====== has Quanty managed to symmetrize the resulting wavefunction w.r.t symmetry of the input problem? ====== ;;# asked by [[mailto:jun.physics@gmail.com|John Lee]] (2023/11/09 21:27) ;;# == == For example, I have the following Hamiltonian containing kinetic operator only on a single site with 7 orbits occupied with 10 electrons in a spin unpolarized state. t_1_1 = -3.8728227 t_2_2 = -3.8728226 t_3_3 = -3.8728227 t_4_4 = -3.8192251 t_5_5 = -3.8192251 t_6_6 = -3.8192251 t_7_7 = -3.8106235 As one can see, the problem has two sets of 3-fold degenerate states and should have identical occupation on orbital 1,2,3 as well as on orbital 4,5,6. But Quanty returns a ground state as, WaveFunction: Wave Function QComplex = 0 (Real==0 or Complex==1) N = 1 (Number of basis functions used to discribe psi) NFermionic modes = 14 (Number of fermions in the one particle basis) NBosonic modes = 0 (Number of bosons in the one particle basis) # pre-factor Determinant 1 1.000000000000E+00 11111111001100 Thus, orbital 1,2,3 are fully occupied, which is expected, while orbital 4,5,6 has sharply different occupations. While this still gives the lowest total energy, it obviously does not fulfill the symmetry inherent in the kinetic Hamiltonian to work with. ~~DISCUSSION|Answers~~