IV. SPIN MECHANISMS OF MAGNETORESISTANCE FOR ACCEPTOR STATES We shall start from the mechanism of spin magnetoresistance first suggested in [8] which seems to be especially important for acceptor dopants. It is related to the fact that inter- ference can occur only if spin states of the final states for both tunneling paths coincide. For 3-cite configuration we discuss it means that the initial and intermediate centers should have the same spin projections (we remind that for destructive interference in question the energy of intermediate center should be negative, i.e. at the equilibrium this center should be occupied). For the case of acceptor states corresponding to the lower Hubbard band the corresponding configuration is in our case ˜ A $^{0}$− ˜ A $^{0}$− A − where the role of intermediate center is played by the site ˜ A $^{0}$. Since the hole has spin 3/2, we have 4 projections of the spin and thus the probability for two sites ˜ A 0 to have the same spin projections is P ( H = 0) = 1 / 4. However in strong magnetic field the site spins are aligned and spin does not affect the (destructive) interference, that is in this case P ( H → ∞ ) = 1. Thus an increase of the magnetic field leads to an enhancement of destructive interference which means positive magnetoresistance which was noted in [8]. In a presence of the states representing the upper Hubbard band (in our case of A + and A 0 states) the situation is somewhat more complicated. In particular, it is related to the fact that the spin structure of doubly occupied A + center is more complex than for single occupied site. In particular, the total spin of A + center is 2 (see [14]) and the possible spin states of A + center are the following: where Ψ (1 , 2) s - are the wave functions characterized by given spin projections of the two holes. Basing on these considerations one can show that for the destructive interference involving 12 | J = 2 ,J$_{z}$ = − 2 〉 = 1 2 Ψ (1) − 1 / $_{2}$Ψ (2) − 3 / 2 − 1 2 Ψ (1) − 3 / $_{2}$Ψ (2) − 1 / 2 | J = 2 ,J$_{z}$ = − 1 〉 = 1 2 Ψ (1) 1 / $_{2}$Ψ (2) − 3 / 2 − 1 2 Ψ (1) − 3 / $_{2}$Ψ (2) 1 / 2 | J = 2 ,J$_{z}$ = 0 〉 = 1 4 Ψ (1) 3 / $_{2}$Ψ (2) − 3 / 2 + 1 4 Ψ (1) 1 / $_{2}$Ψ (2) − 1 / 2 − 1 4 Ψ (1) − 3 / $_{2}$Ψ (2) 3 / 2 − 1 4 Ψ (1) − 1 / $_{2}$Ψ (2) 1 / 2 | J = 2 ,J$_{z}$ = 1 〉 = 1 2 Ψ (1) 3 / $_{2}$Ψ (2) − 1 / 2 − 1 2 Ψ (1) − 1 / $_{2}$Ψ (2) 3 / 2 | J = 2 ,J$_{z}$ = 2 〉 = 1 2 Ψ (1) 3 / $_{2}$Ψ (2) 1 / 2 − 1 2 Ψ (1) 1 / $_{2}$Ψ (2) 3 / 2