# Example of an injective matrix product state (MPS)

I am struggling to understand what is an injective matrix product state (MPS). From the definition, it is said that an injective MPS $$|M(A)\rangle$$is one where the tensor $$A$$ has a projector $$P(A)$$ with a left inverse. I try to think of the example with the GHZ state with $$A^{[0]} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, A^{[1]} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$ but it's difficult to visualize the left inverse since it's a 4-rank tensor. Is there a procedure to break it down and visualize it, as to understand why it is called "injective" and verify if one is?

• The GHZ is not an injective MPS. (Nor is the MPS tensor rank-4.) Commented Jul 30 at 18:53

An MPS tensor $$A^{[s]}$$ is called injective if the map $$X\mapsto\sum_s\mathrm{tr}[XA^{[s]}]\,\lvert s\rangle$$ is injective.1

In particular, the example you give is not an injective tensor -- e.g., if you set $$X=\begin{pmatrix}1 &0\\0&1\end{pmatrix}$$, it is mapped to $$0$$. (In fact, you can already see this from the fact that $$s$$ can take only two values -- for the map to be injective it would have to take at least $$4$$ values.)

1 This map is sometimes denoted as $$P(A)$$ -- note that unlike what you say in your question, $$P(A)$$ is in general not a projector, even though in some important examples it is, and the "Projected" in "Projected Entangled Pair States", i.e., PEPS, refers to that.

The term injective comes from linear transformations. Recall that an injective linear map $$T : V \rightarrow W$$ is one that for every $$w$$ in $$W$$, there is at most one $$v$$ in $$V$$ such that $$T(v)=w$$. If T is injective then we can define a projection $$P : W \rightarrow U$$ such that $$P$$ is a linear map and $$\mathrm{ker}(P) \cap \mathrm{im}(T)=\{0\}$$, and $$P(T)$$ is both injective and surjective, thus it's invertible.

For the case of the GHZ state over 2 qubits, we have: $$|\mathrm{GHZ}\rangle = \frac{1}{\sqrt{2}}\sum_{\{s_1,s_2\}\in\{0,1\}} \mathrm{Tr} (A^{[s_1]} A^{[s_2]}) |s_1s_2\rangle$$

We can write the open boundary condition tensor for this state as follows:

$$\mathrm{GHZ}_\mathrm{OBC} = \frac{1}{\sqrt{2}}\sum_{\{s_1,s_2\}\in\{0,1\}} A^{[s_1]} A^{[s_2]}$$

Which is a 4-tensor. We can write out this 4-tensor as (up to normalization): $$A_{{s_1}{s_2}}^{{a_1}{a_2}} = \delta_{s_1}^0\delta_{s_2}^0\delta_{a_1}^0\delta_{a_2}^0 + \delta_{s_1}^1\delta_{s_2}^1\delta_{a_1}^1\delta_{a_2}^1$$

We can construct a projector that contracts two of the indices: $$(PA)_{s_2}^{a_2} = A_{{s_1}{s_2}}^{{s_1}{a_2}}$$

Using Einstein notation. You can see that this projector results in the identity tensor, which is its own inverse.

• What does this show? The MPS in the question is not injective. Commented Jul 30 at 18:53
• @NorbertSchuch you are welcome to provide your own answer. Commented Aug 1 at 16:50
• I just wanted to explain my downvote, and point out to other readers that the answer, as it is, is plain wrong: Both what it shows and the notion of injectivity used are incorrect. Commented Aug 1 at 18:36
• @NorbertSchuch As it stands, your comment is unhelpful and low-effort. You have not explained why it's 'wrong' nor what you believe the right answer is. Commented Aug 2 at 19:59