<![CDATA[The Large-Number Limit for Reaction Networks (Part 2)]]>I’ve been talking a lot about ‘stochastic mechanics’, which is like quantum mechanics but with probabilities replacing amplitudes. In Part 1 of this mini-series I started telling you about the ‘large-number limit’ in stochastic mechanics. It turns out this is mathematically analogous to the ‘classical limit’ of quantum mechanics, where Planck’s constant \hbar goes to zero.

There’s a lot more I need to say about this, and lots more I need to figure out. But here’s one rather easy thing.

In quantum mechanics, ‘coherent states’ are a special class of quantum states that are very easy to calculate with. In a certain precise sense they are the best quantum approximations to classical states. This makes them good tools for studying the classical limit of quantum mechanics. As \hbar \to 0, they reduce to classical states where, for example, a particle has a definite position and momentum.

We can borrow this strategy to study the large-number limit of stochastic mechanics. We’ve run into coherent states before in our discussions here. Now let’s see how they work in the large-number limit!

Coherent states

For starters, let’s recall what coherent states are. We’ve got k different kinds of particles, and we call each kind a species. We describe the probability that we have some number of particles of each kind using a ‘stochastic state’. For starters, this is a formal power series in variables z_1, \dots, z_k. We write it as

\displaystyle{\Psi = \sum_{\ell \in \mathbb{N}^k} \psi_\ell z^\ell }

where z^\ell is an abbreviation for

z_1^{\ell_1} \cdots z_k^{\ell_k}

But for \Psi to be a stochastic state the numbers \psi_\ell need to be probabilities, so we require that

\psi_\ell \ge 0

and

\displaystyle{ \sum_{\ell \in \mathbb{N}^k} \psi_\ell = 1}

Sums of coefficients like this show up so often that it’s good to have an abbreviation for them:

\displaystyle{ \langle \Psi \rangle = \sum_{\ell \in \mathbb{N}^k} \psi_\ell}

Now, a coherent state is a stochastic state where the numbers of particles of each species are independent random variables, and the number of the ith species is distributed according to a Poisson distribution.

Since we can pick ithe means of these Poisson distributions to be whatever we want, we get a coherent state \Psi_c for each list of numbers c \in [0,\infty)^k:

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

Here I’m using another abbreviation:

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

If you calculate a bit, you’ll see

\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} }

Thus, 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.

What are the main properties of coherent states? For starters, they are indeed states:

\langle \Psi_c \rangle = 1

More interestingly, they are eigenvectors of the annihilation operators

a_i = \displaystyle{ \frac{\partial}{\partial z_i} }

since when you differentiate an exponential you get back an exponential:

\begin{array}{ccl} a_i \Psi_c &=& \displaystyle{ \frac{\partial}{\partial z_i} \frac{e^{c \cdot z}}{e^c} } \\ \\ &=& c_i \Psi_c \end{array}

We can use this fact to check that in this coherent state, the mean number of particles of the ith species really is c_i. For this, we introduce the number operator

N_i = a_i^\dagger a_i

where a_i^\dagger is the creation operator:

(a_i^\dagger \Psi)(z) = z_i \Psi(z)

The number operator has the property that

\langle N_i \Psi \rangle

is the mean number of particles of the ith species. If we calculate this for our coherent state \Psi_c, we get

\begin{array}{ccl} \langle a_i^\dagger a_i \Psi_c \rangle &=& c_i \langle a_i^\dagger \Psi_c \rangle \\ \\ &=& c_i \langle \Psi_c \rangle \\ \\ &=& c_i \end{array}

Here in the second step we used the general rule

\langle a_i^\dagger \Phi \rangle = \langle \Phi \rangle

which is easy to check.

Rescaling

Now let’s see how coherent states work in the large-numbers limit. For this, let’s use the rescaled annihilation, creation and number operators from Part 1. They look like this:

A_i = \hbar \, a_i

C_i = a_i^\dagger

\widetilde{N}_i = C_i A_i

Since

\widetilde{N}_i = \hbar N_i

the point is that the rescaled number operator counts particles not one at a time, but in bunches of size 1/\hbar. For example, if \hbar is the reciprocal of Avogadro’s number, we are counting particles in ‘moles’. So, \hbar \to 0 corresponds to a large-number limit.

To flesh out this idea some more, let’s define rescaled coherent states:

\widetilde{\Psi}_c = \Psi_{c/\hbar}

These are eigenvectors of the rescaled annihilation operators:

\begin{array}{ccl} A_i \widetilde{\Psi}_c &=& \hbar a_i \Psi_{c/\hbar} \\ \\ &=& c_i \Psi_{c/\hbar} \\ \\ &=& c_i \widetilde{\Psi}_c \end{array}

This in turn means that

\begin{array}{ccl} \langle \widetilde{N}_i \widetilde{\Psi}_c \rangle &=& \langle C_i A_i \widetilde{\Psi}_c \rangle \\ \\ &=& c_i \langle C_i \widetilde{\Psi}_c \rangle \\ \\ &=& c_i \langle \widetilde{\Psi}_c \rangle \\ \\ &=& c_i \end{array}

Here we used the general rule

\langle C_i \Phi \rangle = \langle \Phi \rangle

which holds because the ‘rescaled’ creation operator C_i is really just the usual creation operator, which obeys this rule.

What’s the point of all this fiddling around? Simply this. The equation

\langle \widetilde{N}_i \widetilde{\Psi}_c \rangle = c_i

says the expected number of particles of the ith species in the state \widetilde{\Psi}_c is c_i, if we count these particles not one at a time, but in bunches of size 1/\hbar.

A simple test

As a simple test of this idea, let’s check that as \hbar \to 0, the standard deviation of the number of particles in the state \Psi_c goes to zero… where we count particle using the rescaled number operator.

The variance of the rescaled number operator is, by definition,

\langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle - \langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle^2

and the standard deviation is the square root of the variance.

We already know the mean of the rescaled number operator:

\langle \widetilde{N}_i \widetilde{\Psi}_c \rangle = c_i

So, the main thing we need to calculate is the mean of its square:

\langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle

For this we will use the commutation relation derived last time:

[A_i , C_i] = \hbar

This implies

\begin{array}{ccl} \widetilde{N}_i^2 &=& C_i A_i C_i A_i \\ \\ &=& C_i (C_i A_i + \hbar) A_i \\ \\ &=& C_i^2 A_i^2 + \hbar C_i A_i \end{array}

so

\begin{array}{ccl} \langle \widetilde{N}_i^2\widetilde{\Psi}_c \rangle &=& \langle (C_i^2 A_i^2 + \hbar C_i A_i) \Psi_c \rangle \\ \\ &=& c_i^2 + \hbar c_i \end{array}

where we used our friends

A_i \Psi_c = c_i \Psi_c

and

\langle C_i \Phi \rangle = \langle \Phi \rangle

So, the variance of the rescaled number of particles is

\begin{array}{ccl} \langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle - \langle \widetilde{N}_i \widetilde{\Psi}_c \rangle^2 &=& c_i^2 + \hbar c_i - c_i^2 \\ \\ &=& \hbar c_i \end{array}

and the standard deviation is

(\hbar c_i)^{1/2}

Good, it goes to zero as \hbar \to 0! And the square root is just what you’d expect if you’ve thought about stuff like random walks or the central limit theorem.

A puzzle

I feel sure that in any coherent state, not only the variance but also all the higher moments of the rescaled number operators go to zero as \hbar \to 0. Can you prove this?

Here I mean the moments after the mean has been subtracted. The pth moment is then

\langle (\widetilde{N}_i - c_i)^p \; \widetilde{\Psi}_c \rangle

I want this to go to zero as \hbar \to 0.

Here’s a clue that should help. First, there’s a textbook formula for the higher moments of Poisson distributions without the mean subtracted. If I understand it correctly, it gives this:

\displaystyle{ \langle N_i^m \; \Psi_c \rangle = \sum_{j = 1}^m {c_i}^j \; \left\{ \begin{array}{c} m \\ j \end{array} \right\} }

Here

\displaystyle{ \left\{ \begin{array}{c} m \\ j \end{array} \right\} }

is the number of ways to partition an m-element set into j nonempty subsets. This is called Stirling’s number of the second kind. This suggests that there’s some fascinating combinatorics involving coherent states. That’s exactly the kind of thing I enjoy, so I would like to understand this formula someday… but not today! I just want something to go to zero!

If I rescale the above formula, I seem to get

\begin{array}{ccl} \langle \widetilde{N}_i^m \; \widetilde{\Psi}_c \rangle &=& \hbar^m \langle N_i^m \Psi_{c/\hbar} \rangle \\ \\ &=& \hbar^m \; \displaystyle{ \sum_{j = 1}^m \left(\frac{c_i}{\hbar}\right)^j \left\{ \begin{array}{c} m \\ j \end{array} \right\} } \end{array}

We could plug this formula into

\langle (\widetilde{N}_i - c_i)^p \; \widetilde{\Psi}_c \rangle = \displaystyle{ \sum_{m = 0}^p \, \binom{m}{p} \; \langle \widetilde{N}_i^m \; \widetilde{\Psi}_c \rangle \, (-c_i)^{p - m} }

and then try to show the result goes to zero as \hbar \to 0. But I don’t have the energy to do that… not right now, anyway!

Maybe you do. Or maybe you can think of a better approach to solving this problem. The answer must be well-known, since the large-number limit of a Poisson distribution is a very important thing.

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