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 Incoming links: Free-electron Fermi circle in a square lattice

Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section A / 5 / a / Solution 2026-10-20

For a free-electron Fermi circle in a square lattice, include two spin states per wavevector. A square cell of side

a

supplies

z / a^{2}

electrons per unit area; counting the disk of occupied states gives

\frac{z}{a ^{2}} = \frac{2 π k _{F}^{2}}{( 2 π ) ^{2}}, k_{F} = \frac{2 π z}{a} .

The first Brillouin zone is

∣ k_{x} ∣, ∣ k_{y} ∣ \leq π / a

. For

z = 1

, $k_{F} = 2.507/ a$ , so the circular Fermi surface lies wholly inside this square and encloses half its area. The figure uses the stated free-electron approximation; the lattice symmetry alone does not prescribe an exact circular surface.

Figure 1.
Free-electron circles and the square Brillouin zone
. Original state-counting construction. Dashed squares are first-zone boundaries; the right-hand circle is shown in the extended-zone representation. Details.

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