## Charge density of an infinite 1D system

Hi there. Long time no see. I hope you're all well.

1. The problem statement, all variables and given/known data

An infinite 1D system has electron plane waves occupying states 0 <= E <= E_F. At time t=0, a potential step is introduced such that V=0 for x<0 and V=V' for x>0. What is the electron density when the system reaches equilibrium again?

2. Relevant equations

The initial (unperturbed) electron density, in atomic units, is $$n(x) = \int_{0}^{k_{F}} \frac{dk}{\pi}$$ where $$k_{F} = \sqrt{2E_{F}}$$

3. The attempt at a solution

Well, when the pertubation is switched on the wavenumbers for x<0 are unchanged while those for x>0 are given by $$k = \sqrt{2(E - V'}$$. The initial occupancy for x>0 is $$V' < E < E_{F}+V'$$. When in equilibrium, the left and right sides must be energetically equal. Since the initial energy difference is V', and the system is symmetric about x=0, I'm figuring that the final occupancies will be:

$$0 < E < E_{F} + \frac{V'}{2}$$ for x < 0
$$V' < E < E_{F} + \frac{V'}{2}$$ for x > 0

in atomic units. The equation for the ground state depends on $$\sqrt{V'}$$, but looking at a graph the difference between n(x) on the left and right sides is just V'. So clearly I'm using the wrong equation. Anyone know the right one?
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Is there something wrong with my wording here? Please tell me if there is and I will amend the question. I could do with sussing this in the next week. Cheers... EHI