 Solution

When $\tilde{b}=0$, integration by parts makes the kinetic term $-\tilde{a}{\partial }_{\mu }{V}_{\nu }{\partial }^{\mu }{V}^{\nu }$. The Lorentzian contraction over the field index gives the time component and the three spatial components opposite kinetic-energy signs. Reversing the sign of $\tilde{a}$ merely exchanges which component is a ghost field; no nonzero choice makes all four energies positive. At $\tilde{a}=0$ there is no healthy kinetic term. Thus the Hamiltonian is never positive definite.