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 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 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 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 We write it as

where is an abbreviation for

But for to be a **stochastic state** the numbers need to be probabilities, so we require that

and

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

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 th 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 for each list of numbers

Here I’m using another abbreviation:

If you calculate a bit, you’ll see

Thus, the probability of having things of the th species is equal to

This is precisely the definition of a **Poisson distribution** with mean equal to

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

More interestingly, they are eigenvectors of the annihilation operators

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

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

where is the creation operator:

The number operator has the property that

is the mean number of particles of the th species. If we calculate this for our coherent state we get

Here in the second step we used the general rule

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:

Since

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

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

These are eigenvectors of the rescaled annihilation operators:

This in turn means that

Here we used the general rule

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

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

says the expected number of particles of the th species in the state is *if we count these particles not one at a time, but in bunches of size *

### A simple test

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

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

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

We already know the mean of the rescaled number operator:

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

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

This implies

so

where we used our friends

and

So, the variance of the rescaled number of particles is

and the standard deviation is

Good, it goes to zero as 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 Can you prove this?

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

I want this to go to zero as

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:

Here

is the number of ways to partition an -element set into 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

We could plug this formula into

and then try to show the result goes to zero as 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.