The Tight-binding model for electronic band structure

Like the free-electron-gas model, the tight-binding model belongs in the independent-electrons framework.

This model is basically the only time that we see (at the level of the Struttura della Materia course) the actual calculation of an electronic band structure, that takes explicitly into account the presence of the periodic lattice potential.

Contrary to the free-electron picture, the tight-binding model describes the electronic states starting from the limit of isolated-atom orbitals.

This simple model gives good quantitative results for bands derived from strongly localized atomic orbitals, which decay to essentially zero on a radius much smaller than the next neighbor half-distance in the solid.

For the more interesting bands, the conduction bands, the results of tight-binding are usually in rather poor agreement with experiment. As we shall see, tight binding could be systematically improved by including additional levels/bands, so that the accuracy of the calculated bands increases, at the expense of the simplicity and transparency of the model.

Here we shall apply the general formalism to the simplest case of an isolated s band.

The starting point of this model is the decomposition of the total single-electron Hamiltonian into:

H = Hat + ΔU(r) ,

where Hat contains the kinetic energy plus the potential of a single ion (the one placed at R=0, say), and ΔU(r) is the potential generated by all the others ions in the lattice except for the one already considered. The eigenfunctions of the atomic problem satisfy the Schrödinger equation

H φn(r) = En φn(r) ,

where n represents collectively a full set of (orbital) atomic quantum numbers. We then expand a generic state localized around the atom as

φ(r) = sumn bn φn(r)

[notice that this state is not quite generic, as the continuum states of the atom are left out of the sum]. We than make a combination of such localized states, with the symmetry of the lattice:

ψ(r) = sumR exp(i k·R) φ(r-R)

[Exercise: verify that this state has the Bloch property ψ(r+R) = exp(i k·R) ψ(r) ].

For fixed k, we plug this state into the Schrödinger equation:

H ψ(r) = [Hat + ΔU(r)] ψ(r) = E(k) ψ(r)

This differential equation maps to a matrix equation by multiplication on the left by φm*(r) and integration over the r variable:

sumn A(k)mn bn = E(k) sumn B(k)mn bn

This, for each fixed k, is a generalized eigenvalue problem for the matrix A(k), with a "metric" matrix B in place of the identity. E(k) represents the eigenvalue, corresponding to the eigenvector b (of course, also b is k-dependent, in general). The "energy" matrix above is:

A(k)mn = sumR exp(i k·R) int φm*(r) H φn(r-R) dr

The "overlaps" matrix above is:

B(k)mn = sumR exp(i k·R) int φm*(r) φn(r-R) dr

where the sums over R extend over all the lattice points. This is conventiently rearranged (using the decomposition H = Hat + ΔU(r), the fact that φm*(r) are (left) eigenstates of Hat and the orthonormality of φn(r) and φm(r) ) into:

sumn C(k)mn bm = (E(k) - Em) sumn B(k)mn bn

where the new matrix C(k) contains what is left of the Hamiltonian, i.e.

C(k)mn= sumR exp(i k·R) int φm*(r) ΔU(r) φn(r-R) dr

Now the eigenvalue (E(k) - Em) of the generalized secular problem measueres the displacement of the "band" energy E(k) with respect to the original atomic value Em.

The size of this matrix eigenvalue problem is clearly as large as the number of eigenstates of the atomic problem, i.e. infinite. It is therefore necessary to do some approximation here. In particular, one could hope that all the off-diagonal matrix elements of the matrices at the right side of this eq. could be neglected for some given level m. This cannot work for atomic degenerate levels (such as p, d, f... orbitals) where the couplings between the degenerate levels form the main part of the Hamiltonian, the one that resolves the degeneracy: for the case of d-degenerate levels, one has to solve at least a d-dimensional matrix problem (at each value of k).

The only case where it sometimes makes sense forgetting the interaction with all levels with nneqm is with atomic s (l=0) nondegenerate levels. In this approximation, the matrix equation becomes a 1×1 trivial problem, for which one takes bm=1 and all the other bnneqm=0, and only the "m" energy equation:

C(k)mm = (E(k) - Em) B(k)mm

remains, with solution:

E(k) = Em + C(k)mm/B(k)mm

Now, in the infinite R-sums of the definitions of the B(k) and C(k) matrices, it is convenient to separate the contribution from R=0, the contribution from R in the first shell around the origin (first or nearest neighbors), the contribution of the second shell around the origin (second neighbors), etc. The R=0 contribution to B(k) is 1, and that to C(k) is

χ = int ΔU(r) |φm(r)|² dr

which is a negative quantity, reflecting the attraction that the "other" nuclei produce on the band electron, which was not there when the atom was isolated. We shall indicate the Rneq0 contributions to B(k) and C(k) as

β(R)= int φm*(r) φm(r-R) dr
γ(R)= int φm*(r) ΔU(r) φm(r-R) dr

respectively. Note that Accordingly, one can replace the complex exponentials with cosines, and rearrange the solution for E(k) to:

E(k) = Em + [χ + sumRneq0 cos(k·R) γ(R)] / [1 + sumRneq0 cos(k·R) β(R)]

This E(k) gives the tight-binding band structure in terms of a set of parameters β(R), χ and γ(R). We also have an explicit recipe to compute these parameters in terms of overlap integrals at different sites.

Due to the exponential decay of the atomic wave functions at large distance, both the overlap integrals β(R) and the energy integrals γ(R) become exponentially small for large distance R between the centers of the atoms. It therefore makes sense to ignore all the integrals outside some Rmax, which would bring in only negligible corrections to the bandstructure E(k). One may obtain a band structure depending on a minimal number of parameters by making further rather radical approximations:

In this approximation the band energy simplifies further to:

E(k) = Em + χ + γ sumR(NN) cos(k·R)

where γ indicates the value of γ(R) for the nearest neighbors. In this approximation, the band is determined by 2 parameters only: Em + χ, which tunes the band mean energy, and γ, which sets the band width.

[Further details for this model can be studied, for example, in the Ashcroft-Mermin textbook "Solid State Physics" (Chap. 10).]

Exercises

  1. Write out explicitly the s-band tight-binding E(k) for a 1-dimensional lattice of spacing a, in the approximations outlined above. Draw it out in the first Brillouin zone (-π/a, π/a), assuming that the original atomic level position is at -6.3eV, that the on-site potential integral χ=-0.7eV and the NN overlap integral is γ=-1.2eV. How would that change upon inclusion of second-near neighbor overlap γ'=0.4eV? How would the band change upon further inclusion of nearest neighbor orthogonalization correction β=0.15? Draw the three bands.
    RESULT:
    nearest neighbor only: E(k) = Em + χ + 2 γ cos(k a) = -7eV -2.4eV cos(k a)
    all corrections: E(k) = Em + [χ + 2 γ cos(k a) + 2 γ' cos(2 k a)]/ [1 + β cos(k a)]
  2. In 3 dimensions E(k) is a function of the 3 variables (kx,ky,kz). Write out explicitly the s-band tight-binding E(k) for a 3-dimensional simple-cubic (SC) lattice of spacing a, in the simplest NN approximation (no β correction) outlined above. Draw a slice of the band in the first Brillouin zone for kx in (-π/a, π/a), ky=0, kz=0. On the same (kx,E(k)) plot, draw another slice of the band for ky changing to π/a (all the rest left unchanged). What is the total width of this band?
    RESULT: E(k) = Em - χ - 2 γ [cos(kx a) + cos(ky a) + cos(kz a)]. Total bandwidth = 12 γ
  3. Write out explicitly the s-band tight-binding E(k) for a 3-dimensional fcc (face centered cubic) lattice of spacing a, taking into account the overlap integrals to the 12 nearest neighbor sites. Expand E(k) around k=0 and show that (as happens in all cubic cases) the dispersion is isotropic for small k, and similar to that of free fermions.
    NOTE: The first Brillouin zone of the fcc lattice has a nontrivial shape. This does not affect the resolution of this specific exercise, but it is interesting to see the region of k-space containing all independent k points where the computed E(k) is defined.
  4. Write out explicitly the s-band tight-binding E(k) for a 2-dimensional square lattice of spacing a, taking into account the overlap integrals to the 4 nearest neighbor sites. Fill the states according to the 0-temperature Fermi distribution, so that 1 (non-interacting spin-½) electron per site is present, like in, e.g., in an hypothetic square-lattice Na. What is the resulting shape of the Fermi "surface" (i.e. Fermi line) in the (kx,ky) plane? Compare to the Fermi "surface" of the free electrons dispersion.
    HINT: use the symmetry of the band around its energy center, and the fact that an s-band can accommodate 2 (spin-½) electrons at each k-point (thus a total 2×(number of sites) electrons in the completely filled band).
  5. The benzene molecule C6H6 can be thought of as a ring of 6 sites, on each of which a C atom is placed. Each C atom shares one electron in delocalized pz (out-of plane) orbitals, which can be roughly described with the tight-binding s-band model (negative γ, including only nearest neighbor C), in the non-interacting-electron approximation. The excitation energy to the lowest excited state is observed 2.0 eV.
    1. Draw a single-particle energy diagram for these delocalized electronic states.
      RESULT: allowed independent k = n 2 π/(6 a), with n=0, ±1, ±2, +3.
    2. Fill the 6 lowest spin-orbital states with the electrons provided by the C atoms: compute the ground-state total energy.
      RESULT: EGS = 8 γ
    3. Draw the lowest excited 6-electron state, by promoting one electron to the lowest empty state. Compute the energy of this 1st excited state and its excitation energy referred to the ground state. By comparison to the experimental data, determine γ.
      RESULT: E1st ex = 6 γ, ExcE1st ex = -2 γ, whence γ = -1.0 eV
    4. Determine the excitation energies referred to the ground state of the 2nd and 3rd excited states.
      RESULT: ExcE2st ex = 3.0 eV, ExcE3st ex = 4.0 eV

Notes

  1. The sign of γ(R) is determined by the relative signs of the tails of the wavefunctions φm(r) and φm(r-R) in the overlapping region. For s-orbitals, γ(R) has the same sign as ΔU(r), i.e. negative. As a consequence, s-band tigtht binding places the bottom of the band at k=0 (similarly to a free-electron-like dispersion, see ex. 3).

    However, for p bands one can see qualitatively that the overlapping tails of the wavefunctions φm(r) and φm(r-R) can have opposite sign (the p wavefunction changes sign across some plane passing in the nucleus), thus γ(R)>0 in that case. As a consequence, a p band shows typically a band maximum at k=0, with negative curvature of E(k). This does not apply to the π-band of benzene in the exercise above, as the nodal plane where φm(r) changes sign is parallel to the line joining neighboring atoms.

  2. The tight-binding method is an approximate method for computing bandstructures. The approximation involved is a truncation of the basis.

    Another standard elementary technique is the perturbative method: the starting point of the free-electron parabolic dispersion is perturbed by a periodic potential, assumed to be "weak". A qualitative understanding of how the bands change and how gaps open can be gathered by a perturbative analysis. Exact bands are then obtained if the plane-waves basis is used in a full diagonalization approach as in current electronic-structure codes.

    Another popular toy bandstructure scheme is the Kronig-Penney model.

    The best thing about these methods is that they all give the same qualitative result: a band spectrum. This is not a random coincidence: it reflects the discrete translational invariance of the periodic potential, the symmetry enforcing Bloch's theorem.

Comments and debugging are welcome!


created: 11 Jan 2002 last modified: 16 Jan 2012 by Nicola Manini