Effective mass

The energy depending on the velocity contribution is given by an integral (7.30), which can be easily calculated

$$\Delta E^{2D}(V) =\frac{{g_{i}}^2n_{2D}(2m)^3V^2}{2\hbar^4} \int_... ...\frac{k dk}{(2\pi)^2} = \frac{{g_{i}}^2n_{2D}m}{4\pi^2\hbar^2c^2} \frac{V^2}{2}$$ (7.36)

From this result we infer the effective mass

$$m^*={g_{i}}^2n_{2D} m/(4\pi^2\hbar^2c^2)$$ (7.37)