$\sigma_z/2$ or $(\sigma_z +I)/2$ both satisfy Eq. (\[eqn:cnd\]). This arbitrariness causes a shift in the first cumulant. However, only the operator $(\sigma_z+I)/2$ will give a distribution in which all moments are equal, since this matrix has the form $$(\sigma_z+I)/2 = \left( 
\begin{array}{cc} 1 & 0  \\ 
0 & 0
\end{array}
\right),$$ and is equal to itself when raised to any power.

In the case of the RM model we first write the Hamiltonian with the parameter $K$ explicitly as $$\begin{aligned}
\hat{H}(K) &= -J \exp(i K) \sum_{j=1}^{L/2}  c_{j,A}^\dagger c_{j,B} + \mbox{H.c.}\\ 
& \nonumber
         -J' \exp(i K) \sum_{j=1}^{L/2}  c_{j,B}^\dagger c_{j+1,A}  + \mbox{H.c.} + \hat{H}_{\Delta}. \nonumber\end{aligned}$$ The operator $\partial_K \hat{H}(K)$ is the current, $$\begin{aligned}
\partial_K \hat{H}(K) &= -i J \exp(i K) \sum_{j=1}^{L/2}  c_{j,A}^\dagger c_{j,B} + \mbox{H.c.}\\ 
& \nonumber
         -i J' \exp(i K) \sum_{j=1}^{L/2}  c_{j,B}^\dagger c_{j+1,A}  + \mbox{H.c.}. \nonumber\end{aligned}$$ We now write a form for the operator $\hat{O}$ as $$\hat{O} = \sum_{j=1}^{L/2} x_j c_{j,A}^\dagger c_{j,A} + 
y_j c_{j,B}^\dagger c_{j,B}.$$ Evaluating the commutator gives $$\begin{aligned}
i [\hat{H}(K),\hat{O}] = i \sum_{j=1}^{L/2}(y_j - x_j) J \exp(iK) c_{j,A}^\dagger c_{j,B} + \mbox{H.c.} \nonumber \\  
i \sum_{j=1}^{L/2}(x_{j+1} - y_j) J' \exp(iK) c_{j+1,A}^\dagger c_{j,B} + \mbox{H.c.} \hspace{1cm}\end{aligned}$$ For the case $J'=0$ we can chose $x_j=0$ and $y_j=1$, so that $i[\hat{H}(K),\hat{O}]$ corresponds 