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

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Past exam of the quantum condensed matter course of the University of Cambridge / 2012 / 1 / b / Solution by

Codex  0 2026-10-18

For a hydrogenic donor state, dielectric screening replaces

ϵ_{0}

by

ϵ_{r} ϵ_{0}

and the electron mass by the band effective mass. The characteristic orbital radius is therefore

a^{*} = \frac{4 π ϵ _{0} ϵ _{r} ℏ ^{2}}{m ^{*} e ^{2}} = ϵ_{r} \frac{m _{e}}{m ^{*}} a_{0} .

Using

a_{0} = 0.0529177 nm

gives

a^{*} = 120 a_{0} = 6.35 nm .

A donor-state overlap estimate compares this radius with the mean dopant spacing

d \sim N_{D}^{- 1/3}

. Taking

d \sim a^{*}

yields

N_{D} \sim (a^{*})^{- 3} = 3.91 \times 1 0^{24} m^{- 3} = 3.91 \times 1 0^{18} c m^{- 3} .

There is no sharp geometrical onset: hydrogenic wavefunctions have exponential tails. If “overlap” is defined by neighbouring orbital diameters touching,

d \sim 2 a^{*}

, the corresponding estimate is

4.88 \times 1 0^{23} m^{- 3}

, smaller by a factor of eight. These are conventions for the onset of appreciable orbital overlap, not an exact critical density for a metal-insulator transition.

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