 Solution

ID: past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2012/1/c/solution

\frac{1}{m ^{*} ( k )} = \frac{1}{ℏ ^{2}} \frac{d ^{2} E}{d k ^{2}} .

With the stated sign convention and

t > 0

, the band minimum is at

k = 0

. Expanding the cosine, or differentiating twice, gives

E (k) = E_{0} - 2 t + t a^{2} k^{2} + O (k^{4}), \frac{d ^{2} E}{d k ^{2}}_{0} = 2 t a^{2} .

ℏ^{2} k^{2} / (2 m_{e})

therefore requires

t = \frac{ℏ ^{2}}{2 m _{e} a ^{2}} = 6.78 \times 1 0^{- 20} J = 0.423 eV .

The hopping integral has units of energy. A negative

t

would move the minimum to the edge of the Brillouin zone and give the same positive bottom-of-band effective mass with

∣ t ∣

in this expression.

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