Example of an injective matrix product state (MPS)

Question

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?

Answer 1

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 $X=\begin{pmatrix}0 &1\\1&0\end{pmatrix}$, 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.)

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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.

Answer 2

One thing to add to Norbert Schuch's answer is that injectivity of an MPS is determined by whether or not you can "block" the matrices/tensors together to reach an injective mapping; looking at a single matrix is sometimes not enough.

In the way you wrote it down, blocking would just be matrix multiplication of different $A^{[0]}$ and $A^{[1]}$, but it's easy to see for the GHZ state that this will never lead you to a trivial kernel, i.e. an injective mapping.

$$ A^{s_1s_2s_3...s_n}= A^{s_1} \; \; ; \; s_i=s_j \; \;\forall i,j $$ $$ A^{s_1s_2s_3...s_n}= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \; \; ; \; \text{otherwise} $$

Answer 3

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.