Revealing the topological phase diagram of ZrTe5 using the complex strain fields of microbubbles

Centre for Energy Research, Institute of Technical Physics and Materials Science

Zoltán Tajkov

outline

Topological Insulators

Why are they important?

Results

Topological Insulators

$$ E $$

Insulator?

Topology?

$$ \frac{1}{2\pi}\int K \mathrm{d}A = 2(1-g) $$

Topological Insulators

$$ \frac{1}{2\pi}\int K \mathrm{d}A = 2(1-g) $$

Su-Schrieffer-Heeger

H_{\mathrm{fin.}}=\begin{pmatrix} 0 & v & 0 & 0 & 0 & 0 \\ v & 0 & w & 0 & 0 & 0 \\ 0 & w & 0 & v & 0 & 0 \\ 0 & 0 & v & 0 & w & 0 \\ 0 & 0 & 0 & w & 0 & v \\ 0 & 0 & 0 & 0 & v & 0 \end{pmatrix}

H_{\mathrm{per.}}=\begin{pmatrix} 0 & v & 0 & 0 & 0 & w \\ v & 0 & w & 0 & 0 & 0 \\ 0 & w & 0 & v & 0 & 0 \\ 0 & 0 & v & 0 & w & 0 \\ 0 & 0 & 0 & w & 0 & v \\ w & 0 & 0 & 0 & v & 0 \end{pmatrix}

H(k)=\begin{pmatrix} 0 & v+w\mathrm{e}^{ik} \\ v+w\mathrm{e}^{-ik} & 0 \end{pmatrix}

E(k)=\pm\sqrt{v^2+w^2+2vw\cos{(k)}}

J. K. Asbóth, et. al., vol. 909 of Lecture Notes in Physics. Springer International Publishing, 1st ed., 2016.

Topological Insulators

$$ \frac{1}{2\pi}\int K \mathrm{d}A = 2(1-g) $$

Su-Schrieffer-Heeger

H(k)=\begin{pmatrix} 0 & v+w\mathrm{e}^{ik} \\ v+w\mathrm{e}^{-ik} & 0 \end{pmatrix}

E(k)=\pm\sqrt{v^2+w^2+2vw\cos{(k)}}

H(k)=d_x(k) \mathbf{\sigma}_x + d_y(k) \mathbf{\sigma}_y + d_z(k) \mathbf{\sigma}_z

\mathbf{d}(k)=\begin{pmatrix} d_x(k) \\ d_y(k) \\ d_z(k) \end{pmatrix} = \begin{pmatrix} v+w\cos{(k)} \\ w\sin{(k)} \\ 0 \end{pmatrix}

\nu = \frac{1}{2\pi}\int \left( \tilde{\mathbf{d}}(k) \times \frac{\mathrm{d}}{\mathrm{d}k} \tilde{\mathbf{d}}(k) \right)_z\mathrm{d}k

vinding number