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Hole energy levels of an "artificial atom" - Spherical Si Quantum Dot (6-band k.p)

Author: Stefan Birner

-> 3DsphericSiQD_d5nm_6bandkp_nn3.in / *_nnp.in - spherical Si QD with diameter 5 nm - input file for the nextnano³ and nextnano++ software

These input files are included in the latest version.

Hole energy levels of an "artificial atom" - Spherical Si Quantum Dot (6-band k.p)

-> 3DsphericSiQD_d5nm_6bandkp_nn3.in - spherical Si QD with diameter 5 nm

Here, we want to calculate the energy spectrum of a spherical Si quantum dot of radius 2.5 nm.

We assume that the barriers at the QD boundaries are infinite.
The potential inside the QD is assumed to be 0 eV.
We use a grid resolution of 0.25 nm.
We solve the 6-band k.p Schrödinger equation for the hole eigenstates.

The following 6-band k.p parameters are used:

   L = -6.8           ! [Burdov] V.A. Burdov, JETP 94, 411 (2002)
   M = -4.43          ! [Burdov])
   N = -8.61          ! [Burdov])
   Deltaso = 0.044 eV

 6x6kp-parameters = -6.8d0 -4.43d0 -8.61d0 ! L, M, N [hbar^2/2m]
                     0.044d0               ! Deltasplit-off [eV]   spin-orbit splitting energy

These L, M, N parameters correspond to the following Luttinger parameters:
   gamma1 = 4.22
   gamma2 = 0.395
   gamma3 = 1.435
 

The following figure shows the hole eigenenergy spectrum of the Si QD (diameter = 5 nm) calculated with a 6-band k.p Hamiltonian.

For comparison, we also display the energy spectrum where we assumed zero spin-orbit splitting energy.
In this case there is a six-fold symmetry. Spin-orbit splitting reduces this degeneracy to 4 and 2.
In general, each state is two-fold degenerate due to spin.

(Note: We used a cuboidal shaped quantum cluster although we could have used a spherical quantum cluster to reduce the size of the 6-band k.p Hamiltonian matrix that has to be solved for the eigenstates.
If m grid points can be excluded from the quantum region due to an optimal choice of quantum region shape, the dimension N of the 6-band k.p matrix is reduced by N-6m.)

Following the paper of [Burdov], one can calculate the ground state energy for this particular system from the L and M parameters:

E₁ = - h_bar² pi² / ( 2 m_h R² ) = -0.314 eV    (using m_h = 0.192 m₀ as [Burdov] where he uses incorrect k.p parameters: In his definition L must be -5.8 and M = -3.43.)
E₁ = - h_bar² pi² / ( 2 m_h R² ) = -0.254 eV    (using m_h = 0.237 m₀ as V.A. Belyakov, V.A. Burdov, J. Phys.: Condens. Matter 20, 025213 (2008))
The latter is in much better agreement to our calculations.

m_h is given by:
m_h = 3 m₀ / (  L      + 2  M    ) = 3 m₀ / ( -6.8 + 2 * (-4.43)) = -0.192 m₀     [Burdov]
m_h = 3 m₀ / ( (L+1) + 2 (M+1)) = 3 m₀ / ( -5.8 + 2 * (-3.43)) = -0.237 m₀    [Belyakov/Burdov]
The latter definition is consistent to our implementation of the k.p Hamiltonian.

The discrepancy of these equations arises because there are two different definitions of the L,M parameters available in the literature.
More details...

nextnano³ vs. nextnano++ comparison

The following figure compares the nextnano³ results with the nextnano++ results.
The results of both simulators are in excellent agreement.

Additional comment for experts

For this particular geometry, the eigenvalues are highly degenerate, not only due to spin, but also due to geometry.
This might cause problems for certain eigenvalue solvers as they might miss some of these degenerate eigenvalues.
So the software should be used with care.
In our case, the 'chearn' eigenvalue solver (Arnoldi method that uses Chebyshev polynomials as preconditioner) missed some degenerate eigenvalues. So probably one has to adjust some eigenvalue solver parameters to increase the accuracy.
For this reason it is of great advantage if any numerical software has redundancy in terms of several eigensolvers where one can choose from in order to check results for consistency and accuracy, as well as performance.