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documentation:language_reference:functions:blockbanddiagonalize [2024/09/25 16:33] – Sina Shokri | documentation:language_reference:functions:blockbanddiagonalize [2024/09/25 16:36] (current) – Sina Shokri | ||
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The function // | The function // | ||
$$ M = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} $$ | $$ M = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} $$ | ||
- | Now, we can linearly combine the second and third basis vectors, such that we get a single | + | Now, assuming that $M$ corresponds to a tight-binding Hamiltonian defined on some basis, we can linearly combine the second and third basis orbitals, such that we get a single |
$$ U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} $$ | $$ U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} $$ | ||
Now, transforming the matrix $ M $ using the unitary matrix $ U $ results in: | Now, transforming the matrix $ M $ using the unitary matrix $ U $ results in: | ||
$$ M' = U M U^{T} = \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & 0 \\ | $$ M' = U M U^{T} = \begin{pmatrix} 0 & \frac{1}{\sqrt{2}} & 0 \\ | ||
+ | In the new representation, | ||
+ | The basis orbital and not with the third one. | ||
The function // | The function // | ||
### | ### |