Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Dispersion relation in tight binding model

  1. Nov 28, 2016 #1
    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?
     
  2. jcsd
  3. Dec 3, 2016 #2
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dispersion relation in tight binding model
  1. Tight-binding model (Replies: 4)

  2. Tight binding model (Replies: 4)

Loading...