Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 3

Question 1

We have the band structure

E(k_{x}, k_{y}) = −8γ + 4γ cos \frac{k_{x}a}{2} (cos \frac{k_{x}a}{2} + cos \frac{\sqrt{3}k_{y}a}{2})

for the 2D triangular lattice in the single–band nn TB model. We expand E around the Γ–point to second order in k, using cos α ≃ 1 − α^{2}/2 for α ≪ 1:

E ≃ −8γ + 4γ (1 − \frac{k_{x}^{2}a^{2}}{8} ) (1 − \frac{k_{x}^{2}a^{2}}{8} + 1 − \frac{3k_{y}^{2}a^{2}}{8} )
≃ −8γ + 4γ (2 − \frac{3k_{x}^{2}a^{2}}{8} − \frac{3k_{y}^{2}a^{2}}{8} )
= −\frac{3}{2}γa^{2} (k_{x}^{2} + k_{y}^{2})
= −\frac{3}{2}γa^{2}k^{2}

Hence, the inverse effective mass tensor is diagonal and isotropic, i.e., with both diagonal elements equal to

1/m^{∗} = \frac{3|γ|a^{2}}{h̄^{2}}

since we had assumed γ < 0. Anisotropy would require different coefficients in front of k_{x}^{2} and k_{y}^{2}. A non–diagonal inverse effective mass tensor would require a term in E proportional with k_{x}k_{y}.

Question 2

a) We take the hints given in the text and expand E to lowest order in deviations from k_{K}:

cos \frac{3k_{x}a}{2} = cos (π + \frac{3}{2}ε_{x}) = − cos \frac{3}{2}ε_{x}
≃ −1 + \frac{9}{8}ε_{x}^{2}
cos \frac{\sqrt{3}k_{y}a}{2} = cos (\frac{π}{3} + \frac{\sqrt{3}}{2}ε_{y})

= cos \frac{π}{3} cos \frac{\sqrt{3}}{2}ε_{y} − sin \frac{π}{3} sin \frac{\sqrt{3}}{2}ε_{y}
≃ \frac{1}{2} · (1 − \frac{3}{8}ε_{y}^{2}) − \frac{\sqrt{3}}{2} · \frac{\sqrt{3}}{2}ε_{y}
= \frac{1}{2} − \frac{3}{4}ε_{y} − \frac{3}{16}ε_{y}^{2}

The expression under the square root becomes, to second order in ε_{x} and ε_{y},

1 + 4 cos \frac{3k_{x}a}{2} cos \frac{\sqrt{3}k_{y}a}{2} + 4 cos^{2} \frac{\sqrt{3}k_{y}a}{2}

≃ 1 + 4 (−1 + \frac{9}{8}ε_{x}^{2}) (\frac{1}{2} − \frac{3}{4}ε_{y} − \frac{3}{16}ε_{y}^{2}) + 4 (\frac{1}{2} − \frac{3}{4}ε_{y} − \frac{3}{16}ε_{y}^{2})^{2}
≃ \frac{9}{4} (ε_{x}^{2} + ε_{y}^{2})
= \frac{9}{4}a^{2}(Δk)^{2}

Here, Δk = k − k_{K} = (ε_{x}x̂ + ε_{y}ŷ)/a. Hence,

E^{±} ≃ ±\frac{3}{2}|γ|aΔk,

and the constant is C = 3|γ|a/2.

b) As in Exercise 1, Question 3, we have a constant density of states in 2D,

D_{2}(k) = \frac{A}{2π^{2}} ,

which means that the number of states with |k − k_{K}| less than a given Δk is

N_{2}(Δk) = π(Δk)^{2} · D_{2} = \frac{A(Δk)^{2}}{2π} .

With the linear dispersion found in a), we have

N_{2}(E) = \frac{A}{2π} (\frac{E}{C} )^{2} ,

and therefore

D_{2}(E) = \frac{dN_{2}}{dE} = \frac{A}{π} \frac{E}{C^{2}} = \frac{4AE}{9πγ^{2}a^{2}} .

c) Photons are massless particles, so the energy–momentum relation for photons is simply E = pc. For electrons in a crystal, we know that the wavenumber, measured relative to the energy band minimum, and multiplied by h̄, is the crystal momentum. More precisely,

p = h̄Δk,

so we may write for the graphene electrons at the Fermi level, near the conduction band minimum,

E = \frac{3}{2}|γ|aΔk = v_{F} p,

with Fermi velocity

v_{F} = \frac{3|γ|a}{2h̄}.

The nearest neighbour distance in graphene is a ≃ 1.4 Å, so if γ ∼ 3 eV, the velocity of the electrons is

v_{F} ∼ \frac{3 · 3 · 1.6 · 10^{−19} · 1.4 · 10^{−10}}{2 · 1.05 · 10^{−34}} ≃ 10^{6} m/s.