I of course mean boundary conditions, even if the autocorrect of my mail program renames it ....
On Sep 7, 2014, at 13:50, Matthias Troyer troyer@phys.ethz.ch wrote:
With open bondage conditions momentum is not a well defined quantum number
On Sep 7, 2014, at 13:49, hamid mosadegh hamid.mosadegh@gmail.com wrote:
thanks for your answer. I use DMRG with open boundary condition. what about this code?
On Sun, Sep 7, 2014 at 1:25 PM, Matthias Troyer troyer@phys.ethz.ch wrote: What code do you use? If you have a periodic system and use the diagonalization codes then you will already get the spectra momentum resolved.
On Sep 7, 2014, at 9:35, hamid mosadegh hamid.mosadegh@gmail.com wrote:
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
-- H. Mosadeq Shahr-e-Kord University (SKU), Shahr-e-Kord , Iran