Calculating the inelastic quasiparticle lifetime of a screened quantum fluid

Question

I've been studying "Lifetime of a quasiparticle in an electron liquid", by Qian and Vignale. Much of it makes sense, but there is a detail in the calculation of the exchange term that doesn't make sense to me. Eqn. 23 gives

$$\frac{2\pi me^2 }{pqk_s\sqrt{k_s^2+4k_F^2-q^2}} $$

This follows from Eqn. 69 in the appendix of the article (omitting the Heaviside theta functions):

$$ \frac{2\pi me^2}{pq\sqrt{[p^2+k^2+k_s^2-{\bf k}\cdot {\bf q}]^2-[k^2-({\bf k}\cdot {\bf \hat{q}})^2][4p^2-q^2] }} $$

Equating the two, this tells me that

$$k_s\sqrt{k_s^2+4k_F^2-q^2}=\sqrt{[p^2+k^2+k_s^2-{\bf k}\cdot {\bf q}]^2-[k^2-({\bf k}\cdot {\bf \hat{q}})^2][4p^2-q^2] } $$

The authors obtain Eqn. 23 from Eqn. 69 by assuming that $p\sim k\sim k_F$ and ${\bf k}\cdot {\bf q}\sim -\frac{q^2}{2}$. The r.h.s. of the above then becomes

\begin{align} &\sqrt{[p^2+k^2+k_s^2-{\bf k}\cdot {\bf q}]^2-[k^2-({\bf k}\cdot {\bf \hat{q}})^2][4p^2-q^2] }\notag\\ \approx& \sqrt{[2k_F^2+k_s^2+q^2/2]^2-1/4[4k_F^2-q^2]^2 } \notag\\ &=\sqrt{4k_F^4+4k_F^2k_s^2+2k_F^2q^2+k_s^4+k_s^2q^2+q^4/4-4k_F^2+2k_F^2q^2-q^2/4 }\notag\\ &=\sqrt{4k_F^2k_s^2+k_s^4+k_s^2q^2+4k_F^2q^2 }\notag\\ &=k_s\sqrt{4k_F^2+k_s^2+q^2+4k_F^2q^2/k_s^2} \end{align}

which is clearly different from the authors' Eqn. 23. Is there an approximation they invoke that they do not mention? I checked my result numerous times and it appears to be mathematically sound. Specifically, Qian & Vignale's denomainator differs from mine in the sign of $q^2$.

EDIT: Fixed a minor typo pointed out by @vin92. The solution, however, still doesn't match that of Qian and Vignale.

Answer 1

I have contacted one of the authors of this work, and I found out that there is a small typo in the appendix. Eqn. 68 in the Appendix of the arXiv version (Eqn. A2 in the PRB version), which reads

$$ A_3=2\pi e^2 \int_{-1}^1 dx \delta\left( \omega +\frac{pqx}{m}+\frac{q^2}{2m} \right)\frac{1}{\sqrt{ (p^2+k^2+k_s^2+2pk\cos\theta x)^2-4(pk\sin\theta)^2(1-x^2) }} $$

should be $$ A_3=2\pi e^2 \int_{-1}^1 dx \delta\left( \omega +\frac{pqx}{m}+\frac{q^2}{2m} \right)\frac{1}{\sqrt{ (p^2+k^2+k_s^2-2pk\cos\theta x)^2-4(pk\sin\theta)^2(1-x^2) }} $$

Redoing my previous calculation, we find that

\begin{align} &\sqrt{[p^2+k^2+k_s^2+{\bf k}\cdot {\bf q}]^2-[k^2-({\bf k}\cdot {\bf \hat{q}})^2][4p^2-q^2] }\notag\\ \approx& \sqrt{[2k_F^2+k_s^2-q^2/2]^2-1/4[4k_F^2-q^2]^2 } \notag\\ &=\sqrt{4k_F^4+4k_F^2k_s^2-2k_F^2q^2+k_s^4-k_s^2q^2+q^4/4-4k_F^2+2k_F^2q^2-q^4/4 }\notag\\ &=\sqrt{4k_F^2k_s^2+k_s^4-k_s^2q^2}\notag\\ &=k_s\sqrt{4k_F^2+k_s^2-q^2} \end{align}

This leads to Eqn. 24 in the text, a major result in this paper which, from the above calculation, is shown to be correct.