# Polarization of free electron in two dimensions

We consider the polarziation of free electron gas in 2D,

$$$\Pi_0(q, \Omega_n) =-S\int\frac{d^2 \vec k}{{(2\pi)}^2} \frac{n(\epsilon_{\vec{k}})-n(\epsilon_{\vec{k}+\vec{q}})}{i\Omega_n+\epsilon_{\vec{k}}-\epsilon_{\vec{k}+\vec{q}}}\\$$$

where $n(\vec k) = 1/(e^{\beta(\epsilon_{\vec k}-\mu)}+1)$, $\epsilon_{\vec k}=k^2/(2m)-\mu$, and $S$ is the spin factor. It is expressed as

$$$\Pi_0(q, \Omega_n)=-S\int_0^{\infty} \frac{mkdk}{2\pi^2} n(\epsilon_k) \int_{0}^{2\pi} d\theta \left[ \frac{1}{i2m\Omega_n-2kq \cos\theta-q^2}-\frac{1}{i2m\Omega_n-2kq\cos \theta+q^2}\right]$$$

## Static limit $\Omega_n=0$

For real $a,b$, the integral $\int_0^{2\pi}\frac{1}{a-b\cos \theta}=\frac{2\pi}{\sqrt{a^2-b^2}}$ and $\int_0^{2\pi}\frac{1}{a+b\cos \theta}=C\frac{2\pi}{\sqrt{a^2-b^2}}$ where $C=1$ for $a>b$ and $C=-1$ for $a<b$. Hence, we have

$$$\Pi_0(q, 0)= -S\int_0^{q/2}\frac{mkdk}{2\pi^2} n(\epsilon_k) \frac{4\pi}{\sqrt{q^4-4k^2q^2}}$$$

At zero temperature,

• for $q<2k_F$, $$$\Pi_0(q, 0)=-S\int_0^{q/2} \frac{m}{\pi q} \frac{dk^2}{\sqrt{q^2-4k^2}}=-\int_0^1\frac{mS}{4\pi}\frac{dx}{\sqrt{1-x}}=-\frac{mS}{2\pi} \,;$$$
• for $q>2k_F$, $$$\Pi_0(q, 0)=-S\int_0^{k_F} \frac{m}{\pi q} \frac{dk^2}{\sqrt{q^2-4k^2}}=-\int_0^{4k_F^2/q^2}\frac{mS}{4\pi}\frac{dx}{\sqrt{1-x}}=-\frac{mS}{2\pi}\left( 1-\sqrt{1-\frac{4k_F^2}{q^2}}\right) \,.$$$

## Zero temperature

$$$\Pi_0(q, \Omega_n)=-\frac{mS}{2\pi} \left[1-\frac{2k_F}{q}{\rm Re}\sqrt{\left(\frac{q}{2k_F}+ i \frac{m\Omega_n}{qk_F}\right)^2-1} \right] \,.$$$

For $q\to 0$ and $\Omega_n \neq 0$, $\Pi_0(q, \Omega_n) \to 0$.