Charge density of an infinite 1D system

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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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