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I have just started to read about DMRG and MPS.

It is said that in case of simple 1D chain with spins states $|\uparrow\rangle$; $|\downarrow\rangle$ and any state in the complete Hilbert space of such a system could be written as :

$|\Psi\rangle=\sum\limits_{i_1,...,i_n} C^{i_1,...,i_n}|i_1,...,i_n\rangle$

Where each index runs over the local basis at each site. This presents a complexity of the order $2^N$ and by writing it in the form of matrix product states we would making a simplification a mean field simplification and reduce the complexity to $2N$.

$C^{i_1,...,i_n}=C^{i_1}C^{i_2}...C^{i_n}$

EDIT:

A General case:

Suppose I have Heisenberg chain kind of a 1D Lattice problem with N sites, where each site lives on a Hilbert space of dimensionality D. Given the fact that the dimensionality of Hilbert space of the entire system scales exponentially as $D^N$ because of the entanglement, it becomes a very complex problem to solve.

Questions:

  1. How does writing the system in terms of MPS reduce the complexity of the problem and bring it down to $ND$ ?
  2. What is the link between using MPS ansatz and mean field approximation?
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    $\begingroup$ Can you add some details about what you want to understand? Otherwise "it has less parameters" seems a valid answer. $\endgroup$ – Norbert Schuch Feb 1 at 9:40
  • $\begingroup$ I want to understand the of how the complexity for $ C^i's $ scales down and what really is the basis for MPS and how does link up to mean field theory. $\endgroup$ – EverydayFoolish Feb 1 at 13:12
  • $\begingroup$ Using singular value decomposition on wave function expansion coefficients, you can proove MPS representation for arbitrary many body wave function. In this step, if I am correct you dont reduce any complexity. I think you further need to assume/approximate some value bond dimensions. If you impose bond dimensions to be unity i. e., matrices of MPS are scalars, then you will have simple standard mean field ansatz. $\endgroup$ – Sunyam Feb 1 at 15:09
  • $\begingroup$ I don't get it, I find that MPS is a way to reorganize and rewrite the coefficients. Please correct me if I am wrong about this. $\endgroup$ – EverydayFoolish Feb 10 at 15:52
  • $\begingroup$ I really want to understand what is the advantage of talking about t-DMRG in terms of MPS ansatz rather than traditional terms in which it was first formulated. $\endgroup$ – EverydayFoolish Feb 10 at 15:57

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