
