Effective mass

The energy term (7.30), which depends on the velocity ${\bf V}$ can be trivially calculated by using of the integral equality $\int \frac{dx}{(x^2+a^2)^2} =\frac{1}{2a^3}\mathop{\rm arctg}\nolimits \frac{x}{a}+\frac{x}{2a^2(x^2+a^2)}$. The result of the integration is $\Delta E^{1D}(V) = \frac{{g_{i}}^2 nV^2}{4\hbar c^3}$ and the effective mass is given by

$$m^*={g_{i}}^2n_{1D}/2\hbar c^3.$$ (7.42)

It can be expressed in terms of the particle-particle $a$ and particle-impurity $b$ scatering lengths $m^* = \frac{1}{\sqrt{32n_{1D}a}}\left(\frac{a}{b}\right)^2m$