Many aromatic hydrocarbons have doubly or multiply degenerate lowest unoccupied molecular orbitals (LU-MOs) consisting of 2pπ-orbitals. The exchange interaction between electrons on LUMOs in these dinegative ions tends to be considered as ferromagnetic (the spintriplet state) by the Hund's rule coupling. Indeed, the triplet ground state has been observed for the dinegative ions of triphenylbenzene, decacyclene, and so on. 1 Benzene (Bz) has doubly degenerate LUMOs. Theoretically, the electronic state of Bz dianion (Bz 2− ) has been studied by quantum chemical calculations. 2, 3 It has been reported that the ground state is the triplet state, and the energy difference between the singlet state and triplet state is about 0.1 eV 2 or about 1 eV. 3 However, since the many body effects have not been fully treated and the energy difference is tiny, it has not yet been determined whether the ground state is the spin triplet state or the spin singlet state. In addition, the ground state of Bz 2− has not been experimentally studied.In the present paper, we discuss the ground state of Bz 2− on the basis of the numerical diagonalization method of an effective model of π orbitals. Then, we estimate the amplitude of the exchange coupling between two electrons on LUMOs. As a result, we find that the ground state becomes the spin singlet state, and the exchange coupling between LUMOs can be antiferromagnetic.The model Hamiltonian for Bz 2− is given as follows:wherewhere χ's are atomic 2pπ orbitals of Bz; t π , U , U ′ , and J represent the transfer integral, the intra-and interorbital Coulomb interaction and the ferromagnetic exchange coupling, respectively. This effective model is an extended Hubbard model (t-U -V -J model). It is well known that the exchange coupling, J, plays a fundamental role in the property of the ground state. 4, 5The amplitude of U and t π have been estimated by the quantum chemical calculation: U ∼ 17 eV and t π ∼ 2.5eV. 6, 7 Thus, we choose U/t π = 7.0. The other parameters have also been estimated by the same quantum chemical calculations: U ′ ∼ 9 eV (U ′ /t π ∼ 3.6) and J ∼ 1 eV (J/t π ∼ 0.4). However, since we neglect many other interactions (for example, the second nearest Coulomb interactions), we regard U ′ and J as variable parameters.By diagonalizing the kinetic energy term H 0 , six molecular orbitals, φ 1 ∼ φ 6 , are given as follows: 7The eigen energy increases from φ 1 to φ 6 , while φ 2 (φ 4 ) and φ 3 (φ 5 ) are degenerate. We call φ 4 (φ 5 ) LUMO. In Fig. 1, we show the phase diagram for the ground state in the U ′ /t π -J/t π plane within manifolds of S = 2, S = 1, and S = 0. When J/t π is increased, the ground state changes from S = 1 to S = 0, and changes from S = 0 to S = 2. The S = 2 state is such that the spins of four holes become ferromagnetic by the ferromagnetic exchange coupling J. The characters of electronic states of the S = 1 state and S = 0 state are discussed later. Figure 2 shows the number of electrons in the molecular orbital φ i (eq.(4) ∼ eq.(9)) as a function of J/t...