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/sci/ - Science & Math


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File: 45 KB, 500x324, 500px-Electron_orbitals.svg.png [View same] [iqdb] [saucenao] [google]
2326349 No.2326349 [Reply] [Original]

I skimmed through a textbook on particle physics, especially the chapter on bound states, like atoms. There's one thing I didn't quite get, though.

If the potential around a nucleus is spherically symmetrical (the Coulomb potential), then how come atomic orbitals aren't? Is it because of spin? (On other hand doesn't spin just lead to splitting energy levels?)

Pic related it's some orbitals.

>> No.2326416

>skimmed
That's your problem right there.

>> No.2326439

Orbitals are wavefunctions, they depend on various numbers leading from the inability to put more than two electrons in the same place at the same energy.
It's beyond your comprehension- you need a good understanding of the wavefunction and quantum mechanics.

>> No.2326461

Nodes, dude. Nodes.

>> No.2326497

>>2326416
I read it carefully enough to follow the solution for the radial component of the wavefunction. The author just states the solutions for the angular components though. (And fuck they are messy.)

The author also focuses more on the energy levels of hydrogen than on the shapes of the orbitals; there isn't a single plot of the orbitals but plenty of energy diagrams. (Quantum mechanics predicting hyperfine splitting gave me a small nerdgasm, shit's cool.)
>>2326439
The wavefunction is not really beyond my grasp, I know how to solve PDE (well, not as well as I'd like to). I also know where the quantum numbers come from. (Boundary conditions can only be met by some functions, eigenfunctions et cetera) I haven't taken any actual courses in quantum mechanics, but this is more of a mathematical problem, I think.

The problem involves a potential that is spherically symmetric, so I can't quite get it into my head mathematically that the solutions aren't spherically symmetric too. And if they aren't spherically symmetric, then which plane is the preferred? Look at the 4f orbital for example. How do you know which orientation that orbital has? It looks like it's rotationally symmetric around one axis, which led me to think of spin, since it's like angular momentum.

It's something with angular momentum isn't it?

>> No.2326501

>>2326497
>a potential that is spherically symmetric

That is an approximation son.

>> No.2326518

>>2326497
Angular momentum isn't the only part of an electron. If you think back to an s orbital, that is purely spherical.
But then you get p orbitals, which is an s orbital with a node, so contains two opposite-phase lobes.
D and f have weirder shapes because of increased node count, and the addition of several other quantum numbers which changes their shape.
My knowledge on these things is limited to their use in organic chemistry, I'm afraid.

>> No.2326541
File: 74 KB, 312x449, 80514710.jpg [View same] [iqdb] [saucenao] [google]
2326541

>>2326497
You say you did the math, and you can see the solutions. Yet you don't believe them? WTF?

The potential is only specially symmetrical (approx) with respect to charge. However you have all sorts of other degrees of freedom that effect the wavefucntions of the electrons.

All the wavefucntions have some sort of symmetery around the potential, this is due to the spherically symmetric potentail of the ponetial. That is what the symmetery of the potential contributes to the wavefucntion symmetry.

If you are really intrested in that shit, you need to know some Group Theory, and how to apply it to QM.

>> No.2326543

Ultimately, it's because of the exclusion principle.

Also, even though the orbitals aren't spherical, in the absence of an external field they don't have any orientation relative to the outside world. So the odds of finding an electron in any particular region relative to the nucleus depends only upon the distance from that region to the nucleus... which is what you'd expect. Only once you start putting in several electrons, and interacting with outside things, do you notice orbital shapes, and then it comes down to electron pair repulsion, which should be intuitive.

>> No.2326551

>>2326497
>It's something with angular momentum isn't it?

Angular mometum is directly in the calculations dumbshit! How the fuck can you say you looked at that shit, when you missed the angular mometum? WTF?

>> No.2326562

>>2326518
Maybe the picture is misleading and should be animated. I remembered Wikipedia comparing orbitals to standing waves on a drum (I once read an explanation of quantization of orbitals as requiring that the wavefunction "fit together" if you go an entire lap around the atom) so I looked at
http://commons.wikimedia.org/wiki/Category:Drum_vibration_animations
and thought maybe it's like that? The 3d and 4f orbitals would behave something like mode21, which has two orthogonal oscillations, the orbital itself oscillates.

I guess that makes me understand how electrons can have angular momentum, actually... Thanks!

Quantum mechanics sure is weird.
>>2326501
For a hydrogen atom it's a pretty damn good approximation. Actually, wouldn't Noether's theorem imply that if you want angular momentum to be conserved, the potential has to be spherically symmetric? Well there's also a correction for the proton moving slightly, which shifts and splits energy levels (which is super cool).

See, I didn't skim *that* much!

>> No.2326580
File: 125 KB, 567x1040, orbitals.png [View same] [iqdb] [saucenao] [google]
2326580

>>2326551
Not in the book I read. Pic related it's the book I read.

Like I said, the author doesn't talk about the shape of orbitals.

>>2326541
Well I didn't do the math myself, but I did look at the author's solution. And yes, I can see the solutions, but I don't disbelieve them. I just don't quite understand how solutions to what seems to me to be a spherically symmetric problem aren't symmetric too.

Then again there are those d<span class="math">\theta[/spoiler]s in the laplace operator...

I really can't wait until I get to take actual group theory courses and such...

>> No.2326599
File: 1.13 MB, 1200x1600, 1294553702581.jpg [View same] [iqdb] [saucenao] [google]
2326599

>>2326562
>Noether's theorem imply that if you want angular momentum to be conserved, the potential has to be spherically symmetric?

Nope, you are confusing shit. Noethers theorm implies that angular mometum is conserved around a gravitationally symmetric body. It has nothing to do with the symmetry of the charge.

>> No.2326614
File: 60 KB, 700x525, 1292529644904.jpg [View same] [iqdb] [saucenao] [google]
2326614

>>2326580
You are dealing with different types of symmetries corresponding to different properties.

You are confusing the shit out of them, assuming that differnt types of symmetries are related in a very simple way.

You assume a probablistic spatial symmetry should be the same as a symmetery of charge? Why?

>> No.2326641

>>2326599
Right.

In general, isotropy (physics look the same no matter which direction you point) leads to conserved angular momentum. In a similar sense, a potential with no angular dependence leads to a constant one-body angular momentum (not just total system momentum).

Independence from time means conserved energy, and independence from location means conserved momentum.

>> No.2326646

>>2326599
Noethers theorm implies that angular mometum is conserved around a gravitationally symmetric body.
In my understanding it's far more general than that.
>It has nothing to do with the symmetry of the charge.
If the electric field due to the charge somehow enters into the Lagrangian, wouldn't it?

>>2326614
Well in electromagnetism if there's a symmetry in the source of the electric field (the charge distribution), then the electric field inherits that symmetry. I guess what I'm thinking is that the potential plays the same role as the charge distribution; it sets the physical situation. On the other hand, a symmetry in current distribution doesn't generate the same symmetry in magnetic fields...

Also, why are you talking about symmetry of charge? How does that make sense in a hydrogen nucleus (a proton)?

My electromagnetism teacher stressed very much that if there is no preferred axis or plane in a problem, you would expect the solution to have a symmetry. If you have a charged sphere the electric field only depends on distance, if you have a current in a straight wire the magnetic field makes circles around it and so on. That's why I'm confused when I look at atomic orbitals, because the atom should be the same in every direction right?

The only preferred axis I can think of in an atom is angular momentum (the nucleus's spin, right?). The orbitals do look like they have a rotational symmetry around an axis, would that be the direction of the spin?

>> No.2326652

>>2326646
>That's why I'm confused when I look at atomic orbitals, because the atom should be the same in every direction right?
Not sure where you get that from. Clearly, atoms are NOT spherically symmetric, except for special cases.

For instance, bulk magnetism is impossible for spherically symmetric atoms.

>> No.2326653

>>2326652
Noble gases aren't a bad approximation, IIRC.

>> No.2326669

>>2326652
Well, what about hydrogen then?

Obviously bigger atoms are somewhat of a clusterfuck...

>> No.2326685

>>2326669
1s is spherically symmtric, and IIRC the proton is pretty good too.

The problem is instrinsic magnetic moments. Once you consider those, I don't think even atomic hydrogen is spherically symmetric.

http://en.wikipedia.org/wiki/Magnetic_moment#Two_kinds_of_magnetic_sources

>The net magnetic moment of any system is a vector sum of contributions from one or both types of sources. For example, the magnetic moment of an atom of hydrogen-1 (the lightest hydrogen isotope, consisting of a proton and an electron) is a vector sum of the following contributions: (1) the intrinsic moment of the electron, (2) the orbital motion of the electron around the proton, (3) the intrinsic moment of the proton. Similarly, the magnetic moment of a bar magnet is the sum of the intrinsic and orbital magnetic moments of the unpaired electrons of the magnet's material.

>> No.2326688

>>2326580
You're already on the right track. The majority of the posts in thread are utter crap.

1.) Even for a totally spherically symmetric situation of an V = V(r) portential, the wave functions are simply NOT spherically symmetric anymore

2.) Why?

3.) Because of the motherfucking Laplace operator (second derivative) in our Schrödinger equation. If you do the coordinate transformation to spherical coordinates correct, you see that the usual simple form of (dx^2+dy^2+dz^2) isn't recovered anymore.

4.) The angular part (both phi and theta, azimuthal and polar angles) of the Schrödinger equation for an H-like atom is solved using spherical harmonics, and if you strictly do the math, you'll eventually see that the spherical symmetry of the potential doesn't translate to a sph. symmetrical wave function.

The lack of total sph. symmetry here has nothing to do with Pauli exclusion principle, it's simply a nature of our equation.


Also: for non-freaky atoms, the symmetrical Coulomb potential is still a good approximation.

>> No.2326690
File: 9 KB, 251x165, 1292785461516.jpg [View same] [iqdb] [saucenao] [google]
2326690

>>2326646

>the electric field due to the charge somehow enters into the Lagrangian, wouldn't it?

Yes, But that shit is way over your head bro.
In any case, The symmetry known as Gauge invariance in the Lagragian will lead to the conservation of electric charge, by Nothers theorem.

>why are you talking about symmetry of charge?

Isn't that what you were talking about? Sperical symmetric potential? in charge?...DURRRR? It wouldnt make fuckin sense if you were talking about mass.

Dude you are fuckin all over the place. If you have a coherent question, ask. Try and clean up your train of thought, so others can understand.

>> No.2326699

>>2326641
>isotropy (physics look the same no matter which direction you point) leads to contrived angular momentum

fix-ed dat fer yer boss

>> No.2326709
File: 1018 KB, 1365x1242, Hydrogen Wave Functions.png [View same] [iqdb] [saucenao] [google]
2326709

>>2326685
>>2326669
>>2326652
Pic related?

>> No.2326719 [DELETED] 

>>2326709
Yes, but I don't think those functions include magnetic moments.

>> No.2326728

>>2326688
It must fall out of the equation somehow, yes...

But then there seems to be a preferred axis. Is that the spin axis?

Excellent example of a crap post here, by the way >>2326690

>> No.2326733

>>2326709
Yes, but I don't think those functions include intrinsic magnetic moments (spin).

>> No.2326737

>>2326699
WTF? TROLLIN?

>> No.2326777

name of textbook?

>> No.2326781

>>2326733
of course they don't .. in the simple H-like Hamiltonian, there is no term that would affect spin, hence you wouldn't see any difference ..

>> No.2326787

>>2326728
>But then there seems to be a preferred axis

WHAT? why? What is your reasoning?

>> No.2326804

>>2326787
Wouldn't that be the opposite of spherical symmetry? And the 3d and 4f orbitals look like they were rotated around an axis.

>> No.2326908
File: 242 KB, 1440x900, 1294572150666.jpg [View same] [iqdb] [saucenao] [google]
2326908

>>2326804
Ohh, ic your problem. You physically think that the axises in Quantum mech are like classical mech.....LOL.

The direction of the shit in basic QM is arbitrary in space, it all depends on where you chose your z axis to be. It is esentially the way you choose your spin-quantization.

The Heisenberg uncertainty principle tells us that it is not possible to measure position and mometum simultaneously with arbitrary precision, due to noncommutating operators. Therefore, there are limits to what can be known or measured about a particle's angular momentum. It turns out that the best that one can do is to simultaneously measure both the angular momentum vector's magnitude and its component along one axis. The one axis is usually denoted "z". Hence, we define spin as quantized intrinstic angular mometum units along "z".

"z" is a perffered axis we choose to help us define our measurments. It comes from our defintion of intrinsic spin. It doesn't have to correspond to a cartesian coordinate "z".

Make sense now?

>> No.2326922

>>2326908
Seconded. If you have to pick an axis to call "Z" (like in experimental observation of the Zeeman effect) it doesn't matter how that axis is oriented in space.

>> No.2326924
File: 26 KB, 619x352, 127629679242bb2.jpg [View same] [iqdb] [saucenao] [google]
2326924

>>2326908

>> No.2326972

Imagine finding this thread on /b/

>> No.2327020

... read the first 10 posts, they were utter crap, no motivation to read the rest to taylor an answer.
Funny how there's only a handful people on sci that know about orbitals even in a basic sense. (Also fuck that Noether misunderstanding crap all the time.)

>> No.2327132

>>2326908
Even though your answers are good this is supposed to be a work-safe board. Not hating on Kari Sweets or anything, it's just not ideal for us browsing from work.

>> No.2327140

>>2327132
That's not the reason there are worksafe boards you know.
Worksafe is a new addition to 4chan, believe it or not. They were created so that every thread on every board wouldn't become a porn thread- because really, NOTHING on 4chan is worksafe, and you're a fool if you think otherwise, nigger.

>> No.2327160
File: 49 KB, 425x408, karina_hart4.jpg [View same] [iqdb] [saucenao] [google]
2327160

>>2327132
WORKSAFE THIS! BITCHES!!!!
If you are at work browsing 4chan, you have much bigger problems then a couple of nuddie pics on your screen.

>> No.2327170
File: 56 KB, 610x403, 1294669580430.jpg [View same] [iqdb] [saucenao] [google]
2327170

>>2327020
No one could figureout what OP was talking about.

Finally solved the problem here
>>2326908

>> No.2327180

>>2327020
Kaza?