Dispersion relation in tight binding model

  • #1
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Hamiltonian of tight binding model in second quantization is given as [tex]H = -t \sum_{<i,j>} a_i^{\dagger} a_j[/tex]
After changing basis it is [tex]H = \sum_{\vec{k}} E_{\vec{k}} a_{\vec{k}}^{\dagger} a_{\vec{k}}[/tex]
where [tex]E_{\vec{k}} = -t \sum_{\vec{b}} e^{i \vec{k} \cdot \vec{b}}[/tex]
where [itex]\vec{b}[/itex] is a nearest neighbour shift vector.

I've used this relation to calculate [itex]E_{\vec{k}} [/itex] for a square lattice, where nearest neighbours are points [itex](a,0)[/itex], [itex](-a,0)[/itex], [itex](0,a)[/itex], [itex](0,-a)[/itex]. The result is [tex] E_{\vec{k}} = -t[e^{i(k_x a + 0)} + e^{i(- k_x a + 0)} + e^{i(0 + k_y a)} + e^{i(0 - k_y a)}] = -2t (cos k_x a + cos k_y a) [/tex]

For a traingular lattice I set point (0,0) in the middle so neighbours are [itex](a/2,a \sqrt{3}/2)[/itex], [itex](-a/2,-a \sqrt{3}/2)[/itex], [itex](a/2,-a \sqrt{3}/2)[/itex], [itex](-a/2,a \sqrt{3}/2)[/itex], [itex](a,0)[/itex], [itex](-a,0)[/itex]. The result is [tex] -2t (cos k_x a + cos (k_x \frac{a}{2} +k_y \frac{\sqrt{3}}{2}a) + cos (k_x \frac{a}{2} -k_y \frac{\sqrt{3}}{2}a))[/tex]

Are these calculations correct?
 

Answers and Replies

  • #2
18,256
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Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
 

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