<![CDATA[Coherence for Solutions of the Master Equation]]>

<![CDATA[Coherence for Solutions of the Master Equation]]>guest post by Arjun Jain

I am a master’s student in the physics department of the Indian Institute of Technology Roorkee. I’m originally from Delhi. Since some time now, I’ve been wanting to go into Mathematical Physics. I hope to do a PhD in that. Apart from maths and physics, I am also quite passionate about art and music.

Right now I am visiting John Baez at the Centre for Quantum Technologies, and we’re working on chemical reaction networks. This post can be considered as an annotation to the last paragraph of John’s paper, Quantum Techniques for Reaction Networks, where he raises the question of when a solution to the master equation that starts as a coherent state will remain coherent for all times. Remember, the ‘master equation’ describes the random evolution of collections of classical particles, and a ‘coherent state’ is one where the probability distribution of particles of each type is a Poisson distribution.

If you’ve been following the network theory series on this blog, you’ll know these concepts, and you’ll know the Anderson-Craciun-Kurtz theorem gives many examples of coherent states that remain coherent. However, all these are equilibrium solutions of the master equation: they don’t change with time. Moreover they are complex balanced equilibria: the rate at which any complex is produced equals the rate at which it is consumed.

There are also non-equilibrium examples where coherent states remain coherent. But they seem rather rare, and I would like to explain why. So, I will give a necessary condition for it to happen. I’ll give the proof first, and then discuss some simple examples. We will see that while the condition is necessary, it is not sufficient.

First, recall the setup. If you’ve been following the network theory series, you can skip the next section.

Reaction networks

Definition. A reaction network consists of:

• a finite set $S$ of species,

• a finite set $K$ of complexes, where a complex is a finite sum of species, or in other words, an element of $\mathbb{N}^S,$

• a graph with $K$ as its set of vertices and some set $T$ of edges.

You should have in mind something like this:

where our set of species is $S = \{A,B,C,D,E\},$ the complexes are things like $A + E,$ and the arrows are the elements of $T,$ called transitions or reactions. So, we have functions

$s , t : T \to K$

saying the source and target of each transition.

Definition. A stochastic reaction network is a reaction network together with a function $r: T \to (0,\infty)$ assigning a rate constant to each reaction.

From this we can write down the master equation, which describes how a stochastic state evolves in time:

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

Here $\Psi(t)$ is a vector in the stochastic Fock space, which is the space of formal power series in a bunch of variables, one for each species, and $H$ is an operator on this space, called the Hamiltonian.

From now on I’ll number the species with numbers from $1$ to $k,$ so

$S = \{1, \dots, k\}$

Then the stochastic Fock space consists of real formal power series in variables that I’ll call $z_1, \dots, z_k.$ We can write any of these power series as

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

where

$z^\ell = z_1^{\ell_1} \cdots z_k^{\ell_k}$

We have annihilation and creation operators on the stochastic Fock space:

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

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

and the Hamiltonian is built from these as follows:

$\displaystyle{ H = \sum_{\tau \in T} r(\tau) \, ({a^\dagger}^{t(\tau)} - {a^\dagger}^{s(\tau)}) \, a^{s(\tau)} }$

John explained this here (using slightly different notation), so I won’t go into much detail now, but I’ll say what all the symbols mean. Remember that the source of a transition $\tau$ is a complex, or list of natural numbers:

$s(\tau) = (s_1(\tau), \dots, s_k(\tau))$

So, the power $a^{s(\tau)}$ is really an abbreviation for a big product of annihilation operators, like this:

$\displaystyle{ a^{s(\tau)} = a_1^{s_1(\tau)} \cdots a_k^{s_k(\tau)} }$

This describes the annihilation of all the inputs to the transition $\tau.$ Similarly, we define

$\displaystyle{ {a^\dagger}^{s(\tau)} = {a_1^\dagger}^{s_1(\tau)} \cdots {a_k^\dagger}^{s_k(\tau)} }$

and

$\displaystyle{ {a^\dagger}^{t(\tau)} = {a_1^\dagger}^{t_1(\tau)} \cdots {a_k^\dagger}^{t_k(\tau)} }$

The result

Here’s the result:

Theorem. If a solution $\Psi(t)$ of the master equation is a coherent state for all times $t \ge 0,$ then $\Psi(0)$ must be complex balanced except for complexes of degree 0 or 1.

This requires some explanation.

First, saying that $\Psi(t)$ is a coherent state means that it is an eigenvector of all the annihilation operators. Concretely this means

$\Psi (t) = \displaystyle{\frac{e^{c(t) \cdot z}}{e^{c_1(t) + \cdots + c_k(t)}}}$

where

$c(t) = (c_1(t), \dots, c_k(t)) \in [0,\infty)^k$

and

$z = (z_1, \dots, z_k)$

It will be helpful to write

$\mathbf{1}= (1,1,1,...)$

so we can write

$\Psi (t) = \displaystyle{ e^{c(t) \cdot (z - \mathbf{1})} }$

Second, we say that a complex has degree $d$ if it is a sum of exactly $d$ species. For example, in this reaction network:

the complexes $A + C$ and $B + E$ have degree 2, while the rest have degree 1. We use the word ‘degree’ because each complex $\ell$ gives a monomial

$z^\ell = z_1^{\ell_1} \cdots z_k^{\ell_k}$

and the degree of the complex is the degree of this monomial, namely

$\ell_1 + \cdots + \ell_k$

Third and finally, we say a solution $\Psi(t)$ of the master equation is complex balanced for a specific complex $\ell$ if the total rate at which that complex is produced equals the total rate at which it’s destroyed.

Now we are ready to prove the theorem:

Proof. Consider the master equation

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

Assume that $\Psi(t)$ is a coherent state for all $t \ge 0.$ This means

$\Psi (t) = \displaystyle{ e^{c(t) \cdot (z - \mathbf{1})} }$

For convenience, we write $c(t)$ simply as $c,$ and similarly for the components $c_i$ . Then we have

$\displaystyle{ \frac{d\Psi(t)}{dt} = (\dot{c} \cdot (z - \mathbf{1})) \, e^{c \cdot (z - \mathbf{1})} }$

On the other hand, the master equation gives

$\begin{array}{ccl} \displaystyle {\frac{d\Psi(t)}{dt}} &=& \displaystyle{ \sum_{\tau \in T} r(\tau) \, ({a^\dagger}^{t(\tau)} - {a^\dagger}^{s(\tau)}) \, a^{s(\tau)} e^{c \cdot (z - \mathbf{1})} } \\ \\ &=& \displaystyle{\sum_{\tau \in T} c^{t(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)}) e^{c \cdot (z - \mathbf{1})} } \end{array}$

So,

$\displaystyle{ (\dot{c} \cdot (z - \mathbf{1})) \, e^{c \cdot (z - \mathbf{1})} =\sum_{\tau \in T} c^{t(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)}) e^{c \cdot (z - \mathbf{1})} }$

As a result, we get

$\displaystyle{ \dot{c}\cdot z -\dot{c}\cdot\mathbf{1} = \sum_{\tau \in T} c^{s(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)}) }.$

Comparing the coefficients of all $z^\ell,$ we obtain the following. For $\ell = 0,$ which is the only complex of degree zero, we get

$\displaystyle { \sum_{\tau: t(\tau)=0} r(\tau) c^{s(\tau)} - \sum_{\tau\;:\; s(\tau)= 0} r(\tau) c^{s(\tau)} = -\dot{c}\cdot\mathbf{1} }$

For the complexes $\ell$ of degree one, we get these equations:

$\displaystyle { \sum_{\tau\;:\; t(\tau)=(1,0,0,\dots)} r(\tau) c^{s(\tau)} - \sum_{\tau \;:\;s(\tau)=(1,0,0,\dots)} r(\tau) c^{s(\tau)}= \dot{c_1} }$

$\displaystyle { \sum_{\tau\; :\; t(\tau)=(0,1,0,\dots)} r(\tau) c^{s(\tau)} - \sum_{\tau\;:\; s(\tau)=(0,1,0,\dots)} r(\tau) c^{s(\tau)} = \dot{c_2} }$

and so on. For all the remaining complexes $\ell$ we have

$\displaystyle { \sum_{\tau\;:\; t(\tau)=\ell} r(\tau) c^{s(\tau)} = \sum_{\tau \;:\; s(\tau)=\ell} r(\tau) c^{s(\tau)} }.$

This says that the total rate at which this complex is produced equals the total rate at which it’s destroyed. So, our solution of the master equation is complex balanced for all complexes $\ell$ of degree greater than one. This is our necessary condition. █

To illustrate the theorem, I’ll consider three simple examples. The third example shows that the condition in the theorem, though necessary, is not sufficient. Note that our proof also gives a necessary and sufficient condition for a coherent state to remain coherent: namely, that all the equations we listed hold, not just initially but for all times. But this condition seems a bit complicated.

Introducing amoebae into a Petri dish

Suppose that there is an inexhaustible supply of amoebae, randomly floating around in a huge pond. Each time an amoeba comes into our collection area, we catch it and add it to the population of amoebae in the Petri dish. Suppose that the rate constant for this process is 3.

So, the Hamiltonian is $3(a^\dagger -1).$ If we start with a coherent state, say

$\displaystyle { \Psi(0)=\frac{e^{cz}}{e^c} }$

then

$\displaystyle { \Psi(t) = e^{3(a^\dagger -1)t} \; \frac{e^{cz}}{e^c} = \frac{e^{(c+3t)z}}{e^{c+3t}} }$

which is coherent at all times.

We can see that the condition of the theorem is satisfied, as all the complexes in the reaction network have degree 0 or 1.

Amoebae reproducing and competing

This example shows a Petri dish with one species, amoebae, and two transitions: fission and competition. We suppose that the rate constant for fission is 2, while that for competition is 1. The Hamiltonian is then

$H= 2({a^\dagger}^2-a^\dagger)a + (a^\dagger-{a^\dagger}^2)a^2$

If we start off with the coherent state

$\displaystyle{\Psi(0) = \frac{e^{2z}}{e^2}}$

we find that

$\displaystyle {\Psi(t)=e^{2(z^2-z)2+(z-z^2)4} \; \Psi(0)}=\Psi(0)$

which is coherent. It should be noted that the chosen initial state

$\displaystyle{ \frac{e^{2z}}{e^2}}$

was a complex balanced equilibrium solution. So, the Anderson–Craciun–Kurtz Theorem applies to this case.

Amoebae reproducing, competing, and being introduced

This is a combination of the previous two examples, where apart from ongoing reproduction and competition, amoebae are being introduced into the dish with a rate constant 3.

As in the above examples, we might think that coherent states could remain coherent forever here too. Let’s check that.

Assuming that this was true, if

$\displaystyle{\Psi(t) = \frac{e^{c(t)z}}{e^{c(t)}} }$

then $c(t)$ would have to satisfy the following:

$\dot{c}(t) = c(t)^2 + 3 -2c(t)$

and

$c(t)^2=2c(t)$

Using the second equation, we get

$\dot{c}(t) = 3 \Rightarrow c = 3t+ c_0$

But this is certainly not a solution of the second equation. So, here we find that initially coherent states do not remain remain coherent for all times.

However, if we choose

$\displaystyle{\Psi(0) = \frac{e^{2z}}{e^2}}$

then this coherent state is complex balanced except for complexes of degree 1, since it was in the previous example, and the only new feature of this example, at time zero, is that single amoebas are being introduced—and these are complexes of degree 1. So, the condition of the theorem does hold.

So, the condition in the theorem is necessary but not sufficient. However, it is easy to check, and we can use it to show that in many cases, coherent states must cease to be coherent.

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