# I Charge density that minimizes total energy

1. Mar 10, 2016

### aaaa202

I am currently interested in a system, where I have a big box of charges for which the density goes to zero at the boundary. What I wanted to do is try to derive what charge density will minimize the total electrostatic energy in the box.
Now a cubic box gave me some complicated so I switched the problem to be that of a sphere of radius R, for which I want to find the charge density that minimizes the total electrostatic energy. Mathematically I am therefore trying to solve the problem of finding a ρ(r) that minimizes:

E = 1/4πε0 4π∫0Rρ(r) V(ρ(r)) dr
with the boundary conditions:
4π∫0Rρ(r)dr = ρ0
ρ(R)=0.

Now V(r) I can calculate using the symmetry of the sphere, but it will depend on ρ(r), which explains the notation above.
Is it possible to use variational calculus or other means to find ρ(r) such that E is minimized? And is it possible for other geometries?

2. Mar 10, 2016

### BvU

Physically, repelling charges will want to be as far apart as possible. That's why charge is on the surface on a conductor.

3. Mar 10, 2016

### aaaa202

Yes but in my problem there is the boundary condition that the density should vanish on the surface.

4. Mar 10, 2016

In your question, you need to define the types of electrical charges=if they are positive charges, what is their atomic number?, etc., and for an atom, the positive charges need neutrons to hold them together or they would repel each other. If they are negative charges, most likely they are necessarily electrons. Put them together and you create a solid, but besides just simple electrostatic effects, quantum mechanical principles often play a role in the solid that emerges.

5. Mar 10, 2016

### aaaa202

Well this is just a conductor on which we have a put a charge density of electrons.

6. Mar 10, 2016

### BvU

Well, that is rather artificial. The charge will nevertheless want to be as far apart as possible.

7. Mar 10, 2016

The excess charge will go to the surface and arrange itself in such a manner that there is zero electric field everywhere inside the conductor and also zero electric fields parallel to the outer surface. Any electric field will be perpendicular to the surface and external to the surface.

8. Mar 10, 2016

### aaaa202

I don't think it is artificial. Actually it is not really a conductor I am solving but rather electrons trapped inside a heterostructure. The wavefunctions vanish at the boundary which means that the electron density also must vanish at the boundary. The reason I did a sphere was for simple calculations.
So I guess the problem I stated is not solvable?

9. Mar 10, 2016

### BvU

If this is on a QM scale you should look at it as a QM problem. Doesn't make it easier.
Conductor, trapping mechanism/potential, fermion wavefunctions, ....

10. Mar 10, 2016

### aaaa202

I have solved it numerically quantum mechanically. But I wanted to do some classical calculations for intuitions on the solutions.

11. Mar 10, 2016

### BvU

Can you trust the numerical results ?

12. Mar 10, 2016

### aaaa202

Well that is the question. I wanted to get some insight by considering the classical solutions..

13. Mar 10, 2016

### BvU

Well, good hunting ! ( time here)....

14. Mar 11, 2016

### BvU

Your problem kept me busy . Is there a square well for the trapping potential or is that something electrostatic as well ?

Any connection with this 1D thread by you ? I have hard time getting a complete picture of what's going on....

15. Mar 11, 2016

### aaaa202

Yes it is a 1d square well for which I solve the Schrödinger equation numerically (in the other directions the electrons are assumed to be free electrons). I then calculate the electrostatic potential from Poissons equation, plug back into the Schrödinger equation and keep reiterating until a self-consistent solution is reached.

16. Mar 11, 2016

### BvU

What does your SE look like ?

17. Mar 11, 2016

### aaaa202

Okay might as well get the full story. My system considers a box with a semiconductor coated by a thin metallic layer. The semiconductor is much longer in the x- and y-direction than the z-direction such that we only considers electrostatic effects in this direction.
Now the metal sets the Fermi level in the heterostructure and determines the number of states that are filled up in the conduction band of the semiconductor. The electron density in the z-direction obeys:

ρ(z) ∝ ∑nklψ(z)l2

And inside the semiconductor I solve the Schrödinger equation:
2/2m* d2ψ/dz2 - eφ(z)ψ(z) = Eψ(z)
Here φ(z) is the electrostatic potential from the charge distribution ρ(z), which can be found by inverting Poissons equation ∇2 = -ρ/ε
The problem is solved and reiterated until a self-consistent solution is found. I should also note that the effect of the metal is of course that it screens the potential in the semiconductor. Therefore I have placed image charges inside the metal, which I use when I calculate the potential.

Now my question was to get an intuition on how the solutions should look like. Right now I get that the electrons migrate to the sides of the junction as shown on the attached picture, which shows an example of a self consistent solution. I guess that makes sense in terms of what you'd expect classically (charges flow to the surface), but I guess I wanted to do a proper calculation with the boundary condition that the charge densities must still vanish at the boundary to vacuum, which is clear from the expression for ρ(z) (now this is not true at the boundary to the metal, where the density explodes, but I chose to ignore that and look at the simpler, classical problem).

#### Attached Files:

• ###### SelfconsistentSolution.png
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18. Mar 11, 2016

### BvU

Interesting enterprise ! So you mix QM on a 100 nm scale with a continuous charge distribution that follows from the calculated wave functions. That does avoid the singularities from two electrons in one place, but if it's representative is questionable. And I really have no idea either.

@vanhees71 ?

19. Mar 11, 2016