# Derivation of expected value of a function from master equation Glauber Model (Ising Model)

Let $$\sigma =(\sigma_1, \sigma_2,...,\sigma_N)$$ where $$\sigma_i =\pm \sigma_i$$. Let also $$\sigma^i =(\sigma_1, \sigma_2,...,-\sigma_i, ...,\sigma_N)$$. Given the master equation:

$$\begin{equation} \frac{d}{dt}P(\sigma,t) = \sum_{i=1}^N\left[ w_i(\sigma^i)P(\sigma^i,t) - w_i(\sigma)P(\sigma,t)\right] \end{equation}$$

where $$w_i(\sigma)$$ is the transition rate of $$\sigma_i \rightarrow -\sigma_i$$ that depends on the whole configuration $$\sigma$$. I want to find

$$\begin{equation} \frac{d}{dt}\left< f(\sigma)\right> = \sum_{i=1}^N\left< \{f(\sigma^i) - f(\sigma)\}w_i(\sigma)\right> \end{equation}$$

So far I've got to: ($$\left< f(\sigma)\right> = \sum_\sigma f(\sigma)P(\sigma,t)$$)

$$\begin{equation} \frac{d}{dt}\left< f(\sigma)\right> = \sum_\sigma\sum_{i=1}^N\left\{ f(\sigma)w_i(\sigma^i)P(\sigma^i,t) - f(\sigma)w_i(\sigma)P(\sigma,t) \right\} \end{equation}$$

Then:

$$\begin{equation} \frac{d}{dt}\left< f(\sigma)\right> = \sum_{i=1}^N\left\{ \sum_{\sigma}\left[f(\sigma)w_i(\sigma^i)P(\sigma^i,t)\right] - \left \right\} \end{equation}$$

Now I don't see how that first member of the rhs is going to turn into $$\left< f(\sigma^i) w_i(\sigma)\right>$$ (&) which is what I would need at this point. I consider the possibility I'm not attacking the problem the right way, so (&) didn't need to be true, but I still can't figure out what I am missing. Any help or ideas would be extremeley apreciatted.

Thanks,

PD: $$f(\sigma)$$ and $$w_i(\sigma)$$ are completely arbitrary

• The mapping $\sigma\mapsto\sigma^i$ defines a bijection on the set of configurations. Therefore, you can just change the variable of summation from $\sigma$ to $\sigma^i$, which immediately yields what you want. Apr 30, 2020 at 18:07
• @YvanVelenik I thought about this, but if I change the variable to $\sigma^i$ then for the terms which had $\sigma^i$ before, I have $(\sigma^i)^i$ which is different from $\sigma^i$. A
– CMB
Apr 30, 2020 at 18:26
• $(\sigma^i)^i=\sigma$, which is what you want. Apr 30, 2020 at 18:27
• @YvanVelenik What I want to say is: let $\sigma \rightarrow \sigma^i$, then $f(\sigma)w_i(\sigma^i)P(\sigma^i) \rightarrow f(\sigma^i)w_i((\sigma^i)^i)P((\sigma^i)^i) = f(\sigma^i)w_i(\sigma)P(\sigma)$
– CMB
Apr 30, 2020 at 18:31

Since $$(\sigma^i)^i = \sigma$$, the transformation $$\sigma\mapsto\sigma^i$$ is an involution. In particular, changing the summation variable from $$\sigma$$ to $$\tau = \sigma^i$$, you get $$\sum_\sigma f(\sigma) w_i(\sigma^i) P(\sigma^i,t) = \sum_{\tau} f(\tau^i) w_i(\tau) P(\tau,t),$$ which is just the expectation of the function $$\sigma\mapsto f(\sigma^i) w_i(\sigma)$$ with respect to the measure $$P(\cdot,t)$$, as you want.