 Solution

ID: past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2021/section-b/4/d/solution

For

∣ k_{i} a ∣ ≪ 1

, expand each cosine to second order:

E (k) = E_{0} + 6 t - t a^{2} (k_{x}^{2} + k_{y}^{2} + k_{z}^{2}) + O (t a^{4} k^{4}) .

To this order a fixed energy fixes

k_{x}^{2} + k_{y}^{2} + k_{z}^{2}

(m^{*- 1})_{ij} = \frac{1}{ℏ ^{2}} \frac{\partial ^{2} E}{\partial k _{i} \partial k _{j}} = - \frac{2 t a ^{2}}{ℏ ^{2}} δ_{ij}, m^{*} = - \frac{ℏ ^{2}}{2 t a ^{2}} .

For bonding hopping

t < 0

this is a positive electron mass at the band bottom. If

t > 0

k = 0

is a maximum with negative electronic mass, or positive electron hole mass

ℏ^{2} / (2 t a^{2})

. The spherical shape is a quadratic approximation; cubic anisotropy appears in the fourth-order terms.

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