Unitary Coupled Cluster Operator In quantum computing, we always want to deal with unitary evolution operators. That's why the traditional definition of the Coupled Cluster ansatz is modified to the so-called Unitary Coupled Cluster, in which the exponent of the cluster operator is changed to
$$
\operatorname{e}^T \to \operatorname{e}^{T-T^\dagger}
$$
Generally, there are two ways of making an anti-Hermitian operator out of an arbitrary operator $T$:
$$
\begin{alignedat}{4}
&1.\quad &&T \to &&(T-T^\dagger)\\
&2.\quad &&T \to i&&(T+T^\dagger)
\end{alignedat}
$$
I'm wondering if anyone could explain the reasoning for choosing the first option for UCC?
 A: first of all, thank you very much for the question, it made me think about something I never thought of before. Here I try to sketch what my conclusion is.
First, if we call:
\begin{align}
& U=T-T^{\dagger} \\
& V=i\left(\tilde{T}+\tilde{T}^{\dagger}\right)
\end{align}
and we take the Hermitian conjugate of this operators we get:
\begin{align}
& U^{\dagger}=T^{\dagger}-T \\
& V^{\dagger}=-i\left(\tilde{T}^{\dagger}+\tilde{T}\right)
\end{align}
so you can see that both $U$ and $V$ are correctly antihermitian and so $e^{U}$ and $e^{V}$ are correctly unitary. 
Still, in quantum chemical problems, one usually neglects effects as spin-orbit coupling so that the Hamiltonian is completely real and real eigenfunctions can be found. Working with real wavefunction makes two-electron integrals much easier to compute and avoids calculations to become too messy. 
In the end, you just want to use real orbitals which means that the unitary operator you want to use is real itself. Then, in U it is completely useless to have a complex part in the operator T while in V it is completely useless to have a real part in the operator $\tilde{T}$. But then, as both U and V must be real and they must also be equal such that:
\begin{align}
& T-T^{\dagger}=i\left(\tilde{T}+\tilde{T}^{\dagger}\right)
\end{align}
implying that:
\begin{align}
& T=i\tilde{T}
\end{align}
So the two forms you used are completely equivalent but have a different definition of the operator T and then you just use the first form because it is more intuitive.
I hope the answer was clear enough, let me know if it is not.
