Tight-binding equations for a semi-infinite graphene sheet

February 16th, 2012 — 11:00pm

It’s quite straightforward how to solve TB-equations for an infinite graphene sheet (see Castro Neto et al., Rev. Mod. Phys. 2009):

E_{\mu k}\alpha (k,n) = -t[(1+e^{ika})\beta (k,n)+\beta(k,n-1)] ,
E_{\mu k}\beta (k,n) = -t[(1+e^{-ika})\alpha (k,n)+\alpha(k,n+1)] .

One should search a solution (for both \alpha and \beta ) which is proportional to \exp [ i\mu a n ] . It will automatically give a well known expression for a graphene dispersion relation E_{\mu k} . Now to cut a graphene sheet and make it semi-infinite we consider a boundary condition

E_{\mu k}\alpha (k,0) = -t[(1+e^{ika})\beta (k,0)

How to find a solution in this case ? It seems that the usual trick with a standing wave instead of a running wave doesn’t work…

Comment » | S-Journal

Introduction

March 13th, 2011 — 6:46pm

The concept of this journal can be perfectly expressed by just one phrase, which was very popular in BSU when I was studying there. They said: “I have explained this topic so well that I even have understood it myself !”. It’s a joke, but the truth is that the journal is created mainly for educational and self-educational purposes, to spotlight certain relevant topics which people often face on their scientific path. The other point is to learn the art of R. Feynman to explain complicated things in an easy way. The art which is almost forgotten by theorists in the 21st century :)

Comment » | S-Journal

LaTeX test

March 5th, 2011 — 9:53pm

Despite its simplicity, the model is very important from theoretical point of view. It’s analytically solvable in both scattering and ME approaches and provides a basis for discussion and comparison. The atom of impurity plays a role of the molecular bridge, which contains one electronic level with energy \varepsilon_0, available for tunneling, and no internal degrees of freedom. Changing the notation of the basis vector |n=0\rangle to |d\rangle, bridge’s Hamiltonian reads

H_s = |d\rangle \varepsilon_0 \langle d| .

Leads, as always in this work, are presented by two half-infinite tight binding chains (HITBC). Their Hamiltonian in k-representation reads

H_l = \sum\limits_{k\alpha} |k\alpha\rangle \varepsilon_{k\alpha} \langle k\alpha| ,

where \alpha=l,r is used to distinguish left and right lead. Chemical potential \mu_{\alpha} is in general different for different leads. When we speak about a voltage U, applied to the junction, we assume \mu_l=+\frac{V}{2} and \mu_r=-\frac{V}{2}.

Comment » | S-Journal

     

↑ Back to top