compute $\Tr (\hat{K}^2)$. In this case one finds that the exact $A(n)$ may be well approximated by its ray-dynamical analogue $A_{\rm  cl}(n)$ even without averaging and that the result above then also holds for individual maps. (Of course, the trace operation itself performs an average over basis states even if no further averaging is performed.)

The variance of the correlation function $\hat{\Gamma}$
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To get an equivalent expression for the variance of the full correlation function of the stationary wave field, $\hat{\Gamma}$, we start from Eq. (\[G0\]), giving $$\hat{\Gamma} = \sum_{n_1, n_2=0}^\infty v ^{n_1+n_2}\,  \hat{U}^{n_1} \hat{\Gamma}_0 \hat{U}^{-n_2}$$ from which we obtain (using cyclic permutability of the trace again) $$\begin{aligned}
{\rm Tr} \,\hat{\Gamma}^2 &=& \sum_{n_1,\ldots,n_4 =0}^\infty v^{n_1+n_2+n_3+n_4} {\rm Tr} \left[\hat{U}^{n_1} \hat{\Gamma}_0 \hat{U}^{-n_2} \hat{U}^{n_3} \hat{\Gamma}_0 \hat{U}^{-n_4}\right]\nonumber \\ \label{qsum}
&=& \sum_{n_1,\ldots,n_4 =0}^\infty v^{n_1+n_2+n_3+n_4} {\rm Tr} \left[\hat{U}^{n_1-n_4} \hat{\Gamma}_0 \hat{U}^{-n_2+n_3} \hat{\Gamma}_0 \right].\end{aligned}$$ We argue next that, after averaging, $$\langle {\rm Tr}\, [\hat{U}^n \hat{\Gamma}_0 \hat{U}^{-n'} \hat{\Gamma}_0 ] \rangle = {\delta}_{nn'} \langle A(n)\rangle$$ due to the phase difference that one necessarily finds in contributions to $\hat{U}^n$ and $\hat{U}^{n'}$ when $n\ne n'$. On average, the sum in (\[qsum\]) is then dominated by terms with $n_1 -n_4 = n_2-n_3 = n$ and, after reordering, we may write $$\begin{aligned}
\langle{\rm 