Two coupled system of which one has dephasing I'm reading Environment-assisted quantum transport by Rebentrost at all, where they deal with a system of 2 sites that hosts 1 excitation. They describe this in terms of two states: $\vert 1 \rangle$ denotes an excitation at site 1, $\vert 2 \rangle$ denotes an excitation at site 2. They are coupled, given by the Hamiltonian
\begin{equation}
H = \frac{\Delta}{2} \left(\vert1\rangle\langle1\vert\ - \vert2\rangle\langle2\vert\right) + \frac{V}{2}\left( \vert1\rangle\langle2\vert\ + \vert2\rangle\langle1\vert\ \right)
\end{equation}
where $\Delta$ is the energy mismatch between the sites and $V$ is the coupling strength. 
The next ingredient is that both sites couple to an uncorrelated (between the sites) white noise bath, leading to a pure dephasing $\gamma_\phi$ for both states. They write that the master equation is then given by
\begin{equation}
\dot{\rho} = \gamma_\phi\sum_m\left(A_m \rho A_m^\dagger  - \frac{1}{2} A_m A_m^\dagger \rho  - \frac{1}{2} \rho A_m A_m^\dagger \right)
\end{equation}
with $A_m = \vert m \rangle \langle m \vert$. 
I don't see why this is true. Isn't pure dephasing normally with $\sigma_z$, which here would be $\vert1\rangle\langle1\vert\ - \vert2\rangle\langle2\vert$? The above would miss the minus sign, right? 
Assuming I am just not reading the above correctly, this system has a Lindblad term $\sqrt{\gamma_\phi} \left(\vert1\rangle\langle1\vert\ - \vert2\rangle\langle2\vert \right)$. Would it then be the case that, starting from an excitation at site 1, the dephasing leads to population at site 2? I suppose so, because the Hamiltonian is not diagonal; we're not along the z-axis of the Bloch sphere and the dephasing thus leads to rotations between states 1 and 2. Is this correct, and can we quantify it?
What I am then also interested in is a slight change to the above; I figured it is not worth a new question as I think the analysis is very similar. What changes in this picture if only site 1 is coupled to a bath, so that $\gamma_{\phi 2}$ is zero? Would we still end up at site 2, starting from site 1? 
 A: It might help you to note that a pure dephasing Lindblad term can be written in many ways. Suppose we write a general dissipator as $$\mathcal{D}[A]\rho = A\rho A^\dagger - \frac{1}{2}A^\dagger A \rho - \frac{1}{2} \rho A^\dagger A.$$
Then a straightforward computation demonstrates that $\mathcal{D}[\sigma^z] = 4\mathcal{D}[\Pi^\uparrow ]$, where $\Pi^\uparrow$ is the projector onto the excited/spin-up state, i.e. $\Pi^\uparrow = (1 + \sigma^z)/2$. Rebentrost et al. are simply using the second form, where the Lindblad operator is given by the projector onto the excited state of the relevant site, i.e. $\Pi^\uparrow_{1} = \lvert 1 \rangle \langle 1 \rvert $ and $\Pi^\uparrow_{2} = \lvert 2 \rangle \langle 2\rvert $. The form given in the paper describes local dephasing acting independently on each site.
Amusingly, in the single-excitation subspace your proposed Lindblad operator is actually equivalent to local dephasing acting on just one site. This can be shown by exactly the same argument as above, since $\Pi^\uparrow_1 = (1 + \Pi^\uparrow_1 - \Pi^\uparrow_2)/2$, and therefore $\mathcal{D}[\Pi^\uparrow_1 - \Pi^\uparrow_2] = 4\mathcal{D}[\Pi^\uparrow_1]$. 
Finally, note that your interpretation of how the dephasing works is not at all correct. Whatever the choice of Lindblad operator above, the dephasing terms do not directly affect the population of either site, since the Lindblad operators commute with the site populations $\Pi_{1,2}^\uparrow$. The dephasing terms instead affect the coherences, i.e. the matrix elements $\langle 1\rvert \rho\lvert 2\rangle$, causing them to decay exponentially. Of course, the dephasing indirectly affects the populations, since the coherences and populations are coupled by the interaction term proportional to $V$ (this is the whole point of the paper).
