Nice!

I've been a bit absorbed in blogging about this stuff and am just catching up with your work.

We can why there's a 1-parameter family of solutions of the master equation in this case using Noether's theorem. We've got a reversible reaction

$$A + 2 B \leftrightarrow 3 A $$

and if we write

$$ A + 2 B = 3 A $$

we can solve and get

$$ B = A $$

so the*overall effect* of this reaction is to turn an $A$ into a $B$. This means that the number of $A$'s, plus the number of $B$'s, is conserved. So, we get a conserved quantity

$$ O = N_A + N_B = a_A^\dagger a_A + a_B^\dagger a_B$$

and we therefore can apply any function of $O$ to an equilibrium solution $\Psi$ of the master equation and get another equilibrium solution:

$$ H \Psi = 0 \implies H (f(O) \Psi) = 0 $$

since Noether's theorem says

$$ [H , O] = 0 $$

so

$$ [H, f(O)] = 0 $$

so

$$ H f(O) \Psi = f(O) H \Psi = 0 $$

This kind of argument is fairly general. We just need to find a conserved quantity!

I've been a bit absorbed in blogging about this stuff and am just catching up with your work.

We can why there's a 1-parameter family of solutions of the master equation in this case using Noether's theorem. We've got a reversible reaction

$$A + 2 B \leftrightarrow 3 A $$

and if we write

$$ A + 2 B = 3 A $$

we can solve and get

$$ B = A $$

so the

$$ O = N_A + N_B = a_A^\dagger a_A + a_B^\dagger a_B$$

and we therefore can apply any function of $O$ to an equilibrium solution $\Psi$ of the master equation and get another equilibrium solution:

$$ H \Psi = 0 \implies H (f(O) \Psi) = 0 $$

since Noether's theorem says

$$ [H , O] = 0 $$

so

$$ [H, f(O)] = 0 $$

so

$$ H f(O) \Psi = f(O) H \Psi = 0 $$

This kind of argument is fairly general. We just need to find a conserved quantity!