# k.p dispersion in bulk GaAs (strained / unstrained)¶

Input files:
• bulk_kp_dispersion_GaAs_nnp.in

• bulk_kp_dispersion_GaAs_nnp_strained.in

Scope:

We calculate $$E(k)$$ of strained and unstrained $$GaAs$$.

## Band structure of bulk $$GaAs$$¶

Input file: bulk_kp_dispersion_GaAs_nnp.in

We want to calculate the dispersion $$E(k)$$ from $$|k|$$ = 0 nm-1 to $$|k|$$ = 1.0 nm-1 along the following directions in k space:

•  to 

•  to 

We compare 6-band and 8-band k.p theory results. We calculate $$E(k)$$ for bulk $$GaAs$$ at a temperature of 300 K.

### Bulk dispersion along  and along ¶

quantum{
region{
...
bulk_dispersion{
lines{ # set of dispersion lines along crystal directions of high symmetry
name = "lines"
position{ x = 5.0 }
k_max = 1.0
spacing  = 0.01
shift_holes_to_zero = yes
}

path{ # dispersion along arbitrary path in k-space
name = "user_defined_path"
position{ x = 5.0 }
point{ k = [0.7071, 0.7071, 0.0] }
point{ k = [0.0, 0.0, 0.0] }
point{ k = [1.0, 0.0, 0.0] }
spacing  = 0.01
shift_holes_to_zero = yes
}
}
}
}


We calculate the pure bulk dispersion at position x = 5 nm. In our case this is $$GaAs$$, but it could be any strained alloy. In the latter case, the k.p Bir-Pikus strain Hamiltonian will be diagonalized. The grid point at position{ x = 5.0 } must be located inside a quantum region. shift_holes_to_zero = yes forces the top of the valence band to be located at 0 eV. How often the bulk k.p Hamiltonian should be solved can be specified via spacing. To increase the resolution, just increase this number. We use two direction in k space, i.e. from  to  and from  to . In the latter case the maximum value of $$|k|$$ is

$k_\mathrm{max} = \sqrt{0.7071^2+0.7071^2} = 1.0$

Note that for values of $$|k|$$ larger than 1.0 nm-1, k.p theory might not be a good approximation anymore.

The results of the calculation can be found in the folder bias_00000\Quantum\Bulk_dispersions. Figure 2.5.8.1 visualizes the results. Figure 2.5.8.1 Bulk k.p dispersion in $$GaAs$$: $$E(k)$$ along  and .¶

The split-off energy of 0.341 eV is identical to the split-off energy as defined in the database:

...
valence_bands{ delta_SO = 0.341 } # [eV] Vurgaftman1
...


If one zooms into the holes and compares 6-band vs. 8-band k.p, one can see that 6-band and 8-band coincide for $$|k|$$ < 1.0 nm-1 for the heavy and light hole but differ for the split-off hole at larger $$|k|$$ values, see Figure 2.5.8.2. Figure 2.5.8.2 Bulk k.p dispersion in GaAs: $$E(k)$$ along  and  - Comparision between 6x6 and 8x8 k.p¶

### 8-band k.p vs. effective-mass approximation¶

Now we want to compare the 8-band k.p dispersion with the effective-mass approximation. The effective mass approximation is a simple parabolic dispersion which is isotropic (i.e. no dependence on the k vector direction). For low values of k ($$|k|$$ < 0.4 nm-1) it is in good agreement with k.p theory, see Figure 2.5.8.3. Figure 2.5.8.3 Bulk k.p dispersion in $$GaAs$$: $$E(k)$$ along  and  - Comparision between 8x8 k.p and effective-mass approximation¶

## Band structure of strained $$GaAs$$¶

Input file: bulk_kp_dispersion_GaAs_nnp_strained.in

Now we perform these calculations again for $$GaAs$$ that is strained with respect to $$In_{0.2}Ga_{0.8}As$$. The $$InGaAs$$ lattice constant is larger than the $$GaAs$$ one, thus $$GaAs$$ is strained tensely. The changes that we have to make in the input file are the following:

strain{
pseudomorphic_strain{}
}

run{
strain{}
}


As substrate material we take $$In_{0.2}Ga_{0.8}As$$ and assume that $$GaAs$$ is strained pseudomorphically (pseudomorphic_strain{}) with respect to this substrate, i.e. $$GaAs$$ is subject to a biaxial strain. Due to the positive hydrostatic strain (i.e. increase in volume or negative hydrostatic pressure) we obtain a reduced band gap with respect to the unstrained $$GaAs$$. Furthermore, the degeneracy of the heavy and light hole at $$k= 0 is lifted, see :numref:fig-1D-kp-dispersion-bulk-GaAs-kp-bandedges-strained$$. Now, the anisotropy of the holes along the different directions  and  is very pronounced. There is even a band anti-crossing along . (Actually, the anti-crossing looks like a “crossing” of the bands but if one zooms into it (not shown in this tutorial), one can easily see it.) Note: If biaxial strain is present, the directions along $$x$$, $$y$$ or $$z$$ are not equivalent anymore. This means that the dispersion is also different in these directions (, , ). Figure 2.5.8.4 Bulk k.p dispersion in $$GaAs$$ strained with respect to $$In_{0.2}Ga_{0.8}As$$ : $$E(k)$$ along  and .¶

If one zooms into the holes and compares 6-band vs. 8-band k.p, one can see that the agreement between heavy and light holes is not as good as in the unstrained case where 6-band and 8-band k.p lead to almost identical dispersions, compare Figure 2.5.8.5. Figure 2.5.8.5 Bulk valence band k.p dispersion in $$GaAs$$ strained with respect to $$In_{0.2}Ga_{0.8}As$$ : $$E(k)$$ along  and  - Comparision between 6x6 and 8x8 k.p approximation.¶

Note that in the strained case, the effective-mass approximation is very poor.

## Analysis of eigenvectors¶

(preliminary)

Using the Voon-Willatzen-Bastard-Foreman k.p basis one obtains the following output for the eigenvectors at the Gamma point, $$k$$ = ($$k_x$$, $$k_y$$, $$k_z$$) = 0.

Example: The x_up component contains a complex number. Here, we show the square of X_up. This gives us information on the strength of the coupling of the mixed states.

   eigenvalue  S+     S-     HH     LH     LH     LH     SO     SO
1           0      1.0    0      0      0      0      0      0
2           1.0    0      0      0      0      0      0      0
3           0      0      0      1.0    0      0      0      0
4           0      0      0      0      1.0    0      0      0
5           0      0      0      0      0      1.0    0      0
6           0      0      1.0    0      0      0      0      0
7           0      0      0      0      0      0      0      1.0
8           0      0      0      0      0      0      1.0    0

eigenvalue  S+          S-          X+          Y+          Z+          X-          Y-          Z-
1           1.0         0           0           0           0           0           0           0
2           0           1.0         0           0           0           0           0           0
3           0           0           0           0           0.5         0.5         0           0
4           0           0           0           0           0.166       0.166       0.666       0
5           0           0           0.5         0           0           0           0           0.5
6           0           0           0.166       0.666       0           0           0           0.166
7           0           0           0           0           0.333       0.333       0.333       0
8           0           0           0.333       0.333       0           0           0           0.333


+: spin up, -: spin down

• The electron eigenstates are 2-fold degenerate, i.e. have the same energy, and are decoupled from the holes.

 1 $$| S \downarrow{}\rangle$$ $$\;$$ 2 $$| S \uparrow{}\rangle$$ $$\;$$
• The hole eigenstates are 4-fold (heavy and light holes) and 2-fold degenerate (split-off holes).

 3 $$\left|\frac{3}{2}, \frac{3}{2}\right\rangle$$ $$\;$$ hh spin up $$\frac{1}{\sqrt{2}}\left| (X +iY) \uparrow{}\right\rangle$$ 4 $$\left| \frac{3}{2}, \frac{1}{2}\right\rangle$$ $$\;$$ lh $$\frac{1}{\sqrt{6}}\left| (X + iY) \downarrow{} \right\rangle - \sqrt{\frac{2}{3}} \left| Z \uparrow{}\right\rangle$$ 5 $$\left|\frac{3}{2}, -\frac{1}{2}\right\rangle$$ $$\;$$ lh $$\frac{1}{\sqrt{6}}\left| (X - iY) \uparrow{} \right\rangle - \sqrt{\frac{2}{3}} \left| Z \downarrow{}\right\rangle$$ 6 $$\left|\frac{3}{2}, -\frac{3}{2}\right\rangle$$ $$\;$$ hh spin down $$\frac{1}{\sqrt{2}}\left| (X - iY) \downarrow{} \right\rangle$$ 7 $$\left|\frac{1}{2}, \frac{1}{2}\right\rangle$$ $$\;$$ s/o split $$\frac{1}{\sqrt{3}}\left| (X + iY) \downarrow{} \right\rangle - \frac{1}{\sqrt{3}} \left| Z \uparrow{}\right\rangle$$ 8 $$\left|\frac{1}{2}, -\frac{1}{2}\right\rangle$$ $$\;$$ s/o split $$\frac{1}{\sqrt{3}}\left| (X - iY) \downarrow{} \right\rangle - \frac{1}{\sqrt{3}} \left| Z \downarrow{}\right\rangle$$
$$\frac{1}{\sqrt{2}}$$ = 0.707 $$\rightarrow{}$$ $$\left(\frac{1}{2}\right)^2$$ = 0.5
$$\frac{1}{\sqrt{3}}$$ = 0.577 $$\rightarrow{}$$ $$\left(\frac{1}{3}\right)^2$$ = 0.333
$$\frac{1}{\sqrt{6}}$$ = 0.408 $$\rightarrow{}$$ $$\left(\frac{1}{6}\right)^2$$ = 0.166