Yes, I know it. But I use OBC because it is better in DMRG calculations.

However, if I use PBC the code determine the momentum of states.

On Sun, Sep 7, 2014 at 4:22 PM, Matthias Troyer <troyer@phys.ethz.ch> wrote:

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
>

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

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