I've been trying to better understand matrix product states (in order to implement them in code in the near future), so I'm considering small examples. I was wondering if I could get some clarification on how to express states as an MPS for this specific case.

In the 2 site Ising model, we have a Hamiltonian in the form:

$$ H = -(\sigma^z_1 \sigma^z_2 + \sigma^z_2 \sigma^z_1) - h (\sigma^x_1 + \sigma^x_2) $$

where I have set the value of the interaction coefficient to 1. When $h=0$, I would expect my ground state to be some superposition of $| \uparrow \uparrow \rangle$ and $| \downarrow \downarrow \rangle$.

If I'm understanding this reference properly (page 3), if I want to express $| \psi \rangle = | \uparrow \uparrow \rangle + | \downarrow \downarrow \rangle$ (currently not normalized) as a matrix, my goal is to obtain two matrices that multiply to make:

$$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

The solution is to pick $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

So my questions:

  1. Are matrix product states unique? It seems that $A$ is effectively $\begin{bmatrix} | \uparrow \uparrow \rangle & 0 \\ 0 & | \downarrow \downarrow \rangle \end{bmatrix} $, but I could easily change this to be $\begin{bmatrix} | \downarrow \downarrow \rangle & 0 \\ 0 & |\uparrow \uparrow \rangle \end{bmatrix} $

  2. Physically, how do I interpret $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $? What do the matrix elements mean?

  3. In terms of actual implementation in a tensor network, can I express this MPS as a (2, 2, 2) tensor? Where the first two indices indicate that we have $2\times 2$ matrices, and the 3rd index shows that we have two of these $2 \times 2$ matrices?

I'm guessing I have a lot of misconceptions here.

  • $\begingroup$ What does your matrix A here represent? $\endgroup$ Jul 14, 2020 at 0:33
  • $\begingroup$ I'm not sure if this is correct, but I thought A would be the full representation of the ground state? $\endgroup$
    – Jlee523
    Jul 14, 2020 at 5:23
  • 1
    $\begingroup$ @Jlee523 The Ising ground state does not have an exact MPS representation (except at h=0 or infinity). $\endgroup$ Jul 14, 2020 at 16:34

1 Answer 1


To create a matrix product state for a spin system, you need to specify two matrices for every site, as illustrated on the Wikipedia page.

enter image description here

Let's try to make a MPS for your system. Your system has two sites, so we'll have a total of four matrices $A_1^{\downarrow},A_1^{\uparrow},A_2^{\downarrow},A_2^{\uparrow}$. The relationship between the set of matrices $A_i^\sigma$ and the amplitudes of a wavefunction is that the amplitude of the state $|\uparrow\uparrow\rangle$ is given by $\text{Tr}(A_1^\uparrow A_2^\uparrow)$, and similarly for all the other possible amplitudes.

You want to create the state $|\uparrow\uparrow\rangle+|\downarrow\downarrow\rangle$, so you need four matrices $A_i^\sigma$ such that

  • $\text{Tr}(A_1^\uparrow A_2^\uparrow)=1$

  • $\text{Tr}(A_1^\uparrow A_2^\downarrow)=0$

  • $\text{Tr}(A_1^\downarrow A_2^\uparrow)=0$

  • $\text{Tr}(A_1^\downarrow A_2^\downarrow)=1$

If you like, you can also try to make the matrices $A_1^\sigma$ and $A_2^\sigma$ equal to each other, although this is not necessary to define an MPS. One set of matrices that work is $$ A_1^\uparrow=A_2^\uparrow = \frac{1}{\sqrt 2}\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right) \qquad A_1^\downarrow=A_2^\downarrow = \frac{1}{\sqrt 2}\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right). $$ (although note that the matrices in an MPS are not unique, many different sets of matrices can give the same state, and there is no direct physical interpretation of the elements of the matrices in an MPS)

  • $\begingroup$ [1 0;0 0] and [0 0;0 1] would be far better choices in the light of the Ising model, as they would give the GHZ state for any system size. $\endgroup$ Jul 14, 2020 at 16:35

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