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Two dirac cones are described by two bi-spinors of different chirality

  1. Jun 20, 2009 #1
    Is is right to say that the two dirac cones are described by two bi-spinors of different chirality?...

    Is it right to say that each of the dirac cones contains quasi-particles of different helicity (electrons of positive elicity and of holes negative elicity for one dirac cone and the opposit for the other) if we consider holes as "quasiparticles of negative energy" (so not changing sign to the momentum)?...

    I find books quite confusing on this point...
  2. jcsd
  3. Jun 20, 2009 #2
    Re: graphene

    Usually one writes low-energy Hamiltonians for graphene as [tex]H_K=\sigma \cdot k [/tex] and [tex]H_{K'}=\sigma_x k_x - \sigma_y k_y [/tex]. These are the Hamiltonians for the two valleys in the sublattice basis. The first component of the two-component spinors refers to, say, sublattice A and the second to sublattice B. One can, however, use just as well the valley-isotropic representation in which both Hamiltonians are identical. Anyway, these two unequivalent points in the Brillouin zone give rise to the valley degree of freedom (referred to as isospin in some papers).

    By diagonalizing the low-energy Hamiltonian in either valley one gets eigenenergies [tex]\epsilon_{sk} = sv_F |k| [/tex], where [tex]s=\pm 1[/tex]. The eigenbasis could be called "chiral" basis. In this basis, the first component of the spinors refers to "electrons" and the lower component to "holes". It is just a simple unitary transformation from one basis to another to make the Hamiltonian diagonal.

    I'm not quite sure why these two-component creatures are actually spinors, perhaps someone more familiar with QFT can shed some light on this.
  4. Jun 20, 2009 #3
    Re: graphene

    Thank you for the answer :smile:

    very interesting this "valley isotropic" representation...

    what you say gives me a little suggestion that what I previously intended for "helicity" is relabled by these authors with the term "chirality", which term I used to call something else (what I used to call chirality here becomes simply the valley index if I understood well, I mean the one distinguishing the two dirac points)....

    can you tell me something that could help me clarify better the situation?... in this preprint


    you see picture two part 1 at page 4... I wanted to ask you is this picture is referred to states around the K point... and if it is correct that, considering also electrons in K' point I should reverse the direction of all the pseudospin vectors ...

    if it is this way, I wonder: why doesn't the book specify that? Is it because processes between the two valleys do not happen and so I can fix one valley and work only with that? Furthermore, pseudospin is always conserved if two valleys interact?
  5. Jun 21, 2009 #4
    Re: graphene

    The "valley isotropic representation" is quite often used by Beenakker, take a look at e.g. his colloquium on Klein tunneling: http://arxiv.org/abs/0710.3848 , Eq. (5).

    Note that the term "pseudospin" is used (in the paper you referred to) to describe the degree of freedom originating from the two interprenating sublattices, not from the two valleys! So it is just as you said: when intervalley scattering can be neglected, one can work with one valley. Intervalley scattering requires short-range disorder, for example, and is quite often neglected in theoretical papers.

    If one has to take the actual spin as well as the valley index into account, there are eight components in the spinor.

    So let us now fix the quasi-particles to be at one valley. The conservation of pseudospin means, in my opinion, the conservation of helicity, or chirality. (These two terms are exactly the same thing for massless fermions, and are used interchangeably. Chirality is more common.) I mean that if an electron is turned into a hole in a np-junction, for example, one has to also revert the sign of momentum to keep the helicity the same.

    Remember that helicity is defined [tex] \Sigma = \sigma \cdot p/|p| [/tex], so the eigenstates of the Hamiltonian are eigenstates of helicity. The eigenstates are [tex] \psi_{sk} = [1,s \cdot \exp(i \phi_k)] [/tex], multiplied by a plane-wave in the position representation.

    You should take a look at the original papers by Ando, 1998 and Ando, 2002 if you really want to understand the absence of backscattering in clean graphene. But this is not necessary if you just want to understand Klein tunneling at an introductory level.
    Last edited: Jun 21, 2009
  6. Jun 21, 2009 #5
    Re: graphene

    I'll certanily have a look at that paper... not now because in a few hours I'm gonna catch my train... just a first question... the world I underlined in red is it a typo? I would have written "pseudospin" instead of helicity... in fact (as far as I've understood), if we fix the K valley, there electrons have helicity positive whereas holes have negative helicity, so that changing the sign of the momentum of a hole (actually pseudo-momentum, since we talk about brillouin zones) doesn't affect its helicity but only the direction of the pseudospin... is it right all that?
  7. Jun 21, 2009 #6
    Re: graphene

    I'm sorry for my mistake, I'm glad you noticed it! You're correct, helicity is not conserved but pseudospin is. To make this explicit, consider an electron hitting a symmetric np-junction perpendicularly. The state left of the barrier is
    [tex] \psi_1 (x) = [1,+1] \exp(ikx) [/tex]
    and on the right
    [tex] \psi_2 (x) = [1,-\exp(i\pi)] \exp(-ikx) = [1,1] \exp(-ikx). [/tex]
    So now helicities are
    [tex] \Sigma \psi_1 = + \psi_1, \ \Sigma \psi_2 = - \psi_2 [/tex],
    whereas the pseudospins are
    [tex] \sigma_x \psi_1 = + \psi_1, \ \sigma_x \psi_2 = + \psi_2 [/tex].
    So flipping the momentum requires flipping the helicity to conserve pseudospin. Thanks, I think we are on the right track now!

    Any reflected wave would of the form
    [tex] \psi_r (x) = [1,\exp(i\pi) ] \exp(-ikx) [/tex],
    giving the helicity and pseudospin
    [tex] \Sigma \psi_r = + \psi_r, \ \sigma_x \psi_r = - \psi_r [/tex].
    For normal incidence the reflected part vanishes (Klein tunneling), but for other angles the transmission probability is [tex]T=\cos^2 \phi [/tex].
    Last edited: Jun 21, 2009
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