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The Motivation is That:In the Tensor Network method, they say 'time evolution MPS(Matrix Product State) work quite well in 1 Dimension'.

but as I think any 2D could be expressed by 1D

for example in the 2x2 Square lattice J1J2 isingmodel (2D) to

L = 4 1D lattice.

What is the difference between 1D 2D?

.

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  • $\begingroup$ The difference is in the scaling. For a 2-dim system with N spins, you need to contract many more bonds compares to a 1-dim. $\endgroup$
    – proton
    Commented Aug 17, 2023 at 6:35
  • $\begingroup$ I think you should be more detailed/specific. $\endgroup$
    – Norbert Schuch
    Commented Aug 17, 2023 at 6:52
  • $\begingroup$ @proton But any 2D representation could be mapped on 1D. If I try to calculate (fixed lattice) 2D Isingsystem , I Could unfold from 2D to 1D and then I can calculate '1D' Ising system by using MPS is there have any boundary for the between 1D and 2D for calculating some fixed lattice ? $\endgroup$
    – Cha
    Commented Aug 17, 2023 at 7:18
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    $\begingroup$ Why is this question closed? It is perfectly clear and a very reasonable confusion... $\endgroup$
    – AccidentalFourierTransform
    Commented Aug 22, 2023 at 13:15
  • $\begingroup$ @AccidentalFourierTransform I don't know... but your answer was very clear to me thank you! $\endgroup$
    – Cha
    Commented Aug 22, 2023 at 16:25

2 Answers 2

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The resulting '1D' Ising system has long-range interactions (while the initial 2D system had only short-range interactions). Approximating the state of the system by a MPS works well for chains with short-range interactions.

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More generally, there is no notion of dimension for finite lattices. Only a sequence of lattices, of growing size, can be ascribed a dimension. For a fixed lattice, there is no way to define the range of interactions and the dimension of the lattice. (Of course, if the lattice is very large, there is an intuitive notion of dimension but it is only approximate and there is no precise mathematical definition. In more technical terms, a finite set has no dimension).

Your argument precisely shows this. In a somewhat formal sense, all finite lattices are zero-dimensional (they are just good old quantum mechanics).

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