2.4.2.4 Herring-Vogt Transformation

The analytical band structure given by (2.77) represents an anisotropic dispersion law. However, it is possible to transform [21] the coordinate system to obtain a spherical energy surface which is more convenient to work with. The orientation of the coordinate system is specified so that the tensors $m_{\alpha\beta}^{-1}$ for the valleys located along the $\Delta$ axes have the form:

$$m^{-1}_{x}= \begin{bmatrix}\frac{1}{m_{Xl}} & 0 & 0\\ 0 & \frac{1... ...}} & 0 & 0\\ 0 & \frac{1}{m_{Xt}} & 0\\ 0 & 0 & \frac{1}{m_{Xl}} \end{bmatrix}.$$ (2.78)

The transformation $\vec{w}=T\vec{k}$ is given by matrices $T$ of the form:

$$T_{x}= \begin{bmatrix} \sqrt{\frac{m_{0}}{m_{Xl}}} & 0 & 0\\ 0 &... ...\frac{m_{0}}{m_{Xl}}} & 0\\ 0 & 0 & \sqrt{\frac{m_{0}}{m_{Xt}}} \end{bmatrix},$$

$$T_{z}= \begin{bmatrix} \sqrt{\frac{m_{0}}{m_{Xt}}} & 0 & 0\\ 0 &... ...\frac{m_{0}}{m_{Xt}}} & 0\\ 0 & 0 & \sqrt{\frac{m_{0}}{m_{Xl}}} \end{bmatrix}.$$ (2.79)

The dependence of energy on the wave-vector $\vec{w}$ is now isotropic

$$\epsilon(\vec{k})(1+\alpha\epsilon(\vec{k}))=\frac{\hbar^{2}\vec{w}^{2}}{2m_{0}}.$$ (2.80)

For the valleys located at the $L$ points the tensor $m_{\alpha\beta}^{-1}$ is not diagonal in the coordinate system chosen above. The right hand side of (2.77) can be expressed in terms of a rotation transformation $D$ as follows:

$$\frac{\hbar^{2}}{2}\vec{k}^{T}D^{T}\begin{bmatrix}\frac{1}{m_{Ll}... ... 0\\ 0 & \frac{1}{m_{Lt}} & 0\\ 0 & 0 & \frac{1}{m_{Lt}} \end{bmatrix}D\vec{k},$$ (2.81)

with matrix $D$ defined as:

$$T= \begin{bmatrix}\cos\alpha\cos\beta & \sin\alpha\cos\beta & \si... ...ha & 0\\ -\cos\alpha\sin\beta & -\sin\alpha\sin\beta & \cos\beta \end{bmatrix},$$ (2.82)

where the angles $\alpha$ and $\beta$ specify the longitudinal direction of an $L$ valley as shown in Fig. 2.7. This longitudinal orientation is chosen to be $x$ axis of the rotated coordinate system.

Figure 2.7: Angels $\alpha$ and $\beta$ with respect to the coordinate system which diagonalizes $m^{-1}$.

Therefore the transformation matrices in this case are equal to $SD$, where the matrix $S$ is given as

$$S=\begin{bmatrix}\sqrt{\frac{m_{0}}{m_{Ll}}} & 0 & 0\\ 0 & \sqrt{\frac{m_{0}}{m_{Lt}}} & 0\\ 0 & 0 & \sqrt{\frac{m_{0}}{m_{Lt}}} \end{bmatrix}.$$ (2.83)

For example, for the $[111]$ orientation the angles are $\alpha=\frac{\pi}{4}$, $\sin\beta=\frac{1}{\sqrt{3}}$ and the matrix $\bf {D}$ equals

$$D=\begin{bmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{... ...c{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & \frac{\sqrt{2}}{\sqrt{3}} \end{bmatrix}.$$ (2.84)

Analogous expressions can easily be obtained for other orientations:

$$D_{1}=\begin{bmatrix}-\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \... ...rac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{\sqrt{2}}{\sqrt{3}} \end{bmatrix}$$ (2.85)

where indices $1,2$ and $3$ specify orientations $[\overline{1}11]$, $[11\overline{1}]$, and $[1\overline{1}1]$, respectively. S. Smirnov: