The polarization states can be effectively defined in terms of the Stokes operators $\hat{S}_i$ for $i=[0,3]$
$\hat{S}_0 = \hat{a}^{\dagger}_{x}\hat{a}_{x}+\hat{a}^{\dagger}_{y}\hat{a}_{y}$
$\hat{S}_1 = \hat{a}^{\dagger}_{x}\hat{a}_{x}-\hat{a}^{\dagger}_{y}\hat{a}_{y}$
$\hat{S}_2 = \hat{a}^{\dagger}_{x}\hat{a}_{y}+\hat{a}^{\dagger}_{y}\hat{a}_{x}$
$\hat{S}_3 = i\left(\hat{a}^{\dagger}_{y}\hat{a}_{x}-\hat{a}^{\dagger}_{x}\hat{a}_{y} \right)$
These quantum-mechanical observables (Hermitian operators) are defined analogously to their classical counterparts. The annihilation and the creation operators $\hat{a}_{x}$ and $\hat{a}^{\dagger}_{x}$ of the mode (given by the subscript $x$) satisfy the usual commutation relations.
The last three of these operators can also be viewed equivalently to the Pauli matrices $\sigma_z$, $\sigma_x$, and $\sigma_y$ respectively.
To answer your question directly, it is the operator $\hat{S}_1$ whose eigenstates are the horizontal and vertical polarization states $|H\rangle$ and $|V\rangle$.
To visualize this from an experimental perspective, think of a photon counting setup and the fact that $\hat{S}_1 = \hat n_x - \hat n_y$ essentially. And to make a connection to the comments in the other answer, $\hat{S}_1 \equiv \sigma_z = |H\rangle\langle H| - |V\rangle\langle V| = 2|H\rangle\langle H| - I$,
if polarization states were chosen to represent the basis vectors of the Pauli matrices.
Edit on 04/01/2015: I saw a comment asking for some references on the topic. Unfortunately I didn't have time to respond then and now I can't see that comment anymore. Still thought of writing down a couple of works that deal with the topic in a fairly introductory manner.