# The Azimuth Project Experiments in multiple equilibrium states

We have so far always first considered the rate equation and then talked about the master equation and given heuristic arguments on how they are different. I wanted to start from the master equation and show how the rate equation arises from this. That’s exactly what we are going to do in this post.

## The master equation

We can also arrive at a master equation. Now say we have a probability,

$\psi_{i_1...i_k}$ of having $i_1$ things in state 1, $i_k$ things in state $k$ etc. We write

$\psi = \sum \psi_{i_1...i_k}z_1^{i_1}\cdots z_k^{i_k}$

and $a_i^\dagger \psi = z_i \psi$, $a_i\psi = \frac{\partial}{\partial z_i} \psi$ then the master equation says

$\frac{d}{dt}\psi = H\psi$

Each transition corresponds to an operator. There is a term $a_1^{\dagger n_1}\cdots a_k^{\dagger n_1} a_1^{m_1}\cdots a_k^{m_k}$

In general, the master equation is written as

$H = \sum_{\tau \in T} r(\tau) (a_1^{\dagger n_1(\tau)}\cdots a_k^{\dagger n_k(\tau)}a_1^{m_1(\tau)}\cdots a_k^{m_k(\tau)}- N_1^{\underline m_1}\cdots N_k^{\underline m_k})$

We will attempt to solve $H\Psi = 0$, by trying

$\Psi = \prod_{i=1}^k \sum_{n_i = 0}^\infty \frac{\alpha_i^{n_i}}{n_i!} z_i^{n_i} = \prod_{i=1}^k e^{\alpha_i z_i}$
$(a_1^{\dagger n_1(\tau)}\cdots a_k^{\dagger n_k(\tau)}) \Psi = (z_1^{n_1(\tau)}\cdots z_k^{n_k(\tau)})\Psi$
$(a_1^{m_1(\tau)}\cdots a_k^{m_k(\tau)}) \Psi = (\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)}) \Psi$
$(N_1^{\underline m_1}\cdots N_k^{\underline m_k})\Psi = (a_1^{\dagger m_1(\tau)} a_1^{m_1(\tau)})\cdots (a_k^{\dagger m_k(\tau)} a_k^{m_k(\tau)})\Psi = (\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)})(z_1^{m_1(\tau)}\cdots z_k^{m_k(\tau)})\Psi$

So the master equation acting on $\Psi$ becomes

$H\Psi = \sum_{\tau \in T} r(\tau)(\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)})\left(z_1^{n_1(\tau)}\cdots z_k^{n_k(\tau)} - z_1^{m_1(\tau)}\cdots z_k^{m_k(\tau)}\right) \Psi$

We will assume that

$H\Psi = 0$

From which it follows that

$\frac{d}{d z_i}H\Psi = 0$

We will make use of the compact notation for the Hamiltonian as

$H = \sum_{\tau} r(\tau)\alpha^{m(\tau)}(z^{n(\tau)}-z^{m(\tau)})$
$\frac{d}{d z_i}H\Psi = \frac{d H}{d z_i}\Psi + \frac{d\Psi}{dz_i}H$

This becomes

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i z^{n'(\tau)}-m_i z^{m'(\tau)}) \Psi + \alpha_i H \Psi=0$

but from the assumption, the second term vanishes. Hence,

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i z^{n'(\tau)}-m_i z^{m'(\tau)}) \Psi=0$

and we let

$z_1=z_2=z_3=...=z_k=1$

which yields a solution for non-zero $\Psi$ as

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i -m_i)=0$

which gives a solution to the rate equation, as asserted. Creation ex nihilo, something from nothing.

## Some notation

Consider the nth derivative of the product of $\Psi$ and $H$.

$\left(\frac{d}{d z_i}\right)^n H\Psi = \sum_{l+m=n}H^{(l)}\Psi^{(m)}$

and so $H^{(l)}$ and $\Psi^{(m)}$ become short hand for the lth and mth derivative respectively. In the case we consider here, this takes the simpler form as

$\sum_{l+m=n}H^{(l)}\Psi^{(m)} = \left(\sum_{l+m=n} \alpha_i^{(m)} H^{(l)}\right)\Psi$

And so we are left with the scalar $\alpha_i$ to the mth power times the lth derivative of the Hamiltonian operator. We will evaluate this at

$z_1=z_2=z_3=...=z_k=1$

Yielding

$H^{(l)} = \sum_{\tau}r(\tau)\alpha^{m(\tau)}\left(\Pi_{q=0}^l(n_i-q) - \Pi_{q=0}^l(m_i-q)\right)$

Note that the term $\Pi_{q=0}^l(n_i-q)$ will vanish iff we count over a value where $n_i=q$. The same is true for the term $\Pi_{q=0}^l(m_i-q)$.

## This is the moment

We will consider the number operator $N$ acting $l$ times on $H\Psi$

$N^l H \Psi=$

From above we know that

$H\Psi = \sum_{\tau} r(\tau)\alpha^{m(\tau)}(z^{n(\tau)}-z^{m(\tau)})\Psi$

and also that

$N^q (H \Psi) = \sum_{l+m=q}(N^l H)(N^m \Psi)= \sum_{l+m=q}\alpha_i^m (N^l H)\Psi$

For now we will consider each term $(N^l H)\Psi$ separately.

$(N^l H)\Psi = \sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i^l z^{n(\tau)} - m_i^l z^{m(\tau)})\Psi$

When we set $z_1=...=z_k=1$ we arrive at

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i^l(\tau) - m_i^l(\tau))$
• (Result) For $l=1$ we recover that the mean vanishes for appropriate choice of $\alpha^{m(\tau)}$, given that the rate equation vanishes with appropriate choice of $X^{m(\tau)}$.

It is a straight forward application of the factor theorem to show that $(n_i-m_i)$ is a factor of $(n_i^l(\tau) - m_i^l(\tau))$ so we have

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i(\tau) - m_i(\tau))\text{Poly}((n_i(\tau), m_i(\tau))$

We know from the rate equation that

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i(\tau) - m_i(\tau)) = 0$

## Example of multiple equilibrium states

For the general terms and definitions used in this project see

Here we consider examples from Table 1 in the following paper.

P. M. Schlosser and M. Feinberg, A theory of multiple steady states in isothermal homogeneous CFSTRs with many reactions, Chemical Engineering Science 49 (1994), 1749–1767.

• By ‘just one equilibrium state’, I presume they mean just a 1-parameter family of equilibrium states. After all, if $x_A, x_B$ is an equilibrium solution of the rate equation, so is $c x_A, c x_B$ for any $c \ge 0$.

### The chemical reaction network

The example we consider here is from Table 1 (7), giving the chemical reaction network defined by

$2 A + B \to 3 A$
$3 A \to 2 A + B$

The input and output functions then give.

$m(\tau_1)=\{2,1\}$
$n(\tau_1)=\{3,0\}$

and for $\tau_2$

$m(\tau_2)=\{3,0\}$
$n(\tau_2)=\{2,1\}$

### The Stochastic Petri Net ### The chemical rate equation

The general form of the rate equation is

$\frac{d}{d t}X_i=\sum_{\tau\in T}r(\tau)X^{m(\tau)}(n_i(\tau)-m_i(\tau))$

The population of species $A$ ($B$) will be denoted $X_1(t)$ ($X_2(t)$) and the rate of the reaction $\tau_1$ ($\tau_2$) as $r_1$ ($r_2$).

$\frac{d}{d t}X_1 = r_1 X_1^2 X_2 - r_2 X_1^3$

and

$\frac{d}{d t} X_2 = r_1 X_1^2 X_2 + r_2 X_1^3$

### The master equation

The general form of the master equation is

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

In our case, this becomes

$H = r_1 (a_1^\dagger a_1^\dagger a_1^\dagger a_1 a_1 a_2 - a_1^\dagger a_1^\dagger a_2^\dagger a_1 a_1 a_2) + r_2 (a_1^\dagger a_1^\dagger a_2^\dagger a_1 a_1 a_1 - a_1^\dagger a_1^\dagger a_1^\dagger a_1 a_1 a_1 )$

### Proof that if the rate equation vanishes then so does the mater equation

Since the rate equation is known to vanish from the Deficiency Zero Theorem, we have that

$\frac{d}{d t}X_1 = 0 = r_1 X_1^2 X_2 - r_2 X_1^3$

and

$\frac{d}{d t} X_2 = 0 = r_1 X_1^2 X_2 + r_2 X_1^3$
• (Algebraic independence) Let
$f_1,...,f_m$

be polynomials in

$F[x_1,...,x_n]$

are called algebraically independent if there is no non-zero polynomial

$A \in F[y_1,...,y_m]$

such that

$A(f_1,...,f_m)=0$

then its called the annihilating polynomial.

We consider a wave function of the form

$\Psi := e^{b z_1} e^{c z_2}$

We then calculate $H\Psi$. We find solutions where this vanishes, for non-zero $\Psi$ as

$(r_1 b^2 c - r_2 b^3)z_1^3 + (r_2 b^3 - r_1 b^2 c)z_1^2 z_2$

For this to vanish, each of the terms must vanish separately. Hence we arrive at a system of equations

$r_1 b^2 c - r_2 b^3 = 0$
$r_2 b^3 - r_1 b^2 c = 0$

In this case, and in every case I’ve considered so far, a solution of one of these equations, gives a solution of the second.

category: experiments