# Angular part of the wavefunction!

1. Jan 6, 2012

### Chemist20

Hello,

This question is related to wavefunctions and their radial and angular parts.
I know how to draw the radial part, the RDF but how would you draw the angular part?

Thank you!

2. Jan 6, 2012

### ChmDudeCB

Are you referring to the hydrogen atom?

3. Jan 7, 2012

### Chemist20

Im referring to any of the spherical harmonics, including the one for the hydrogen! ;)

4. Jan 7, 2012

5. Jan 7, 2012

6. Jan 7, 2012

### cgk

No, they are not:
1. They are only the angular part (they have only two angles (or alternatively one normalized vector) as parameter!).
2. They are not necessarily wave functions. They are the solutions to the angular part of a Laplace equation with spherically symmetric potential.

There are also "solid harmonics", which actually have an radial part and are three-dimensional functions (unlike the spherical harmonics, which are defined only on the sphere), but normally you don't encounter them.
Why do you think spherical harmonics are total wave functions?

7. Jan 7, 2012

### Chemist20

I'm a bit lost now...
Okay, the link takes you to a google page were the shape of different orbitals appear. That's the shape of the orbital itself and so has to be total wavefunction right?

8. Jan 7, 2012

### nonequilibrium

Well, they might remind you of orbital drawings (which is not too silly as they're used in them), but they aren't really orbital pictures when you compare them, are they?

Anyway, those drawings of the amplitude of the spherical harmonics are exact, whereas the orbital drawings of hydrogen are an approximation, right? The latter uses a .9 probability area to draw the wave functions (?) That being said, I'm wondering how orbitals are drawn: what is the method/algorithm? What exactly are we seeing in those drawings of orbitals? I can't seem to find a clear source on it. My quantum physics professor even told me about it the wrong way...

9. Jan 7, 2012

### tom.stoer

No, I just googled for "spherical harmonics"

10. Jan 7, 2012

### nonequilibrium

He means that the pictures that show up remind him of the pictures of orbitals.

11. Jan 7, 2012

### tom.stoer

Yes, I understand that and that's not unreasonable b/c spherical harmonics are part of the wave function -and b/c the usual 'drawings' of orbitals focus on the spherical part, not on the radial one.

12. Jan 7, 2012

### nonequilibrium

Just wondering: do you know how orbitals are drawn? Where and how does the 90% come in?

13. Jan 7, 2012

### tom.stoer

An orbital is a region of space (not necessarily connected) which contains the electron with a certain fixed probability. What is drawn are just the outer surfaces. For higher shells this is not correct, the orbitals should look more like onions.

Don't know; b/c it looks nice?

14. Jan 7, 2012

### nonequilibrium

Outer surfaces of what? Of for example a region where there's a 90% probability? But it seems there would be an uncountable number of such regions you could pick. Which one is chosen?

15. Jan 7, 2012

### tom.stoer

Define surfaces of constant probability density p; these closed surfaces define several regions (not necessarily connected); the union of all these regions is called 'orbital'. Integrating the probability density over all these regions yields a probability P(p). Now adjust p such that P(p)=90%.

Due to the radial oscillations the structure of the orbital (the different surfaces) can look like an onion; but of course you can only see one such surface, the inner structure is hidden.

16. Jan 7, 2012

### nonequilibrium

Ah interesting. So when you say "Integrating the probability density over all these regions yields a probability P(p)." do you mean integrating the volume enclosed in each of those closed surfaces?

17. Jan 7, 2012

### tom.stoer

For one region V: integrating the probability density over V yields a probability P

$$\int_{V:\, |\psi|^2 = p\text{ on }\partial V} dV\,|\psi(x)|^2 = P$$

18. Jan 7, 2012

### nonequilibrium

But if you're doing a volume integral restricted to a surface, isn't that zero? After all, if the probability distribution is bounded, you can bound the whole integral by a constant times the volume measure of a surface, which is zero.

19. Jan 8, 2012

### tom.stoer

I do not integrate over a surface but over a volume defined by the surface. It's like saying that "integrating over a disc r² ≤ 1" is defined by "integrating over a two-dim. region defined by a circle r² = 1".

20. Jan 8, 2012

### nonequilibrium

Oh, no problem then, I misunderstood your notation (my fault)!

So you end up with (possibly) different surfaces, each with a (possibly) different (constant) value of probability density on its surface, such that each of these gives a (say) probability of 90% when integrating the volume enclosed by the surface? At least this is what your explicit integral is implying to me.

But how can you then get an onion? If one surface B is enclosing another smaller one A, and the volume integral inside of A is .9, then surely the volume integral inside of B (enclosing the volume inside of A) must be bigger than .9,
so I'm probably again misinterpreting something you've said.