• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,726 comments
1,470 users with positive rep
818 active unimported users
More ...

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

+ 3 like - 0 dislike

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.

asked Jul 12, 2016 in General Physics by anna v (2,005 points) [ no revision ]
recategorized Jul 12, 2016 by Dilaton

2 Answers

+ 2 like - 0 dislike

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 Arnold Neumaier (15,787 points) [ no revision ]

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.

+ 0 like - 0 dislike

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 Vladimir Kalitvianski (102 points) [ revision history ]
edited Jul 12, 2016 by Vladimir Kalitvianski

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights