I don't understand physics. How do atomic orbitals work?

September 21, 2009 7:39 PM

I don't understand physics. How do atomic orbitals work?

I don't know anything about modern physics beyond the "I watched a trippy special on the Discovery Channel" level Consider this situation: I have one atom of hydrogen. Bound to it is a single electron in the 1s orbital. This electron has a negative charge. As I understand it, the orbital describes a probability density function: if I sample a point in the physical space described by the orbital, the value of that point in associated probability density function is the probability that I will encounter the electron there when I look for it (bonus side question: so if I integrate across the three dimensions of the orbital do I end up with 1?).

Let's assume I set up the following apparatus: I have my atom. At a fixed point from my atom, I have a very tiny electric field meter. This meter will record the strength of electric fields you put it in. I start recording the field strength. What do I see?

a) an oscillating value: the electron being a charged particle creates an electric field that falls off with distance to the meter. When the electron is closest to the meter, it registers the strongest field, and when it is further, it registers the weakest. The rest of the time it careens through points in between (bonus side question if this is the answer: is the output of the meter smooth, continuous, differentiable, etc., because the electron is careening through the physical space the orbital is defined over, or is it just bliz-blaz-bloz because the electron is teleporting to wherever it feels like?)

b) a fixed value: the field the meter would pick up if a point charge equal to the electron was fixed at some point (i don't know, maybe the closest point?) to the meter.

c) a different fixed value: something like the average charge. because quantum mechanics is mad trippy, the charge meter picks up, at all times, the sum of the charge it would detect from a point charge at each point in the orbital over the volume of the orbital.

d) something completely different.

e) Moo. The poster's understanding of the atom is so far wrong that this question cannot be answered.

(Super-bonus question: I'm not in school or anything, but I still want to learn some basics about how these things work. Is this physics or chemistry? Personally, I'm more interested in going up from here into how these work in groups than down into what makes up the various components at this point. Can you recommend any resources whereby I can teach myself? On Amazon, its easy to identify the "Grand Textbook for People Taking Multiple Years of This With Professors, TAs, and Problem Sets" and the "Kind of Trippy Pop Science Book That A Professor Wrote So He Could Get On The Daily Show", but what is in between? I took a lot of math, hard science, and engineering in undergrad, but no physics beyond classical mechanics and no chem so I don't know any of this stuff. I don't want to read a book thats like "Did you know that Quantum Mechanics begins with a 'Q'? That is a very unusual letter, my friends!" but I also don't want to read a book thats like "Page 1: As you can clearly see from the following Lagrangian Chapters 1-10 as listed in the table of contents can be left as an exercise to the reader so your authors have decided to proceed from the beginning of Chapter 11. More specifically, its the eighth page of Chapter 11 so it begins mid-sentence."

posted by jeb to science & nature (28 answers total)

if I sample a point

e) Moo: you can't sample a point.

You can sample a region, but how you go about doing that and what you are measuring gets complicated quickly.

You can probably teach yourself a great deal of quantum physics, but you should approach it from an experimental perspective rather than a philosophical perspective. Think about life at the human scale, and then start thinking about measuring things outside of the human scale (both larger and smaller). Think and study about how you would go about doing that. Quantum physics is just measuring and studying really tiny things. No one really knows why quantum mechanics works the way it does, but they do have a lot of data to study.

IANAQP
posted by b1tr0t at 7:47 PM on September 21, 2009

Actually, (A) might be closest to reality, except that your data points will probably bounce around randomly. I'm really not a quantum physicist, so I'll leave the answer as an exercise for the reader.
posted by b1tr0t at 7:51 PM on September 21, 2009

If you're willing to do a bit of work, it sounds like you'd have the background you need to read Griffiths's quantum book, which is really wonderful. It's a textbook, for sure, but the explanations are very clear; it might be a good companion to whatever else you end up reading.
posted by you're a kitty! at 7:52 PM on September 21, 2009

e) Moo: you can't sample a point.

You can sample a region, but how you go about doing that and what you are measuring gets complicated quickly.

You can probably teach yourself a great deal of quantum physics, but you should approach it from an experimental perspective rather than a philosophical perspective

that's interesting, but (a) I'm doing my best here to describe an experiment, rather than a philosophical "what does this really mean" approach. Also I'm really only trying to understand this so I can understand chemistry, so even philosophically, the experimental results at this level are what I'm actually interested in. (b), I'm introducing that point-sample part where I'm trying to explain what I understand the definition of the orbital to be. Even if we ignore that bit, does that affect my actual question about the electric field measurement from a point next to the atom? Can I not sample that point either? If so, let's assume I modify my experiment so that my meter is actually measuring the average of a field over a very small sphere. Much smaller than the atom. Is that possible?
posted by jeb at 8:04 PM on September 21, 2009

Let's assume I set up the following apparatus

One of the fundamental insight of quantum physics is that you cannot build such an apparatus. This is known as the Uncertainty Principle. The more you know about the location, the less you know about the value; the more you know about the value, the less you know about the location. Thus, the cloud that you see surrounding the nucleus in illustrations of an atom is in some ways a field, but it's also in some ways a distribution of probabilities.

You can know the probabilities. You can't know the actual location of an electron.
posted by alms at 8:24 PM on September 21, 2009 [2 favorites]

What you seem to be confused by in this setup is what physicists call the Measurement Problem. It's valid to say that if you set up your little tiny field-meter and an atom, looked at the field-meter's reading, and then repeated the experiment many, many times, the distribution of your electric field readings would match what you'd expect if the electron "randomly selected" a position in space every time you measured it, where the probabilities of this "selection" are given by the square of the wave function.

The problem is that according to the orthodox interpretation of quantum mechanics (the Copenhagen interpretation), the wavefunction is irrevocably changed by the act of measuring it. This means that once you read a value off of your field-meter, you've changed the experimental setup. Any measurements you do after this point (without "resetting" the system, as above) are essentially measurements of different physical situation.

So as to the possibilities you mention: (A) is false, since once you've made a measurement on a system you've almost certainly changed the system — the idea of the value "jittering around" isn't really meaningful. (B) and (C) are false, since you won't always get the same result. The most direct answer to your question is (E), with a bit of (D) mixed in. Don't feel bad, though — the question of "what really happens when we measure a quantum system" is one that brilliant physicists have wrestled with for going on a century now.
posted by Johnny Assay at 8:24 PM on September 21, 2009

Your picture of a hydrogen atom is quite accurate - you've got the probability distribution concept down pat.

bonus side question: so if I integrate across the three dimensions of the orbital do I end up with 1?

You sure do - that's what it means to have "one electron".

The answer to your electric field measuring question is a bit complicated. Here's the thing: When you measure a property of a system, you ALWAYS interact with it. So, if you're picturing a passive electric-field meter quietly measuring the electric field created by the electron while the electron goes on about its business, forget about it - such a device is inherently impossible. The electric field meter will, in a sense, CAUSE the electron to exist at a particular place; i.e. when you measure the electron's position, the electron's wave function collapses to an eigenstate of position. In simple terms, every time you measure where the electron is, that probability density function collapses to a single spike at a position.

It turns out that when you measure something using the electromagnetic force, your measurement apparatus will interact with the electron via a photon. That's getting into quantum electrodynamics.

Now, when QM was a young upstart field, people weren't sure if maybe the electron really had a definite position all along, but WE didn't know where it was until you measured it, or if the electron really DIDN'T HAVE a definite position until you measured it. A clever fellow named John Bell demonstrated that it is possible to distinguish between those two theories with a physical experiment, and since then, it has been proven beyond any reasonable doubt that, as weird as it is, the electron really doesn't have a definite position - it's a probability cloud. Don't feel bad if you have trouble believing it yourself - it even made Einstein a bit queasy.

Phew. I hope that helps a bit.

As for a good layman's text, Griffith's Intro to QM is quite reasonably written - it's probably near the top of the range of accessibility for someone who isn't studying physics.

Also, I highly recommend reading the Feynman Lectures, which, if you haven't heard of them, are the gold (nay, the diamond) standard in college physics lecturing. Someone said "If you can't explain it to a college freshman, you don't understand it yourself." It might have been Feynman who said that, but google isn't backing me up on that. Anyhow, that's sort of the philosophy behind his lectures.

Good luck - QM is very weird, amazing, trippy, sometimes well-nigh unbelievable, but it is the most accurate theory humanity has ever created, and it's the way our world works*!

* Disclaimer: may not apply under certain extreme circumstances or in relation to some elementary forces, or in some cosmic jurisdictions. No refunds. Due to the probabilistic nature of the theory, liability may or may not be waived for personal injury.
posted by Salvor Hardin at 8:29 PM on September 21, 2009 [6 favorites]

Oh, and

(bonus side question: so if I integrate across the three dimensions of the orbital do I end up with 1?).

Yes. This is called normalization of the wave function, and it's essentially how the overall scale of the wave function is defined.
posted by Johnny Assay at 8:29 PM on September 21, 2009

Is this question in response to the recent news of the atomic orbital photographs? [1, 2] Note that they got this image by pulling many different electrons through the screen to generate the probability distribution.

Your question reads like you're trying to apply your "macro" electrodynamics understanding to quantum mechanics, which as the above replies show, isn't too helpful.
posted by FuManchu at 8:33 PM on September 21, 2009

You can actually directly measure the orbitals, and here are some pictures that show it. This is pretty recent, coming out in Phys Rev B soon. Also they don't claim to be exactly measuring the E-field, rather just probability density of e-. The linked summary (and presumably the article, once it's available) have more details.

But you're certainly looking at s and p orbitals here, as you can tell by the shapes. Nifty!
posted by nat at 8:36 PM on September 21, 2009

One of the fundamental insight of quantum physics is that you cannot build such an apparatus.

Huh. Man, I really don't get this. So I knew that was true for the electron, but that's true for the proton as well? I was hoping in my apparatus that the meter was at a fixed point relative to the proton, and the electron cloud would do its bit in the general area.

Don't feel bad, though — the question of "what really happens when we measure a quantum system"

Double-Huh. I really thought this would be like a basic, who-cares-what-REALLY-happens type of problem.

So like, doesn't whatever would happen in this non-doable experiment have implications for real-world, observable-every-day chemistry? Like..I have a water molecule. The electrons are more probably in the oxygen atom's orbital relative to either of the electrons' orbitals, so like...shouldn't the charges of the hydrogen ends of the molecule change? Shouldn't their electronegativity, etc. all the things that govern chemistry depend on how this works? Really what made me ask this question is this bio textbook I'm reading never mentions this. (So far, I'm only the very beginning, presumably they are oversimplifying stuff since its meant for n00bs).

It's valid to say that if you set up your little tiny field-meter and an atom, looked at the field-meter's reading, and then repeated the experiment many, many times, the distribution of your electric field readings would match what you'd expect if the electron "randomly selected" a position in space every time you measured it, where the probabilities of this "selection" are given by the square of the wave function.

If I did this a zillion times, what would be the average of these values? If I follow what you are saying, you can't just let the meter run and read the output, but if you somehow could do the experiment a zillion times, the average reading of the meter would be like (c), the measured charge * the distance falloff * the probability of that position.
posted by jeb at 8:38 PM on September 21, 2009

Oh, I forgot to include the fun controversy about the "wavefunction collapse". The "wavefunction collapse" is a phenomenological theory. Wavefunction collapse cannot be derived from anything - it just seems to be what happens when you measure a property of a system. Weird! Now, there are two mainstream ideas about what is happening when a quantum mechanical wavefunction collapses.

The Copenhagen interpretation: It's just what it sounds like - the probability function just snaps into a infinitesimally skinny spike when you measure it. Dunno why, but that's how it works.

The many-worlds interpretation: Every time you measure (read: interact with) a wavefunction, the entire universe splits into an infinite number of universes for each possibly measurement value; i.e. when you measure the electron's position, an infinite number of universes are created, in each of which the electron is measured to be in a different place.

Which one is crazier? I don't know - they're both pretty weird. Luckily, or maybe unluckily, no one has thought of a way in which the two interpretations can be physically distinguished from one another; i.e. there is no test that one can run to determine whether wavefunction collapse is just literally wavefunction collapse, or a splitting of the universe. Most physicists throw up their hands and say, "Interesting issue, but it's not our problem if you can't test it experimentally. Talk to the philosophers."

Sorry - you didn't specifically ask about this, but I think it's a cool problem with weird implications.
posted by Salvor Hardin at 8:40 PM on September 21, 2009

Also, I highly recommend reading the Feynman Lectures, which, if you haven't heard of them, are the gold (nay, the diamond) standard in college physics lecturing. Someone said "If you can't explain it to a college freshman, you don't understand it yourself."

I'll definitely check them out, my friend has a copy, but I was under the impression that it was the gold standard of college physics lecturing...to college physics grad students. I also heard that quote attribute to Feynman, but with a more aggressive formulation ("eight-year old").

Is this question in response to the recent news of the atomic orbital photographs? [1, 2] Note that they got this image by pulling many different electrons through the screen to generate the probability distribution.

No, but that is super cool eh? The question came up when I was trying to understand some chem background in this bio textbook I'm reading. I got sidetracked trying to figure out which ochem book to order, wikipedia...zillion clicks later...

Your question reads like you're trying to apply your "macro" electrodynamics understanding to quantum mechanics, which as the above replies show, isn't too helpful.

Perhaps it is also coming through that I do not have a macro or any other size electrodynamics understanding.

Anyway thanks everyone so far this is so awesome. I'm definitely going to get that griffiths book and I'll take a look at the Feynman Lectures...probably both over my head. I'm getting the distinct feeling that this whole side-project of trying to understand some basic biology is going to end up in a slender, yellow, shockingly expensive Springer-Verlag math book, just like everything else.
posted by jeb at 8:46 PM on September 21, 2009

I recommend Feynman's QED, which is mostly about the interaction of electrons and photons... basically the stuff you are asking about. It does better than any other book I've read to answer for the non-physicist what this wave + particle stuff is all about.
posted by zompist at 8:54 PM on September 21, 2009

If you are trying to understand quantum as a means towards understanding chemistry (and thence bio), I really rather liked McQuarrie's "Quantum Chemistry." As its name implies, it's aimed towards chemists learning quantum, not hardcore theoretical physicists. If you're more or less OK with multivariable calculus and are willing to pick up a few bits of differential equations and linear algebra, you can get quite a lot out of it.

Incidentally, one of Heisenberg's arguments regarding his Uncertainty Principle involved an imaginary tiny microscope rather like the one you're hypothesizing.

I have a water molecule. The electrons are more probably in the oxygen atom's orbital relative to either of the electrons' orbitals, so like...shouldn't the charges of the hydrogen ends of the molecule change? Shouldn't their electronegativity, etc. all the things that govern chemistry depend on how this works?

One way of thinking of this is that you have changed where the electrons are likely to be. You're not any closer to being able to say with certainty "they are here" or "this is their momentum", but run the calculations (this gets much, much harder when we get more complicated than single hydrogen atoms), and you'll get a different set of probabilities than you did for the lone hydrogen atom. Whether we know exactly where the electrons are, or what their momentum is, rules about electron pairing, electron shielding, etc. still apply, and help govern how the probabilities change when atoms get close to each other and bind.

It's a huge headache, but at the same time, it helps explain what on earth orbitals are, and how they interact. Molecular orbital diagrams started to make much more sense after I took quantum. You start to get an understanding for why certain orbitals are or are not likely to interact, and for what on earth phases really are (where by phases I mean those shaded versus unshaded or + versus - lobes of orbitals you might have seen in organic chemistry books.)
posted by ubersturm at 9:02 PM on September 21, 2009

As I understand it, the probability density is not even throughout a given orbital. Check out this graph. It gives the probability of finding an electron at various distances from the nucleus (l=0 indicates an s orbital, l=1 indicates a p orbital). Notice that the various orbitals are not straight vertical lines but broad hills. Technically, I think the hills are infinitely broad, but the probability drops off to unmeasurable insignificance very quickly.

Nobody knows what the inside of an atom is like. Thousands of scientist far smarter than you or I have banged their heads against walls trying to figure it out. Eventually everyone just kind of gave up trying to interpret QM. Now they just accept that it's true and that we don't know what it means.

As far as measurements go, the Heisenberg uncertainty principle prohibits anything useful. You can't measure a quantum system without disrupting it; i.e., causing the wavefunction to collapse. So, assuming your very tiny electric field meter were possible, you might be able to take a single snapshot of the electric field inside a given atom, but you couldn't take a continuous measurement over time. And that picture would be of a regular boring old Newtonian system of fixed points, not a fancy cloud of QM weirdness. There really is no cheating.

Now I will seemingly contradict myself: visible orbitals! They basically took a huge number of snapshots and blended them together. You could do that with your tiny electric field meter too, but it wouldn't really answer your question.
posted by dephlogisticated at 9:14 PM on September 21, 2009

Dammit. First link should go here.
posted by dephlogisticated at 9:15 PM on September 21, 2009

A good place to start on the wave-particle duality is the double slit experiment. Several people have mentioned the wave function collapse because (as far as I know) no one knows if an electron is actually at a specific place at a specific time, or just seems to be when you try to interact with it in a particle-like way.

Although the double slit experiment strongly suggests that photons (or electrons, or other quantum stuff) are really waves, a photodetector (such as your digital camera in very low light) makes it seem like they are particles. If photons were always wavelike, you would expect digital camera images to produce very smooth and blurry images in low light, as the photon "splashes" across several photosites. Instead, you see that photons land in a particular bucket which becomes bright, while the adjacent bucket is dark.
posted by b1tr0t at 9:29 PM on September 21, 2009 [1 favorite]

So like, doesn't whatever would happen in this non-doable experiment have implications for real-world, observable-every-day chemistry? Shouldn't their electronegativity, etc. all the things that govern chemistry depend on how this works?

Yes, pretty much. The electron orbitals are fundamental to the interaction of atoms (bonding). This should be covered in a decent first year chem textbook (and probably won't be covered in a pure QM textbook).

alms: One of the fundamental insight of quantum physics is that you cannot build such an apparatus. This is known as the Uncertainty Principle.

The uncertainty principle makes it impossible to know the position and momentum of the electron simultaneously. We don't need to know the momentum of the electron to know the electrostatic field it generates. So, theoretically, we could measure the electrostatic field in some region over some period of time and then be able to say (probabilistically) that the electron was located in some region.

With those measurements, you could to this:

If I did this a zillion times, what would be the average of these values? If I follow what you are saying, you can't just let the meter run and read the output, but if you somehow could do the experiment a zillion times, the average reading of the meter would be like (c), the measured charge * the distance falloff * the probability of that position.

A couple points: If you want to start talking about the electromagnetic field, then you have an uncertainty problem, because the magnetic field depends on the momentum of the electron and you can't know both the momentum and position of the electron. Secondly, electrons don't really act like simple point charges. The electromagnetic field is much weirder than that and to really answer your question, you'd need to get into what exactly an electromagnetic field is.
posted by ssg at 9:43 PM on September 21, 2009

I was under the impression that it was the gold standard of college physics lecturing...to college physics grad students

The Feynman Lectures are based on the lectures he gave to Caltech undergrads. Who were, granted, probably as smart as grad students elsewhere...
posted by nicwolff at 10:13 PM on September 21, 2009

jeb: So like, doesn't whatever would happen in this non-doable experiment have implications for real-world, observable-every-day chemistry? Like..I have a water molecule.

Yes. And as a matter of fact, some of the behavior of water is much easier to understand when you consider negative electrical charge as a field. Or especially when you get into things like aromatic rings where that electron field gets spread out over the entire ring system.
posted by KirkJobSluder at 4:45 AM on September 22, 2009

KirkJobSluder has it. See resonance. For an aromatic ring, there are two possible electron configurations. One can imagine that the ring switches rapidly between the two states, or one can imagine that ring exists in a continuous average of the two states. The situation is somewhat analogous to the classical/quantum picture.

Molecular dynamics often involves this kind of thing, because reactions happen so rapidly and continuously on that scale. For example, one can conceptually treat a volume of liquid H₂O as an equivalent volume of separate ions (OH^- and H₃O⁺). Water molecules are continuously trading hydrogen cations (H⁺) with each other, such that there is always a small amount of charged species present. H₂O is only the average species, and 7 is only the average pH.
posted by dephlogisticated at 7:42 AM on September 22, 2009

RESONANCE! that is the word I needed to find here. Ok, that is actually what I wanted to know. I actually started out wondering about aromatic rings, I just used the water example because I thought it would be easier. But yeah, that's what I was wondering about, the double lines in the lewis diagram can be rotated and everything is the same. In fact, they are rotated and not-rotated, right? So the reason I was wondering about the charge-measurement apparatus is I was wondering if like...do these rings or other molecules flap away as their charges shift (I'm picturing like a scallop swimming) or do the charges stay balanced at some kind of average and the ring lays flat?

One can imagine that the ring switches rapidly between the two states, or one can imagine that ring exists in a continuous average of the two states. The situation is somewhat analogous to the classical/quantum picture.

Uhhh...one can? But doesn't one of these have to actually happen, or do we end up with some kind of macro-scale wavefunction here? Like I'd think that the shape of a molecule would totally depend on this, and the shapes of the molecules matter a lot e.g. for binding.

Ok, this QM discussion is very interesting and I ordered Feynman's QED (Bringing that up reminded me that one of my beloved high school physics teachers said that book was amazing too). But 'resonance' is really what I was wondering about so at least I know what the appropriate search terms are now. From the resonance wikipedia page: "The ozone molecule is represented by two resonance structures in the top of scheme 2. In reality the two terminal oxygen atoms are equivalent and the hybrid structure is drawn on the right with a charge of -1/2 on both oxygen atoms and partial double bonds."

Thanks everyone.
posted by jeb at 8:12 AM on September 22, 2009

Some random guy wrote this guide up. It may have what you're looking for.
http://www.phys.uu.nl/~thooft/theorist.html
posted by fairykarma at 9:24 AM on September 22, 2009

Molecules do indeed flap around, at least ones that are permitted to do so by their bonds. Take an amine group, for instance. A tertiary amine (NR₃) has three carbon bonds and one pair of electrons, which gives the group a pyramidal shape. But the electron pair is (relatively) free moving, so the group can form either an upward-pointed pyramid, or a downward-pointed pyramid. At room temperature, in fact, an amine group will rapidly flap up and down, such that the observed NMR structure comes out as something like an average of the two (IIRC).

An aromatic group, however, has double bonds, and double bonds don't allow for the same freedom of movement. Well, not quite double bonds (due to resonance), but the equivalent of something halfway between a single and a double bond, and that's enough to keep the ring structurally stable, so it doesn't do any flapping.

As for which way you look at it—bonds that switch back and forth or bonds that are evenly spread out—it helps to keep in mind the time scales on the quantum level. If the bonds do indeed switch back and forth, they will be doing so at such a high rate of speed that, for all practical purposes, the electrons might as well be evenly spread out. Quantum mechanics would say that the probability distribution around the ring is even at each carbon. Classical mechanics would say that a given electron can be in only one place at one time, and thus the bonds must alternate. As for which is correct, well, it goes back to same old argument. We just don't know and can't say for sure.
posted by dephlogisticated at 9:46 AM on September 22, 2009

Classical mechanics would say that a given electron can be in only one place at one time, and thus the bonds must alternate. As for which is correct, well, it goes back to same old argument. We just don't know and can't say for sure.

I'll disagree with this because the classical mechanical view of electrons often can't explain the physical properties of resonant molecules, or chemical systems that take advantage of phenomena such as quantum tunneling.
posted by KirkJobSluder at 1:36 PM on September 22, 2009

I'll disagree with this because the classical mechanical view of electrons often can't explain the physical properties of resonant molecules, or chemical systems that take advantage of phenomena such as quantum tunneling.

You're right—I was mistakenly conflating classical mechanics with local realism. My point still stands, though, because there is no proven interpretation of QM. Some interpretations assume locality, while others do not. As an example, the Bohm interpretation has particles (such as electrons) existing in a definite position at all times, while the Copenhagen interpretation says that they aren't localized in space until they are measured. Yet the two interpretations are said to be empirically equivalent.

Still, it certainly seems like all interpretations of QM require you to give up at least one cherished notion. You can have a straight timeline, a single universe, locality, or determinism, but you can’t have all four.
posted by dephlogisticated at 9:46 PM on September 28, 2009

zompist-- QED is probably the best "explaining some stuff" book I've ever read. thanks a ton for the rec.

If anyone finds this thread in the future, READ QED! Actually, it still won't really help you understand the original question here, but on the other hand, it will.

QED is amazing. Even apart from the physics, just as a work of teaching, its incredible.
posted by jeb at 12:07 PM on October 1, 2009