Charge density of an infinite 1D system

Click For Summary
SUMMARY

The discussion focuses on determining the electron density of an infinite one-dimensional system after a potential step is introduced. The initial electron density is given by the equation n(x) = ∫_{0}^{k_{F}} (dk/π), where k_{F} = √(2E_{F}). Upon introducing the potential step, the wavenumbers for regions x<0 remain unchanged, while those for x>0 are modified to k = √(2(E - V')). The equilibrium conditions lead to final energy occupancies that differ on either side of the potential step, necessitating a reevaluation of the initial equations used to describe the system.

PREREQUISITES
  • Understanding of quantum mechanics, specifically electron wave functions
  • Familiarity with potential energy concepts in one-dimensional systems
  • Knowledge of Fermi energy and its implications in electron density calculations
  • Proficiency in mathematical integration techniques in atomic units
NEXT STEPS
  • Review the derivation of electron density equations in quantum mechanics
  • Study the effects of potential steps on wave functions in one-dimensional systems
  • Learn about the implications of Fermi energy in electron distributions
  • Explore advanced topics in quantum mechanics, such as perturbation theory
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying electron behavior in one-dimensional systems and potential energy effects. This discussion is also beneficial for anyone involved in theoretical physics or materials science.

El Hombre Invisible
Messages
691
Reaction score
0
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 n(x) = \int_{0}^{k_{F}} \frac{dk}{\pi} where k_{F} = \sqrt{2E_{F}}


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&#039;}. The initial occupancy for x>0 is V&#039; &lt; E &lt; E_{F}+V&#039;. 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 &lt; E &lt; E_{F} + \frac{V&#039;}{2} for x < 0
V&#039; &lt; E &lt; E_{F} + \frac{V&#039;}{2} for x > 0

in atomic units. The equation for the ground state depends on \sqrt{V&#039;}, 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?
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
Replies
4
Views
2K