# Hole energy levels of an “artificial atom” - Spherical Si Quantum Dot (6-band k.p)¶

Input files:
• 3DsphericSiQD_d5nm_6bandkp_nnp.in

Scope:

In this tutorial, we calculate the energy spectrum of a spherical Si quantum dot of radius 2.5 nm.

Output Files:
• bias_00000\Quantum\energy_spectrum_qr_6band_kp6_00000.dat

## Introduction¶

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:

 kp_6_bands{
L = -6.8           # [Burdov] V.A. Burdov, JETP 94, 411 (2002)
M = -4.43          # [Burdov]
N = -8.61          # [Burdov]
}


These L, M, N parameters correspond to the following Luttinger parameters:

• $$\gamma_1$$ = 4.22

• $$\gamma_2$$ = 0.395

• $$\gamma_3`$$ = 1.435

## Results¶

Figure 2.4.8.7 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

The nextnano++ tool only allows a cuboidal shaped quantum region, thus we can’t employ a spherical quantum region that would reduce the dimension of the 6-band k.p Hamiltonian matrix and thus the overall execution time.

Following the paper of [Burdov2002], one can calculate the ground state energy for this particular system from the L and M parameters:

$E_1 = - \frac{\hbar^2 \pi^2}{2 m_h R^2} = -0.314 eV$

using $$m_h$$ = 0.192 $$m_0$$ as [Burdov2002], where he uses incorrect k.p parameters: In his definition L must be -5.8 and M = -3.43.

$E_1 = - \frac{\hbar^2 \pi^2}{2 m_h R^2} = -0.254 eV$

using $$m_h$$ = 0.237 $$m_0$$ as [BelyakovBurdov2008]. The latter is in much better agreement to our calculations. $$m_h$$ is given by:

$m_h = \frac{3 m_0}{L+2M} = \frac{3 m_0}{ -6.8 + 2 \cdot (-4.43)} = -0.192 m_0$

in [Burdov2002] and

$m_h = \frac{3 m_0}{ (L+1) + 2 (M+1)} = \frac{3 m_0}{ -5.8 + 2 * (-3.43)} = -0.237 m_0 \quad$

in [BelyakovBurdov2008]. 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.

## Comparison of nextnano³ and nextnano++¶

Figure 2.4.8.8 compares the nextnano³ results with the nextnano++ results. The results of both simulators are in excellent agreement.