Dear All
What is the momentum of , E_1(N_{\uparrow},(N_{\downarrow} ,L), the lowest-excitation energy in the Hubbard model in the S=0 sector.

I means, I want to evaluate the velocity of charge excitation from

E_1(N_{\uparrow},(N_{\downarrow} ,L)- E_0(N_{\uparrow},(N_{\downarrow} ,L)

where E_1(N_{\uparrow},(N_{\downarrow} ,L) is the lowest-excitation energy with

momentum k=2\pi/L and total spin S=0.

If i run 

MODEL="fermion Hubbard"
CONSERVED_QUANTUMNUMBERS="Nup,Ndown"
Ndown_total=10;Nup_total=10
NUMBER_EIGENVALUES=2

what is the momentum of first excitation. is it possible to determine or change it?

thanks