  The lifetime of two-vibron bound states in the overtone region of a
one-dimensional anharmonic molecular lattice is investigated. The
anharmonicity, introduced within an attractive Hubbard Hamiltonian for bosons,
is responsible for the formation of bound states which belong to a finite
linewidth band located below the continuum of two-vibron free states. The decay
of these bound states into either bound or free states, is described by
considering the coupling between the vibrons and a thermal bath formed by a set
of low frequency acoustic phonons. The relaxation rate is expressed in terms of
the spectral distribution of the vibron/phonon coupling and of the two-vibron
Green operator which is calculated exactly by using the number states method.
The behavior of the two-vibron bound states relaxation rate is analyzed with a
special emphasis on the influence of the anharmonicity. It is shown that the
rate exhibits two distinct regimes depending on the thermal bath dimension.
When the bath dimension is equal to unity, the rate increases with the
anharmonicity and the decay of the two-vibron bound states into the other bound
states appears as the main contribution of the rate. By contrast, when the bath
dimension is equal to 2 and 3, the rate decreases as the anharmonicity
increases indicating that the two-vibron bound states decay into the two-vibron
free states continuum.
