How does the complexity in Matrix Product states ansatz drop from $D^N$ to $ND$?

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?
• Can you add some details about what you want to understand? Otherwise "it has less parameters" seems a valid answer. – Norbert Schuch Feb 1 at 9:40
• 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. – EverydayFoolish Feb 1 at 13:12
• 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. – Sunyam Feb 1 at 15:09
• 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. – EverydayFoolish Feb 10 at 15:52
• 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. – EverydayFoolish Feb 10 at 15:57