Determining Orbital Shapes from Schrodinger's Equation

In summary: Solving for the wave function involves Legendre polynomials or something like that. And from that you can get a function who's...Solving for the wave function involves Legendre polynomials or something like that. And from that you can get a function who's...Solving for the wave function involves Legendre polynomials or something like that. And from that you can get a function who's...The wave function of a particle in an n-dimensional space.
  • #1
Bladibla
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  • #2
You can not deduce the shape of the orbital from just the radial part of the wave function. You also need the angular parts. Then for a fixed value of [itex]|\psi| ^2 [/itex], you can plot [itex] (r, \theta, \phi) [/itex] that give you that value, and this 3D plot will give you the shape of the orbitals.

PS : The question is certainly not "dumb" and this is often misunderstood by a lot of people.

If I find the time, I'll see if I can show how a 2p orbital looks like a dumbell...or if you'd like me to do this, say so.
 
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  • #3
If I find the time, I'll see if I can show how a 2p orbital looks like a dumbell...or if you'd like me to do this, say so.
You can do that without much math.Wow.Do it please.
 
  • #4
actually, i think it DOES require a lot of maths. But he is (i think) clever than most of us and can do it..

It would really appreciative if you could show us, although if you don't want to, you don't have to.
 
  • #5
The trick is in being able to draw polar plots roughly, from the spherical harmonics.

Consider the state, n=2, l=1, m=0
The polar part of the wavefunction is given by :
[tex] |Y_1^0|^2 = \frac{3}{4 \pi} cos^2 \theta [/tex]

Ignoring the constant term, the functional behavior is [itex] cos ^2 \theta [/itex], which we want to plot against [itex]\theta [/itex]

Let's draw the 2D version of this plot (or you can simply graph this using Mathematica or a calculator that can do polar plots). Draw the X (theta = 90, -90) and Z (theta = 0, 180) axes, and draw a large number of lines, all passing through the origin - like the spokes on a bicycle wheel. These lines represent the different values of theta. Now on each line, place a point at a distance (from the origin) given by [itex]cos^2 \theta [/itex]. Finally, join all these points, neighbor to neighbor. Remember, the X-axis represents [itex] \theta = 90 [/itex] and the Z axis represents [itex] \theta = 0 [/itex]

At [itex]\theta =90,~ cos^2 \theta = 0[/itex]. So, the points along the horizontal spokes are at the origin. At [itex] \theta [/itex] increases, or decreases, [itex] cos ^2\theta[/itex] increases till it reaches maxima at [itex] \theta = 0, 180 [/itex]. So, for spokes above and below the X-axis, the points move farther and farther out, reaching a maximum at the +/-Z-axis. Join these points and you'll find it looks like a vertical dumbell oriented along the Z-axis. So, clearly this is the [itex]2p_z[/itex] orbital. While this is not all of it (you must now combine the radial part with the polar part), and the best way involves using some 3D plot software, I can't really do that here.

For the s-orbitals, the total wavefunction is something like [itex] |\psi _n(r, \theta \phi)|^2 ~=~A r^{2n}e^{-2r/na_0} [/itex]. Since these orbitals have no polar ([itex]\theta[/itex]) or azimuthal ([itex]\phi[/itex]) dependence, they are spherically symmetric, and all the "equipotential" surfaces are spheres.

If this was too confusing, I'll try and attach a picture, when I find a little more time.

PS : Not more clever...maybe a little more experienced, that's all ! :biggrin:
 
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  • #6
Gokul43201 said:
The trick is in being able to draw polar plots roughly, from the spherical harmonics.

Consider the state, n=2, l=1, m=0
The polar part of the wavefunction is given by :
[tex] |Y_1^0|^2 = \frac{3}{4 \pi} cos^2 \theta [/tex]
Can you explain how you got that .
Thanks.
 
  • #7
poolwin2001 said:
Can you explain how you got that .
Thanks.

Looking for a link to spherical harmonics, I found this excellent page, which not only gives you the general formula, it also provides a list of values of [itex]Y_l^m[/itex] for different (l,m), but especially nice, is that it also has beautiful 3D plots of the orbitals.

But anyway, you'll find that [itex]Y_1^0 = \frac{1}{2} \sqrt {\frac{3}{\pi}} ~cos \theta [/itex].

If you want to know why the polar part of the 2p orbital is given by this equation, you'll have to go through the entire process of learning how to solve the Schrodinger equation in different geometries.
 
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  • #8
Gokul43201 said:
Looking for a link to spherical harmonics, I found this
Thanks
Gokul43201 said:
If you want to know why the polar part of the 2p orbital is given by this equation, you'll have to go through the entire process of learning how to solve the Schrodinger equation in different geometries.
Something to do this summer .I hope! :biggrin:
 
  • #9
Doesn't solving for the wave function involve legendre polynomials or something like that? And from that you can get a function who's plot will give the shape of the oribitals?
 
  • #10
Yes, that's roughly right.

The polar part of the solution to the Time Independent SE in spherical co-ordinates is indeed something very much like the Legendre Polynomials. To be precise, they are the Associated Legendre Polynomials.

The function I used above for the p_z orbital is just the Associated Legendre Polynomial, [itex]P_1^0(cos \theta) [/itex].
 
  • #11
Thanks for the detailed explanation.

I was always curious how scientists got the answer for that..And i always thought our chem teachers were being too dogmatic about this kinda stuff.
But now i understabd why (pretty complicated stuff!)

At roughly what age did you learn this stuff?
 
  • #12
Your chem teachers were dogmatic about this stuff because they had no idea about it themselves. You learn this in Physics, not chemistry.

Only theoretical/atomic chemists will know anything about Quantum Mechanics and solutions of the commonly occurring partial differential equations.

I learned this stuff pretty late - my first year of Grad School - because my major in College was Engineering. But I think most physics majors will learn this stuff towards the end of college.
 
  • #13
I see.

I personally don't do Physics, but am very, very curious about quantum mechanics. Do you reckon its possible to learn QM without doing physics (in high school, this is)

My presumption would be that we don't learn about quantum mechanics in high school anyway. But in uni..
 
  • #14
You might be able to learn the basics, but without a knowledge of physics, it would have no ...aroma.

It would be like learning to play music simply by memorizing a sequence of keys - that would be meaningless.
 
  • #15
Gokul43201 said:
Only theoretical/atomic chemists will know anything about Quantum Mechanics and solutions of the commonly occurring partial differential equations.

Partly right,my friend.I won't go and quote Feynman saying no one will ever understand QM,but i just want to say I've never met (and honeslty never will) a chemist,a "theorist" kind (:tongue2:) with whom to discuss mathematical and axiomatic structure of Copenhagen/textbook version of QM.They probably know a bit about the works of Schroedinger,Heisenberg,Born,Dirac,Bethe & Salpeter (the famous book),Ritz,Legendre,Heitler,London,Hueckel and Pauling and names like von Neumann,Weyl,Stone,Borel are totally unfamiliar or probably unknown.

Generalizations are always dangerous... :wink:

Gokul43201 said:
Your chem teachers were dogmatic about this stuff because they had no idea about it themselves. You learn this in Physics, not chemistry.

It's true for most of the physics teachers as well.


Daniel.

PS.PROBABLY (it's QM,right?? :wink:) none of chemistry "theorists" thinks about Whittaker functions and Gauss confluent hypergeometric functions when seeing the PDE of the (the most simple model) of Hydrogen atom.
 
  • #16
Bladibla said:
I see.
I personally don't do Physics, but am very, very curious about quantum mechanics. Do you reckon its possible to learn QM without doing physics (in high school, this is)
My presumption would be that we don't learn about quantum mechanics in high school anyway. But in uni..

You don't do physics in HS??Too bad.The simplicity of the exposure and a talented proffessor would have made you love this science of nature.
Decent understanding of QM would require strong knowledge of both mathematics and physics.As a theoretical physicist,i would emphasize the role of mathematics.Without proper knowledge of advanced mathematics,it's impossible to understand the axiomatical structure,i.e.the 6 (my count) principles.

In school,HS that is,the results of quantum mechanins regarding atomic structure are presented and students are required to them for granted.There are some universities in which the course is being taught sloppy and u'll have to search through (maybe old and dusty) books for the answers to your questions.

Daniel.

PS.Like many chapters of physics,quantum mechanics is just about combining advanced mathematical notions,an axiomatic structure and a lot of imagination and patience...
 
  • #17
Well the deal goes like this.
I currently do Biology , chemistry, Maths and DT(AS level). I do them all for a year, then i can drop 1 subject.(For the other 3 subjects to carry on to A level)

But i CAN drop 1 and then just do AS physics. This probably sounds gobblygoop for you people who don't know about the system here in UK, but i think it just might make me do physics..

I am thinking of doing physical chemistry, as it does both of the science branches. But as i said, no physics = :(
 
  • #18
Bladibla said:
Well the deal goes like this.
I currently do Biology , chemistry, Maths and DT(AS level). I do them all for a year, then i can drop 1 subject.(For the other 3 subjects to carry on to A level)
But i CAN drop 1 and then just do AS physics. This probably sounds gobblygoop for you people who don't know about the system here in UK, but i think it just might make me do physics.
I am thinking of doing physical chemistry, as it does both of the science branches. But as i said, no physics = :(

I'm only 200Km away from Britain and i find what u're telling me to be very curious.I mean the "dropping" science disciplines,it's really weird and i should say illogical at the same time... :rolleyes: Anyway,i assume it's only up to to take the WISE decision.U have to chose among preferences,maybe,and,if u've chosen "physical chemistry",then that should be it...I wish u only luck...

Take as it a good thing:physical chemistry involves an "itsy-bitsy" amount of Quantum Mechanics,at least in the latters's results in atomic and molecular structure.

Daniel.
 
  • #19
dextercioby said:
I'm only 200Km away from Britain and i find what u're telling me to be very curious.I mean the "dropping" science disciplines,it's really weird and i should say illogical at the same time... :rolleyes: Anyway,i assume it's only up to to take the WISE decision.U have to chose among preferences,maybe,and,if u've chosen "physical chemistry",then that should be it...I wish u only luck...

Take as it a good thing:physical chemistry involves an "itsy-bitsy" amount of Quantum Mechanics,at least in the latters's results in atomic and molecular structure.

Daniel.

Well, i was thinking of dropping Design technology.. :blushing:
 
  • #20
Honestly, I have never taken a physics course so far in my university career and have taken a hardcore class on quantum chemistry (Physical Chemistry II) and did extremely well (although I am a math major too, which helped a lot). P-Chem is extremely instense if you don't have a very decent math background. P-chem is the bridge between chemistry and physics. I wouldn't write P-chem off as having an itsy bitsy amount of QM, I think it really depends on how it is taught.
 
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  • #21
gravenewworld said:
Honestly, I have never taken a physics course so far in my university career and have taken a hardcore class on quantum chemistry (Physical Chemistry II) and did extremely well (although I am a math major too, which helped a lot). P-Chem is extremely instense if you don't have a very decent math background. P-chem is the bridge between chemistry and physics. I wouldn't write P-chem off as having an itsy bitsy amount of QM, I think it really depends on how it is taught.

Thanks for the word of experience of physical chemistry. I am wouldn't say I am crap at maths, but not the work genius either. But i am willing to work hard..
 

1. What is Schrodinger's equation and how does it relate to orbital shapes?

Schrodinger's equation is a mathematical equation that describes the behavior of quantum particles, such as electrons, in a given system. It relates to orbital shapes because it provides a way to calculate the probability of finding an electron in a specific location around a nucleus, which determines the shape of the electron's orbital.

2. How do you determine orbital shapes from Schrodinger's equation?

To determine orbital shapes from Schrodinger's equation, one must solve the equation for a specific system, such as a hydrogen atom. This involves using mathematical techniques, such as separation of variables, to obtain a set of solutions, or wavefunctions, which describe the possible states of the electron in that system. The square of the wavefunction gives the probability density of finding the electron in a particular location, and the shape of this probability distribution corresponds to the shape of the orbital.

3. Can Schrodinger's equation be used to determine the shapes of all orbitals?

Yes, Schrodinger's equation can be used to determine the shapes of all orbitals, including s, p, d, and f orbitals. However, the complexity of the equation increases with the number of electrons and the energy level of the orbital, making it more difficult to solve for higher energy orbitals.

4. How does the shape of an orbital affect its energy?

The shape of an orbital affects its energy by determining the distance of the electron from the nucleus and the amount of electron-electron repulsion. For example, s orbitals have a spherical shape and the electrons are closer to the nucleus, resulting in lower energy levels. P orbitals, on the other hand, have a dumbbell shape and the electrons are further from the nucleus, resulting in higher energy levels.

5. Are there any limitations to using Schrodinger's equation to determine orbital shapes?

Yes, there are limitations to using Schrodinger's equation to determine orbital shapes. The equation assumes a stationary system, which does not account for the movement of electrons. It also does not take into account the effects of relativity or electron-electron interactions, which can affect the accuracy of the results. Additionally, the equation only applies to non-relativistic particles, so it cannot be used to determine the shapes of orbitals for particles traveling at high speeds.

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