• Support PF! Buy your school textbooks, materials and every day products Here!

Charge density of an infinite 1D system

  • #1
Hi there. Long time no see. I hope you're all well.

Homework Statement



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?


Homework 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]


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?
 

Answers and Replies

  • #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
 

Related Threads for: Charge density of an infinite 1D system

Replies
9
Views
973
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
1
Views
539
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
6K
Replies
2
Views
909
Replies
8
Views
8K
Top