In this paper by G. D. Mahan, he obtains the following electron Hamiltonian in a nearest-neighbour tight binding scheme: (page 2 of the paper, top of the right column)

\begin{align} H_0 &= J_0 \sum_{k,\alpha}{\left(C_{A,k\alpha}^{\dagger}C_{B,k\alpha}^{\vphantom{\dagger}}\gamma^{\vphantom{\dagger}}_{\vphantom{A}\alpha}(k) + C_{B,k\alpha}^{\dagger}C^{\vphantom{\dagger}}_{A,k\alpha}\gamma^{*\vphantom{\dagger}}_{\vphantom{A}\alpha}(k)\right)} \end{align}

where the $A$ and $B$ indices refer to the two atoms in the graphene unit cell, the $C$'s and $C^\dagger$'s are electronic destruction and creation operators and

\begin{align} \gamma_\alpha(k) &= |\gamma_\alpha(k)| \mathrm{e}^{i\xi(k,\alpha)} \end{align}

is some complex scalar, the precise definition of which is irrelevant to this question (but is in the paper).

Now, Mahan claims the eigenvalues of the above Hamiltonian are

\begin{align} E^{(\pm)}(k,\alpha) &= \pm J_0 |\gamma_\alpha(k)|. \end{align}

Unfortunately, I don't immediately see how this follows. A bit further in the paper Mahan does diagonalize the above Hamiltonian but then seems to use the eigenvalues as already given to rewrite this diagonalization instead of deriving them from it, as I would do.

So my question is: how did Mahan obtain the eigenvalues for the above Hamiltonian without diagonalizing it? I am looking for a mathematical derivation but perhaps there is a simple argument I'm missing that makes this evident? Or did he in fact diagonalize $H_0$?


1 Answer 1


For each value of $k$, the above hamiltonian has energy contributions for an electron hopping from an $A$ to a $B$ site and vice versa. If we wanted to write the hamiltonian in matrix form (in a basis corresponding to $A$ and $B$ sites), it would look like

\begin{equation} H_0 = \sum_{k,\alpha}\left(\begin{array}{c c}0& J_0\gamma_{k,\alpha}\\ J_0\gamma_{k,\alpha}^* & 0\end{array}\right) \end{equation}

and we can diagonalize each term to get $\epsilon_{k,\alpha} = \pm J_0 |\gamma_{k,\alpha}|$.

  • $\begingroup$ Alright, so actual diagonalization it is. Thanks, I was wondering if maybe it was obvious in some other way, but I guess not. Though I should probably have been able to see the result of this diagonalisation without having to actually perform it. Oh well, takes a while for the fingerspitzengefühl to return I guess. :) Thanks again! $\endgroup$
    – Wouter
    Feb 16, 2015 at 17:51

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