# Electron Orbital

1. Sep 16, 2010

### mishima

Hello,

I'm a bit confused learning about how an electron can be found anywhere within the probability density. What exactly are the bounds of the probability density? Could the electron (with an insanely low probability) be found light years away from the nucleus?

2. Sep 16, 2010

### zhermes

The bounds depend mostly on the potential energy environment around the particle (e.g. electron). For example: for a hydrogen atom, the potential energy is mostly determined by the coulomb force
$$U \propto - \frac{1}{r}$$
or for a 'square well' (like an electron between capacitor plates) the potential is zero between the plates, and some number outside the plates.

The only time (generally) that the electron probability density will be zero, is in regions with (or separated off by) and infinite potential. In the hydrogen atom example, the potential is infinite only at the origin (r = 0), and thus there is zero probability density there. Everywhere else in the universe (of that hypothetical and ideal hydrogen atom), even light years away, there is some probability of finding the electron. Within very short distances, however, the probability becomes so small that you would never statistically expect to find the electron there in the entire age of the universe.... but it would still be, 'possible.'

3. Sep 16, 2010

### alxm

To answer the OP: Well, yes, in principle. But it does drop off exponentially, and becomes insignificantly low very quickly.

Err, no. Electronic orbitals (with the exception of 1s orbitals which have no nodes) have lots of nodes where there are quite finite potentials. And infinite potential does not mean a zero probability density, because that singularity may be canceled by the kinetic energy term. Hence:
Is also wrong. The ground-state hydrogen electron (1s) has a quite finite probability of being at the nucleus, not only that, it's the most likely place - a volume element at r=0 has the highest possible probability density. You're confusing the probability density with the radial probability distribution, which is not the same thing. The latter is zero at r=0 because a sphere of radius zero has no surface area.

4. Sep 16, 2010

### zhermes

I think you're missing the forest a little here.
Yes, absolutely, entirely correct. But the only place, where generally and systematically the probability will tend to be zero... is at infinite potentials (e.g. infinite square well). Nodes do pop up all over the place in bound systems, but I don't find that point to be at all illustrative to the general concept of delocalization. I should have been more clear, my apologies to the OP.

Right again, I mis-spoke, I meant probability, not probability density. But, alxm, I don't see how you can say there is a finite probability at r=0--for just the reason you state, the volume is zero.

5. Sep 16, 2010

### alxm

But this isn't the case for (generalized) Coulomb potentials, which are the only potentials of interest in this context.

I said 'a volume element', I didn't refer to the volume of a single point. Choose a box or sphere of whatever volume you like. If you want to maximize the density in it, it must be centered at x=y=z=0. The 1s wave function is exp(-r), and the density distribution merely the square of that.

The density distribution is n(x,y,z), the radial density distribution is this function integrated over the surface of a sphere of radius r. That will obviously go to zero as r->0, regardless of what the density distribution happens to be.