Screening in the electron gas (physical discussion)
If one places an impurity charge \(Q\) in an interacting electron gas, how do the electrons rearrange themselves to account for its presence? Well, one would na"ively expect that the electrons would feel a force due to the Coulomb potential energy
\[\begin{equation} V(\x) = \frac{Q}{4\pi\epsilon_0 |\x|} \end{equation}\]Even when we take into account the Thomas-Fermi screening, as was done in the lecture, the potential is modified to
\[\begin{equation} V(\x) = \frac{Qe^{-k_{\text{TF}}|\x|}}{4\pi\epsilon_0 |\x|} \end{equation}\]We see that it is much weaker, but it is still repulsive. Let us consider the following cartoon (see the following \href{https://gravityandlevity.wordpress.com/2009/06/02/friedel-oscillations-wherein-we-learn-that-the-electron-has-a-size}{blog post}, where I learned this explanation from). Consider the interacting electron gas. Pictorially, we draw the electron gas as a bunch of negative charges moving in a uniformly charged positive background.
Now, suppose we have an external charge \(+Q\) sitting in an electron gas. Note that an electron gas has a uniformly charged positive background. Then, we have something like the following Figure. We see that the negative charges will be drawn to it.
However, the picture that we have just discussed is incomplete. We remember that, in quantum mechanics, electrons are described by wave functions, and in the electron gas electron wave functions look like \(\psi_{\k}(\x) = \frac{1}{\sqrt{\V}}\sin(\k\cdot\x)\). This wave function has a charge density which looks like \(|\psi_{\k}(\x)|^2\). Thus, an electron with wave vector \(\k\) has size \(\lambda/2 = \frac{\pi}{k}\). Note that the size is halved because we take the modulus squared to get the actual charge density of the electron.
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Now to modify our cartoon, we need to draw the electrons as ovals with length \(\lambda/2\). Doing this, we have that
The key thing to notice here is that the size of the electron is \(\lambda/2 = \pi/k\), but since only low-energy electrons contribute, \(k = k_F\). Thus the oscillation should repeat itself with period \(\lambda_F/2 = \pi/k_F\). Thus, we should have something like \(\cos(2\pi r/(\pi/k_F)) = \cos(2k_Fr)\). The rate at which the decay occurs, we cannot predict unless we do some math. Let us see if we are at least correct about the oscillations.