I'm trying to figure out what's going on in this paper here between lines 10 and 11. But I'll give a brief rundown of what's going on.
We are trying to compute the trace of the exponential operator $e^{A}$ where $A = a c + b c^{\dagger}$ knowing the commutation relation $[c,c^{\dagger}] = 1$.
The paper proceeds by exploiting the fact that
$$ e^{ac + bc^{\dagger}} = e^{bc^{\dagger} + ac} $$
Then using the Baker-Campbell-Hausdorff formula, it follows that: $$ e^{-\frac{1}{2} ab} e^{ac}e^{bc^{\dagger}} = e^{\frac{1}{2} ab} e^{bc^{\dagger}} e^{ac} $$
Then they consider the trace of the operators:
$$ Tr(e^{ac}e^{bc^{\dagger}}) = e^{ab}Tr(e^{bc^{\dagger}}e^{ac}) $$
Then they seem to claim that given a pair of inverse operators $UU^{-1} = 1$ it follows solely from the invariance of the trace under cyclic permutations that $Tr(AB) = Tr(UAU^{-1}B)$ so that:
$$ Tr(e^{ac}e^{bc^{\dagger}}) = e^{ab}Tr(Ue^{ac}U^{-1}e^{bc^{\dagger}}) $$
But proceeding from the third displayed equation and using the cyclic property, it seems that:
$$ e^{ab}Tr(e^{bc^{\dagger}}e^{ac}) = e^{ab}Tr(Ue^{bc^{\dagger}}U^{-1}Ue^{ac}U^{-1}) \\ = e^{ab}Tr(Ue^{ac}U^{-1}Ue^{bc^{\dagger}}U^{-1}) $$
So it seems that for the fourth displayed equation to be true, it must be that
$$ Ue^{bc^{\dagger}}U^{-1} = e^{bc^{\dagger}}$$
Which requires that $[e^{bc^{\dagger}}, U^{-1}] = 0$
But given the form of $U= e^{\beta ( c^{\dagger} c + \frac{1}{2} )}$, $[e^{bc^{\dagger}}, U] \ne 0$.
Is there something I'm missing here, or am I misinterpreting the author's justification (I've seen this justification made in several different places)?