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A question on the existence of Dirac points in graphene?

+ 5 like - 0 dislike
36 views

As we know, there are two distinct Dirac points for the free electrons in graphene. Which means that the energy spectrum of the 2$\times$2 Hermitian matrix $H(k_x,k_y)$ has two degenerate points $K$ and $K^{'}$ in BZ.

According to the von Neumann-Wigner theorem (no-crossing theorem): To make two eigenvalues of a Hermitian matrix (depending on some independent real parameters) cross, generally speaking, we need to change at least 3 parameters. But in the 2D graphene case, the variation of only 2 parameters $k_x,k_y$ can cause the energy levels cross.

So I want to know whether there are some physical or mathematical reasons for the existence of Dirac points in graphene.


This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

asked Apr 5, 2013 in Theoretical Physics by Kai Li (975 points) [ revision history ]
retagged Mar 25, 2014 by dimension10

2 Answers

+ 2 like - 0 dislike

Wikipedia says:

The eigenvalues of a Hermitian matrix depending on $N$ continuous real parameters cannot cross except at a manifold of $N-2$ dimensions.

Since the Hamiltonian has $N=2$ parameters ($k_x$, $k_y$), the crossing manifold has a dimension $N-2=0$, which is a point. So it's, in principle, allowed for graphene to have degenerate states (there are also a lot of other degenerate states if you look at the whole bandstructure). This is definitely only a necessary, not a sufficient condition (e.g. one could look at bilayer graphene which does not have this degeneracy).

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Rafael Reiter
answered Apr 5, 2013 by Rafael Reiter (20 points) [ no revision ]
Most voted comments show all comments
Thanks a lot for your answer. Yeah, the criterion you present just gives a possibility of degeneracy. But what I most concern here is that the underlying mechanism to cause degeneracy in 2D graphene.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
And the von Neumann-Wigner theorem also says that:For the real symmetric matrix case, the minimal number of real parameters we need to tune to make level crossing reduces to 2. So I want to know the 2D graphene case whether has something to do with the real symmetric matrix case in von Neumann-Wigner theorem ?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
1) I would say symmetry - I never came up with a better explanation. 2) No, because the matrix is not real. The matrix you need to solve to obtain the bandstructure is complex and Hermitian.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Rafael Reiter
Maybe, but can you specify what kinds of symmetries make degeneracy possible?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
How about this explanation:As we know, there are at most $n^2$ independent real parameters for a $n\times n$ Hermitian matrix , and if we want use von Neumann-Wigner theorem, the $n\times n$ Hermitian matrix which we deal with should has $n^2$ independent real parameters, then the theorem works.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
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So in 2D graphene case, can we say that broken inversion symmetry commonly results gapped spectrum ?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
I'm not going to/can't answer that here - you should open another question so that other people read this too.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Rafael Reiter
+ 1 like - 0 dislike

Your use of the no-crossing idea is correct - we do not expect level crossings in two dimension to appear unless protected by symmetry. The symmetries in this case are the symmetries of the honeycomb lattice and time reversal. The protection of level crossings by symmetry is ubiquitous in solid-state.

I should add that the existence of these Dirac point is actually slightly more robust than would be implied by simple symmetry arguments. The band structure will still have Dirac cones if one applies any perturbation that does not violate parity, time reversal and is not extremely strong[1]. This is because of the interplay of the Berry's curvature and the Dirac point, which I could find a reference for if you would like.


[1] Extremely strong means that if I imagined increasing the strength of this perturbation up from zero it would drag the Dirac cones from the $K$, $K'$ points into each other. This would mean a perturbation of energy about the bandwidth, which is several electron-volts.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user BebopButUnsteady
answered Apr 23, 2013 by BebopButUnsteady (325 points) [ no revision ]
@ BebopButUnsteady Thanks for your brilliant answer. Yeah, please show me some relevant references, I'd like to have a look, thank you.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy
@ BebopButUnsteady And I can't imagine that this phenomena yet has something to do with Berry's curvature?

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user K-boy

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