# Why do we use matrix product states?

Given a many body $$\vert\psi\rangle$$, we can express it in terms of a matrix product state. That is,

$$\vert\psi\rangle = \sum_{i,j..k}\psi_{i,j..k}\vert i,j..k\rangle$$

can be rewritten as

$$\vert\psi\rangle = \sum_{i,j..k}A^{}_i A^{}_j..A^{[k]}_k\vert i,j..k\rangle$$.

It's not clear to me why this is an improvement, even if the size of the $$A$$ matrices are bounded since one replaced scalars $$\psi_{i,j..k}$$ with a matrix product instead. I haven't understood the claim for why as $$k$$ gets very large, the matrix product representation is often said to scale better, provided the state has no long range correlations.

Could someone give an example of a state where the MPS representation is more concise than the coefficients based representation?

## 1 Answer

Take a product state $$|\psi\rangle=|+,+,\dots,+\rangle$$, on $$N$$ spins. Then, to write it in the first form takes a tensor $$\psi_{i,j,\dots}$$ with $$2^N$$ non-zero elements. For $$N=100$$, there is no way this fits in your computer.

On the other hand, as an MPS, this can be written with $$A_i^{[s]}=\frac{1}{\sqrt{2}}$$ for all $$s$$, so you only need $$2N$$ coefficients, which easily fits in your memory.

• A follow up question, specific to this example: The number of nonzero coefficients required to represent a pure state depends on the choice of basis - here you have $2^N$ because you chose the computational basis. Is the advantage of MPS representation that one doesn't need to care about the basis anymore and the dimensions are purely determined by the amount of entanglement? – user1936752 Jan 12 '19 at 20:03