AutoMPO with products of three/four terms operators

Question

Hi all,

I am trying to implement the AKLT Hamiltonian for the Spin-2 case.
Having computed with Mathematica the power (Si * Sj)^3 and (Si * Sj)^4 (that are contained in the Hamiltonian) as a function of the operators S+ and S-, I have terms of the form

ampo += 1.0, "Sz*S+*S+",indices_prime[i],"Sz*S-*S-",indices_prime[i+1];

and

ampo += 1.0, "S+*S-*Sz*Sz",indices_prime[i],"S-*S+*Sz*Sz",indices_prime[i+1];

with products of three or four operators on each site. I compile the code, but when I try to run it, I obtain the following error:

"Operator S- name not recognized".

Actually, the code seems to recognise the operator "S-", because, if I put only two operators for each site (for instance, "S+S-" and "S-S+"), it works. I think the error it is due to the fact that I am using more than two operators in the product on each site.

Any advice?

Thank you very much.

Comments

Hi, thanks for the question. One question before I try to find the issue: what is the value type for the array or vector "indices_prime"? Are those integers or ITensor Index types? Thanks -

---

Hi Miles,

thank you for your precious help. In the array "indices_prime" I have normal integers (1,2,3...N).

Best.

---

Hi, so the following case worked for me:

    int N = 10;
    auto sites = SpinTwo(N);

    auto ampo = AutoMPO(sites);
    for(int j = 1; j < N; ++j)
        {
        ampo += "Sz*S+*S+",j,"Sz*S-*S-",j+1;
        ampo += "S+*S-*Sz*Sz",j,"S-*S+*Sz*Sz",j+1;
        }
    auto H = toMPO(ampo);

so then perhaps the issue you are seeing is because of the specific values of indices_prime[i] and indices_prime[i+1] ? Could you print out or test all of these values and make sure they are between 1 and N, inclusive?

Otherwise I think I will need more information or a minimal example code reproducing the bug in order to investigate it.

Thanks,
Miles

---

Hi Miles,

thank you very much. In the end, I solved the problem: there was a typo in the string of operators. Now it works for any number of operators.


Thank you very much for your precious help.

Best.

---

Great. Thanks for following up! It’s useful to know so we can catch any potential bugs.

Feel free to ask us if you need any more help

Miles

Answer 1

(This question was answered in the above comments thread.)