What an oxymoron. You know, how “quantum leap” in colloquial speech means “lots of progress” but quantum actually means small. Heh.
For those of you still eagerly awaiting for the post with a word count perhaps exceeding the total verbiage of the rest of my blog combined, this is unfortunately not that post. Notably, it is a more quantum post, both in size and by content. The longpost does not have significant quantum mechanics material.
Also, post restriction broken.
Disclaimer: I don’t claim to actually have formal knowledge of quantum mechanics. I won’t even claim to have any knowledge of quantum mechanics! So please be aware of any ramifications in using this potentially inaccurate information in serious applications, such as studying for a test, writing a research paper, or other life-changing events. Notably, much of this stuff was derived from a combination of websites and personal intuition.
Disclaimer 2: A lot of this stuff is review if you’ve already taken some sort of chemistry class. I’m sorry for giving you potentially inaccurate or redundant information. In addition, a lot of this stuff may be more focused on competitive-math-oriented subjects, so I’m going to repeat that information too. So that might be redundant as well.
It was a cool and rainy (if my memory holds on the weather that day) afternoon, about one year ago. In Honors Chemistry, we had been learning about orbitals. Now, the most important part about electron orbitals is not the naming. In fact, that probably holds for 90% of everything else, in that naming is generally not terribly important.
However, naming can be interesting, despite its lack of importance. In this case, orbital naming. No, not the s-orbital, p-orbital, d-orbital, f-orbital naming (or else this blog post would be: sharp, principal, diffuse, fundamental. Ok this blogpost is done.), but the naming of each individual orbital.
We start with the s-orbital. I’m actually unsure what it’s named. Probably the 1-orbital. Since the names tend to be subscripts below the s/p/d/f designation, I think the only shape – that is, a sphere – would merely be an s orbital, sans subscript. Well, that was boring.
We could enlarge our orbital shell. The nth orbital shell can hold n supertypes (supertype being each individual letter/block) of orbitals. Yes, I’m being really nonrigorous with my naming. And that’s because naming is not important.
We call the s-orbital in shell number #2 the s2 orbital, the s-orbital in shell number #7 the s7 orbital, and so on. It turns out these orbitals are just larger spheres, with a bunch of inner spheres inside them. In a similar vein other larger orbitals are merely simple expansions, so I won’t delve too deeply into that subject.
The next one, the p-orbitals, is pretty straightforward as well. There are three types, the x, y and z. They are shaped like two-lobed thingies. I will now proceed to add a picture to this blog. This is unprecedented. Wow!
In other words, the p_x points in the x direction, the p_y points in the y direction, and the p_z points in the z direction. So far, pretty intuitive.
It’s interesting too, as all the p orbitals fit into a nice little compartament, along with the (not shown) s2 orbital.
Next, with some theory-crafting, although I’m guessing it’s well-explained already. Electrons, as you probably know, have several quantum numbers:
The principal quantum number, n, dictates the shell number.
The orbital quantum number, l, dictates what type of orbital it is. You know, the s, p, d, f, g, h, i, j, k, l, m, n, o, etc. thing.
The magnetic quantum number, m_l, dictates what shape of orbital it is. For instance, p -> p_x, p_z, p_y.
The spin quantum number, m_s, dictates which spin the electron is. As electrons are fermions, they have half-integer spins, i.e. either +1/2 or -1/2.
They are constrained by the following:
Notably, this gives us 2l-1 total orbital types per orbital. Moreover, this also produces the fact that there are 2n^2 electrons maximum in each electron shell. This also furthermore explains why the periodic table has 2 elements, 8, 8, 18, 18, 32, 32, … elements per row, in conjunction with the Aufbau principle. I always thought it was “Aufbau’s principle” but “Aufbau” means “filling-up” in German and therefore cannot really be in possession of a principle.
When l=0, we get the p_z orbital. When l=1, we get the p_x orbital. Or the p_y orbital. I’m not sure. The other one would be l=-1.
At this point, you ask yourself: why is p_z in the middle? Let’s keep going on.
Come on, it’s not that complicated… yet.
We have five orbitals: d_xy, d_xz, d_yz all look like they have four lobes.
But what’s this? There’s ANOTHER d-orbital with four lobes? Yeah, d_(x^2-y^2). Are you even allowed to put such an expression in the subscript? Oh well.
Then there’s this fifth d-orbital that doesn’t have four lobes. Instead it has three balls stacked on top of each other, d_(z^2). Yay complications.
First, can you see any patterns? Unless you’re ridiculously good at inference, probably not. It took me up to the f-orbitals to figure out some meaningful trend. Why is this? Technically, I cheated here. The groups I picked are rather misleadingly grouped. As hinted in the earlier section, something is special with the z-direction. It becomes even more painfully obvious with the d_(z^2) somewhat sticking out. A few trends that you may notice, on the other hand: all subscripts are of degree l, so they’re all quadratic terms here.
Let’s reorganize the d-orbitals by the z-exponent.
I guess it might be somewhat a stretch to say that these clouds easily fit, along with s and p orbitals it would share. Although, I suppose the picture itself definitely does not fit the margins set by my blog. Darn.
So I’ll give instructions, by request, on how to view these oversized images. Open a new tab or window. Right-click the picture and copy the URL, and then paste it onto the tab. Your browser may have more functionality like “open image in new tab” when you right-click, so there may be more direct alternatives.
Yeah, d_(z^2) looks kinda like a normal p_z orbital with a donut in the middle. How confusing. Also, d_xy isn’t aligned to axes unlike p_x or p_y (while d_(x^2-y^2) is!) What gives?
At this point, the “Modern Chemistry” book stops here, leaving most students content and happy that they don’t have to learn more in chemistry. But this is rather unsatisfying. Exactly what causes this sort of behavior? What would the f-orbitals look like, and for that matter, the g-orbitals and beyond? How are they named, because they sure don’t look too consistent to me.
Fun news! For the f-orbitals, two of them are named f_(x^3-3xy^2) and f_(y^3-3yx^2)! This obviously suggests some sort of pairing mechanism, and indeed this is the basis of the m_l being positive or negative. The other orbitals are named:
f_(zx^2-zy^2), f_(xyz), f(xz^2), f(yz^2), and f(z^3). The homogeneity condition still holds. The z-exponent thing also holds. (2 per z-exponent except when there’s only z^l.) Let’s try dividing out all the z’s and see what we get:
p: “1”, x, y
d: 1, x, y, xy, x^2-y^2
f: 1,x,y,xy,x^2-y^2,x^3-3xy^2, y^3-3yx^2
The pattern is pretty clear now, no?
Meanwhile, I’ll also try to describe the f-orbitals:
The two new f-orbitals have six lobes each. The ones with z^1 exponent have 8 lobes each, the ones with z^2 exponent have 6 lobes each, and the ones (or rather, I should say, one) with z^3 exponent has 4 lobes, all stacked on top of each other.
Since tabling seems to be a useful idea, we perform this again:
p: 2 / 2
d: 3 / 4 / 4
f: 4 / 6 / 8 / 6
We know that those z^n orbitals all seem to have balls stacked on top of each other; it seems rather logical that the next one will have n+1 lobes, preserving the pattern.
But if we investigate the xz^(n-1) orbitals, for instance, we see that ALL of these double-lobes are stacked together. We can see the same thing start to form in the xyz^(n-2) and (x^2-y^2)z^(n-2) orbitals as well, with quad-lobes stacked instead.
And all of this stacking is iterated every time we multiply by z. With induction, we can probably name every single orbital…
Except for the orbitals with no z exponent to begin with.
Let’s take a look:
They all have one layer, unsurprisingly (given that their z-exponent is zero). Starting from the p, d, f, orbitals, we see that the number of lobes increases by 2. They are equally spaced; anyone with decent math contest exposure should probably think about roots of unity at this point. [Root of unity: any complex number x satisfying the equation x^n=1 is known as a nth root of unity] In fact, out of curiosity, let’s factor the d and f orbitals:
x^2-y^2 = (x-y)(x+y)
When are these equations zero? For the former, (x-y)(x+y)=0 => x=y or x=-y. Look at the gaps between the lobes. Sure enough, these gaps exactly correspond with the zeroes! So the lobes can be seen as when the equation of the orbital is not zero. In fact, if we use the definition that these electron clouds are merely probability distributions (it could be possible that an errant electron is somewhere else; it’s just very unlikely), then we can see that the regions where the equation has a high magnitude exactly correspond with the regions where the electrons are “likely” to be found. I would also guess that this is how two electrons can share the same orbital, with such parity.
This makes naming future orbitals rather simple. We’re looking at every other root of unity of the 2(m_l) roots of unity. You may have noticed, especially with the f-orbitals, that the coefficients somewhat follow the binomial theorem. And this is no coincidence; if you are somewhat familiar with math contests, root of unity filters can easily kill certain problems. For example, this one:
What is nC0+nC2+nC4+…+nCn? (n is even)
If n were odd, that would be strange as 0,2,4… tends to not have any odd numbers in this sequence. It would also be easier, assuming that I typoed nCn as “nC(n-1)”, where you could easily invoke a symmetry argument.
However, there is another way.
Recall that the Binomial Theorem states that (x+y)^n = nC0 * x^n + nC1 * x^(n-1)y + … nCn * y^n. By plugging in (1+1)^n, we get the well-known fact that nC0+nC1+nC2+…+nCn = 2^n.
However, how would we extract out every other term? We can experiment and try (1-1)^n. Ok, that’s pretty obviously 0, but what exactly is it by the Binomial Theorem? It is equal to nC0-nC1+nC2-…+nCn. Still not quite there yet, but if we add the two up we get
2(nC0+nC2+…+nCn) = 2^n so our desired result is 2^(n-1).
It is a more complicated path to compute nC0+nC3+nC6+…+nCn (n is divisible by 3). Sure, you could do so combinatorially (really!) but it’s not much fun. Roots-of-unity filters are better in this case. If you don’t do math contests regularly, try this exercise out first. It’s admittedly quite hard if you haven’t seen this before.
Darn this isn’t a forum.
Ok, so let the two roots of x^2+x+1=0 be w0 and w1. (They should be omega’s but oh well.) One property that you might notice is that w0^2=w1 and viceversa (w1^2=w0). You might think that property, that two numbers are squares of each other, is rather weird (or you may point out 0 and 1 being squares of themselves, but that’s slightly different), but it’s easily seen:
w0=w1^2=w0^4 => w0^4-w0 = w0*(w0-1)(w0^2+w0+1). So if w0^2+w0+1=0, it satisfies that property. Yes, w0 and w1 are complex numbers, easily seen with the discriminant of x^2+x+1 being negative.
The insight: Consider (1+w0)^n and (1+w1)^n, along with (1+1)^n. By the Binomial Theorem, only the terms divisible by 3 align with each other (giving you 1+1+1), while everything else cancels out, as x^2+x+1=0.
You get something like (2^n+(1+w0)^n+(1+w1)^n)/3 as your final answer. In fact, the fact that n is divisible by 3 was not even necessary for this problem. You could clean up 1+w0 and 1+w1 in this case (they turn out to become sixth roots of unity) but I’ll leave the algebra for later.
That was somewhat of a tangent, sorry.
At any rate, what do the formulas remind you of? We have x^2-y^2, and xy. We have x^3-3xy^2 and 3x^2y-y^3. (If you think I’m scaling up and down arbitrarily, you’d be right, because as it turns out constant multiples don’t really affect anything.) Notably, they appear to “look” like expansions of (x+y)^n. However, there’s a catch. Both (x+y)^n and (x-y)^n have positive coefficients of x^2 and y^2 for instance. We wouldn’t expect x^3 and 3xy^2 to alternate signs either.
Insight: Expand out (x+iy)^n. The real part spits out one part of the orbital, and the imaginary part spits out the other. I guess motivation for this could be wanting to get opposite signs for x^2 and y^2 😛
However, it doesn’t answer a few important questions:
Question 1: How does this factor out into alternating roots of unity?
First, if you didn’t know, the n roots of unity have the form cis(2pi*k/n), because x^n=1 =>(r*cis(th))^n=r^n*cis(th*n)=cis(0)
cis(th*n)=cis(0) only when th*n=2k*pi => th = 2k*pi/n.
[cis notation: cos theta + i sin theta, i.e. polarizing rectangular complex numbers. Polar coordinates are easy to multiply as it’s just multiplying magnitudes and adding angles. Thus, [r*cis(t1)]*[s*cis(t2)]=rs*cis(t1+t2).]
So, this is equivalent to either Re((x+iy)^n)=0 or Im((x+iy)^n)=0. It may be more useful to convert x+iy into r*cis notation here (r^n doesn’t really affect things to be honest).
Re(r^n*cis(n*theta))=0 => cos(n*theta)=0 => n*theta = pi/2, 3pi/2, 5pi/2, …
In other words, theta should equal to (2k-1)*pi/2n, i.e. the odd 2nth roots of unity. Similar logic for Im, just use sin instead. Therefore, (x+iy) must have an angle measure equal to the 2nth roots of unity, so it must be a scalar multiple of a root of unity. Hence the gaps are where they are supposed to be. Yay.
Question 2: But I want a nicer formula!
Well, that’s not a question.
But sure. Do note that the polynomial looks like normal binomial expansion, except with every other term…
Wait this sounds like that root of unity filter I talked about earlier! It might not be so tangential of a tangent after all!
But wait. It’s alternating signs! Let’s root of unity filter again, with different filters.
(1+i)^n => 1 i -1 -i 1…
(1-i)^n => 1 -i -1 i 1…
This seems to work; the i’s cancel out, and every other term alternate signs as we want.
So, the proper polynomial would be ((x+iy)^n+(x-iy)^n)/2.
Now, this explanation is only satisfactory to a certain extent. For instance, it still doesn’t explain why the d_(z^2) orbital has that strange donut in the middle. Well, now is the time.
Actually, pop quiz because I feel like giving a pop quiz. And “since I’m such a nice person”* (*cue Bellotti) this pop quiz is entirely optional!
Here are some g-orbitals. Please name them, and give the magnetic quantum numbers of each. Have a nice day.
(Note: the weird gray sphere is the atom itself.)
a: z^4, m_l=0
b: z(x^3-3xy^2) OR z(y^3-3yx^2), m_l=+/-3
c: z^2(x^2-y^2) OR xyz^2, m_l = +/-2
d: x^4-6x^2y^2+y^4 OR xy(x^2-y^2), m_l = +/-4]
It was a fine day, knowing what was the approximate shape of the orbitals and knowing every single name of every orbital! But then I stumbled upon this site http://www.uky.edu/~holler/html/equations.html and it was once again a state of agony. (As it turns out, the functions on the right are the wavefunctions themselves. I suppose armed with that information one could accurately construct models for all of the orbitals. However, the associated text to these equations, http://archive.org/stream/introductiontoqu031712mbp#page/n141/mode/2up, is supposedly an “Introduction to Quantum Mechanics”. It was also written in 1935, perhaps a decade after the leading scientists formulated quantum mechanics itself :P)
I have no idea how that’s an introduction. Either they had really smart people back then where even the average layperson knew triple integrals, or I guess it’s not really an introduction. But I digress.
Let’s try to tackle the donut problem. Why would there be a donut there? My best guess is that it’s sort of like a standing wave. You have positive and negative regions, but they just swap with each other, over and over again.
The gold regions represent positive and the red regions represent negative. Or maybe it’s the other way around. Either way, they alternate ad nauseum.
Also, there appears to be a sort of rotational effect as well, centered on the z-axis (of course, what other axis is so special?). Meh time to learn about sigma pi delta phi bonding.
I guess I’ll end this post with an anticlimactic link: http://www.falstad.com/qmatom/
Enjoy playing with that app.