# Calculating the inelastic quasiparticle lifetime of a screened quantum fluid

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.

• Can you explain the difference between $\hat{\boldsymbol{q}}$ and $\boldsymbol{q}$? I just took a quick look but can it be that you messed up in the second line of your developement? Since you say that $k^2-(\boldsymbol{k}\cdot \hat{\boldsymbol{q}})^2\approx \dfrac{1}{4}[4k_F^2-q^2]$ but it should be $\dfrac{1}{4}[4k_F^2-q^4]$. – vin92 May 22 at 9:25
• @vin92 $\hat{\bf q}$ is the unit vector. When you derive Eqn. 69 in the paper, this value comes up (I believe that's correct because I followed that portion of the calculation assuming this, and I got the correct answer). Using the information supplied, we then find the term in the second line of my derivation. Also, assuming that you are correct and it is a $q^4$ term, I do not believe this would solve the discrepancy. – Joshuah Heath May 22 at 12:55
• Ok thanks for the notation. Are you sure you expanded $-1/4[4k_F^2-q^2]^2$ correctly? I get $-1/4[4k_F^2-q^2]^2=-4k_F^2+2k_F^2q^2-q^4/4$. – vin92 May 22 at 13:25
• @vin92 Thank you for pointing that out. I have corrected that issue in the main part of the text; it seems as if it modifies it slightly, but the main issue is still there; namely, the sign of the -q^2 term (assuming the last term in the square root of my expression can be approximated to zero). – Joshuah Heath May 22 at 15:03
• Yes I thought the same, I'm sorry I don't have the time to look deep in the article but could it be that there is a relation between $k_s,k_F$ and $q$, then it may be posssible to approximate the term $q^2[1+2k_F^2/k_s^2]=-q^2$. – vin92 May 22 at 15:52

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