I'm following a calculation done by a guy who's done it a bit different than what I've done before (used nearest neighbour vectors and a DFT instead of what I will show below), I'm not quite sure how to invert this expression he gives.

We are looking at formulating the tight binding picture for graphene, using the convention $ \mathbf{r} = l \mathbf{a_1} + j \mathbf{a_2}$ for the position of the unit cell, and the number $s = 1,2$ for the intra-cell atom. State 1 is located at position $\mathbf{r}$. On-site potential is $\epsilon_0$ and hopping potential is $t_0$ as usual.

The Hamiltonian of this picture is written in a basis of localised orbitals $\left| \mathbf{r} , s\right>$.

$$\begin{align} \hat{H} &= \sum_\mathbf{r} \biggl(\sum_{s=1,2} \left|\mathbf{r},s\right> \epsilon_0 \left<\mathbf{r},s\right|\biggr) \\ &\quad + \left|\mathbf{r},1\right> t_0 \bigl(\left<\mathbf{r},2\right| + \left<\mathbf{r} - \mathbf{a_1},2\right| + \left<\mathbf{r} - a_2 , 2\right|\bigr) \\ &\quad + \left|\mathbf{r},2\right> t_0 \bigl(\left<\mathbf{r},1\right| + \left<\mathbf{r} + \mathbf{a_1},1\right| + \left<\mathbf{r} + a_2 , 1\right|\bigr)\end{align}$$

To make this a bit clearer without drawing a diagram, the top line obviously refers to on-site potential and the second and third to the nearest neighbour hopping (i.e. electron at atom 1 hops to atom 2 either in same unit-cell or the one at $r-a_1$ or $r-a_2$).

This is where it starts to get hazy for me, the next step is that using Bloch's Theorem the eigenvectors for the Hamiltonian are given by:

$$ \left|\mathbf{k}, \alpha\right> = \frac{1}{\sqrt{N_C}} \sum_\mathbf{r} \sum^2_{s=1} e^{i \mathbf{k} \cdot \mathbf{r}} A_s \left|\mathbf{r},s\right>$$

where $A_s$ are coefficients, $N_C$ normalising factor. When I have done the TBM method for graphene in the past, I used a discrete Fourier Transform to get an expression for the state in momentum space, inverted it and put it back into the original Hamiltonian, giving an expression for the Hamiltonian in k-space. I would guess it is a similar technique in this case and my main question would be how to invert the above expression if this is the case? The sum of s is confusing me when I try to do this!


1) How exactly do we arrive at this using Bloch's theorem? I know in the TBM we look for eigenfunctions which are a linear combination of orbitals, and that the expression does look kind of similar to that. Is the above expression in reciprocal space?

2) The $\alpha$ is later used to refer to the positive or negative (conduction or valence) bands, is it included here for convenience and continuity to later or is there a way to infer that from the Hamiltonian or unit cell already? It is not doing anything in the expression at the moment, but is it used somehow when we invert it (maybe like a k-space equivalent of s=1 or 2)?

3) I guess the result of applying that equation to the Hamiltonian is that you get a 2x2 matrix of the Hamiltonian in k-space, diagonalise this and find the energy eigenvalues?

Would really appreciate some help on any of this, I have been looking online for stuff all day but everyone does it differently, leaves stuff out and uses different notation!

  • $\begingroup$ I think maybe this should get a homework tag --- just because the necessary knowledge for the OP to understand is so minimal that it's hard to justify a long answer. $\endgroup$ – genneth Apr 12 '12 at 18:26
  • $\begingroup$ The thing you need to keep in mind is that there are 2 atoms per lattice point --- which is where the sum over $s$ comes from. Applying Bloch's theorem is really just Fourier transform by another name. You've got points 2 and 3 covered already. $\endgroup$ – genneth Apr 12 '12 at 18:28
  • $\begingroup$ @genneth: the criterion for the homework tag is that the intended value of the question is not the solution itself, but the demonstration of the method used. I don't think that applies here. user8469, is this actually a homework problem or self-study exercise? (also, would you like me to merge your two accounts?) $\endgroup$ – David Z Apr 12 '12 at 23:41
  • $\begingroup$ Hey thanks for the reply, I know where the s comes from but my problem is trying to see how to invert the expression like he has done it. I have done the TBM calculation before and when I invert the transform there has never been this sum over s since I ignored the unit cell from the outset and just did NN for an atom and the c.c. My only issue with this is how to handle the sum over s when inverting the transform, and how to handle the alpha. $\endgroup$ – Josh Apr 13 '12 at 14:37
  • $\begingroup$ Hi David, this is a self study exercise but it could be treated as "homework" in that my question is more about the methodology employed when inverting the expression for the eigenvectors |k,a> if that makes sense? $\endgroup$ – Josh Apr 13 '12 at 14:39

1) The Bloch theorem comes from the fact that the group of translations is Abel, thus its representations are defined by number which is called $\mathbf{k}$. It means that when you translate (by let's say vector $\mathbf{a}$) the wavefunction with given $\mathbf{k}$ it is multiplied by exponent $e^{i\mathbf{ka}}$ (more or less by definition), which gives you exactly this form of the wavefunction.

2) $\alpha$ enumerates the solutions of the Hamiltonian. Different solutions have different set of tight-binding coefficients $A_s$.

3) Yes. You should put wavefunction in this form into real-space Hamiltonian and get 2x2 matrix parametrized by $\mathbf{k}$ which gives you energies and coefficients $A_s$ if you solve this eigenproblem.

I recommend you to read some book with a chapter on tight-binding (e.g. "Fundamentals of Semiconductors" by Yu, Cardona).


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