Chemical Orbitals from Eigenstates

A small puzzle I recently set for myself was finding out how the hydrogenic orbital eigenstates give rise to the S- P- D- and F- orbitals in chemistry (and where s, p, d and f came from).

The reason this puzzle is important to me is that many of my interests sort of straddle how to go from the angstrom scale to the nanometer scale. There is a cross-over where physics becomes chemistry, but chemists and physicists often look at things very differently. I was not directly trained as a P-chemist; I was trained separately as a Biochemist and a Physicist. Remarkably, the Venn diagrams describing the education for these pursuits only overlap slightly. When Biochemists and Molecular Biologists talk, the basic structures below that are frequently just assumed (the scale here is >1nm), while Physicists frequently tend to focus their efforts toward going more and more basic (the scale here is <1 Angstrom). This leads to a clear non-overlap in the scale where chemistry and P-chem are relevant (~1 angstrom). Quite ironically, the whole periodic table of the elements lies there. I have been through P-chem and I’ve gotten hit with it as a Chemist, but this is something of an inconvenient scale gap for me. So, a cat’s paw of mine has been understanding, and I mean really understanding, where quantum mechanics transitions to chemistry.

One place is understanding how to get from the eigenstates I know how to solve to the orbitals structuring the periodic table.

1200px-periodic_table_chart

This assemblage is pure quantum mechanics. You learn a huge amount about this in your quantum class. But, there are some fine details which can be left on the counter.

One of those details for me was the discrepancy between the hydrogenic wave functions and the orbitals on the periodic table. If you aren’t paying attention, you may not even know that the s-, p-, d- orbitals are not all directly the hydrogenic eigenstates (or perhaps you were paying a bit closer attention in class than I was and didn’t miss when this detail was brought up). The discrepancy is a very subtle one because often times when you start looking for images of the orbitals, the sources tend to freely mix superpositions of eigenstates with direct eigenstates without telling why the mixtures were chosen…

For example, here are the S, P and D orbitals for the periodic table:

ao

This image is from http://www.chemcomp.com. Focusing on the P row, how is it that these functions relate to the pure eigenstates? Recall the images that I posted previously of the P eigenstates:

P-orbital probabiltiy densityorbital21-1 squared2

In the image for the S, P and D orbitals, of the Px, Py and Pz orbitals, all three look like some variant of P210, which is the pure state on the left, rather than P21-1, which is the state on the right. In chemistry, you get the orbitals directly without really being told where they came from, while in physics, you get the eigenstates and are told somewhat abstractly that the s-, p-, d- orbitals are all superpositions of these eigenstates. I recall seeing a professor during an undergraduate quantum class briefly derive Px and Py, but I really didn’t understand why he selected the combinations he did! Rationally, it makes sense that Pz is identical to P210 and that Px and Py are superpositions that have the same probability distribution as Pz, but are rotated into the X-Y plane ninety degrees from one another. How do Px and Py arise from superpositions of P21-1 and P211? P21-1 and P211 have identical probability distributions despite having opposite angular momentum!

Admittedly, the intuitive rotations that produce Px and Py from Pz make sense at a qualitative level, but if you try to extend that qualitative understanding to the D-row, you’re going to fail. Four of the D orbitals look like rotations of one another, but one doesn’t. Why? And why are there four that look identical? I mean, there are only three spatial dimensions to fill, presumably. How do these five fit together three dimensionally?

Except for the Dz^2, none of the D-orbitals are pure eigenstates: they’re all superpositions. But what logic produces them? What is the common construction algorithm which unites the logic of the D-orbitals with that of the P-orbitals (which are all intuitive rotations).

I’ll actually hold back on the math in this case because it turns out that there is a simple revelation which can give you the jump.

As it turns out, all of chemistry is dependent on angular momentum. When I say all, I really do mean it. The stability of chemical structures is dependent on cases where angular momentum has tended in some way to cancel out. Chemical reactivity in organic chemistry arises from valence choices that form bonds between atoms in order to “complete an octet,” which is short-hand for saying that species combine with each other in such a way that enough electrons are present to fill in or empty out eight orbitals (roughly push the number of electrons orbiting one type of atom across the periodic table in its appropriate row to match the noble gases column). For example, in forming the salt crystal sodium chloride, sodium possesses only one electron in its valence shell while chlorine contains seven: if sodium gives up one electron, it goes to a state with no need to complete the octet (with the equivalent electronic completion of neon), while chlorine gaining an electron pushes it into a state that is electronically equal to argon, with eight electrons. From a physicist stand-point, this is called “angular momentum closure,” where the filled orbitals are sufficient to completely cancel out all angular momentum in that valence level. As another example, one highly reactive chemical structure you might have heard about is a “radical” or maybe a “free radical,” which is simply chemist shorthand for the situation a physicist would recognize contains an electron with uncancelled spin and orbital angular momentum. Radical driven chemical reactions are about passing around this angular momentum! Overall, reactions tend to be driven to occur by the need to cancel out angular momentum. Atomic stoichiometry of a molecular species always revolves around angular momentum closure –you may not see it in basic chemistry, but this determines how many of each atom can be connected, in most cases.

From the physics, what can be known about an orbital is essentially the total angular momentum present and what amount of that angular momentum is in a particular direction, namely along the Z-axis. Angular momentum lost in the X-Y plane is, by definition, not in either the X or Y direction, but in some superposition of both. Without preparing a packet of angular momentum, the distribution ends up having to be uniform, meaning that it is in no particular direction except not in the Z-direction. For the P-orbitals, the eigenstates are purely either all angular momentum in the Z-direction, or none in that direction. For the D-orbitals, the states (of which there are five) can be combinations, two with angular momentum all along Z, two with half in the X-Y plane and half along Z and one with all in the X-Y plane.

What I’ve learned is that, for chemically relevant orbitals, the general rule is “minimal definite angular momentum.” What I mean by this is that you want to minimize situations where the orbital angular momentum is in a particular direction. The orbits present on the periodic table are states which have canceled out angular momentum located along the Z-axis. This is somewhat obvious for the homology between P210 and Pz. P210 points all of its angular momentum perpendicular to the z-axis. It locates the electron on average somewhere along the Z-axis in a pair of lobes shaped like a peanut, but the orbital direction is undefined. You can’t tell how the electron goes around.

As it turns out, Px and Py can both be obtained by making simple superpositions of P21-1 and P211 that cancel out z-axis angular momentum… literally adding together these two states so that their angular momentum along the z-axis goes away. Px is the symmetric superposition while Py is the antisymmetric version. For the two states obtained by this method, if you look for the expectation value of the z-axis angular momentum, you’ll find it missing! It cancels to zero.

It’s as simple as that.

The D-orbitals all follow. D320 already has no angular momentum on the z-axis, so it is directly Dzz. You therefore find four additional combinations by simply adding states that cancel the z-axis angular momentum: D321 and D32-1 symmetric and antisymmetric combinations and then the symmetric and antisymmetric combinations of D322 and D32-2.

Notice, all I’m doing to make any of these states is by looking at the last index (the m-index) of the eignstates and making a linear combination where the first index plus the second gives zero. 1-1 =0, 2-2=0. That’s it. Admittedly, the symmetric combination sums these with a (+) sign and a 1/sqrt(2) weighting constant so that Px = (1/sqrt(2))(P21 + P21-1) is normalized and the antisymmetric combination sums with a (-) sign as in Py = (1/sqrt(2))(P211 – P21-1), but nothing more complicated than that! The D-orbitals can be generated in exactly the same manner. I found one easy reference on line that loosely corroborated this observation, but said it instead as that the periodic table orbitals are all written such that the wave functions have no complex parts… which is also kind of true, but somewhat misleading because you sometimes have to multiply by a complex phase to put it genuinely in the form of sines for the polar coordinate (and as the polar coordinate is integrated over 360 degrees, expectation values on this coordinate, as z-axis momentum would contain, cancel themselves out; sines and cosines integrated over a full period, or multiples of a full period, integrate to zero.)

Before I wrap up, I had a quick intent to touch on where S-, P-, D- and F- came from. “Why did they pick those damn letters?” I wondered one day. Why not A-, B-, C- and D-? The nomenclature emerged from how spectral lines appeared visually and groups were named: (S)harp, (P)rincipal, (D)iffuse and (F)undamental. (A second interesting bit of “why the hell???” nomenclature is the X-ray lines… you may hate this notation as much as me: K, L, M, N, O… “stupid machine uses the K-line… what does that mean?” These letters simply match the n quantum number –the energy level– as n=1,2,3,4,5… Carbon K-edge, for instance, is the amount of energy between the n=1 orbital level and the ionized continuum for a carbon atom.) The sharpness tends to reflect the complexity of the structure in these groups.

As a quick summary about structuring of the periodic table, S-, P-, D-, and F- group the vertical columns (while the horizontal rows are the associated relative energy, but not necessarily the n-number). The element is determined by the number of protons present in the nucleus, which creates the chemical character of the atom by requiring an equal number of electrons present to cancel out the total positive charge of the nucleus. Electrons, as fermions, are forced to occupy distinct orbital states, meaning that each electron has a distinct orbit from every other (fudging for the antisymmetry of the wave function containing them all). As electrons are added to cancel protons, they fall into the available orbitals depicted in the order on the periodic table going from left to right, which can be a little confusing because they don’t necessarily purely close one level of n before starting to fill S-orbitals of the next level of n; for example at n=3, l can equal 0, 1 and 2… but, the S-orbitals for n=4 will fill before D-orbitals for n=3 (which are found in row 4). This has purely to do with the S-orbitals having lower energy than P-orbitals which have lower energy than D-orbitals, but that the energy of an S-orbital for a higher n may have lower energy than the D-orbital for n-1, meaning that the levels fill by order of energy and not necessarily by order to angular momentum closure, even though angular momentum closure influences the chemistry. S-, P-, D-, and F- all have double degeneracy to contain up and down spin of each orbital, so that S- contains 2 instead of 1, P- contains 6 instead of 3, and D- from 10 instead of 5. If you start to count, you’ll see that this produces the numerics of the periodic table.

Periodic table is a fascinating construct: it contains a huge amount of quantum mechanical information which really doesn’t look much like quantum mechanics. And, everybody has seen the thing! An interesting test to see the depth of a conversation about periodic table is to ask those conversing if they understand why the word “periodic” is used in the name “Periodic table of the elements.” The choice of that word is pure quantum mechanics.

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Powerball Probabilities

If you’ve read anything else in this blog, you’ll know I write frequently about my playing around with Quantum Mechanics. As a digression away from a natural system that is all about probabilities, an interesting little toy problem I decided to tackle is figuring out how the “win” probabilities are determined in the lottery game Powerball.

Powerball is actually quite intriguing to me. They have a website here which details by level all the winners across the whole country who have won a Powerball prize in any given drawing. You may have looked at this chart at some point while trying to figure out if your ticket won something useful. A part of what intrigues me about this chart is that it tells you in a given drawing exactly how much money was spent on Powerball and how many people bought tickets. How does it tell you this? Because probability is an incredibly reliable gauge of behavior with big samples sizes. And, Powerball quite willingly lays all the numbers out for you to do their book keeping for them by telling you exactly how many people won… particularly at the high-probability-to-win levels which push into the regime of Gaussian statistics. For big samples, like millions of people buying powerball tickets, where N=big, the errors on average values become relatively insignificant since they go as sqrt(N). And, the probabilities reveal what those average values are.

The game is doubly intriguing to me because of the psychological component that drives it. As the pot becomes big, people’s willingness to play becomes big even though the probabilities never change. It suddenly leaps into the national consciousness every time the size of the pot becomes big and people play more aggressively as if they had a greater chance of winning said money. It is true that somebody ultimately walks away with the big pot, but what’s the likelihood that somebody is you?

But, as a starter, what are the probabilities that you win anything when you buy a ticket? To understand this, it helps to know how the game is set up.

As everybody knows, powerball is one of these games where they draw a bunch of little balls printed with numbers out of a machine with a spinning basket and you, as the player, simply match the numbers on your ticket to the numbers on the balls. If your ticket matches all the numbers, you win big! And, as an incentive to make people feel like they’re getting something out of playing, the powerball company awards various combinations of matching numbers and adds in multipliers which increase the size of the award if you do get any sort of match. You might only match a number or two, but they reward you a couple bucks for your effort. If you really want, you can pick the numbers yourself, but most people simply grab random numbers spat out of a computer… not like I’m telling you anything you don’t already know at this point.

One of the interesting qualities of the game is that the probabilities of prizes are very easy to adjust. The whole apparatus stays the same; they just add or subtract balls from the basket. In powerball, as currently run, there are two baskets: the first basket contains 69 balls while the second contains 26. Five balls are drawn from the first basket while only one, the Powerball, is drawn from the second. There is actually an entire record available of how the game has been run in the past, how many balls were in either the first or second baskets and when balls were added or subtracted from each. As the game has crossed state lines and the number of players has grown, the number of balls has also steadily swelled. I think the choice in numbering has been pretty careful to make the smallest prize attainably easy to get while pushing the chances for the grand prize to grow enticingly larger and larger. Prizes are mainly regulated by the presence of the Powerball: if your ticket manages to match the Powerball and nothing else, you win a small prize, no matter what. Prizes get bigger as a larger number of the other five balls are matched on your ticket.

The probabilities at a low level work almost exactly as you would expect: if there are 26 balls in the powerball basket, at any given drawing, you have 1 chance in 26 of matching the powerball. This means that you have 1 chance in 26 of winning some prize as determined by the presence of the powerball. There are also prizes for runs of larger than three matching balls drawn from the main basket, which tends to push the probabilities of winning anything to a slightly higher frequency than 1 in 26.

For the number savvy this begins to reveal the economics of powerball: an assured win by these means requires you to spend, on average, $48. That’s 26 tickets where you are likely to have one that matches the powerball. Note, the prize for matching that number is $4. $44 dollars spent to net only $4 is a big overall loss. But, this 26 ticket buy-in is actually hiding the fact that you have a small chance of matching some sequence of other numbers and obtaining a bigger prize… and it would certainly not be an economic loss if you matched the powerball and then the 5 other balls, yielding you a profit in the hundreds of millions of dollars (and this is usually what people tell themselves as they spend $2 for each number).

The probability to win the matched powerball prize only, that is to match just the powerball number, is actually somewhat worse than 1 in 26. The probability is attenuated by the requirement that you hit no matches on any other of the five possible numbers drawn.

Finding the actual probability is as follows: (1/26)*(64/69)*(63/68)*(62/67)*(61/66)*(60/65). If you multiply that out and invert it, you get 1 hit in 38.32 tries. The first number is, of course, the chances of hitting the powerball, while the other five are the chance of hitting numbers that aren’t picked… most of these probabilities are naturally quite close to 1, so you are likely to hit them, but they are probabilities that count toward hitting the powerball only.

This number may not be that interesting to you, but lots of people play the game and that means that the likelihood of hitting just the powerball is close to Gaussian. This is useful to a physicist because it reveals something about the structure of the Powerball playing audience on any given week: that site I gave tells you how many people won with only the powerball, meaning that by multiplying that number by 38.32, you know how many tickets were purchased prior to the drawing in question. For example, as of the August 12 2017 drawing, 1,176,672 numbers won the powerball-only prize, meaning that very nearly 38.32*1,176,672 numbers were purchased: ~45,090,071 numbers +/- 6,715, including error (notice that the error here is well below 1%).

How many people are playing? If people mostly purchase maybe two or three numbers, around 15-20 million people played. Of course, I’m not accounting for the slavering masses who went whole hog and dropped $20 on numbers; if everybody did this, 4.5 million people played… truly, I can’t really know people’s purchasing habits for certain, but I can with certainty say that only a couple tens of millions of people played.

The number there reveals quite clearly the economics of the game for the period between the 8/12 drawing and the one a couple days prior: $90 million was spent on tickets! This is really quite easy arithmetic since it’s all in factors of 2 over the number of ticket numbers sold. If you look at the total prize pay-out, also on that page I provided, $19.4 million was won. This means that the Powerball company kept ~$70 million made over about three days, of which some got dumped into the grand prize and some went to whatever overhead they keep (I hear at least some of that extra is supposed to go into public works and maybe some also ends up in the Godfather’s pocket). Lucrative business.

If you look at the prize payouts for the game, most of the lower level prizes pay off between $4 and $7. You can’t get a prize that exceeds $100 until you match at least 4 balls. Note, here, that the probability of matching 4 balls (including the powerball) is about 1 in 14,494. This means, that to assure yourself a prize of $100, you have to spend ~$29,000. You might argue that in 14,494 tickets, you’ll win a couple smaller prizes ($4 prizes are 1 in 38, 1 in 91, and $7 prizes are 1 in 700 and 1 in 580) and maybe break even. Here’s the calculation for how much you’ll likely make for that buy-in: $4*(14,494*(1/38 + 1/91)) + $7*(14,494*(1/700 + 1/580))… I’ve rounded the probabilities a bit… =$2482.65. For $29,000 spent to assure a single $100 win, you are assured to win at most $2500 from lesser winnings for a total loss of $27,500. Notice, $4 on a $44 loss is about 10%, while $2500 on $27,500 is also about 10%… the payoff does not improve at attainable levels! Granted, there’s a chance at a couple hundred million, but the probability of the bigger prize is still pretty well against you.

Suppose you are a big spender and you managed to rake up $29,000 in cash to dump into tickets, how likely is it that you will win just the $1 million prize? That’s five matched balls excluding the powerball. The probability is 1 in 11,688,053. By pushing the numbers, your odds of this prize have become 14,500/11,688,053, or about 1 chance in 800. Your odds are substantially improved here, but 1 in 800 is still not a wonderful bet despite the fact that you assured yourself a fourth tier prize of $100! The grand prize is still a much harder bet with odds running at about 1 in 20,000, despite the amount you just dropped on it. Do you just happen to have $30,000 burning a hole in your pocket? Lucky you! Lots of people live on that salary for a year.

Most of this is simple arithmetic and I’ve been bandying about probabilities gleaned from the Powerball website. If you’re as curious about it as me, you might be wondering exactly how all those probabilities were calculated. I gave an example above of the mechanical calculation of the lowest level probability, but I also went and figured out a pair of formulae that calculate any of the powerball prize probabilities. It reminded me a bit of stat mech…

prob without powerball

prob with powerball

number for hits

I’ve colored the main equations and annotated the the parts to make them a little clearer. The final relation just shows how you can see the number of tries needed in order to hit one success, given a probability as calculated with the other two equations. The first equation differs from the second in that it refers to probabilities where you have matched numbers without managing to match the powerball, while the second is the complement, where you match numbers having hit the powerball. Between these two equations, you can calculate all the probabilities for the powerball prizes. Since probabilities were always hard for me, I’ll try to explain the parts of these equations. If you’re not familiar with the factorial operation, this is what is denoted by the exclamation point “!” and it denotes a product string counting up from one to the number of the factorial… for example 5! means 1x2x3x4x5. The special case 0! should be read as 1. The first part, in blue, is the probability relating to either hitting on missing the powerball, where K = 26, the number of balls in the powerball basket. The second part (purple) is the multiplicity and tells you how many ways that you can draw a certain number of matches (Y) to fill a number of open slots (X), while drawing a number of mismatches (Z) in the process, where X=Y+Z. In powerball, you draw five balls, so X=5 and Y is the number of matches (anywhere from 0 to 5), while Z is the number of misses. Multiplicity shows up in stat mech and is intimately related to entropy. The totals drawn (green) is perhaps mislabeled… here I’m referring to the number of possible choices in the main basket, N=69, and the number of those that will not be drawn M = N – X, or 64. I should probably have called it “Main basket balls” or something. The last two parts determine the probabilities related to the given number of hits (Y) (orange) and the given number of misses (Z) (red) and I have applied the product operator to spiffy up the notation. Product operator is another iterand much like the summation operator and means that you repeatedly multiply successive values, much like a factorial, but where the value you are multiplying is produced from a particular range and given a set form. In these, the small script m and n start at zero (my bad, this should be under the Pi) and iterate until they are just less than the number up top (Y – 1 or Z – 1 and not equal to). At the extreme cases of either all hits or all misses, the relevant product operator (either Miss or Hit respectively) must be set equal to one in order to not count it.

This is one of those rare situations where the American public does a probability experiment with the values all well recorded where it’s possible to see the outcomes. How hard is it to win the grand prize? Well, the odds are one in 292 million. Consider that the population of the United States is 323 million. That means that if everybody in the United States bought one powerball number, about one person would win.

Only one.

Thanks to the power of the media, everybody has the opportunity to know that somebody won. Or not. That this person exists, nobody wants to doubt, but consider that the odds of winning are so scant that you not only won’t win, but you pretty likely will never meet anyone who did. Sort of surreal… everything is above board, you would think, but the rarity is so rare that there’s no assurance that it ever actually happens. You can suppose that maybe it does happen because people do win those dinky $4 prizes, but maybe this is just a red herring and nobody really actually wins! Those winner testimonials could be from actors!

Yeah, I’m not much of a conspiracy theorist, but it is true that a founding tenant of the idea of a ‘limit’ in math is that 99.99999% is effectively 100%. Going to the limit where the discrepancy is so small as to be infinitesimal is what calculus is all about. It is fair to say that it very nearly never happens! Everybody wants to be the one who beats the odds, which is why Powerball tickets are sold, but the extraordinarily vast majority never will win anything useful… I say “useful” because winning $4 or $7 is always a net loss. You have to win one of the top three prizes for it to be anywhere near worth anything, which you likely never will.

One final fairly interesting feature of the probability is that you can make some rough predictions about how frequently the grand prize is won based on how frequently the first prize is won. First prize is matching all five of the balls, but not the powerball. This frequency is about once per 12 million numbers, which is about 26 times more likely than all 5 plus the Powerball. In the report on winnings, a typical frequency is about 2 to 3 winners per drawing. About 1 time in 26 a person with all five manages to get the powerball too, so, with two drawings per week and about 2.5 first prize winners per drawing, that’s five winners per week… which implies that the grand prize should be won at a frequency of about once every five to six weeks –every month and a half or so. The average here will have a very large standard deviation because the number of winners is compact, meaning that the error is an appreciable portion of the measurement, which is why there is a great deal of variation in period between times when the grand prize is won. The incidence becomes much more Poissonian and stochastic, and allows some prizes to get quite big compared to others and causes their values to disperse across a fairly broad range. Uncertainty tends to dominate, making the game a bit more exciting.

While the grand prize is small, the number of people winning the first prize in a given week is small (maybe none or one), but this number grows in proportion to the size of the grand prize (maybe 5 or 6 or as high as 9). When the prize grows large enough to catch the public consciousness, the likelihood that somebody will win goes up simply because more people are playing it and this can be witnessed in the fluctuating frequency of the wins of lower level prizes. It breathes around the pulse of maybe 200 million dollars, lubbing at 40 million (maybe 0 to 1 person winning the first prize) and dubbing at 250 million (with 5 people or more winning the first prize).

Quite a story is told if you’re boring and as easily amused as me.

In my opinion, if you do feel inclined to play the game, be aware that when I say you probably won’t win, I mean that the numbers are so strongly against you that you do not appreciably improve your odds by throwing down $100 or even $1,000. The little $4 wins do happen, but they never pay and $1,000 spent will likely not get you more than $100 in total of winnings. It might as well be a voluntary tax. Cherish the dream your $2 buys, but do not stake your well-being on it. There’s nothing wrong with dreaming as long as you understand where to wake up.

(edit 8-24-17)

There was a grand prize winner last night (Wednesday 8-23-17). The outcomes are almost completely as should be expected: the winner is in Massachusetts… the majority of the country’s population is located in states on either the east or west coast, so this is unsurprising. There were 40 match 5 winners, so you would anticipate at least one to be a grand prize winner, which is exactly what happened (1 in 26 difference between 5 with powerball and 5 without). There were about 5.9 million powerball-only winners, so 38.32*5.9 is 226 million total powerball numbers sold in the run-up to last night’s drawing… with grand prize odds of 1 in 292 million, this is approaching parity. This means that more than $452 million was spent since Saturday on powerball lottery numbers (calculation excludes the extra dollar spent on multipliers). About five times as many ticket numbers were sold for this drawing as when I made my original analysis a week ago. With that many tickets sold, there was almost assuredly going to be a winner last night. This is not to say there shouldn’t have been a winner before this –probability is a fickle mistress– but the numbers are such that it was unlikely, but not impossible, for the prize to grow bigger. The last time the powerball was won was on 6-10-17, about two months and thirteen days ago… you can know that this is an unusually large jackpot because this period is longer than the usual period between wins (I had generously estimated 6 weeks based on the guess of 2 match 5 winners per drawing, but I think this might actually be a bit too high).

There was only one grand prize winning number out of 226 million tickets sold (not counting all the drawings that failed to yield a grand prize winner prior to this.) Think on that for a moment.