The Malay Archipelago

17 September, 2011

I live on the fringes of the Malay Archipelago. At least that’s what the naturalist Alfred Russel Wallace called it. In his famous book of that name, he wrote:

The Malay Archipelago extends for more than 4,000 miles in length from east to west, and is about 1,300 in breadth from north to south. It would stretch over an expanse equal to that of all Europe from the extreme west far into Central Asia, or would cover the widest parts of South America, and extend far beyond the land into the Pacific and Atlantic oceans. It includes three islands larger than Great Britain; and in one of them, Borneo, the whole of the British Isles might be set down, and would be surrounded by a sea of forests. New Guinea, though less compact in shape, is probably larger than Borneo. Sumatra is about equal in extent to Great Britain; Java, Luzon and Celebes are each about the size of Ireland. Eighteen more islands are, on the average, as large as Jamaica; more than a hundred are larger than the Isle of Wight; while the isles and islets of smaller size are innumerable.

Wallace claimed to have fond a line running through the Malay Archipelago that separates islands with Asian flora and fauna from those with plants and animals more like those of Austalia. This is now called the ‘Wallace Line’. It runs between Bali (on the western, Asiatic side) and the lesser-known nearby island of Lombok (on the eastern side).

Why does the Wallace Line run right between these two nearby islands? Wallace had a theory: it’s because the ocean between them is very deep! Even when sea levels were much lower—for example during the last ice age—and many islands were connected by land bridges, Bali and Lombok remained separate. Indeed, all the islands on one side of the Wallace Line have been separate from those on the other side for a very, very long time.

How long? Maybe forever, since Australia used to be down near Antarctica. I don’t really know. Do you? I’ve brought Wallace’s book to read with me, but that’s probably not the best way to find out. It should be easy to look up.

However, everything gets more complicated when you examine it carefully. Now the Wallace Line is just one of several important lines dividing biogeographical regions in the Malay Archipelago:

Sundaland is the land shelf containing the Malay Peninsula, Borneo, Java, and Sumatra. Sahul, also known as Australasia or simply Australia, is the land shelf containing Australia and New Guinea. Wallacea is a group of islands between Sundaland and Sahul—islands that haven’t been haven’t been recently connected to either of these land shelves.

The Wallace line is the western boundary of Wallacea, separating it from Sundaland. The Lydekker line, named after Richard Lydekker, is the eastern boundary of Wallacea separating it, from Sahul.

Wallacea is the red, heart-shaped region here:

The blue line through Wallacea is called the ‘Weber line’. Max Carl Wilhelm Weber argued that this line, not the boundary between Bali and Lombok, mark the major boundary between Asiatic and Australian organisms.

I bet the real truth is even more complicated.

Anyway, today my wife and I are going to Lombok. We’ll say there until the 21st, based in Senggigi; then we’ll go to Bali and stay in the town of Ubud until the 25th. So, we’ll have a good chance to see the difference between Asiatic and Wallacean flora and fauna.

15 Comments | biodiversity, biology | Permalink
Posted by John Baez

Network Theory (Part 10)

16 September, 2011

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 \beta, and species 2 turning back into species 1 with rate constant \gamma. So, the rate equation is:

\begin{array}{ccc} \displaystyle{ \frac{d x_1}{d t} } &=& -\beta x_1 + \gamma x_2 \\ \qquad \mathbf{} \\ \displaystyle{ \frac{d x_2}{d t} } &=& \beta x_1 - \gamma x_2 \end{array}

where x_1 and x_2 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

\displaystyle{ \frac{d x_1}{d t} = \frac{d x_2}{d t} = 0 }

Clearly this happens precisely when

\beta x_1 = \gamma x_2

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 \psi_{n_1, n_2}(t) be the probability that we have n_1 things of species 1 and n_2 things of species 2 at time t. We summarize all this information in a formal power series:

\Psi(t) = \sum_{n_1, n_2 = 0}^\infty \psi_{n_1, n_2}(t) z_1^{n_1} z_2^{n_2}

Then the master equation says

\frac{d}{d t} \Psi(t) = H \Psi (t)

where following the general rules laid down in Part 8,

\begin{array}{ccl} H &=& \beta (a_2^\dagger - a_1^\dagger) a_1 + \gamma (a_1^\dagger - a_2^\dagger )a_2 \end{array}

This may look scary, but the annihilation operator a_i and the creation operator a_i^\dagger are just funny ways of writing the partial derivative \partial/\partial z_i and multiplication by z_i, so

\begin{array}{ccl} H &=& \displaystyle{ \beta (z_2 - z_1) \frac{\partial}{\partial z_1} + \gamma (z_1 - z_2) \frac{\partial}{\partial z_2} } \end{array}

or if you prefer,

\begin{array}{ccl} H &=& \displaystyle{ (z_2 - z_1) \, (\beta \frac{\partial}{\partial z_1} - \gamma \frac{\partial}{\partial z_2}) } \end{array}

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 (x_1,x_2) is a complex balanced solution of the rate equation, this recipe gives an equilibrium solution of the master equation:

\displaystyle{ \Psi = \frac{e^{x_1 z_1 + x_2 z_2}}{e^{x_1 + x_2}} }

In other words: whenever \beta x_1 = \gamma x_2, we have

we have

H\Psi = 0

Let’s check this! For starters, the constant in the denominator of \Psi doesn’t matter here, since H is linear. It’s just a normalizing constant, put in to make sure that our probabilities \psi_{n_1, n_2} sum to 1. So, we just need to check that

\displaystyle{ (z_2 - z_1) (\beta \frac{\partial}{\partial z_1} - \gamma \frac{\partial}{\partial z_2}) e^{x_1 z_1 + x_2 z_2} = 0 }

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

\displaystyle{ (z_2 - z_1) (\beta x_1 - \gamma x_2) e^{x_1 z_1 + x_2 z_2} = 0 }

And this is indeed true when \beta x_1 = \gamma x_2 .

So, the theorem works as advertised. And now we can work out the probability \psi_{n_1,n_2} of having n_1 things of species 1 and n_2 of species 2 in our equilibrium state \Psi. To do this, we just expand the function \Psi as a power series and look at the coefficient of z_1^{n_1} z_2^{n_2}. We have

\Psi = \displaystyle{ \frac{e^{x_1 z_1 + x_2 z_2}}{e^{x_1 + x_2}} } = \displaystyle{ \frac{1}{e^{x_1} e^{x_2}} \sum_{n_1,n_2}^\infty \frac{(x_1 z_1)^{n_1}}{n_1!} \frac{(x_2 z_2)^{n_2}}{n_2!} }

so we get

\psi_{n_1,n_2} = \displaystyle{ \frac{1}{e^{x_1}} \frac{x_1^{n_1}}{n_1!} \; \cdot \; \frac{1}{e^{x_2}} \frac{x_1^{n_2}}{n_2!} }

This is just a product of two independent Poisson distributions!

In case you forget, a Poisson distribution says the probability of k 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 \lambda, the Poisson distribution is

\displaystyle{ \frac{1}{e^\lambda} \frac{\lambda^k}{k!} }

and it looks like this for various values of \lambda:

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

Anyway: we’ve seen that in our equilibrium state, the number of things of species i = 1,2 is given by a Poisson distribution with mean x_i. 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 n_1 things of species 1 and n_2 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 \Psi and change it like this: set the probability \psi_{n_1,n_2} equal to 0 unless

n_1 + n_2 = n

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 \Psi_n. Why? Because, as we’ve already seen, no transitions will carry us from one value of n_1 + n_2 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 \Psi_n depends only on n and the ratio x_1/x_2 = \gamma/\beta, not on anything more about the values of x_1 and x_2 in the original solution

\Psi = \displaystyle{ \frac{e^{x_1 z_1 + x_2 z_2}}{e^{x_1 + x_2}} }

Puzzle 2. What is this new solution like when \beta = \gamma? (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.

8 Comments | mathematics, networks, physics | Permalink
Posted by John Baez

NSF Funding for Research in Asia

15 September, 2011

The National Science Foundation has a program called EAPSI, which stands for East Asia and Pacific Summer Institutes. This is a program where US grad students are supported to participate in 8-week research experiences at laboratories in Australia, China, Korea, New Zealand, Singapore, Taiwan and Japan (where you actually get 10 weeks). It runs from June to August. The program provides a $5,000 summer stipend, round-trip airfare to the host location, living expenses abroad, and an introduction to the society, culture, language, and research environment of the host location.

Sounds cool, huh?

The application for the summer of 2012 is now open. It will close at 5:00 pm proposer’s local time on November 9, 2011. Application instructions are available online here. For further information concerning benefits, eligibility, and tips on applying, go here or here.

Since there’s a good chance I’ll be here at the Centre for Quantum Technologies in Singapore for the summer of 2012, I’d love it if you could apply to this program and get a fellowship to work with me. However, I have no idea if this is actually possible! I just learned about this program five minutes ago, and I’ve told you everything I know about it.

Anyway, it sounds like a good thing. I believe the 21st century is the century of Asia. If you’re a grad student in any sort of science, you want to get in on that.

Leave a Comment » | jobs | Permalink
Posted by John Baez

Climate Reality Project

14 September, 2011

The Climate Reality Project is planning a presentation called “24 Hours of Reality” beginning at 7 pm Central Time on September 14th, arguing for the connection between more extreme weather and climate change. “There will be a full-on assault on climate skeptics, exploring where they get their funding from.”

The Washington Post has an interview with Al Gore about this project:

• Brad Plumer, Al Gore: ‘The message still has to be about the reality we’re facing’ , Washington Post, 12 September 2011.

I’ll quote a bit:

Brad Plumer: “An Inconvenient Truth” was basically a primer on global warming—the causes, the problems it creates, the ways we can avert it. So what more is there to add? How will this new presentation be different?

Al Gore: It’s very different—a few of the images are the same, but 95 percent of the slides are completely new. The science linking the increased frequency and severity of extreme weather to the climate crisis has matured tremendously in the last couple of years. Think about the last year, we’ve had floods in Pakistan displacing 20 million people and further destabilizing a nuclear-armed country. We’ve had drought and wildfires in Russia. In Australia you’ve got floods the size of France and Germany combined. Then there’s drought in Texas—out of 254 counties in Texas, 252 are on fire. I’m talking to you from Nashville, where the city lost the equivalent of an entire year’s budget from recent floods—the area has never been flooded like this before, so no one had flood insurance.

That’s the reality we’ve got to focus on. This presentation is a defense of the science and the scientists, against the timeworn claims by deniers.

BP: Now, whenever a natural disaster happens—say, a flood or a wildfire—you typically see scientists quoted in the press saying, “Well, it’s hard to attribute any single event to global warming, although this is the sort of event we should see more of as the planet warms.” As I understand it, this sort of extra-careful hedge is becoming outdated. Scientists actually are making tighter connections between current disasters and climate change, correct?

AG: Yes, that shift in the way scientists describe the linkage is one of the elements of this new slideshow. It’s a subtle but extremely important shift. They used to say that the climate crisis changes the odds of extreme weather events—this was the old metaphor of “loading the dice.” Now, they say there’s not only a greater likelihood of rolling 12s, but we’re actually loading 13s and could soon be rolling 15s and 16s. As scientists like James Hansen [of NASA’s Goddard Institute for Space Studies] and Kevin Trenberth [of the National Center for Atmospheric Research] point out, the changes brought about by man-made global-warming pollution have reached the stage that every event is now being affected by it in some way.

In the last 30 years, for instance, we’ve seen water vapor above the oceans increase by 4 percent, and many storms reach as far as 2,000 miles out to collect water vapor. So when you have a 4 percent increase over such a large area, the storms are now fueled with more water vapor than was the case 30 years ago. That means we’re getting larger downpours. And in drought-prone areas, we’re seeing increasing intervals between downpours, which is one of several reasons why we’re seeing extreme droughts.

BP: Now, you’re talking about presenting the stark facts as a way of persuading people that climate change is a problem. Yet when you look at polls on climate belief, one thing that stands out is that the people most dismissive of global warming tend to be the most confident that they have all the information they need. Doesn’t that suggest there’s a point at which more information doesn’t actually persuade anyone?

AG: Well, that logic hasn’t led deniers to stop pressing the inaccurate disinformation about climate science. And the fact is that quite a few of the large carbon polluters and their allies in the ideological right wing have been spending hundreds of millions of dollars per year to mislead people. Have you read Naomi Oreske’s book Merchants of Doubt? The tobacco companies a few decades ago pioneered this organized disinformation technique to create artificial doubt about the science of their product—they hired actors to dress them up as doctors and had them say, “I’m a doctor, there’s nothing wrong with smoking cigarettes; in fact, it’ll make you feel better.” And some of the same people who took money from tobacco companies to lie about tobacco science are now taking money from large carbon polluters to lie about the reality of the climate crisis.

BP: Okay, but taking that opposition is a given, there’s been a lot of discussion about whether something more is needed to fight it than yet another recital of climate science facts.

AG: Right, you hear a lot of people giving advice on how to talk about climate science—how you need to dress differently or stand on your head and deliver the message in rhyme. And I respect all that, and I hope a lot of people will present the message in their own way. But my message is about presenting the reality. I have faith in the United States and our ability to make good decisions based on the facts. And I believe Mother Nature is speaking very loudly and clearly. We’ve had ten disasters in the United States this year alone costing more than $1 billion and which were climate-related. It’s only a matter of time before reality sinks in, and we need both parties involved. And the only way to get the right answer is to understand the question.

4 Comments | carbon emissions, climate | Permalink
Posted by John Baez

Network Theory (Part 9)

13 September, 2011

jointly written with Brendan Fong

Last time we reviewed the rate equation and the master equation. Both of them describe processes where things of various kinds can react and turn into other things. But:

• In the rate equation, we assume the number of things varies continuously and is known precisely.

• In the master equation, we assume the number of things varies discretely and is known only probabilistically.

This should remind you of the difference between classical mechanics and quantum mechanics. But the master equation is not quantum, it’s stochastic: it involves probabilities, but there’s no uncertainty principle going on.

Still, a lot of the math is similar.

Now, given an equilibrium solution to the rate equation—one that doesn’t change with time—we’ll try to find a solution to the master equation with the same property. We won’t always succeed—but we often can! The theorem saying how was proved here:

• D. F. Anderson, G. Craciun and T. G. Kurtz, Product-form stationary distributions for deficiency zero chemical reaction networks.

To emphasize the analogy to quantum mechanics, we’ll translate their proof into the language of annihilation and creation operators. In particular, our equilibrium solution of the master equation is just like what people call a ‘coherent state’ in quantum mechanics.

So, if you know about quantum mechanics and coherent states, you should be happy. But if you don’t, fear not!—we’re not assuming you do.

The rate equation

To construct our equilibrium solution of the master equation, we need a special type of solution to our rate equation. We call this type a ‘complex balanced solution’. This means that not only is the net rate of production of each species zero, but the net rate of production of each possible bunch of species is zero.

Before we make this more precise, let’s remind ourselves of the basic setup.

We’ll consider a stochastic Petri net with a finite set S of species and a finite set T of transitions. For convenience let’s take S = \{1,\dots, k\}, so our species are numbered from 1 to k. Then each transition \tau has an input vector m(\tau) \in \mathbb{N}^k and output vector n(\tau) \in \mathbb{N}^k. These say how many things of each species go in, and how many go out. Each transition also has rate constant r(\tau) \in [0,\infty), which says how rapidly it happens.

The rate equation concerns a vector x(t) \in [0,\infty)^k whose ith component is the number of things of the ith species at time t. Note: we’re assuming this number of things varies continuously and is known precisely! This should remind you of classical mechanics. So, we’ll call x(t), or indeed any vector in [0,\infty)^k, a classical state.

The rate equation says how the classical state x(t) changes with time:

\displaystyle{ \frac{d x}{d t} = \sum_{\tau \in T} r(\tau)\, (n(\tau)-m(\tau)) \, x^{m(\tau)} }

You may wonder what x^{m(\tau)} means: after all, we’re taking a vector to a vector power! It’s just an abbreviation, which we’ve seen plenty of times before. If x \in \mathbb{R}^k is a list of numbers and m \in \mathbb{N}^k is a list of natural numbers, we define

x^m = x_1^{m_1} \cdots x_k^{m_k}

We’ll also use this notation when x is a list of operators.

Complex balance

The vectors m(\tau) and n(\tau) are examples of what chemists call complexes. A complex is a bunch of things of each species. For example, if the set S consists of three species, the complex (1,0,5) is a bunch consisting of one thing of the first species, none of the second species, and five of the third species.

For our Petri net, the set of complexes is the set \mathbb{N}^k, and the complexes of particular interest are the input complex m(\tau) and the output complex n(\tau) of each transition \tau.

We say a classical state c \in [0,\infty)^k is complex balanced if for all complexes \kappa \in \mathbb{N}^k we have

\displaystyle{ \sum_{\{\tau : m(\tau) = \kappa\}} r(\tau) c^{m(\tau)} =\sum_{\{\tau : n(\tau) = \kappa\}} r(\tau) c^{m(\tau)} }

The left hand side of this equation, which sums over the transitions with input complex \kappa, gives the rate of consumption of the complex \kappa. The right hand side, which sums over the transitions with output complex \kappa, gives the rate of production of \kappa . So, this equation requires that the net rate of production of the complex \kappa is zero in the classical state c .

Puzzle. Show that if a classical state c is complex balanced, and we set x(t) = c for all t, then x(t) is a solution of the rate equation.

Since x(t) doesn’t change with time here, we call it an equilibrium solution of the rate equation. Since x(t) = c is complex balanced, we call it complex balanced equilibrium solution.

The master equation

We’ve seen that any complex balanced classical state gives an equilibrium solution of the rate equation. The Anderson–Craciun–Kurtz theorem says that it also gives an equilibrium solution of the master equation.

The master equation concerns a formal power series

\displaystyle{ \Psi(t) = \sum_{n \in \mathbb{N}^k} \psi_n(t) z^n }

where

z^n = z_1^{n_1} \cdots z_k^{n_k}

and

\psi_n(t) = \psi_{n_1, \dots,n_k}(t)

is the probability that at time t we have n_1 things of the first species, n_2 of the second species, and so on.

Note: now we’re assuming this number of things varies discretely and is known only probabilistically! So, we’ll call \Psi(t), or indeed any formal power series where the coefficients are probabilities summing to 1, a stochastic state. Earlier we just called it a ‘state’, but that would get confusing now: we’ve got classical states and stochastic states, and we’re trying to relate them.

The master equation says how the stochastic state \Psi(t) changes with time:

\displaystyle{ \frac{d}{d t} \Psi(t) = H \Psi(t) }

where the Hamiltonian H is:

\displaystyle{ H = \sum_{\tau \in T} r(\tau) \, \left({a^\dagger}^{n(\tau)} - {a^\dagger}^{m(\tau)} \right) \, a^{m(\tau)} \label{master} }

The notation here is designed to neatly summarize some big products of annihilation and creation operators. For any vector n \in \mathbb{N}^k, we have

a^n = a_1^{n_1} \cdots a_k^{n_k}

and

\displaystyle{ {a^\dagger}^n = {a_1^\dagger }^{n_1} \cdots {a_k^\dagger}^{n_k} }

Coherent states

Now suppose c \in [0,\infty)^k is a complex balanced equilibrium solution of the rate equation. We want to get an equilibrium solution of the master equation. How do we do it?

For any c \in [0,\infty)^k there is a stochastic state called a coherent state, defined by

\displaystyle{ \Psi_c = \frac{e^{c z}}{e^c} }

Here we are using some very terse abbreviations. Namely, we are defining

e^{c} = e^{c_1} \cdots e^{c_k}

and

e^{c z} = e^{c_1 z_1} \cdots e^{c_k z_k}

Equivalently,

\displaystyle{ e^{c z} = \sum_{n \in \mathbb{N}^k} \frac{c^n}{n!}z^n }

where c^n and z^n are defined as products in our usual way, and

n! = n_1! \, \cdots \, n_k!

Either way, if you unravel the abbrevations, here’s what you get:

\displaystyle{ \Psi_c = e^{-(c_1 + \cdots + c_k)} \, \sum_{n \in \mathbb{N}^k} \frac{c_1^{n_1} \cdots c_k^{n_k}} {n_1! \, \cdots \, n_k! } \, z_1^{n_1} \cdots z_k^{n_k} }

Maybe now you see why we like the abbreviations.

The name ‘coherent state’ comes from quantum mechanics. In quantum mechanics, we think of a coherent state \Psi_c as the ‘quantum state’ that best approximates the classical state c. But we’re not doing quantum mechanics now, we’re doing probability theory. \Psi_c isn’t a ‘quantum state’, it’s a stochastic state.

In probability theory, people like Poisson distributions. In the state \Psi_c, the probability of having n_i things of the ith species is equal to

\displaystyle{ e^{-c_i} \, \frac{c_i^{n_i}}{n_i!} }

This is precisely the definition of a Poisson distribution with mean equal to c_i. We can multiply a bunch of factors like this, one for each species, to get

\displaystyle{ e^{-c} \frac{c^n}{n!} }

This is the probability of having n_1 things of the first species, n_2 things of the second, and so on, in the state \Psi_c. So, the state \Psi_c is a product of independent Poisson distributions. In particular, knowing how many things there are of one species says nothing all about how many things there are of any other species!

It is remarkable that such a simple state can give an equilibrium solution of the master equation, even for very complicated stochastic Petri nets. But it’s true—at least if c is complex balanced.

The Anderson–Craciun–Kurtz theorem

Now we’re ready to state and prove the big result:

Theorem (Anderson–Craciun–Kurtz). Suppose c \in [0,\infty)^k is a complex balanced equilibrium solution of the rate equation. Then H \Psi_c = 0.

It follows that \Psi_c is an equilibrium solution of the master equation. In other words, if we take \Psi(t) = \Psi_c for all times t, the master equation holds:

\displaystyle{ \frac{d}{d t} \Psi(t) = H \Psi(t) }

since both sides are zero.

Proof. To prove the Anderson–Craciun–Kurtz theorem, we just need to show that H \Psi_c = 0. Since \Psi_c is a constant times e^{c z}, it suffices to show H e^{c z} = 0. Remember that

\displaystyle{ H e^{c z} = \sum_{\tau \in T} r(\tau) \left( {a^\dagger}^{n(\tau)} -{a^\dagger}^{m(\tau)} \right) \, a^{m(\tau)} \, e^{c z} }

Since the annihilation operator a_i is given by differentiation with respect to z_i, while the creation operator a^\dagger_i is just multiplying by z_i, we have:

\displaystyle{ H e^{c z} = \sum_{\tau \in T} r(\tau) \, c^{m(\tau)} \left( z^{n(\tau)} - z^{m(\tau)} \right) e^{c z} }

Expanding out e^{c z} we get:

\displaystyle{ H e^{c z} = \sum_{i \in \mathbb{N}^k} \sum_{\tau \in T} r(\tau)c^{m(\tau)}\left(z^{n(\tau)}\frac{c^i}{i!}z^i - z^{m(\tau)}\frac{c^i}{i!}z^i\right) }

Shifting indices and defining negative powers to be zero:

\displaystyle{ H e^{c z} = \sum_{i \in \mathbb{N}^k} \sum_{\tau \in T} r(\tau)c^{m(\tau)}\left(\frac{c^{i-n(\tau)}}{(i-n(\tau))!}z^i - \frac{c^{i-m(\tau)}}{(i-m(\tau))!}z^i\right) }

So, to show H e^{c z} = 0, we need to show this:

\displaystyle{ \sum_{i \in \mathbb{N}^k} \sum_{\tau \in T} r(\tau)c^{m(\tau)}\frac{c^{i-n(\tau)}}{(i-n(\tau))!}z^i =\sum_{i \in \mathbb{N}^k} \sum_{\tau \in T} r(\tau)c^{m(\tau)}\frac{c^{i-m(\tau)}}{(i-m(\tau))!}z^i }

We do this by splitting the sum over T according to output and then input complexes, making use of the complex balanced condition:

\displaystyle{ \sum_{i \in \mathbb{N}^k} \sum_{\kappa \in \mathbb{N}^k} \sum_{\{\tau : n(\tau)=\kappa\}} r(\tau)c^{m(\tau)}\frac{c^{i-n(\tau)}}{(i-n(\tau))!} \, z^i = }

\displaystyle{ \sum_{i \in \mathbb{N}^k} \sum_{\kappa \in \mathbb{N}^k} \frac{c^{i-\kappa}}{(i-\kappa)!}\, z^i \sum_{\{\tau : n(\tau) = \kappa\}} r(\tau)c^{m(\tau)} = }

\displaystyle{\sum_{i \in \mathbb{N}^k} \sum_{\kappa \in \mathbb{N}^k} \frac{c^{i-\kappa}}{(i-\kappa)!}\, z^i \sum_{\{\tau : m(\tau) = \kappa\}} r(\tau)c^{m(\tau)} = }

\displaystyle{ \sum_{i \in \mathbb{N}^k} \sum_{\kappa \in \mathbb{N}^k} \sum_{\{\tau : m(\tau) = \kappa\}} r(\tau)c^{m(\tau)}\frac{c^{i-m(\tau)}}{(i-m(\tau))!}\, z^i }

This completes the proof! It’s just algebra, but it seems a bit magical, so we’re trying to understand it better.

I hope you see how amazing this result is. If you know quantum mechanics and coherent states you’ll understand what I mean. A coherent state is the "best quantum approximation" to a classical state, but we don’t expect this quantum state to be exactly time-independent when the corresponding classical state is, except in very special cases, like when the Hamiltonian is quadratic in the creation and annihilation operators. Here we are getting a result like that much more generally… but only given the "complex balanced" condition.

An example

We’ve already seen one example of the theorem, back in Part 7. We had this stochastic Petri net:

We saw that the rate equation is just the logistic equation, familiar from population biology. The equilibrium solution is complex balanced, because pairs of amoebas are getting created at the same rate as they’re getting destroyed, and single amoebas are getting created at the same rate as they’re getting destroyed.

So, the Anderson–Craciun–Kurtz theorem guarantees that there’s an equilibrium solution of the master equation where the number of amoebas is distributed according to a Poisson distribution. And, we actually checked that this was true!

Next time we’ll look at another example.

56 Comments | mathematics, networks, probability | Permalink
Posted by John Baez

Fool’s Gold

12 September, 2011

My favorite Platonic solids are the regular dodecahedron, with 12 faces and 20 corners:

and the regular icosahedron, with 12 corners and 20 faces:

They are close relatives, with all the same symmetries… but what excites me most is that they have 5-fold symmetry. It’s a theorem that no crystal can have 5-fold symmetry. So, we might wonder whether these shapes occur in nature… and if they don’t, how people dreamt up these shapes in the first place.

It’s widely believed that the Pythagoreans dreamt up the regular dodecahedron after seeing crystals of iron pyrite—the mineral also known as ‘fool’s gold’. Nobody has any proof of this. However, there were a lot of Pythagoreans in Sicily back around 500 BC, and also a lot of pyrite. And, it’s fairly common for pyrite to form crystals like this:

This crystal is a ‘pyritohedron’. It looks similar to regular dodecahedron—but it’s not! At the molecular level, iron pyrite has little crystal cells with cubic symmetry. But these cubes can form a pyritohedron:


(By the way, you can click on any of these pictures for more information.)

You’ll notice that the front face of this pyritohedron is like a staircase with steps that go up 2 cubes for each step forwards. In other words, it’s a staircase with slope 2. That’s the key to the math here! By definition, the pyritohedron has faces formed by planes at right angles to these 12 vectors:

\begin{array}{cccc} (2,1,0) & (2,-1,0) & (-2,1,0) & (-2,-1,0) \\ (1,0,2) & (-1,0,2) & (1,0,-2) & (-1,0,-2) \\ (0,2,1) & (0,2,-1) & (0,-2,1) & (0,-2,-1) \\ \end{array}

On the other hand, a regular dodecahedron has faces formed by the planes at right angles to some very similar vectors, where the number 2 has been replaced by this number, called the golden ratio:

\displaystyle {\Phi = \frac{\sqrt{5} + 1}{2}}

Namely, these vectors:

\begin{array}{cccc} (\Phi,1,0) & (\Phi,-1,0) & (-\Phi,1,0) & (-\Phi,-1,0) \\ (1,0,\Phi) & (-1,0,\Phi) & (1,0,-\Phi) & (-1,0,-\Phi) \\ (0,\Phi,1) & (0,\Phi,-1) & (0,-\Phi,1) & (0,-\Phi,-1) \\ \end{array}

Since

\Phi \approx 1.618...

the golden ratio is not terribly far from 2. So, the pyritohedron is a passable attempt at a regular dodecahedron. Perhaps it was even good enough to trick the Pythagoreans into inventing the real thing.

If so, we can say: fool’s gold made a fool’s golden ratio good enough to fool the Greeks.

At this point I can’t resist a digression. You get the Fibonacci numbers by starting with two 1’s and then generating a list of numbers where each is the sum of the previous two:

1, 1, 2, 3, 5, 8, 13, …

The ratios of consecutive Fibonacci numbers get closer and closer to the golden ratio. For example:

\begin{array}{ccl} 1/1 &=& 1 \\ 2/1 &=& 2 \\ 3/2 &=& 1.5 \\ 5/3 &=& 1.6666.. \\ 8/5 &=& 1.6 \\ \end{array}

and so on. So, in theory, we could use these ratios to make cubical crystals that come closer and closer to a regular dodecahedron!

And in fact, pyrite doesn’t just form the 2/1 pyritohedron I showed you earlier. Sometimes it forms a 3/2 pyritohedron! This is noticeably better. The 2/1 version looks like this:

while the 3/2 version looks like this:

Has anyone ever seen a 5/3 pyritohedron? That would be even better. It would be quite hard to distinguish by eye from a true regular dodecahedron. Unfortunately, I don’t think iron pyrite forms such subtle crystals.

Okay. End of digression. But there’s another trick we can play!

These 12 vectors I mentioned:

\begin{array}{cccc} (\Phi,1,0) & (\Phi,-1,0) & (-\Phi,1,0) & (-\Phi,-1,0) \\ (1,0,\Phi) & (-1,0,\Phi) & (1,0,-\Phi) & (-1,0,-\Phi) \\ (0,\Phi,1) & (0,\Phi,-1) & (0,-\Phi,1) & (0,-\Phi,-1) \\ \end{array}

besides being at right angles to the faces of the dodecahedron, are also the corners of the icosahedron!

And if we use the number 2 here instead of the number \Phi, we get the vertices of a so-called pseudoicosahedron. Again, this can be made out of cubes:



However, nobody seems to think the Greeks ever saw a crystal shaped like a pseudoicosahedron! The icosahedron is first mentioned in Book XIII of Euclid’s Elements, which speaks of:

the five so-called Platonic figures which, however, do not belong to Plato, three of the five being due to the Pythagoreans, namely the cube, the pyramid, and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus.

So, maybe Theaetetus discovered the icosahedron. Indeed, Benno Artmann has argued that this shape was the first mathematical object that was a pure creation of human thought, not inspired by anything people saw!

That idea is controversial. It leads to some fascinating puzzles, like: did the Scots make stone balls shaped like Platonic solids back in 2000 BC? For more on these puzzles, try this:

• John Baez, Who discovered the icosahedron?

But right now I want to head in another direction. It turns out iron pyrite can form a crystal shaped like a pseudoicosahedron! And as Johan Kjellman pointed out to me, one of these crystals was recently auctioned off… for only 47 dollars!

It’s beautiful:

So: did the Greeks ever seen one of these? Alas, we may never know.

For more on these ideas, see:

• John Baez, My favorite numbers: 5.

• John Baez, Tales of the dodecahedron: from Pythagoras to Plato to Poincaré.

• John Baez, This Week’s Finds in Mathematical Physics, "week241" and "week283".

• Ian O. Angell and Moreton Moore, Projections of cubic crystals, International Union of Crystallography.

To wrap up, I should admit that icosahedra and dodecahedra show up in many other places in nature—but probably too small for the ancient Greeks to see. Here are some sea creatures magnified 50 times:

And here’s a virus:


The gray bar on top is 10 nanometers long, while the bar on bottom is just 5 nanometers long.

The mathematics of viruses with 5-fold symmetry is fascinating. Just today, I learned of Reidun Twarock‘s recent discoveries in this area:

• Reidun Twarock, Mathematical virology: a novel approach to the structure and assembly of viruses, Phil. Trans. R. Soc. A 364 (2006), 3357-3373.

Most viruses with 5-fold symmetry have protein shells in patterns based on the same math as geodesic domes:

 

But some more unusual viruses, like polyomavirus and simian virus 40, use more sophisticated patterns made of two kinds of tiles:

They still have 5-fold symmetry, but these patterns are spherical versions of Penrose tilings! A Penrose tiling is a nonrepeating pattern, typically with approximate 5-fold symmetry, made out of two kinds of tiles:

To understand these more unusual viruses, Twarock needed to use some very clever math:

• Thomas Keef and Reidun Twarock, Affine extensions of the icosahedral group with applications to the three-dimensional organisation of simple viruses, J. Math. Biol. 59 (2009), 287-313.

But that’s another story for another day!


27 Comments | biology, chemistry, mathematics | Permalink
Posted by John Baez

Network Theory (Part 8)

9 September, 2011

Summer vacation is over. Time to get back to work!

This month, before he goes to Oxford to begin a master’s program in Mathematics and the Foundations of Computer Science, Brendan Fong is visiting the Centre for Quantum Technologies and working with me on stochastic Petri nets. He’s proved two interesting results, which he wants to explain.

To understand what he’s done, you need to know how to get the rate equation and the master equation from a stochastic Petri net. We’ve almost seen how. But it’s been a long time since the last article in this series, so today I’ll start with some review. And at the end, just for fun, I’ll say a bit more about how Feynman diagrams show up in this theory.

Since I’m an experienced teacher, I’ll assume you’ve forgotten everything I ever said.

(This has some advantages. I can change some of my earlier terminology—improve it a bit here and there—and you won’t even notice.)

Stochastic Petri nets

Definition. A Petri net consists of a set S of species and a set T of transitions, together with a function

i : S \times T \to \mathbb{N}

saying how many things of each species appear in the input for each transition, and a function

o: S \times T \to \mathbb{N}

saying how many things of each species appear in the output.

We can draw pictures of Petri nets. For example, here’s a Petri net with two species and three transitions:

It should be clear that the transition ‘predation’ has one wolf and one rabbit as input, and two wolves as output.

A ‘stochastic’ Petri net goes further: it also says the rate at which each transition occurs.

Definition. A stochastic Petri net is a Petri net together with a function

r: T \to [0,\infty)

giving a rate constant for each transition.

Master equation versus rate equation

Starting from any stochastic Petri net, we can get two things. First:

• The master equation. This says how the probability that we have a given number of things of each species changes with time.

• The rate equation. This says how the expected number of things of each species changes with time.

The master equation is stochastic: it describes how probabilities change with time. The rate equation is deterministic.

The master equation is more fundamental. It’s like the equations of quantum electrodynamics, which describe the amplitudes for creating and annihilating individual photons. The rate equation is less fundamental. It’s like the classical Maxwell equations, which describe changes in the electromagnetic field in a deterministic way. The classical Maxwell equations are an approximation to quantum electrodynamics. This approximation gets good in the limit where there are lots of photons all piling on top of each other to form nice waves.

Similarly, the rate equation can be derived from the master equation in the limit where the number of things of each species become large, and the fluctuations in these numbers become negligible.

But I won’t do this derivation today! Nor will I probe more deeply into the analogy with quantum field theory, even though that’s my ultimate goal. Today I’m content to remind you what the master equation and rate equation are.

The rate equation is simpler, so let’s do that first.

The rate equation

Suppose we have a stochastic Petri net with k different species. Let x_i be the number of things of the ith species. Then the rate equation looks like this:

\displaystyle{ \frac{d x_i}{d t} = ??? }

It’s really a bunch of equations, one for each 1 \le i \le k. But what is the right-hand side?

The right-hand side is a sum of terms, one for each transition in our Petri net. So, let’s start by assuming our Petri net has just one transition.

Suppose the ith species appears as input to this transition m_i times, and as output n_i times. Then the rate equation is

\displaystyle{ \frac{d x_i}{d t} = r (n_i - m_i) x_1^{m_1} \cdots x_k^{m_k} }

where r is the rate constant for this transition.

That’s really all there is to it! But we can make it look nicer. Let’s make up a vector

x = (x_1, \dots , x_k) \in [0,\infty)^k

that says how many things there are of each species. Similarly let’s make up an input vector

m = (m_1, \dots, m_k) \in \mathbb{N}^k

and an output vector

n = (n_1, \dots, n_k) \in \mathbb{N}^k

for our transition. To be cute, let’s also define

\displaystyle{ x^m = x_1^{m_1} \cdots x_k^{m_k} }

Then we can write the rate equation for a single transition like this:

\displaystyle{ \frac{d x}{d t} = r (n-m) x^m }

Next let’s do a general stochastic Petri net, with lots of transitions. Let’s write T for the set of transitions and r(\tau) for the rate constant of the transition \tau \in T. Let n(\tau) and m(\tau) be the input and output vectors of the transition \tau. Then the rate equation is:

\displaystyle{ \frac{d x}{d t} = \sum_{\tau \in T} r(\tau) (n(\tau) - m(\tau)) x^{m(\tau)} }

For example, consider our rabbits and wolves:

Suppose

• the rate constant for ‘birth’ is \beta,

• the rate constant for ‘predation’ is \gamma,

• the rate constant for ‘death’ is \delta.

Let x_1(t) be the number of rabbits and x_2(t) the number of wolves at time t. Then the rate equation looks like this:

\displaystyle{ \frac{d x_1}{d t} = \beta x_1 - \gamma x_1 x_2 }

\displaystyle{ \frac{d x_2}{d t} = \gamma x_1 x_2 - \delta x_2 }

If you stare at this, and think about it, it should make perfect sense. If it doesn’t, go back and read Part 2.

The master equation

Now let’s do something new. In Part 6 I explained how to write down the master equation for a stochastic Petri net with just one species. Now let’s generalize that. Luckily, the ideas are exactly the same.

So, suppose we have a stochastic Petri net with k different species. Let \psi_{n_1, \dots, n_k} be the probability that we have n_1 things of the first species, n_2 of the second species, and so on. The master equation will say how all these probabilities change with time.

To keep the notation clean, let’s introduce a vector

n = (n_1, \dots, n_k) \in \mathbb{N}^k

and let

\psi_n = \psi_{n_1, \dots, n_k}

Then, let’s take all these probabilities and cook up a formal power series that has them as coefficients: as we’ve seen, this is a powerful trick. To do this, we’ll bring in some variables z_1, \dots, z_k and write

\displaystyle{ z^n = z_1^{n_1} \cdots z_k^{n_k} }

as a convenient abbreviation. Then any formal power series in these variables looks like this:

\displaystyle{ \Psi = \sum_{n \in \mathbb{N}^k} \psi_n z^n }

We call \Psi a state if the probabilities sum to 1 as they should:

\displaystyle{ \sum_n \psi_n = 1 }

The simplest example of a state is a monomial:

\displaystyle{ z^n = z_1^{n_1} \cdots z_k^{n_k} }

This is a state where we are 100% sure that there are n_1 things of the first species, n_2 of the second species, and so on. We call such a state a pure state, since physicists use this term to describe a state where we know for sure exactly what’s going on. Sometimes a general state, one that might not be pure, is called mixed.

The master equation says how a state evolves in time. It looks like this:

\displaystyle{ \frac{d}{d t} \Psi(t) = H \Psi(t) }

So, I just need to tell you what H is!

It’s called the Hamiltonian. It’s a linear operator built from special operators that annihilate and create things of various species. Namely, for each state 1 \le i \le k we have an annihilation operator:

\displaystyle{ a_i \Psi = \frac{d}{d z_i} \Psi }

and a creation operator:

\displaystyle{ a_i^\dagger \Psi = z_i \Psi }

How do we build H from these? Suppose we’ve got a stochastic Petri net whose set of transitions is T. As before, write r(\tau) for the rate constant of the transition \tau \in T, and let n(\tau) and m(\tau) be the input and output vectors of this transition. Then:

\displaystyle{ H = \sum_{\tau \in T} r(\tau) \, ({a^\dagger}^{n(\tau)} - {a^\dagger}^{m(\tau)}) \, a^{m(\tau)} }

where as usual we’ve introduce some shorthand notations to keep from going insane. For example:

a^{m(\tau)} = a_1^{m_1(\tau)} \cdots a_k^{m_k(\tau)}

and

{a^\dagger}^{m(\tau)} = {a_1^\dagger }^{m_1(\tau)} \cdots {a_k^\dagger}^{m_k(\tau)}

Now, it’s not surprising that each transition \tau contributes a term to H. It’s also not surprising that this term is proportional to the rate constant r(\tau). The only tricky thing is the expression

\displaystyle{ ({a^\dagger}^{n(\tau)} - {a^\dagger}^{m(\tau)})\, a^{m(\tau)} }

How can we understand it? The basic idea is this. We’ve got two terms here. The first term:

\displaystyle{ {a^\dagger}^{n(\tau)} a^{m(\tau)} }

describes how m_i(\tau) things of the ith species get annihilated, and n_i(\tau) things of the ith species get created. Of course this happens thanks to our transition \tau. The second term:

\displaystyle{ - {a^\dagger}^{m(\tau)} a^{m(\tau)} }

is a bit harder to understand, but it says how the probability that nothing happens—that we remain in the same pure state—decreases as time passes. Again this happens due to our transition \tau.

In fact, the second term must take precisely the form it does to ensure ‘conservation of total probability’. In other words: if the probabilities \psi_n sum to 1 at time zero, we want these probabilities to still sum to 1 at any later time. And for this, we need that second term to be what it is! In Part 6 we saw this in the special case where there’s only one species. The general case works the same way.

Let’s look at an example. Consider our rabbits and wolves yet again:

and again suppose the rate constants for birth, predation and death are \beta, \gamma and \delta, respectively. We have

\displaystyle{ \Psi = \sum_n \psi_n z^n }

where

\displaystyle{ z^n = z_1^{n_1} z_2^{n_2} }

and \psi_n = \psi_{n_1, n_2} is the probability of having n_1 rabbits and n_2 wolves. These probabilities evolve according to the equation

\displaystyle{ \frac{d}{d t} \Psi(t) = H \Psi(t) }

where the Hamiltonian is

H = \beta B + \gamma C + \delta D

and B, C and D are operators describing birth, predation and death, respectively. (B stands for birth, D stands for death… and you can call predation ‘consumption’ if you want something that starts with C. Besides, ‘consumer’ is a nice euphemism for ‘predator’.) What are these operators? Just follow the rules I described:

\displaystyle{ B = {a_1^\dagger}^2 a_1 - a_1^\dagger a_1 }

\displaystyle{ C = {a_2^\dagger}^2 a_1 a_2 - a_1^\dagger a_2^\dagger a_1 a_2 }

\displaystyle{ D = a_2 - a_2^\dagger a_2 }

In each case, the first term is easy to understand:

• Birth annihilates one rabbit and creates two rabbits.

• Predation annihilates one rabbit and one wolf and creates two wolves.

• Death annihilates one wolf.

The second term is trickier, but I told you how it works.

Feynman diagrams

How do we solve the master equation? If we don’t worry about mathematical rigor too much, it’s easy. The solution of

\displaystyle{ \frac{d}{d t} \Psi(t) = H \Psi(t) }

should be

\displaystyle{ \Psi(t) = e^{t H} \Psi(0) }

and we can hope that

\displaystyle{ e^{t H} = 1 + t H + \frac{(t H)^2}{2!} + \cdots }

so that

\displaystyle{ \Psi(t) = \Psi(0) + t H \Psi(0) + \frac{t^2}{2!} H^2 \Psi(0) + \cdots }

Of course there’s always the question of whether this power series converges. In many contexts it doesn’t, but that’s not necessarily a disaster: the series can still be asymptotic to the right answer, or even better, Borel summable to the right answer.

But let’s not worry about these subtleties yet! Let’s just imagine our rabbits and wolves, with Hamiltonian

H = \beta B + \gamma C + \delta D

Now, imagine working out

\Psi(t) = \displaystyle{ \Psi(0) + t H \Psi(0) + \frac{t^2}{2!} H^2 \Psi(0) + \frac{t^3}{3!} H^3 \Psi(0) + \cdots }

We’ll get lots of terms involving products of B, C and D hitting our original state \Psi(0). And we can draw these as diagrams! For example, suppose we start with one rabbit and one wolf. Then

\Psi(0) = z_1 z_2

And suppose we want to compute

\displaystyle{ H^3 \Psi(0) = (\beta B + \gamma C + \delta D)^3 \Psi(0) }

as part of the task of computing \Psi(t). Then we’ll get lots of terms: 27, in fact, though many will turn out to be zero. Let’s take one of these terms, for example the one proportional to:

D C B \Psi(0)

We can draw this as a sum of Feynman diagrams, including this:

In this diagram, we start with one rabbit and one wolf at top. As we read the diagram from top to bottom, first a rabbit is born (B), then predation occur (C), and finally a wolf dies (D). The end result is again a rabbit and a wolf.

This is just one of four Feynman diagrams we should draw in our sum for D C B \Psi(0), since either of the two rabbits could have been eaten, and either wolf could have died. So, the end result of computing

\displaystyle{ H^3 \Psi(0) }

will involve a lot of Feynman diagrams… and of course computing

\displaystyle{ \Psi(t) = \Psi(0) + t H \Psi(0) + \frac{t^2}{2!} H^2 \Psi(0) + \frac{t^3}{3!} H^3 \Psi(0) + \cdots }

will involve even more, even if we get tired and give up after the first few terms. So, this Feynman diagram business may seem quite tedious… and it may not be obvious how it helps.

But it does, sometimes!

Now is not the time for me to describe ‘practical’ benefits of Feynman diagrams. Instead, I’ll just point out one conceptual benefit. We started with what seemed like a purely computational chore, namely computing

\displaystyle{ \Psi(t) = \Psi(0) + t H \Psi(0) + \frac{t^2}{2!} H^2 \Psi(0) + \cdots }

But then we saw—at least roughly—how this series has a clear meaning! It can be written as a sum over diagrams, each of which represents a possible history of rabbits and wolves. So, it’s what physicists call a ‘sum over histories’.

Feynman invented the idea of a sum over histories in the context of quantum field theory. At the time this idea seemed quite mind-blowing, for various reasons. First, it involved elementary particles instead of everyday things like rabbits and wolves. Second, it involved complex ‘amplitudes’ instead of real probabilities. Third, it actually involved integrals instead of sums. And fourth, a lot of these integrals diverged, giving infinite answers that needed to be ‘cured’ somehow!

Now we’re seeing a sum over histories in a more down-to-earth context without all these complications. A lot of the underlying math is analogous… but now there’s nothing mind-blowing about it: it’s quite easy to understand. So, we can use this analogy to demystify quantum field theory a bit. On the other hand, thanks to this analogy, all sorts of clever ideas invented by quantum field theorists will turn out to have applications to biology and chemistry! So it’s a double win.

42 Comments | networks, physics, probability | Permalink
Posted by John Baez

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