Matrix element of hubbard Hamiltonian using hubbard siteset

+1 vote
asked Mar 2 by JunjieChen (340 points)

I did some simple test about hubbard siteset and use "overlap" function to calculate matrix element of fermion hopping term and I just found some problems.

#include "itensor/all.h"

using namespace itensor;

int main()
    int N=100;
    auto tf=1.0;
    auto Uf=1.0;

auto sites=Hubbard(N);
auto ampo=AutoMPO(sites);
for(int i=1; i<N; i++)
    ampo += -tf, "Cdagup", i, "Cup", i+1;
    ampo += -tf, "Cdagup", i+1, "Cup", i;
    ampo += -tf, "Cdagdn", i, "Cdn", i+1;
    ampo += -tf, "Cdagdn", i+1, "Cdn", i;
for(int i=1; i<=N; i++)
    ampo += Uf, "Nup", i, "Ndn", i;
auto H=IQMPO(ampo);

/************** TEST *************/
auto s1=InitState(sites);
auto s2=InitState(sites);


auto p1=IQMPS(s1);
auto p2=IQMPS(s2);


as shown in the source code, I expect overlap(p2,H,p1)=+1 due to the symmetry of fermion "UpDn" state as described in . But the output is -1. This is puzzling. I also checked the source code of "hubbard.h" in Itensor and I found

if(opname == "Cdn")
    if(opname == "Cdagdn")

and the minus sign in "Cdn" and "Cdagdn" is correct as I expected.

So, What's wrong with it or did I misunderstand the sign convention ?

1 Answer

0 votes
answered Mar 2 by miles (14,880 points)

Hi Junjie,
So just doing it in my head (meaning I could be making a mistake) there's:
1. a minus sign for the process of annihilating the down electron from the UpDn state (with an even number (=0) of electrons to the left of site 1
2. a minus sign from re-creating the down electron on site 2, because there are an odd number of electrons to the left of site 2
3. a minus sign from the -t coefficient of the hopping term where (t=1)

So overall a -1. Hope that helps -

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