0 votes
asked by (300 points)
edited by

I'm trying to measure expected values of Sx, Sy and Sz on each site for given Hamiltonian and graph. My code looks as follows:

#include "itensor/all.h"
using namespace itensor;

int
main(int argc, char* argv[])
{
int Nx = 8; // Nx can be understood as the size of an elementary cell
auto N = atoi(argv[1])*Nx;

auto sites = SpinOne(N,{"ConserveQNs=",false});

auto ampo = AutoMPO(sites);

// [Definitiion of my Hamiltonian]

auto H = toMPO(ampo);

auto state = InitState(sites);

for (int i = 1; i <= N; i++) {
    state.set(i,"Up");
}

auto sweeps = Sweeps(10);
sweeps.maxm() = 50,100,200,300,400;
sweeps.cutoff() = 1E-10;
println(sweeps);

//
// Begin the DMRG calculation
// for the ground state
//

auto [en0,psi0] = dmrg(H,randomMPS(sites),sweeps,{"Quiet=",true});

//
// Print the final energies reported by DMRG
//

printfln("\n      Ground State Energy = %.10f",en0);    

//
// Measuring Sx, Sy & Sz
//

println("Ground state");
println("\nj | Sx | Sy | Sz | S_total |: ");
for( auto j : range1(N) ) {
    // Re-gauge psi0 to get ready to measure at position j
    psi0.position(j);

    auto ket = psi0(j);
    auto bra = dag(prime(ket,"Site"));

    auto Sxjop = op(sites,"Sx",j);
    auto Syjop = op(sites,"Sy",j);
    auto Szjop = op(sites,"Sz",j);

    //take an inner product
    auto sxj = eltC(bra*Sxjop*ket);
    auto syj = eltC(bra*Syjop*ket);
    auto szj = eltC(bra*Szjop*ket);
    println(j, " | ", sxj, " | ", syj, " | ", szj, " | ", sxj + syj + szj, " | ");
}

return 0;

}

When I run the above program I get:

Ground state
j | Sx | Sy | Sz |: 
1 | (2.82726e-12,0) | (0,-2.64747e-28) | (-0.98518,0) |
[...]

However, if I only change the part:

[...]
auto Sxjop = op(sites,"Sx",j);
auto Syjop = op(sites,"Sy",j);
auto Szjop = op(sites,"Sz",j);
[...]

into:

[...]
auto Sxjop = op(sites,"Sx2",j);
auto Syjop = op(sites,"Sy2",j);
auto Szjop = op(sites,"Sz2",j);
[...]

I get:

Ground state
j | Sx^2 | Sy^2 | Sz^2 | S_total^2 |: 
1 | (0.5,0) | (0.5,0) | (1,0) | (2,0) | 
[...]

which makes sense, because I believe the square of total spin satisfies following rules:

$$
S{total}^2 = Sx^2 + Sy^2 + Sz^2
S_{total}^2 = S ( S + 1),
$$

so for the spin @@ S = 1 @@ I should get @@ S_{total}^2 = 2 @@, which holds.

But I have to measure the values Sx, Sy and Sz instead of their squares. Do you have any suggestions?

commented by (460 points)
Hi Ozaru,

Does your Hamiltonian conserve total Sz? Because if it does, even you set {"ConserveQNs=",false} the ground state will still be in one of the Sz sectors so the expectation value of Sx and Sy will always be 0.

Best,

Yixuan

Please log in or register to answer this question.

Welcome to ITensor Support Q&A, where you can ask questions and receive answers from other members of the community.

Formatting Tips:
  • To format code, indent by four spaces
  • To format inline LaTeX, surround it by @@ on both sides
  • To format LaTeX on its own line, surround it by $$ above and below
  • For LaTeX, it may be necessary to backslash-escape underscore characters to obtain proper formatting. So for example writing \sum\_i to represent a sum over i.
If you cannot register due to firewall issues (e.g. you cannot see the capcha box) please email Miles Stoudenmire to ask for an account.

To report ITensor bugs, please use the issue tracker.

Categories

...