# Can one say that, in neutral atoms, l>0 atomic orbitals allow for spill over electric fields?

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I have been involved in a discussion on stack.exchange about neutral atoms and the electric field.

Orbitals with l>0  have a shape , which is the probability density of finding the electron. I have been intuitively assuming in the answer that this shape will create changing electric fields, negative where the probability density of the  electron predominates positive from the nucleus where it does not, and thus a spill over field would appear.

The measured orbitals of the hydrogen atom are symmetric. My second assumption then is that this measurement averages over many not oriented in space atoms  and generates an average symmetry. At an instant in time the hydrogen atom at l=1 has the probability of being asymmetric, given by the orbital distribution.

But thinking about it,  the same is true for the l=0 orbital.

The differences on orbital shape would then only appear if one experimented with the same , oriented atom.

I will be grateful if this "image " I have  been using of orbitals shaping electric fields so that atoms are like small lego blocks that can fit or not, is wrong if the mathematics is done correctly.

recategorized Jul 12, 2016

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Orbitals are distinguished basis functions, reflecting the shape of the basis of spherical harmonics. The individual shapes correspond to fixed values of $l$ and $m$. As far as I can tell, their shapes mean nothing observable but are usually drawn simply as illustrations for the form of spherical harmonics - and to make the abstract wave function stuff look intuitive to chemists.

On the other hand, observed wave functions are just that, hence usually somewhat arbitrary superpositions of the basis states. Your reference on ''measured orbitals'' actually measured $L^2$ which is rotationally invariant. It did not measure $p$ or $d$ orbitals. (At least if I understood the paper correctly.)

answered Jul 18, 2016 by (15,777 points)

Arnold, you contradict yourself: "their shapes mean nothing observable", but "On the other hand, observed wave functions are just that". As I wrote in my answer, there are cases when calculations (corresponding to experimental conditions) deal with these wave functions, so the calculation result is directly expressed via $\psi_{nlm}$. One may write an atomic form-factor in terms of $\psi_{nlm}$; equally one may write an atomic potential $U({\bf{r}})$ in terms of $\psi_{nlm}$ (formula (7) in Atom as a "Dressed" Nucleus). In any case, the result is expressed via a sum (integral is a sum) over "sub-clouds".

It seems the word "orbital" is assigned to wave functions. Would you agree that the complex conjugate square of the orbital as a probability density would have a measurable shape as seen in fig 3.10 here ?

@annav: I agree that if the electron of a single hydrogen atom could be prepared in a pure 2p-state and if the electron charge density of this hydrogen atom could be measured, it would have the shape given in Fig. 3.10. But I believe that both requirements in this statement are unsatisfiable to an experimentally meaningful accuracy.

What one can certainly measure is the charge density of individual metal atoms on a surface. But this includes the joint charge density of all electrons and the nuclei. See my article ''How do atoms and molecules look like?'' from my Theoretical Physics FAQ.

A number of years ago I checked the literature about what was known about the observability of orbitals in more general context. I found (in 2005) that the topic was controversially discussed; see the references at the end of the cited FAQ entry. If you find more recent evidence that improves on the picture I'd be happy if you give references, and would update the description there.

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Yes, it can. Note, you are reasoning in terms of atomic wave functions, as if you could "see" them.

In calculations one may work with non distorted wave functions if the probe projectile is sufficiently fast. Then the first Born approximation applies and you have the cross section expressed via $\psi_{n,l,m}$. In order to avoid angular averaging, one has to prepare equally oriented target atoms.

One funny thing about it is that the projectile will see partially charged sub-clouds similar to quarks (I call them shmarks).

answered Jul 12, 2016 by (102 points)
edited Jul 12, 2016

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