d/dt x1 = alpha(2 c – x1) – 2 b x1^2

It shouldn’t have the factor 2 on the c. Same for the integral that follows. Because: when solve for x2, you get:

x2 = (c – x1) / 2, and when this gets multiplied by 2 alpha, you get:

alpha * (c – x1).

But it doesn’t make a real difference, because c is an unspecified constant.

d/dt x1 + 2 x2 = 0

should say:
d/dt x1 + 2 d/dt x2 = 0

Remaining in the field of continuous markov processes, you have two possible decompositions of the Hamiltonian and : in general the two parts of the first decomposition (symmetric and skew-symmetric, w.r.t. the stationary solution) and the two parts of the second decomposition (irreversible and reversible w.r.t. the time-reversal parities of the variables) do not coincide. In the first decomposition the stationary current is entirely contained in the antisym part: . In the second decomposition the current is separated: and and .

Note that to identify the two parts of the first decomposition you have to know the stationary solution, because symmetrization and antisymmetrization must be done w.r.t. that measure. On the contrary, to identify the two parts of the second decomposition you only have to know the timereversalparities of the variables (I’m still keeping secret this decomposition, but if you wish I can show you that more explicitly, or you can just browse Risken’s book, section 6.4 :-) )

When detailed balance is satisfied, however, the two decompositions exactly coincide, i.e. and . Much better, if detailed balance is satisfied and there is no reversible current (so that ) then is self-adjoint w.r.t. the stationary solution; this happen very often, e.g. in “overdamped” (or aristotelian) dynamics, where you typically have only even variables (if detailed balance is satisfied, i.e. if there are no non-conservative of time-dependent external forces). Nevertheless detailed balance alone is already good because it let you immediately know what are the symmetric and antisymmetric parts of the hamiltonian which makes life much easier (including knowing the stationary solution).

In principle could be self-adjoint even if detailed balance is not satisfied, but I’ve never seen a physical example of that. You should have a vanishing total current but non-vanishing irreversible and reversible parts, i.e. and . That would be strange.

It would be interesting to find the same structure described above also in discrete systems as those you are considering.

This language applies to a continuous space of states as well as for a discrete one, as sketched in Network Theory (Part 12). I’m hoping that your split of the current into ‘reversible’ and ‘irreversible’ parts corresponds to the split of a Dirichlet form into its symmetric and skew-symmetric parts, as sketched in that post and, in vastly more detail, here:

• Zhi-Ming Ma and Michael Röckner, Introduction to the Theory of (Non-Symmetric) Dirichlet Forms, Springer, Berlin, 1992.

Then you can see that is self-adjoint with respect to the stationary solution (as you wrote) if and only the stationary current is zero . The subtle thing is that such a condition is stronger than detailed balance. Indeed one can verify that the total current can be decomposed in two parts, one so-called “reversible” and the other “irreversible”: . This decomposition is based upon the parities, with respect to time-reversal, of the variables (the component of vector ): indeed one may have a master equation describing odd and/or even variables, e.g. velocities and positions respectively. The detailed balance condition is equivalent to , which is less than the previous condition on the total current. When detailed balance is satisfied, the “reversible” current can still be non-zero and so the total current. Imagine for instance a pendulum with non-negligible inertia, so that it is described by position and momentum : its master (Fokker-Planck) equation reduces to the Liouville equation because there is no noise. The system is clearly time-reversible: trajectories in phase space have the same “probabilities” (assuming a uniform measure of initial conditions) of occurring in one direction and in the opposite direction, provided that is changed of sign. Nevertheless in phase-space there is a non-zero current, a flow of probability going circularly in a fixed and not-invertible direction. This current is an example of “reversible” current, in the sense that it does not add irreversibiity to the system. When the phase space is 1-dimensional, the current cannot have a reversible part and therefore detailed-balance and absence of total current are equivalent.

In general however, when detailed balance is satisfied, the total current can be non-zero and therefore the Hamiltonian is not self-adjoint w.r.t. . Detailed balance is still very useful because it allows to identify easily the stationary solution and consequently a particular decomposition of the Hamiltonian in a hermitian and anti-hermitian (always w.r.t. the stationary solution) making easy to study the whole (non-stationary) problem by expanding in right and left eigenfunctions. If detailed balance is not satisfied, things get much more complicate.

For the things I’ve summarized here, my reference book is Risken’s “The Fokker-Planck equation: methods …..” There you can find the definitions of things like irreversible currents etc. which I have not given for brevity. In Risken’s book the connection of Fokker-Planck operator (your Hamiltonian) with a Schroedinger operator as well as with creation-destruction operators is also discussed. What is not discussed is an extension to discrete systems, as the ones you consider. For those systems, I again suggest (if you’re interested in the connection with time-reversibility, entropy production, etc.) to have a look to the review by J. Schnakenberg (Network theory of microscopic and macroscopic behavior of master equation systems, Rev. Mod. Phys. 48, 571–585 (1976) )

p. 225: “where is a function that the probability of being in each state”: missing verb

p. 229: “Because our reaction network is weakly reversible, Theorem 58 there exists”