Diagonalizing Faddeev-Popov Lagrangian $U(1)$ How can we diagonalize the U(1) Faddeev-Popov Lagrangian in a consistent manner.
I can't seem to find any papers on this but I can't believe that they don't exist. 
Any pointers would be greatly appreciated. 
My own attempt
$$S=\int dV -\partial_{\mu} \bar c \partial^\mu c \tag{1}$$
with $\bar c$ antihermitian and $c$ hermitian.
Obviously setting $\bar{c} = i(\phi_1 + \phi_2)$ and $c = \phi_1 - \phi_2$ diagonalized the action to:
$$S=\int dV -i((\partial \phi_1)^2 -(\partial \phi_2)^2) \tag{2}$$
With eg: $$\Pi_1 = \frac{\delta S}{\delta \partial \phi_1} = -i\partial \phi_1 \tag{3}$$ such that the commutation relations become:
$$\{\phi_1(x),-i\partial \phi_1(y)\} = i\delta(x-y)\tag{4}$$
such that $$\{a_1(k), a_1(k')^\dagger\} \sim i \delta(k-k')\tag{5}$$
Such that $$\langle a|a\rangle = \langle 0|aa^\dagger|0\rangle \sim i \tag{6}$$
That I cannot understand since the original theory did not have any states with negative norm... Also if I calculate the Hamiltonian I find that it is imaginairy. 
 A: There's one serious flaw in your entire strategy. Since $\overline{c},\,c$ are fermions, they are Grassmann-number-valued. Thus any complex numbers $w,\,z$ satisfy $(w\overline{c}+zc)^2=0$. You therefore can't rewrite the result in the manner you intended. More precisely, if $\phi=w\overline{c}+zc$ then $$\partial_\mu\phi\partial^\mu\phi=g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi=g^{\mu\nu}\partial_\mu(w\overline{c}+zc)\partial_\nu(w\overline{c}+zc),$$which vanishes because Grassmann numbers anticommute.
A: Here is a slightly different perspective: OP's question seems to (possibly indirectly) inquire whether a quadratic ghost action term of the form $$\bar{c}_i ~M^i{}_j~ c^j,\tag{i}$$ 
where $M^i{}_j$ is a (possible infinite-dimensional) matrix,
can be recast into an action term of the form
$$ \frac{1}{2} C^I ~A_{IJ}~ C^J~? \tag{ii}$$
Here $A_{IJ}$ is an antisymmetric matrix, due to the Grassmann nature of the $C$'s.
Since OP's matrix (or rather operator) $M=\partial_{\mu}\partial^{\mu}$ is symmetric to begin with, this endeavour is a lost cause/impossible.
