Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with crystals
Search options not deleted
user 160881
Crystals are solid material whose constituents, such as atoms, molecules or ions, are *arranged in a highly ordered microscopic pattern*, a crystal lattice that extends with regularity in all directions. Use for all crystallography and ordered structure topics.
1
vote
1
answer
515
views
Momentum Space Representation of the Tight Binding Hamiltonian
I am trying to represent the tight-binding Hamiltonian
\begin{equation}
\hat{H}_{TB} = \sum_{\sigma} \sum_{\alpha,\beta} \sum_{\mathbf{R}_1,\mathbf{R}_2}
t^{\alpha,\beta}_{\mathbf{R}_1,\mathbf{R}_2}
…
- 45
1
vote
Momentum Space Representation of the Tight Binding Hamiltonian
In (2) we can substitute
$t^{\alpha,\beta}_{\mathbf{R}_{1}-\mathbf{R}_{0},\mathbf{R}_{2}-\mathbf{R}_{0}}$. Then since the left hand side of (2) does not depend on $\mathbf{R}_{0}$, if we sum on it w …
- 45