Question about the Hamiltonian MPO

General discussions on the ideas behind the algorithm.
Post Reply
Fabio_Mendez
Posts: 4
Joined: 01 Feb 2022, 01:50

Question about the Hamiltonian MPO

Post by Fabio_Mendez »

Hi, thanks a lot for creating these libraries. I love the simplicity of constructing DMRG codes with them!

I have a question, why do the first Hamiltonian MPO sites always have two rows instead of one? I have worked with the TNT libraries, and we use 1 row and 1 column at the first and last sites in that software. Wouldn't having two rows or columns make the code inefficient for big Hilbert spaces?

Thanks in advance,
User avatar
Johannes
Site Admin
Posts: 413
Joined: 21 Jul 2018, 12:52
Location: TU Munich

Re: Question about the Hamiltonian MPO

Post by Johannes »

MPOs are extensive sums of local terms; thus they take a Jordan block form
\[ W = \begin{pmatrix} 1 & C & D \\ 0 & A & B \\ 0 & 0 & 1 \end{pmatrix} = \text{e.g. for transverse field ising} \begin{pmatrix} 1 & X & - Z \\ 0 & 0 & X \\ 0 & 0 & 1 \end{pmatrix} \]
The first and last index here are special, because they have the identity blocks. TeNPy chooses to keep those identity blocks/indices around by default, just because that allows to easily modify/extract onsite terms and avoid complications for some methods that require them (e.g in calc_H_bond_from_MPO).

In general, this is a tiny bit of extra work for finite MPS, yes, but it's definitely a subleading contribution - even for huge local bond dimensions, it's not gonna give you a substantial speedup of more than a few percent (unless you only have this huge local dimension only at the boundaries, and/or a tiny system).
That being said, if it really bothers you that much, you can easily disable this by setting the option insert_all_Ids=False of the function from_terms when constructing the MPO, presumably in the source code of calc_H_MPO.
Post Reply