How to use AutoMPO with a matrix prefactor?

ITensor C++ Questions
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1

Hi, I want to add a term into my mpo using the AutoMPO function, and the term is a matrix multiplied by a S=\frac{1}{2} operator. Specifically, in my code I include it as:

for (int j=1;j<N;j++){
  mpo += matrix,"Sx",j;
}

and the matrix I want to put in can be site dependent or site independent. I’ve gone through the relevant tutorials and I read that the AutoMPO function only takes in real or complex values as the prefactors, and not a matrix. So when I try to build my project, I encounter an error in the Accumulator class in AutoMPO

no known conversion for argument 2 from ‘std::array<std::array<double, 2>, 2>’ to ‘itensor::Real’ {aka ‘double’}

Can I know what is the best way I can achieve this? Any help would be greatly appreciated! Thanks!

2

Pretty sure this is not possible. I think the solution is to create a matrix of MPOs each of which has scalar coefficients.

3

Ben is correct that this input is not supported. But it makes me want to ask @zomal : what was your goal, mathematically speaking, that led you to want to input a matrix as a coefficient? What would a matrix-valued coefficient mean and what sort of output did you want or expect?

4

Thanks @corbett5 , how can I go about making a matrix of MPOs using ITensor? I cant seem to find any tutorial for that, or maybe I missed it.

Thanks @miles , the model I am using has g tensors terms like

\Sigma_i \; g_u \hat{S_{i}}

Im new to using this model and also actually ITensor and AutoMPO, so I wasnt sure of the best way to deal with this. If you have another idea, Im open to hear them of course!

5

Thanks for the extra info, but that’s still not enough information for me to know why you would need a coefficient to be a matrix. Is g_u a matrix? If so, what does that mean mathematically and what vector space does it live in / act on?

6

Hi @miles , thanks for the help, and sorry for the delay. Actually I realized I made a mistake and do not in fact need a matrix prefactor. Your question of what it means mathematically has led me to realizing that. Thanks so much again!

7

Glad the discussion was helpful :+1: