Atomic Orbitals

[alright1234]

How can a spherical coordinate system represent a non-spherical structure of the atomic orbitals? 

 

[GAHD]

I'm not sure what you're getting at here but I'll take a stab at what I think you're trying to get at.
The rotation in space of a point(or wave) is arbitrary, meaning absolute values of a traditional cubic system are silly. What matters is relative locations, particularly when such things cannot be known with precision. Spherical coordinates do not mean they describe a purely spherical structure.
Here's a quick primer on the system in question. I somehow think you're using English to try to understand math, and being confuse by the imprecise nature of the words in your personal vocabulary. Hopefully this helps.
https://cnx.org/contents/[email protected]:KTzGGLDe@4/Cylindrical-and-Spherical-Coordinates
A little bit of math and you can translate between systems easily, and math results tend to be the same regardless of what base you end up using. The only difference in this case is that absolutes are simplified to an origin point. Based on other interactions I think you're missing a couple of tools: Vector Calculus is probably the main one that will help you the most. From there you can move to tensors.

[exchemist]

It depends what models you are looking at, the following link explains Atomic Orbitals https://en.wikipedia.org/wiki/Atomic_orbital

The orbit an electron may take around a nucleus can be described in spherical coordinates according to the bohr model, but this does not work in real life. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Bohr%27s_Hydrogen_Atom ie it has limitations. 

 

The idea of atomic orbitals comes after the Bohr model, which is still taught.

Yes the Bohr orbital model does not work, as was immediately recognised at the time,  and is only taught today to children as a simplification.

 

But in fact the choice of coordinate system is simply a matter of convenience and is not a function of the choice of model. It is standard to express Schrödinger's equation for the atom in spherical polar coordinates rather than Cartesian (xyz) coordinates, because that elegantly gives one separate solutions for the radial part of the wave function (quantum number n) and the spherical harmonic parts (quantum numbers l and m). In chemistry the behaviour of the radial part and the spherical harmonic parts are both very important, but for quite different reasons.

 

(For any interested readers, the shapes of the non-spherical orbitals result from the shapes of the spherical harmonics. These look just the same as the shapes of the s, p, d , f etc orbital types: https://en.wikipedia.org/wiki/Spherical_harmonics ) 

[ralfcis]

No link to the spdf orbital types. When I took chemistry, the s were spherical but the p were dumbell shaped. Is that real and is that shape caused by spherical harmonics? The dumbells were always depicted as going right through the nucleus.

 

I watched the video. I didn't know there was still a debate about the wave function. I thought it contained all probabilities and it wasn't your measurement that caused the collapse of the wavefunction but it was your measuring device that could only detect one of the probabilities. So you do get that multifood item but if you can only taste hotdogs, that's all you'll be able to get out of the box. That multiworlds idea just turns my stomach. 

 

So it looks like you can go around the restrictions set by quantum physics. By taking enough still shots, you can combine them to see the motion shots. That means I am wrong about how the food box analogy works. You're not limited to only tasting hotdogs. What happens is when you open enough boxes, you will be able to see what percentage of burgers, pizza and hotdogs will pop out of the multifood item.

 

That means there's now a 3rd option of wavefunction collapse. The 1st was the act of measuring that influenced the collapse into the same result as what happens in the double slit experiment. The 2nd is the type of equipment or where you placed the equipment you used to measure only allowed you to see 1 result. The 3rd is now the result is independent of the act of measurement or the equipment used but how many measurements you take. That sure isn't in my quantum physics book (which is the greatest book I ever read (A quantum story by baggot)).

 

PS. So time is the critical missing piece of the puzzle. So measurement over time will eventually reveal all the probabilities. I thought time had been factored out of the wavefunction.

Edited March 24, 2019 by ralfcis

[exchemist]

No link to the spdf orbital types. When I took chemistry, the s were spherical but the p were dumbell shaped. Is that real and is that shape caused by spherical harmonics? The dumbells were always depicted as going right through the nucleus.

 

I watched the video. I didn't know there was still a debate about the wave function. I thought it contained all probabilities and it wasn't your measurement that caused the collapse of the wavefunction but it was your measuring device that could only detect one of the probabilities. So you do get that multifood item but if you can only taste hotdogs, that's all you'll be able to get out of the box. That multiworlds idea just turns my stomach. 

 

So it looks like you can go around the restrictions set by quantum physics. By taking enough still shots, you can combine them to see the motion shots. That means I am wrong about how the food box analogy works. You're not limited to only tasting hotdogs. What happens is when you open enough boxes, you will be able to see what percentage of burgers, pizza and hotdogs will pop out of the multifood item.

 

That means there's now a 3rd option of wavefunction collapse. The 1st was the act of measuring that influenced the collapse into the same result as what happens in the double slit experiment. The 2nd is the type of equipment or where you placed the equipment you used to measure only allowed you to see 1 result. The 3rd is now the result is independent of the act of measurement or the equipment used but how many measurements you take. That sure isn't in my quantum physics book (which is the greatest book I ever read (A quantum story by baggot)).

True Ralf, I didn't bother to link to spdf etc. Here's a link then: http://slideplayer.com/slide/8655418/26/images/18/Orbital+shapes+for+s,+p,+d,+&+f.jpg

 

Yes the p are dumbell-shaped, but they have a node, that is a point of zero amplitude, at the nucleus.

 

I'm going to stay out of all this wavefunction collapse mumbo-jumbo firstly because it is not important to the application of quantum theory to chemistry and secondly because there is all sorts of ghastly woo associated with it.  

[ralfcis]

So the dumbell shape comes from spherical harmonics? So if they used the quantum microscope technique on anything above helium, the pictures they would get would include these dumbells going through the s orbitals? That seems very weird to me.

[exchemist]

So the dumbell shape comes from spherical harmonics? So if they used the quantum microscope technique on anything above helium, the pictures they would get would include these dumbells going through the s orbitals? That seems very weird to me.

Well I don't know if the resolution would enable the shapes of individual orbitals to appear. In atoms with several electrons you will have some in s and some in p. And don't forget too that the p, d etc orbital orientation is not fixed in space unless there is something to align them.

 

But I think there have been pictures of molecules showing π orbitals, i.e. bonding with 2 phases, one above and one below the line of contact of the atoms and a node along the line of contact itself.   

[ralfcis]

Is this the link?   https://en.wikipedia.org/wiki/Schrödinger_equation

Edited March 25, 2019 by ralfcis

[GAHD]

Well I don't know if the resolution would enable the shapes of individual orbitals to appear...

AFAIK there's nothing with that level of resolution, maybe in the next decade if IBM keeps it's nanotech division funded.

There ARE images of complex bonding paths with AFM, eg https://www.nature.com/articles/ncomms8766 or https://www.wired.com/story/new-microscope-shows-the-quantum-world-in-crazy-detail/ ...but individual orbitals are still outside of resolution.

QFM (AKA photoionization microscopy) doesn't seem to have much in the way of published slides. There's hydrogen, helium, lithium, but they don't exactly seem to offer superior insight just from the imaging itself. I'm also not really finding anything direct for any of the first P group, but that might be because I don't have access to AIP https://aip.scitation.org/doi/figure/10.1063/1.2354478

 

 

[tetrahedron]

The electrons in an s orbital don't actually 'orbit' per se. There is no angular momentum. Instead they merely dive down through the nucleus and out the other side along a radial path. The path wanders angularly, so in the end what we have is a densely populated spherical system of such straight pathways along radii.

 

Jess Tauber

[alright1234]

Dear Honorable and beloved Friends and Scholars 

 

It is such an intense pleasure in conversing with such high minded and gifted people of both gender and those in-between. I believe though I may be incorrect that the coordinate system has precedence before the structure that is being depicted. Using an analogy, if you have a pair of pants you would not normally use a pair of pants as a shirt. You could put your arms though the legging but the rest of the pant's structure would not cover your stomach that is indicative of the purpose of the structure of a shirt in normal circumstances. So in conclusion, if you are using a spherical coordinates system the structure depict would inherently be a sphere since the structure of a sphere corresponds with the exact variables orientation of a spherical coordinate system. Back to the analogy, putting pants on your lower body through the legs zipper in the forward position would correspond with a sphere represented with a spherical coordinate system since the segments of the spherical coordinate system are also segments of a sphere. I adamantly wait for you responses and am overwhelmed with joy with the attention that I am getting. Thank you. You the great people of this forum are make life worth living and the living worthy of a life.

 

 

https://en.wikipedia.org/wiki/Spherical_coordinate_system

 

 

Sincerely your

Edited March 25, 2019 by alright1234

[Vmedvil2]

Spherical coordinates in Orbitals are a set of Quantum numbers that depict the orbital of the electrons, it is higher math made easy with the S,P,D,F, orbital system.

[exchemist]

Spherical coordinates are nothing to do with quantum numbers. They simply replace the 3 distance axes x,y, z by a radius and two angles, r, θ, φ, in order to locate points in space.

 

The algebra of Schrödinger's equation will obviously look different expressed in terms of  r, θ, φ from the way it looks in terms of x, y z, but for a system with a spherically symmetrical confining potential, as is the case with an electron confined by attraction towards a +ve atomic nucleus,  r, θ, φ is a better choice of coordinates for solving it. 

 

It is the solutions to this equation that lead to the three quantum numbers: n for the radial part ( r ) and l and m for the two angular parts (θ, φ).  l denotes the number of units of orbital angular momentum possessed by the electron (zero in the case of s orbitals, 1 for p, 2 for d, 3 for f, etc) and m denotes the orientation in space of these relative to one another, e.g. there are 3 p orbitals at right angles to one another.  (It is usual now to identify m as "m(l)", to distinguish it from "m(s)", which is used for the "spin" of the electron - something that does not come out of Schrödinger's equation.) 

Edited March 26, 2019 by exchemist

[alright1234]

The Bohr-de Broglie atomic matter wave is transformed into an electron particle-in-a-box wave that is transformed into a probability wave then represented in a spherical coordinate system to create the structural equations of the atomic orbitals but a spherical coordinate system is being used to derive the structure of an atom using the box normalized probability wave which is physically invalid since a coordinate system is a representation of a structure and cannot be used to construct a structure of an atom or molecule that is depict in the QM atomic orbital equations.

 

 

LOVE

 

Alright1234

[alright1234]

"The algebra of Schrödinger's equation will obviously look different expressed in terms of  r, θ, φ from the way it looks in terms of x, y z, but for a system with a spherically symmetrical confining potential, as is the case with an electron confined by attraction towards a +ve atomic nucleus,  r, θ, φ is a better choice of coordinates for solving it."

 

 

EXCHEM   none of this explain why a spherical coordinate system is used to represent non spherical structures of the atomic orbitals.

[alright1234]

I watched the video. I didn't know there was still a debate about the wave function. I thought it contained all probabilities and it wasn't your measurement that caused the collapse of the wavefunction but it was your measuring device that could only detect one of the probabilities. So you do get that multifood item but if you can only taste hotdogs, that's all you'll be able to get out of the box. That multiworlds idea just turns my stomach. 

 

So it looks like you can go around the restrictions set by quantum physics. By taking enough still shots, you can combine them to see the motion shots. That means I am wrong about how the food box analogy works. You're not limited to only tasting hotdogs. What happens is when you open enough boxes, you will be able to see what percentage of burgers, pizza and hotdogs will pop out of the multifood item.

 

That means there's now a 3rd option of wavefunction collapse. The 1st was the act of measuring that influenced the collapse into the same result as what happens in the double slit experiment. The 2nd is the type of equipment or where you placed the equipment you used to measure only allowed you to see 1 result. The 3rd is now the result is independent of the act of measurement or the equipment used but how many measurements you take. That sure isn't in my quantum physics book (which is the greatest book I ever read (A quantum story by baggot)).

 

___________________________________________________

 

Dear Sir Ralfcis None of this explain why a spherical coordinate system is used to represent non spherical structures of the atomic orbitals.

 

 

Love Alright1234

[Vmedvil2]

Spherical coordinates are nothing to do with quantum numbers. They simply replace the 3 distance axes x,y, z by a radius and two angles, r, θ, φ, in order to locate points in space.

 

The algebra of Schrödinger's equation will obviously look different expressed in terms of  r, θ, φ from the way it looks in terms of x, y z, but for a system with a spherically symmetrical confining potential, as is the case with an electron confined by attraction towards a +ve atomic nucleus,  r, θ, φ is a better choice of coordinates for solving it. 

 

It is the solutions to this equation that lead to the three quantum numbers: n for the radial part ( r ) and l and m for the two angular parts (θ, φ).  l denotes the number of units of orbital angular momentum possessed by the electron (zero in the case of s orbitals, 1 for p, 2 for d, 3 for f, etc) and m denotes the orientation in space of these relative to one another, e.g. there are 3 p orbitals at right angles to one another.  (It is usual now to identify m as "m(l)", to distinguish it from "m(s)", which is used for the "spin" of the electron - something that does not come out of Schrödinger's equation.) 

 Which those are called Quantum Numbers when you do that. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

Edited March 26, 2019 by VictorMedvil