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nextnano3 - Tutorial
next generation 3D nano device simulator
2D Tutorial
Energy dispersion of a cylindrically shaped GaN nanowire
Author:
Stefan Birner 
If you want to obtain the input files that are used within this tutorial, please 
check if you can find them in the installation directory. 
If you cannot find them, please submit a
Support Ticket. 
 -> 2DGaN_nanowire_nn3.in / *_nnp.in - input file for the nextnano3 
and nextnano++ software 
  
 
Energy dispersion of a cylindrically shaped GaN nanowire
In this tutorial we study the electron and hole energy levels of a two-dimensional 
freestanding GaN nanowire of cylindrical shape. 
The tutorial is based on the following paper: 
 
    X.W. Zhang, J.B. Xia 
    Optical properties of GaN wurtzite quantum wires 
    J. Phys.: Condens. Matter 18, 3107 (2006) 
  
	- We assume a cylindrically shaped GaN nanowire 
	(wurtzite structure) that has a radius of 2 nm with infinite barriers so 
	that the wave functions are zero at the nanowire boundary. This assumption is 
	consistent to the above cited paper. The GaN nanowire is shown in 
	red in the following figure. The GaN 
	nanowire is discretized on a mesh with a grid resolution of 0.05 nm.
 
	 
	  
  
 
  
Electrons
	- The following figure shows the electron states as a function of k of the 
	GaN nanowire. It is in excellent agreement with Fig. 1 of the above cited 
	paper. All states are two-fold degenerate due to spin. In addition, the 2nd 
	and 3rd state are degenerate, as well as the 4th and 
	the 5th.
 
	The ground state has quantum number L = 0. For L /= 0, the states are 
	degenerate due to L = +/- 1. 
	The energy levels increase with increasing k as quadratic terms of k 
	(parabolic dispersion). 
	 
	Technical details: We calculated the electron energy levels at kz 
	= 0 with nextnano³ numerically by solving the 2D single-band 
	Schrödinger equation. 
	The parabolic dispersion for kz /= 0 has been calculated 
	analytically using Ei(kz)=Ei + hbar2
	kz2 / (2m*), i.e. not with nextnano³. 
	The eigenvalues for kz = 0 can be found in the following file: 
	Schroedinger_1band/ev_cb1_sg1_deg1.dat 
	 
	  
  
	- The wave function (psi²) of the electron ground state at k = 0 is shown 
	in this figure.
 
	 
	  
  
 
  
Holes
	- The following figures show the ground state wave function (psi²) of the 
	hole (left figure) and the 1st excited hole state (right figure) 
	as calculated within the 6-band k.p approximation at k = 0.
 
	According to the above cited paper, the right figure would be the ground 
	state for GaN nanowires with a radius < 0.7 nm. Because our nanowire has a 
	radius of 2 nm, the ground state wave function is according to the left 
	figure. 
	Following the paper of Zhang and Xia, this means that the probability for 
	electron-hole transitions (e1-h1) is not very high at a radius of 2 nm 
	because the wave functions don't have much overlap and the electron ground 
	state has L = 0, whereas the hole ground state has L = +/- 1 
	(dark exciton effect). 
	 
	   
	 
  
	- The following figure shows the hole states as a function of k of the GaN 
	nanowire as calculated with 6-band k.p theory. It corresponds to Fig. 2 
	and Fig. 3 of the paper of Zhang and Xia. Note that the authors assumed the 
	hole energies to be positive.
 
	All states are two-fold degenerate, i.e. h1=h2, h3=h4, h5=h6, ... 
	The nextnano³ results are a bit different. Several reasons could 
	explain this: 
	- The authors use the "cylindrical approximation" for the 
	k.p parameters. However, the parameters that they are citing are not 
	exactly cylindrical. Thus for our calculations, we had to employ the 
	parameters that they were citing (without making use of the cylindrical 
	approximation). 
	- Our cylinder does not have exactly cylindrical symmetry. It 
	is approximated to be cylindrical by a rectangular grid with a grid 
	resolution of 0.05 nm. 
	- For the k.p parameters that are given in the paper it 
	must hold: 
  A5 = 1/2 (L1 - M1) must be equal to 
  A5 = 1/2 N1 
  but they differ by 0.0064. 
	  
	The data that has been plotted in this figure is contained in this file: 
	Schroedinger_kp/kpar2D_disp_0_1_hl_6x6kp_ev_min001_ev_max030.dat 
	 
	In the input file, one can specify the number of k|| = kz 
	points. 
	 
	
	$quantum-model-holes 
 ... 
 num-kp-parallel = 41 
	 
	Note: 41 points means 16 minutes CPU 
	time (Intel i5, 2015). If one uses only 1, then one 
	only calculates the k.p states at kz=0 and the calculation takes 
	less than a minute. 
  
	- Zhang and Xia used the following 6-band k.p parameters:
 
	 
	 
	Crystal field and spin-orbit splitting energies: 
	 
 Deltacr = 0.021   Deltaso = 0.018 
	 
	 
	"Dresselhaus" parameters: 
	 
	 Zhang/Xia           
	nextnano³ 
 ==========        ============= 
 L = 6.3055  <==>  L1 = -6.3055 - 1 
	= -7.3055   (!) ==> The definition of the 
	k.p Hamiltonians differs. 
 M = 0.1956  <==>  M1 = -0.1956 - 1 
	= -1.1956   (!) ==> The definition of the 
	k.p Hamiltonians differs. 
 N = 0.3813  <==>  M2 = -0.3813 - 1 
	= -1.3813   (!) ==> The definition of the 
	k.p Hamiltonians differs. 
 R = 6.1227  <==>  N1 = -6.1227 
 S = 0.4335  <==>  M3 = -0.4335 - 1 
	= -1.4335   (!) ==> The definition of the 
	k.p Hamiltonians differs. 
 T = 7.3308  <==>  L2 = -7.3308 - 1 
	= -8.3308   (!) ==> The definition of the 
	k.p Hamiltonians differs. 
 Q = 4.0200  <==>  N2 = -4.0200 
	 
	Conversion to "Luttinger" parameters: 
	 
	 A1 = L2 + 1  = -8.3308
	+ 1      = -7.3308       
	(!) ==> The definition of the k.p Hamiltonians differs. 
 A2 = M3 + 1  = -1.4335
	+ 1      = -0.4335       
	(!) ==> The definition of the k.p Hamiltonians differs. 
 A3 = M2 - L2 = -0.3813 + 7.3308 =  6.9495 
 A4 = 1/2 (L1 + M1 - 2M3)        = 
	-2.81705 
 A5 = 1/2 (L1 - M1)              
	= -3.05495      (!) 
	==> inconsistent to -3.06135 
    = 1/2  N1                    
	= -3.06135      (!) 
	==> inconsistent to -3.05495 
 A6 = SQRT(2)/2 N2               
	= -2.84256926 
	 
  
	- cylindrical (axial) approximation
 
	 
      L           
	-  M           - R   
	= 0   (Zhang) 
      L1          
	-  M1          - N1  
	= 0   (nextnano³) 
 ==> (A2+A4+A5-1) - (A2+A4-A5-1) - 2A5 = 0 
	 
      A1 - A2   = -A3 = 2 A4 
      A3 + 4 A5 =       
	SQRT(2) A6 
	 
      Delta2 = Delta3 = 1/3 Deltaso 
  
 
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