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
H. Mosadeq
Shahr-e-Kord University (SKU),
Shahr-e-Kord , Iran