has Quanty managed to symmetrize the resulting wavefunction w.r.t symmetry of the input problem?
asked by 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.
Answers
Hi John Lee,
Setting up a ground state calculation can be a little bit tricky with Quanty, specifically when it comes to the Eigensystem function. Could you please post your entire script so that we can take a look at how you've set up yours?
Dear John,
One other thing to consider is degeneracies. I would suggest for you to calculate not just the lowest state but all 1001 possible states with 10 electrons in 14 spin-orbitals. If two or more states have the same energy than each of these states will brake the symmetry of the Hamiltonian, but the sum of expectation values over these states will obey the Hamiltonian symmetry.
Maurits