# When is a particles charge spread out?

1. Jun 13, 2005

### box

Correct me if i'm wrong but i think in quantum mechanics there are times when the charge of an electron is spread out in the area where its positon is uncertin, such has a ground state hydrogen atom. And there are times when even though the electrons position is unknown over a certin area the charge is not spread over the whole area but is modeled to be in a more certin spot. Such has when you have two protons and one electron that is equaly likely to be orbiting both of them, The protons don't repell each other because the electrons charge is only on one of them and not evenly spread betwen the two. So am I right in saying that sometimes an elecrons charge is spread out over the area were its position is uncertin and somtimes its not? If so what determines if its the former or the latter?

2. Jun 14, 2005

### Dr.Brain

aS PER hEISENBERG'S uNCERTAINITY pRINCIPLE , YOU CANNOT POSSIBLE DTERMINE THE VELOCITY AND POSITION OF AN ELECTRON simultaneously..

Imagine this , suppose you witness a fast moving electron and wth great precision you pin it up against the wall with a pin, before pinning it up , the electron was so fast that you couldnot determine its position and now you pinned it up in hope to determine the position correctly, now when you see the pinned-electron , it spreads itself and looks more like a cloud rather than a particle , this is what happens as per Heisenberg's..

More at :
http://www.doxlab.co.nr/

3. Jun 14, 2005

### dextercioby

Ever heard of overlap integral,or charge exchange integral...?My guess is "no".

Daniel.

4. Jun 14, 2005

### reilly

For nonrelativistic QM, the square of the electron's wave function times the electric charge is the spatial charge density, which in turn is determined by the specific dynamics of the electron's system. For relativistic QM the same is true -- up to a few technical matters.

There are no stable two-proton one or two electron systems. Replace a proton by a neutron, and you are talking helium city, a stable place.

Regards,
Reilly Atkinson

5. Jun 14, 2005

### dextercioby

I think he meant a Hidrogen molecule.That's pretty stable.And it has no neutrons.

Daniel.

6. Jun 15, 2005

### box

This is what I guessed but when reading the Feynman lectures on physics vol 3 chapter 10 he talks about a two state system made up of two separated protons with one electron thats orbiting a different proton in each state. Now when you have a superposition of both states I would guesse the electrons charge would be spread out over both protons (it seemes to me your statment above says this) but then I would think the protons would repel each other. However Feynman says the protons don't repel because one proton is nutrual because of the electron orbiting it. (well they did repel or atract but not because both protons were positive)
Sorry if what I wrote dosn't make alot of sense.

7. Jun 15, 2005

### dextercioby

Since you're talkig about the H_{2} molecule,words won't do.Mathematics should come first.So how about you do some reading & caculations...?The proton may be screened,but not neutral.

Daniel.

8. Jun 15, 2005

### Edgardo

Hello box,

it seems strange that a $H_{2}^{+}$ molecule exists, because as you mentioned one would think that the protons repel each other. However QM calculations indeed show that it is stable only because of that one electron.

Why is this so? You can think of the electron acting as glue between the protons (or as dextercioby said the proton's charge is screened by the electron's negative charge)

9. Jun 15, 2005

### ZapperZ

Staff Emeritus
And this is a good opportunity to add that that single electron in between the two H nucleus is actually in a Schrodinger Cat-type state where it is "localized" at BOTH nucleus simultaneously! [See? All of physics/QM are really connected!] The electron is being spread out over both nucleus in such a way that the overlap of its location from both nucleus causes a bonding state and an antibonding state of the ground state. This is something chemists have seen way before QM, but it had to wait till QM for there to be an explanation for the existence of such a state.

Zz.

10. Jun 15, 2005

### Gokul43201

Staff Emeritus
To answer the original question :
No, that's not right. As explained previously, the charge density at some point $\vec{r}$ scales with the probability of finding the electron in the vicinity of $\vec{r}$.

$$\rho(\vec{r}) = e|\psi({\vec{r})|^2$$