Why is the local coupling constant (vibronic coupling) given in eV?

Question

Should the local coupling constant g not be without a unit, as the Holstein Hamiltonian suggests?

$ H=\underbrace{\sum_{P, Q} h_{P Q}^{(0)} a_P^{\dagger} a_Q}_{\text{Fixed geometry}}+\underbrace{\sum_{P \in \Omega} \sum_\lambda \hbar \omega_\lambda g_{P P}^\lambda\left[b_\lambda^{\dagger}+b_\lambda\right] a_P^{\dagger} a_P}_{\text{Atom displacement}}+\underbrace{\sum_\lambda \hbar \omega_\lambda\left(b_\lambda^{\dagger} b_\lambda+\frac{1}{2}\right)}_{\text{Vibrations}}$

$H$ = Holstein-Peierls Hamiltonian .$P$ or $Q$ = electronic levels
λ = normal mode g = coupling constant h(0) = transfer integral
a† and a = creation and annihilation operator for fermions
b† and b = creation and annihilation operator for bosons ℏω = energy of phonon

To give more context, in Ref. 1 for example, the local coupling constant is given in eV.

References
(1) Shizu, K.; Sato, T.; Tanaka, K. Vibronic Coupling Density Analysis for α-Oligothiophene Cations: A New Insight for Polaronic Defects. Chemical Physics 2010, 369 (2–3), 108–121. https://doi.org/10.1016/j.chemphys.2010.03.014.

Answer 1

The local coupling constant is typically given in units of energy, such as electronvolts (eV), because it represents the strength of the coupling between the electronic and vibrational degrees of freedom in a system (which is thus an 'energy-related' quantity). The Holstein Hamiltonian, which is used to describe the interaction between electrons and phonons in a system, includes the atom displacement term, which represents the coupling between the electronic state $P$ and the vibrational state $\lambda$, with the coupling constant $g^\lambda$ being the coefficient in front of this term for unit equality in the equation.