Effects of diabaticity on fusion of heavy nuclei in the dinuclear model ...

collective velocity the ·
and inversely proportional to the square <strong>of</strong> the static coupling H q, ′
between the diabatic states. A diabatic basis is advantageous if J>0.5 and hence, ∆ > 1.44.
The expression (2.1) is obtained in the linear two-state Landau-Zener model [55] which assumes
that the motion <strong>of</strong> the system in the nona<strong>diabaticity</strong> region is quasi-classical.
In the adiabatic TCSM one can observe the mutual repulsion <strong>of</strong> levels having the same z-
component <strong>of</strong> total angular momentum j z, which prevents crossings. It is due to the noncrossing
rule obtained by Neumann and Wigner [56]. The noncrossing rule reads [55]: the eigenvalues <strong>of</strong>
the Hamiltonian H(q), taken to be functions <strong>of</strong> the coordinate q, do not cross if they belong to
the same irreducible representations <strong>of</strong> the symmetry group <strong>of</strong> the Hamiltonian. The eigenvalues
corresponding to different irreducible representations <strong>of</strong> this symmetry group may cross. The
pro<strong>of</strong> <strong>of</strong> the noncrossing rule is based on the following arguments: Let the Hamiltonian H(q)
have closely lying eigenvalues E1(q0) andE2(q0) corresponding to the eigenfunctions | 1 >0and
| 2 >0 at a certain value q0 <strong>of</strong> the collective coordinate, e.g., relative distance R0 <strong>of</strong> the nuclei.
Treating the Hamiltonian for q values close to q0 as a perturbed H(q0)
=H(q0)+ H(q) ∂H
∂q | q=q0 ·δq = H(q0) (2.3)
+V,
it can be readily seen that the difference between the energy eigenvalues at point q is
∆E12(q) =
+V11] − [E2(q0) +V22]} {[E1(q0) 2 V12 | +4| 2 1/2
(2.4)
,
where V ik =0< i| V | k >0. In order that the terms cross, both terms in the radical <strong>of</strong>
(2.4) must go to zero simultaneously. If | 1 >0 and | 2 >0 belong to different irreducible
representations <strong>of</strong> the H(q) symmetry group, V12 = 0 and (2.4) can be converted to zero by the
proper choice <strong>of</strong> δq. But when | 1 >0 and | 2 >0 belong to the same irreducible representations
<strong>of</strong> the Hamiltonian symmetry group, V12 = 0 and generally speaking both terms in the radical
<strong>of</strong> (2.4) cannot go to zero simultaneouly, since they are functions <strong>of</strong> only one parameter δq. The
noncrossing rule uses a complete symmetry group <strong>of</strong> the Hamiltonian.
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