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}


\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<f(\sigma)w_i(\sigma)\right> \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.


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

  • $\begingroup$ 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. $\endgroup$ Apr 30, 2020 at 18:07
  • $\begingroup$ @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 $\endgroup$
    – CMB
    Apr 30, 2020 at 18:26
  • $\begingroup$ $(\sigma^i)^i=\sigma$, which is what you want. $\endgroup$ Apr 30, 2020 at 18:27
  • $\begingroup$ @YvanVelenik What I want to say is: let $\sigma \rightarrow \sigma^i$, then $f(\sigma)w_i(\sigma^i)P(\sigma^i) \rightarrow f(\sigma^i)w_i((\sigma^i)^i)P((\sigma^i)^i) = f(\sigma^i)w_i(\sigma)P(\sigma)$ $\endgroup$
    – CMB
    Apr 30, 2020 at 18:31

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.