12 Electron-electron interactions, Quantum electronics
12.1 Electron Screening
For a single electron in vacuum Gauss’s law is
∇·E=
−eδ(r)
ε. (174)
If you use the relationship
E=−∇V (175)between the electric fieldEand the electrostatic potentialV and plug it into Gauss’s law, you get the
Poisson equation
∇^2 V =
eδ(r)
ε. (176)
The solution of this differential equation is
V =
−e
4 πr. (177)
If a single electron is put in a metal and fixed, the other mobile electrons will react to the electric field
whereas the uniformly distributed positive ions will stay at their position. Before the single electron
was put in, the electrons where uniformly distributed too. The electrons far away are still where they
were before but now the electrons near the fixed negative charge are repelled. The poisson equation
now looks like
∇^2 V =eδ(r)
ε+
−ρreaction
ε. (178)
Now we want to know howρreactionlooks like. Based on observations we can say that the potential goes
to zero far away from the extra added electron. The reaction goes to zero far away too, therefore we
assume in the simplest approximation thatρreactionis proportional toV. So we can rewrite eqn. (178)
as
∇^2 V−k^2 V =eδ(r)
ε(179)
which is the Helmholtz equation where the solution is well known. The proportionality constant is
typically chosen to bek^2 withkas a wave vector because the dimension of the factor islength^12. The
Helmholtz equation in 3-d is
(
∇^2 −k^2
)
g=δ(r) (180)wheregis the appropriate Green’s function
g=
−exp (−k|r−r′|)
4 π|r−r′|