1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charge density of an infinite 1D system

  1. May 18, 2009 #1
    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 [tex]n(x) = \int_{0}^{k_{F}} \frac{dk}{\pi}[/tex] where [tex]k_{F} = \sqrt{2E_{F}}[/tex]


    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 [tex]k = \sqrt{2(E - V'}[/tex]. The initial occupancy for x>0 is [tex]V' < E < E_{F}+V'[/tex]. 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:

    [tex]0 < E < E_{F} + \frac{V'}{2}[/tex] for x < 0
    [tex]V' < E < E_{F} + \frac{V'}{2}[/tex] for x > 0

    in atomic units. The equation for the ground state depends on [tex]\sqrt{V'}[/tex], 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?
     
  2. jcsd
  3. May 21, 2009 #2
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook