Last time Brendan showed us a proof of the Anderson-Craciun-Kurtz theorem on equilibrium states. Today we’ll look at an example that illustrates this theorem. This example brings up an interesting ‘paradox’—or at least a puzzle. Resolving this will get us ready to think about a version of *Noether’s theorem* relating conserved quantities and symmetries. Next time Brendan will state and prove a version of Noether’s theorem that applies to stochastic Petri nets.

#### A reversible reaction

In chemistry a type of atom, molecule, or ion is called a **chemical species**, or **species** for short. Since we’re applying our ideas to both chemistry and biology, it’s nice that ‘species’ is also used for a type of organism in biology. This stochastic Petri net describes the simplest reversible reaction of all, involving two species:

We have species 1 turning into species 2 with rate constant and species 2 turning back into species 1 with rate constant So, the rate equation is:

where and are the amounts of species 1 and 2, respectively.

#### Equilibrium solutions of the rate equation

Let’s look for *equilibrium* solutions of the rate equation, meaning solutions where the amount of each species doesn’t change with time. Equilibrium occurs when each species is getting created at the same rate at which it’s getting destroyed.

So, let’s see when

Clearly this happens precisely when

This says the rate at which 1’s are turning into 2’s equals the rate at which 2’s are turning back into 1’s. That makes perfect sense.

#### Complex balanced equilibrium

In general, a chemical reaction involves a *bunch* of species turning into a *bunch* of species. Since ‘bunch’ is not a very dignified term, a bunch of species is usually called a **complex**. We saw last time that it’s very interesting to study a strong version of equilibrium: **complex balanced equilibrium**, in which each *complex* is being created at the same rate at which it’s getting destroyed.

However, in the Petri net we’re studying today, all the complexes being produced or destroyed consist of a single species. In this situation, any equilibrium solution is automatically complex balanced. This is great, because it means we can apply the Anderson-Craciun-Kurtz theorem from last time! This says how to get from a complex balanced equilibrium solution of the *rate equation* to an equilibrium solution of the *master equation*.

First remember what the master equation says. Let be the probability that we have things of species 1 and things of species 2 at time . We summarize all this information in a formal power series:

Then the master equation says

where following the general rules laid down in Part 8,

This may look scary, but the **annihilation operator** and the **creation operator** are just funny ways of writing the partial derivative and multiplication by , so

or if you prefer,

The first term describes species 1 turning into species 2. The second describes species 2 turning back into species 1.

Now, the Anderson-Cranciun-Kurtz theorem says that whenever is a complex balanced solution of the rate equation, this recipe gives an equilibrium solution of the master equation:

In other words: whenever , we have

we have

Let’s check this! For starters, the constant in the denominator of doesn’t matter here, since is linear. It’s just a normalizing constant, put in to make sure that our probabilities sum to 1. So, we just need to check that

If we do the derivatives on the left hand side, it’s clear we want

And this is indeed true when .

So, the theorem works as advertised. And now we can work out the probability of having things of species 1 and of species 2 in our equilibrium state . To do this, we just expand the function as a power series and look at the coefficient of . We have

so we get

This is just a product of two independent Poisson distributions!

In case you forget, a Poisson distribution says the probability of events occurring in some interval of time if they occur with a fixed average rate and independently of the time since the last event. If the expected number of events is , the Poisson distribution is

and it looks like this for various values of :

It looks almost like a Gaussian when is large, but when is small it becomes very lopsided.

Anyway: we’ve seen that in our equilibrium state, the number of things of species is given by a Poisson distribution with mean . That’s very nice and simple… but the amazing thing is that these distributions are *independent*.

Mathematically, this means we just *multiply* them to get the probability of finding things of species 1 and of species 2. But it also means that knowing how many things there are of one species says nothing about the number of the other.

But something seems odd here. One transition in our Petri net consumes a 1 and produces a 2, while the other consumes a 2 and produces a 1. The total number of particles in the system never changes. The more 1’s there are, the fewer 2’s there should be. But we just said knowing how many 1’s we have tells us nothing about how many 2’s we have!

At first this seems like a paradox. Have we made a mistake? Not exactly. But we’re neglecting something.

#### Conserved quantities

Namely: the equilibrium solutions of the master equation we’ve found so far are not the only ones! There are other solutions that fit our intuitions better.

Suppose we take any of our equilibrium solutions and change it like this: set the probability equal to 0 unless

but otherwise leave it unchanged. Of course the probabilities no longer sum to 1, but we can rescale them so they do.

The result is a new equilibrium solution, say Why? Because, as we’ve already seen, no transitions will carry us from one value of to another. And in this new solution, the number of 1’s is clearly *not* independent from the number of 2’s. The bigger one is, the smaller the other is.

**Puzzle 1.** Show that this new solution depends only on and the ratio , not on anything more about the values of and in the original solution

**Puzzle 2.** What is this new solution like when ? (This particular choice makes the problem symmetrical when we interchange species 1 and 2.)

What’s happening here is that this particular stochastic Petri net has a ‘conserved quantity’: the total number of things never changes with time. So, we can take any equilibrium solution of the master equation and—in the language of quantum mechanics—`project down to the subspace’ where this conserved quantity takes a definite value, and get a new equilibrium solution. In the language of probability theory, we say it a bit differently: we’re ‘conditioning on’ the conserved quantity taking a definite value. But the idea is the same.

This important feature of conserved quantities suggests that we should try to invent a new version of Noether’s theorem. This theorem links conserved quantities and *symmetries* of the Hamiltonian.

There are already a couple versions of Noether’s theorem for classical mechanics, and for quantum mechanics… but now we want a version for *stochastic* mechanics. And indeed one exists, and it’s relevant to what we’re doing here. We’ll show it to you next time.

You need the ‘s in the right hand side of

.

As for puzzle 1, I reached

So, if we put , we’d have

or something like that.

Nice! There must be some more symmetrical way to express this, probably by using both and instead of

How about

I like ways of writing solutions that leave the symmetries of the problem manifest.

This should remind you of one of those Pascal’s triangle gadgets where a ball can fall from the top to the bottom making a bunch of random binary choices en route. But here the probability of going left (picking ‘1’) is , while the probability of going right (picking ‘2’) is .

That would do.

Then you need to word your puzzles more carefully, and not break the symmetry there:)

True, true. I found that hard to do.

Anyway, I hope the point of the puzzle is clear. We start with a product of independent Poisson distributions to get a probability distribution on pairs . Then we ‘slice’ it, fixing a constant value of . We get a probability distribution on which is exactly like the distribution for flipping a biased coin and getting heads and tails.

Hello, here’s the solution to puzzle 1:

In Puzzle 2, we get:

I hope, I’m correct this time, and haven’t hurried too much. :)

Here are Graham Jones’ answers to the puzzles:

The particular stochastic network considered here, had a conserved quantity n1+n2.

For a general network, how do we identify all the conserved quantities and therefore all the symmetries of the corresponding Hamiltonian?