Consider a ring containing N sites, each of which can either be 'on' or 'off', with equal probability. At $t=0$, the sites are either all 'on' or all 'off'. If we allow each site to switch states (i.e. an 'on' state becomes 'off') at a rate $\mu$ and the possibility of each state to replace one of its neighbour's at a rate $\lambda$, how does the probability that $k$ sites are 'on' vary as a function of time, $t$?
Prior to doing any calculations, we can consider the limits of the problem. If $\mu >> \lambda$ then we will have a distribution that rapidly drifts towards a binomial distribution with $p=0.5$ and $n=N$. Conversely, if $\mu << \lambda$ the distribution will stay as that at $t=0$, with
$$P(k) = \begin{cases} 0.5, & \text{if $k$ is 0 or N} \\ 0, & \text{otherwise} \end{cases}$$
If we initially consider a single site that is 'on' at $t=0$, we essentially have a 2-state system with a switching rate $\mu$ and initial conditions $p(t=0)=1$ the probability that the site is 'on' is: $$p(t) = \frac{1}{2}(1+e^{-2\mu t})$$
We now consider N sites that are able to switch states (for now, ignoring the possibility of replacement) but are all either 'on' or 'off' at $t=0$. Essentially we have a sum of two binomial distributions (due to the symmetry between 'on' and 'off'): $$P(k) = \frac{1}{2} \binom{N}{k} \frac{1}{2^N} \left( (1+e^{-2\mu t})^{k}(1-e^{-2\mu t})^{N-k} + (1-e^{-2\mu t})^{k}(1+e^{-2\mu t})^{N-k} \right)$$
Equivalently: $$P(k) = \frac{1}{2} \binom{N}{k} e^{-N\mu t} \left( \cosh^{k}{\mu t}\sinh^{N-k}{\mu t} + \sinh^{k}{\mu t}\cosh^{N-k}{\mu t} \right)$$
However, I am unsure of the next steps required to incorporate the replacement of neighbouring states into this probability distribution.
EDIT 1:
On reflection, the above approach is not the best way to solve this problem. Rather than consider the number of 'on' sites as our variable of interest, we should focus on the unique states. As an example, consider the case $N=4$. We have 16 separate states (although there is a level of redundancy between states, so there are actually 6 unique states, however I'm not entirely sure how to take into account this multiplicity). We now have a true Markov chain, as the transition probability is only dependent on the current state. I believe that we then have the transition matrix, $T$:
$$ \begin{pmatrix} 8l & m & m & m & m & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ m + 2 l & 4 l & 0 & 0 & 0 & m & m + l & 0 & m + l & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ m + 2 l & 0 & 4 l & 0 & 0 & 0 & m + l & m + l & 0 & 0 & m & 0 & 0 & 0 & 0 & 0\\ m + 2 l & 0 & 0 & 4 l & 0 & 0 & 0 & 0 & m + l & m + l & m & 0 & 0 & 0 & 0 & 0\\ m + 2 l & 0 & 0 & 0 & 4 l & m & 0 & m + l & 0 & m + l & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & m + 2 l & m + 2 l & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & m + 2 l & m + 2 l & 0 & 0 & 0\\ 0 & m + l & 0 & 0 & m + l & 0 & 4 l & 0 & 0 & 0 & 0 & 0 & m + l & m + l & 0 & 0\\ 0 & 0 & m + l & 0 & m + l & 0 & 0 & 4 l & 0 & 0 & 0 & 0 & m + l & 0 & m + l & 0\\ 0 & m + l & 0 & m + l & 0 & 0 & 0 & 0 & 4 l & 0 & 0 & m + l & 0 & m + l & 0 & 0\\ 0 & 0 & m + l & m + l & 0 & 0 & 0 & 0 & 0 & 4 l & 0 & m + l & 0 & 0 & m + l & 0\\ 0 & 0 & 0 & m + 2 l & m + 2 l & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & m + 2 l & m + 2 l & 0\\ 0 & 0 & 0 & 0 & 0 & m & 0 & 0 & m + l & m + l & 0 & 4 l & 0 & 0 & 0 & m + 2 l \\ 0 & 0 & 0 & 0 & 0 & m & m + l & m + l & 0 & 0 & 0 & 0 & 4 l & 0 & 0 & m + 2 l \\ 0 & 0 & 0 & 0 & 0 & 0 & m + l & 0 & m + l & 0 & m & 0 & 0 & 4 l & 0 & m + 2 l \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & m + l & 0 & m + l & m & 0 & 0 & 0 & 4 l & m + 2 l \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & m & m & m & m & 8l\\ \end{pmatrix} $$ We then know that $P_k(t)=P_k(0)e^T$ where $e^T$ is the matrix exponential of T. How do I generalise this process to N states?