# Electrostatic attraction between an electron and a hole

1. Jun 13, 2014

### HeavyMetal

The exciton is defined as a bound state of an electron and an electron hole. From what I've read, this state is described by Coulomb's law. Coulomb's law describes the interaction between two charged particles. So my question is: because an electron hole is not an actual particle, how does an electron feel attraction to this vacancy in space? Also, what makes this force greater than the force of repulsion caused by the neighboring electrons?

2. Jun 14, 2014

### ehild

The hole is a missing electron. Without it, the potential would be constant in the sea of electrons and the free electron would not feel any force as the repulsive forces of the other electrons cancel. A missing electron does not repel the free electron, and the other electrons push it towards that hole. The same situation as if the electron was attracted by the hole.

ehild.

3. Jun 14, 2014

### HeavyMetal

Without what the what potential would be constant? Without the mathematical treatment of a hole, the potential to recombine would be constant? I know a hole wouldn't repel an electron, it would be attracted to it. I was wondering why the neighboring core electrons wouldn't repel said electron. When you say, "the other electrons push it towards that hole," which electrons are you talking about? Core electrons? I need you to be more specific please, I don't know much about this subject!

EDIT: Wait, are we assuming that this hole is in the core, and that there are other valence electrons "sandwiching" the electron and its hole? If so, I understand a bit more what you mean. If so, then I am gathering that this is not actually an attractive force between a particle (the electron) and an imaginary particle (the hole), but rather a repulsive force between real particles, the valence electrons, and a void of space, or the hole. But it is mathematically less challenging to treat this situation as if the hole and the electron are feeling attracted to each other. Y/N? And if this are the case, why in fact is it that all of the core electrons' repulsive forces cancel? I know that there are still only two quantum states per orbital. Is it the fact that when a hole is created, two paired electrons become unpaired, and that vacancy allows the newly unpaired spin to recombine at a later time? This is the repulsion that I meant, by the way, the pairing energy. Not the repulsion from the other core electrons. My guess is that the valence electrons exert stronger repulsive forces than that force which is required to pair the two electrons, so it is still a favored process.

Last edited: Jun 14, 2014
4. Jun 14, 2014

### ZapperZ

Staff Emeritus
The hole is not in the "core". This is still the valence band of a semiconductor we are talking about here. Core levels are "discrete".

Remember that the conduction electrons are moving inside the lattice and a sea of electrons. When there is a hole in this sea of electrons, an electron inside that sea tends to be pushed more towards it, since there's less repulsive forces in that direction than all around it. If you renormalize" the hole, it is the same as putting a positive charge on it. Dealing with the physics and math becomes simpler, because you can now, instead of dealing with huge amount of negative electrons in the valence band, just deal with a small number of positive holes in that band. The physics comes out the same.

This is no different than in the conduction band. I can easily say that it is full of positive holes. When I introduce these electrons in it, I renormalized everything and let those holes have zero charge, causing the electrons to have negative charges, and off I go!

Zz.

5. Jun 14, 2014

### HeavyMetal

So a hole behaves like a particle, and is easier to treat mathematically as a particle. What math is used to treat this, the Schrödinger equation? Just sub in the charge as $+e$ instead of $-e$, but leaving $m_{e}$ unchanged?

And what did ehild mean when he said this?