Charge density of an infinite 1D system

AI Thread Summary
The discussion revolves around determining the electron density in an infinite 1D system after a potential step is introduced. Initially, the electron density is defined by the equation n(x) = ∫(0 to k_F) dk/π, where k_F = √(2E_F). When the potential step is applied, the wavenumbers for x<0 remain unchanged, while those for x>0 are modified by the potential V'. The equilibrium condition requires that the energy distributions on both sides of the potential step be equal, leading to new energy ranges for electron occupancy. The original poster seeks clarification on the appropriate equations to use for this scenario, indicating potential confusion in their calculations.
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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?
 
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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
 
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