Hi, Giampiero! I’m glad you get it and sort of like it. I’m going to resume posts in this series soon: I’ll start by reviewing how to get the rate equation and master equation from a stochastic Petri net, and how the latter reduces to the former in the limit of ‘large particle number’ (as the physicists would say).

The reason people work in ‘label space’ instead of ‘state space’, if I understand what you mean by those terms, is that many different probability distributions $\psi_\ell(t)$ have the same mean values for the number of things of the $\ell$th type, and their time evolution will be different. Only in a certain limit where there are many things of every type can we describe dynamics in terms of these mean values.

]]>OK, i finally had some time to go through this yesterday, and now everything makes sense.

The last formula with the Kronecker’s delta is really what turned all the the lights on for me. I guess i was not really allowing *l’* to be an “independent” variable in *H*, in some sense.

Maybe a part of it is that it does not come natural for me to reason in the “labeling space”, as opposite to the “state space”, of the system (actually i am not yet too sure about what are the advantages of doing so). I guess it’s because I haven’t studied quantum mechanics long enough :)

]]>Greetings John,

Bruce Smith sent me to try to say something constructive (and a while back, I started thinking + searching for an answer, realized how long it was going to be, then forgot to follow-up).

I am pleased to report that (1) you’d be hard pressed to find a chemist with a convincing answer to all other chemists, (2) this is more atomic physics than chemistry, which is why you’d easily stump a chemist on this, (3) I have found a few very eloquent and involved discussions that say “it’s spherically symmetric” and “it’s not spherically symmetric,” (which means the answer is either “neither” or a superposition of the two) and (4) it probably doesn’t matter in the grand scheme because the response of the atom to any kind of external stimuli (such as an H atom) would be effectively instantaneous.

Just going by a combination of Hund’s Rule(s) (max multiplicity = lowest E; largest orbital angular momentum = lowest energy for a given multiplicity; outermost subshell half-filled or less = lowest total angular momentum is lowest in E / outermost subshell more than half-filled = highest total angular momentum is lowest in E), the Aufbau Principle (add e- into a p-shell one-at-a-time, so in nitrogen each p-orbital has 1 e- each), and Unsold’s Theorem (the square of the electronic wavefunction for a completely filled or half-filled sub-shell is spherically symmetric), oxygen is NOT spherically symmetric (well, not spherically symmetric if you were standing on the nucleus). These are the most fundamental “non-computational-quantum-chemical” ways to interpret SINGLE atoms and are all from good olde fashioned spectroscopy back in the day. I note that these are SINGLE atom cases, where the larger the vacuum the better the result. Importantly, these rules break down (or don’t apply) when the atom exists in an external potential (such as an H atom, H nucleus, whatever).

My initial thought (as a computational chemist) was that the “I’m scared that in reality, the oxygen atom’s ground state is a superposition of all possible rotated versions of some charismatic easy-to-draw non-spherically symmetric state” was a bit more formally correct as an *interpretation* of the system. That is, all isolated atoms are spherically symmetric, with the electrons existing in a superposition of all linear combinations of the available orbitals, making the “density” of the atom the “ensemble density” of the linear combinations. More to the point, the potential around an isolated atom would be spherically symmetric (if it is truly isolated), so the electrons have no bias to group in any arrangement one might describe as px, py, pz, whatever. One obtains a differentiation of “orbital arrangements” in an external potential (introducing the laboratory reference frame), and it is the balance of internal electronic repulsion in the presence of this potential and the stabilization of electronic arrangements that come from the presence of the external potential (two H atoms around an oxygen, for instance) that gives one a reference frame for any separation of electron density into “directions” (along bonds, external field, etc.).

All that said, orbitals are a pleasant fiction we use to describe the behavior of electrons in our modern one-electron wavefunction treatment in quantum chemistry. The isolated atom is one of the hardest systems to treat with great accuracy by quantum chemical methods because one must use multireference computational approaches to deal with the “spherical degeneracy” of the electronic states. I’ll take a methane calculation over isolated carbon any day.

Extra credit – once you’ve formed OH, the electron density of the O will attract the second H, which will then adopt the standard H2O geometry discussed above. The 104.45 angle is best explained as “the two lone pairs on the “other side” of the oxygen take up more space because you’ve got four repulsing electrons, so they take up more space, giving the H-O-H bend reduction to take up the slack.

I hope that it any way helps. If not, I’ll be happy to try to attempt a better explanation.

I’m enjoying reading the rest…

]]>Well, I think I’d go with the “simultaneous states of non-spherically symmetric” interpretation because there’s no way to define axes without imposing on the atom. The olde guard atomic physics would argue non-spherically symmetric because you’ve got two unpaired electrons in two p orbitals, but you’d never be able to “dock” along an unfilled orbital in any kind of chemical trajectory (which was how I interpreted the basis of your question in my first pass of your query).

My tweet-friendly would be “non-spherically symmetric locally, spherical enough at a distance, no way to sneak up and ask at any instant.”

]]>Hi, Damian! Thanks for trying to help! I’m still a bit confused, but misery loves company, so I feel much better hearing you say “I’ll take a methane calculation over isolated carbon any day” and scary stuff like:

The isolated atom is one of the hardest systems to treat with great accuracy by quantum chemical methods because one must use multireference computational approaches to deal with the “spherical degeneracy” of the electronic states.

It makes me feel like I wasn’t just being really dumb.

To compress your long answer into a soundbite, I guess you’re saying that a completely isolated unexcited oxygen atom *is* spherically symmetric.

Thanks for all the corrections, David! I’ll fix that stuff. As usual, a more beautiful version of this blog entry, with solutions to the puzzles, is available on my networks page.

So, with a collection of transitions in place in your Petri net you just add these matrices, and all is well since infinitesimal stochastic matrices are closed under addition.

Right! They’re closed under nonnegative linear combinations.

Hmm. If we only keep the condition that the columns sum to zero, and drop the condition that the off-diagonal entries are nonnegative, it seems we get not just a vector space of matrices, but even a Lie algebra . I’m guessing this because if we take the definition of ‘stochastic matrix’, and drop the condition that the entries are nonnegative, keeping the condition that the columns sum to 1, we seem to get a Lie group . This is the Lie group of transformations of that preserve the sum of the entries. The stochastic matrices form a sub-monoid of . And thus the infinitesimal stochastic matrices form a cone in the Lie algebra .

]]>In the final equation, , you ought to have a couple of instances of on the right.

So, with a collection of transitions in place in your Petri net you just add these matrices, and all is well since infinitesimal stochastic matrices are closed under addition.

]]>**Puzzle.** Suppose we have a stochastic Petri net with states and just one transition, whose rate constant is . Suppose the th state appears times as the input of this transition and times as the output. A labelling of this stochastic Petri net is a -tuple of natural numbers saying how many things are in each state. Let be the probability that the labelling is at time . Then the master equation looks like this:

for some matrix of real numbers . What is this matrix?

**Answer.** To compute it’s enough to start the Petri net in a definite labelling and see how fast the probability of being in some labelling changes. In other words, if at some time we have

then

at this time.

Now, suppose we have a Petri net that is labelled in some way at some moment. Then I said the probability that the transition occurs in a short time is approximately:

• the rate constant , times

• the time , times

• the number of ways the transition can occur, which is the product of falling powers Let’s call this product for short.

Multiplying these 3 things we get

So, the *rate* at which the transition occurs is just:

And when the transition occurs, it eats up things in the *i*th state, and produces things in that state. So, it carries our system from the original labelling to the new labelling

So, *in this case* we have

and thus

However, that’s not all: there’s another case to consider! Since the probability of the Petri net being in this new labelling is going up, the probability of it staying in the original labelling must be going down by the same amount. So we must also have

We can combine both cases into one formula like this:

Here the first term tells us how fast the probability of being in the new labelling is going up. The second term tells us how fast the probability of staying in the same labelling is going down.

Note: each column in the matrix sums to zero, and all the off-diagonal entries are nonnegative. That’s good: we now know that this matrix must be ‘infinitesimal stochastic’, meaning it must have these properties!

]]>Jon wrote:

I’m no chemist, but aren’t the orbitals you drew calculated as the excited states of a single electron around an otherwise bare nucleus? Since they don’t consider any interactions between multiple electrons except the exclusion “force”, it feels like they just don’t have it in them to predict bond angles.

Right, they don’t have the ability to predict bond angles. I was only trying to settle the question of whether a lone oxygen atom is spherically symmetric or not. And why? It’s because I told David Corfield this:

If we’re talking about real-world chemistry, rather than ball-and-stick models of molecules, I don’t think a lone oxygen atom really has two distinct ‘slots’ waiting for hydrogens. It’s more like it has a propensity to attach to a hydrogen, which can come in from any direction and stick… and then the resulting hydroxyl radical OH still has a propensity to attach to another hydrogen, which will then move to form a 104.45° angle to the first.

As soon as I wrote this, I started doubting it. I started wondering: is the oxygen atom really spherically symmetric before the hydrogen comes along, so that the hydrogen can attach itself anywhere with equal probability? Or does a hydrogen atom coming up to an oxygen atom attach to one of two predefined ‘slots’ with more or less well-defined locations?

(Of course ‘slots’ is a somewhat naive term for patterns in electron wavefunctions, but never mind.)

Until someone tells me the angular momentum of a lone oxygen atom, I won’t know the answer for sure. But I’m now pretty sure the latter picture is closer to the truth. I believe the 4 electrons in the outermost shell of the oxygen atom *break the spherical symmetry* and in fact pick out two preferred axes, lying at a roughly 90° angle to each other. That’s what those ‘p orbital’ pictures are supposed to suggest. But of course those pictures are only approximate, since (as you note) they neglect inter-electron forces.

Then, when two hydrogens actually attach to the oxygen, they wind up attaching at an angle of 104.45°, for complicated reasons that take a computer simulation to fully fathom. Presumably the repulsion of the hydrogens’ nuclei is largely to blame.

]]>Giampiero wrote:

And i actually wonder why we can’t just propagate the random variable \ell in time, which we should be able to do if we have the deterministic evolution equations (and we do) and the probability distribution (and it looks like we assume it).

There are a couple of reasons:

First, those deterministic evolution equations (the ‘rate equations’) assume the amount of things we have in each state varies *continuously*. For example, the rate equation for this model

might predict that starting out with 10 rabbits now, we will have 10.5 rabbits one week later.

But now we are studying the ‘master equation’, which assumes the number of things we have in each state varies *discretely*. No half-rabbits! The number must be a natural number.

Second, the master equations are really probabilistic, not deterministic. Starting out with 10 rabbits now, they’ll tell us the *probability* that we’ll have 9 or 10 or 11 or some other number of rabbits one week from now.

So the master equation is really answering a different question than the rate equation. It just so happens—we’ll see this later—that in the limit of large numbers, the results we get from the master equation are very close *on average* to the results we get from the rate equation.

I am still at a loss on this.

Okay. If you start with things in the th state, and something happens that eats things in the th state and produces new things in that state, you’re left with

things in the th state.

We can say the same thing faster using more jargon: if our system starts in the labelling , our transition takes it to the labelling .

So, we expect that the matrix entry

will be nonzero if

(Note that we’ve switched the roles of and . In my article I was talking about ; now we’re talking about . It doesn’t really matter as long as we pay attention.)

So, it would be nice to know the formula for in this case. The answer is supposed to be contained in these cryptic comments of mine:

A bit more precisely: suppose we have a Petri net that is labelled in some way at some moment. Then the probability that a given transition occurs in a short time is approximately:

• the rate constant for that transition, times

• the time , times

• the number of ways the transition can occur.

More precisely still: this formula is correct up to terms of order . So, taking the limit as , we get a differential equation describing precisely how the probability of the Petri net having a given labelling changes with time! And this is the

master equation.

But be careful: is

the *only* value of for which is nonzero?

The answer to that may become clearer upon reading part 5. On the other hand, you can also figure it out just by thinking.

Here’s another way to ask the question. The question is, which labellings have their probabilities changing

if at some moment we’re sure the system has the labelling :

We know that

is one such labelling. But is it the only one?

Once we settle this, we can work out exactly what

equals. We know

so once we know which matrix entries are nonzero, we can figure out what they are and we’ll be done!

(By the way, I encourage other folks to join in and help Giampiero solve this puzzle! The equation we’re looking for is called the ‘chemical master equation’, because it’s the basic equation governing chemical reactions. So if you’re already an expert on that, this puzzle might be too easy to be enjoyable. But if you’ve never thought about it, there’s no better way to learn it than to reinvent it.)

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