Why is the s state spherically symmetric?

[KFC]

Hi there,
I am reading something about quantum numbers, there the author introduce the quantum number by solving Schrödinger 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

 

[Doc Al]

It is said that for s state which $\ell=0$, the shape is something like a completely flattened ellipse going through the nucleus.

Where does your reference say that?

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.

Right. s orbitals are spherically symmetric.

 

[KFC]

Where does your reference say that?

Right. s orbitals are spherically symmetric.

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."

 

[Doc Al]

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.

 

[KFC]

Thanks a lot.

 

[vanhees71]

Hi there,
I am reading something about quantum numbers, there the author introduce the quantum number by solving Schrödinger 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

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

 

[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.

 

[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.