Generally band E-k band diagram of a material look like parabolic type. However this is not the case in the TIs. Why bands in TIs are like this (Fig. shown)? Is it due to spin-orbit splitting however generally this type of splitting is seen on the energy axis and not on the k-axis.
Parabolicity is to be expected at k points with full point-group symmetry of a cube (in two dimension, that of a square) [1]. As for the topological insulators, in the mean-field approximation they are in almost all respects similar to ordinary insulators, the main difference being that in topological insulators one has protected edge states, which are insensitive to 'small' perturbations - sufficiently 'large' perturbations give rise to phase transitions, whereby e.g. the bulk is no longer insulating. ■
[1] To appreciate this issue, consider a k point at a regular extremum of a band. Let k0 be such a point. Expand the energy band εl(k) at k = k0. The Linear term is vanishing. The quadratic term is of the form ½ (k-k0)T • F(k0) • (k-k0), where F is a d × d symmetric matrix (equal to the inverse of the effective-mass tensor in atomic units), with d denoting the dimension of the space (d = 1, 2, 3, ...); the symmetric nature of F implies that it has d(d+1)/2 independent elements.
In order for the band to be locally parabolic around k = k0, it is required that F be diagonal with equal diagonal elements - that is, it must be equal to a constant times the d × d unit matrix. While by symmetry F must commute with all d × d matrix representations of the point-group symmetries of the underlying crystal that leave k0 invariant,*) for an insufficiently symmetric k0 the number of restrictions thus imposed on the elements of F are not sufficient for forcing, by symmetry, the off-diagonal elements of F to be vanishing (this amounts to d(d-1)/2 conditions) and its d diagonal elements to be equal (this amounts to d-1 conditions - there are therefore in total d(d+1)/2-1 conditions to be satisfied). In this connection, I note that by Schur's lemma, when F commutes will all elements of a d-dimensional irreducible representation of a group, F is either vanishing or equal to a non-vanishing constant times the d × d unit matrix. For a general k0, this point can in the worst case be invariant only under the identity operation of the point-group operations of the underlying crystal, in which case the d-dimensional representation of this group (consisting of one element, namely the identity operation) is reducible; for a more symmetric k0, the subgroup of point-group operations that leave k0 invariant may not possess a d-dimensional irreducible representation. ■
*) Note that F(k0) is a tensor, so that it transforms as a tensor for transformations of k0.
Parabolicity is to be expected at k points with full point-group symmetry of a cube (in two dimension, that of a square) [1]. As for the topological insulators, in the mean-field approximation they are in almost all respects similar to ordinary insulators, the main difference being that in topological insulators one has protected edge states, which are insensitive to 'small' perturbations - sufficiently 'large' perturbations give rise to phase transitions, whereby e.g. the bulk is no longer insulating. ■
[1] To appreciate this issue, consider a k point at a regular extremum of a band. Let k0 be such a point. Expand the energy band εl(k) at k = k0. The Linear term is vanishing. The quadratic term is of the form ½ (k-k0)T • F(k0) • (k-k0), where F is a d × d symmetric matrix (equal to the inverse of the effective-mass tensor in atomic units), with d denoting the dimension of the space (d = 1, 2, 3, ...); the symmetric nature of F implies that it has d(d+1)/2 independent elements.
In order for the band to be locally parabolic around k = k0, it is required that F be diagonal with equal diagonal elements - that is, it must be equal to a constant times the d × d unit matrix. While by symmetry F must commute with all d × d matrix representations of the point-group symmetries of the underlying crystal that leave k0 invariant,*) for an insufficiently symmetric k0 the number of restrictions thus imposed on the elements of F are not sufficient for forcing, by symmetry, the off-diagonal elements of F to be vanishing (this amounts to d(d-1)/2 conditions) and its d diagonal elements to be equal (this amounts to d-1 conditions - there are therefore in total d(d+1)/2-1 conditions to be satisfied). In this connection, I note that by Schur's lemma, when F commutes will all elements of a d-dimensional irreducible representation of a group, F is either vanishing or equal to a non-vanishing constant times the d × d unit matrix. For a general k0, this point can in the worst case be invariant only under the identity operation of the point-group operations of the underlying crystal, in which case the d-dimensional representation of this group (consisting of one element, namely the identity operation) is reducible; for a more symmetric k0, the subgroup of point-group operations that leave k0 invariant may not possess a d-dimensional irreducible representation. ■
*) Note that F(k0) is a tensor, so that it transforms as a tensor for transformations of k0.
I don't have enough knowledge about that concept but you'd better to check the notion "Dirac cone". Maybe, Wiki would be helpful at first sight:
[https://en.wikipedia.org/wiki/Dirac_cone]
or you may wish to consult this publication given below:
npj Quantum Materials volume 3, Article number: 29 (2018)[https://www.nature.com/articles/s41535-018-0101-8]
and/or you can watch this video:
Topological Insulators, by José Luis Crespo Cepeda, tutor Juan José Palacios [https://www.youtube.com/watch?v=ny0KvKknQig]
Bests.
Bahadır Ozan Aktaş : The question by Vk Singh concerns 'parabolic type' bands. Dirac cones consist of linear energy dispersions and are also not necessarily isotropic. You can find an example of an 'elliptic' Dirac cone on p. 221 in [1] ‒ close to the apex of the cone, the cross-section of the energy dispersion with a constant-energy surface is an ellipse. In graphene one has circular cones (close to the apex of a Dirac cone, the cross-section of the energy dispersion with a constant-energy surface is a circle) because of the symmetry of the apices of the cones - the points K and K' are on the corners of a hexagon (see pp. 175 and 176 in [1]).
Incidentally, with reference to what I have written earlier on this page, for linear energy dispersions associated with Dirac cones in 2D, isotropy of the energy dispersion is enforced by a lower symmetry than D4 because the dispersion is determined by a rank-1 tensor ‒ in contrast, for a locally-quadratic energy dispersion, one has to deal with a rank-2 tensor.
[1] H Aoki, and MS Dresselhaus, Physics of Graphene (Springer, Heidelberg, 2014).
Behnam Farid : Thank you very much for your detailed explanations. Looks like the notion Dirac cone has nothing to do with the question.
Vk Singh : My pleasure.
Best regards to all.
Bahadır Ozan Aktaş : You are welcome. Yes, Dirac cone is at best only tangential to the main question - the main question is about massive excitations, not massless ones.
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