# I Why is the s state spherically symmetric?

1. Sep 8, 2016

### KFC

Hi there,
I am reading something about quantum numbers, there the author introduce the quantum number by solving Schrodinger equation for Hydrogen atom. It gives me an example when the principal quantum number n=4, there are four different sub-level $s, p, d, f$. It also depicts the sublevel for them. It is said that for s state which $\ell=0$, the shape is something like a completely flattened ellipse going through the nucleus. But since $\ell=0$, there is no angular momentum, the electron is not "orbiting" anything; hence, it must be totally radially so s state is spherically symmetric.

It is quite confusing for me. Since it is said at the beginning it is flatten ellipse but later becomes spherically symmetric.

reference: http://web.pdx.edu/~pmoeck/lectures/312/chapter 7 part 3.doc
page 3 and page 8

Last edited: Sep 8, 2016
2. Sep 8, 2016

### Staff: Mentor

Where does your reference say that?
Right. s orbitals are spherically symmetric.

3. Sep 8, 2016

### KFC

Sorry that I give the wrong reference. I just get it corrected. And in the document given by the link, in page 3, note below the caption of FIGURE 7.3, it is said that "for l=0 we have a completely flattened ellipse that goes through the nucleus, this will be later called an s state."

4. Sep 8, 2016

### Staff: Mentor

Those elliptical orbits are remnants of the Bohr-Sommerfeld model of the atom. (Old stuff!) Note the diagram on page 20 of your reference, which shows the s orbitals as spherical.

5. Sep 8, 2016

### KFC

Thanks a lot.

6. Sep 9, 2016

### vanhees71

According to my prejudices, I always warn against physics manuscripts written with Word. This one cements my prejudice. SCNR.

7. Sep 10, 2016

### DrDu

Classically, an orbit with l=0 corresponds to the electron falling straight on or through the nucleus. Quantum mechanically, the orientation of this line becomes completely undetermined.

8. Sep 10, 2016

### vanhees71

Quantum mechanically $l=0$ means that your wave function and thus the probability distribution for the particle's position (and momentum) is spherically symmetric.