This is why I prefer to say, when I put to zero the time-derivative in the l.h.s. of a master equation, that I’m looking for the “stationary solution”, not the “equilibrium” one, reserving the word equilibrium only to stationary solutions which are invariant for time-reversal.

That makes sense. This series has not mentioned ‘detailed balance’, which is indeed different than ‘complex balance’. But Jacob has been thinking about time reversal symmetry lately, so he’ll be interested to read these comments of yours (as was I).

Thanks for the reference to Shnakenberg!

Is there a typo or am I missing something?

Thanks for catching that!

The problem was that I picked a hard notation to remember: the reaction with rate constant reduces the number of atoms. By the time I wrote the last section I’d forgotten this and mentally switched to the more reasonable convention where the bigger is, the more atoms there are in equilibrium, while making bigger makes more molecules.

In fact I wrote down the rate equation based on that other more reasonable convention, and my whole discussion with Greg Egan was based on that!

I like that other convention better, and I’m turning all these articles into a book. So, right now I just rewrote this article to use that other convention.

Of course now there are probably more new typos.

Chemical networks often are not symmetric under time-reversal, because they represent phenomena out of thermal equilibrium. This is why I prefer to say, when I put to zero the time-derivative in the l.h.s. of a master equation, that I’m looking for the “stationary solution”, not the “equilibrium” one, reserving the word equilibrium only to stationary solutions which are invariant for time-reversal (in order to guarantee “consistency” with the idea of thermodynamic equilibrium). This symmetry allows you to map the master equation into a Schroedinger-like equation with a symmetric operator (the hamiltonian). Usually this is not possible and one has to symmetrize the operator (or to find left/right basis etc.). In the master equation the time-reversal symmetry appears as the so-called “detailed balance” which, it seems to me, is different from the complex balance concept. In your post it was perfectly legit to use the word equilibrium, but it comes as a word totally equivalent to

which is not.

Time-reversal is related to the absence of closed loops (indeed in your example the diffusion occurs over a single line). Basically you are at equilibrium if there is only one route connecting a state with another state. If there are more than one route, it is possible to break the time-reversal symmetry by having a stationary solution with a non-zero probability current flowing in your network. There is a very elegant theory of non-equilibrium master equation in terms of closed loops in the chemical network, done by J. Schnakenberg, where the network is decomposed in “fundamental loops” (such that a current in any edge is the sum of currents in the fundamental loops including that edge). You can see

• J. Schnakenberg, Rev. Mod. Phys. 48 (1978), 571.

I have however a problem, because detailed balance means putting to zero each “term” of the sum, if you write the master equation as a sum of terms each one representing the (bi-directional) connection between two states, e.g.

where is a state (node) of the network, which in your notation is a couple . In your notation this should correspond to ask being zero separately the two couples of terms

and

If I use your solution obtained with the ACK theorem I find that it satisfies detailed balance but with which is slightly different than your . Is there a typo or am I missing something?

You can get the whole series on my website.

If we look at the components of in the eigenspaces with definite values for , the symmetry will take each to another vector in . But the relative size of these components won’t be preserved, so I shouldn’t expect to be preserved.

On the other hand, applied to an individual will certainly preserve the definite value of . But in that case, we no longer have .

I think the symmetry maps the equilibrium solution of the master equation associated with the solution of the rate equation to that associated with .

Right!

Clearly the equation

is still satisfied by the new concentrations and .

Right! I feel like rhapsodizing about this a bit:

We’re studying a diatomic gas in equilibrium, assuming fixed rates for the reactions

and

Of course these rates will depend on various things, but we’re treating them as fixed.

The rate equation lets us find the equilibrium in the limit of large numbers of atoms. In this limit, we’ve seen the number of diatomic molecules is proportional to the square of the number of lone atoms. So, the set of equilibria forms half a parabola:

There’s a one-parameter group of symmetries:

that maps this parabola to itself. For example, we can multiply the number of lone atoms by 1.5 and multiply the number of molecules by 2.25.

More surprisingly, this symmetry exists even when we consider small numbers of atoms and molecules, where we treat these numbers as integers instead of real numbers. If we have 3 atoms, we can’t multiply the number of atoms by 1.5. So this is a bit shocking at first!

The trick is to treat the gas stochastically using the master equation rather than deterministically using the rate equation. What our symmetry does is multiply the relative probability of finding our container of gas in a given state by a factor of for each lone atom, and by a factor of for each molecule.

This symmetry commutes with time evolution as given by the master equation. And for probability distributions that are products of Poisson distributions, this symmetry has the effect of multiplying the mean number of lone atoms by , and the mean number of molecules by .

But as you note, this symmetry the property that if we start in a state with a definite total number of atoms (that is, lone atoms plus twice the number of molecules), it will map us to another state with the same total number of molecules!

And if we start in a state with a definite number of lone atoms and a definite number of molecules, the symmetry will leave this state completely unchanged!

These facts sound paradoxical at first, but of course they’re not. They’re just a bit weird.

In fact they’re a lot like this other weird fact. If we take a quantum system and start it off in an eigenstate of energy, it will never change, except for an unobservable phase. Every state is a superposition of energy eigenstates. So you might think that nothing can ever change in quantum mechanics. But that’s wrong: the phases that are unobservable in a single energy eigenstate become observable relative phases in a superposition.

Indeed the math is exactly the same, except now we’re multiplying relative probabilities by positive real numbers, instead of multiplying relative amplitudes by complex numbers!