$G$-injective MPS and symmetry-broken phases

Question

First, a little bit of motivation. I was reading the paper "Matrix Product States and Projected Entangled Pair States" to try to learn more about MPS representations of symmetry broken states. There's a discussion of the GHZ states $\frac{|{0000...}\rangle \pm |{1111...}\rangle}{\sqrt{2}}$ which are the ground states deep in the symmetry-broken phase of the transverse field Ising model (i.e. the ground states of $-\sum_i \sigma^z_i \sigma^z_{i+1} - h\sum_i \sigma^x_i$ for $h=0$). The GHZ states have $A$ matrices of the form $$A^{[0]} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$ and $$A^{[1]} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$

The paper makes the following interesting statement:

Because the corresponding MPS is not injective, the tensors $A^i$ exhibit a non-trivial symmetry $G$ on the virtual level represented by the matrices that commute simultaneously with all $A^i$ ($\sigma^z$ in this GHZ case). MPS with this property are called $G$-injective. Note that this symmetry is dual to the physical symmetry which permutes the blocks (represented by $\sigma^x$ in the case under consideration).

I interpret this as follows: being $G$-injective implies there exists a basis on the bond space for which the $A^i$ matrices are block diagonal. The set of block diagonal matrices commutes with symmetry operators that are themselves block diagonal and proportional to the identity on each block. However, it's not clear to me what is meant by the fact that these symmetry operators that commute with the $A$ matrices are "dual" to physical symmetry operators. It seems like the paper is making a claim that the symmetry operators that permute blocks are physical in some sense and related to the symmetry of the phase.

Indeed, $\prod_i \sigma^x_i$ is the symmetry operator for the transverse field Ising model, and $(\sigma^x)^\dagger A^i \sigma^x$ permutes the $1$ by $1$ blocks of the $A^i$ matrices, so it seems like there is a close connection.

My aim is to try to understand the theory above within a symmetry broken phase but away from any special points within that phase. My main interest in the theory above is that at the special GHZ point above, in the physical basis of $\sigma^x$ eigenstates, $|+\rangle$ and $|-\rangle$, we have that $$A^{[+]} = \frac{1}{\sqrt{2}} I$$ and $$A^{[-]} = \frac{1}{\sqrt{2}} \sigma^z.$$ This is a special form for the $A$ matrices - it makes them proportional to unitary matrices, and I believe this is related to successfully implementing a teleportation protocol via $\sigma^x$ measurement on the GHZ state. Suppose I am in a symmetry broken phase like that given by $-\sum_i \sigma^z_i \sigma^z_{i+1} - h\sum_i \sigma^x_i$ for very small but nonzero $h$. In this basis away from the GHZ point but still in the phase, could I always write the $A$ matrices as $A^{[+]} = I \otimes B^{[+]}$ and $A^{[-]} = \sigma^z \otimes B^{[-]}$? Here, the $B$ matrices are arbitrary matrices to help allow my bond-dimension to grow away from the special GHZ point.

Answer 1

Suppose I am in a symmetry broken phase like that given by $-\sum_i \sigma^z_i \sigma^z_{i+1} - h\sum_i \sigma^x_i$ for very small but nonzero $h$. In this basis away from the GHZ point but still in the phase, could I always write the $A$ matrices as $A^{[+]} = I \otimes B^{[+]}$ and $A^{[-]} = \sigma^z \otimes B^{[-]}$?

The answer is yes, and I thank Frank Pollmann for a useful discussion. Note that in the symmetry broken phase, we can construct two symmetry-broken ground states in the thermodynamic limit. One symmetry broken ground state will be mostly up, with scattered downs, and the other will be mostly down, with scattered ups.

The mostly up state will be $$|\psi_u\rangle = \sum_{i_1,...,i_L} \text{Tr}[A_u^{[i_1]}A_u^{[i_2]}...A_u^{[i_L]}] |i_1 i_2,...,i_L\rangle$$ and the mostly down state will be $$|\psi_d\rangle = \sum_{i_1,...,i_L} \text{Tr}[A_d^{[i_1]}A_d^{[i_2]}...A_d^{[i_L]}] |i_1 i_2,...,i_L\rangle$$

Note that the MPS matrices are related to one another by the symmetry transformation $\prod_i \sigma^x_i$, which interchanges the two symmetry broken ground states: $$A_u^{[0]} = A_d^{[1]}\,\,\text{ and }A_u^{[1]} = A_d^{[0]} $$

The symmetric ground state will be a cat state built out of an MPS with the direct sum of these matrices:

$$|\psi_c\rangle = \sum_{i_1,...,i_L} \text{Tr}[\begin{pmatrix} A_u^{[i_1]} & 0 \\ 0 & A_d^{[i_1]} \end{pmatrix}\begin{pmatrix} A_u^{[i_2]} & 0 \\ 0 & A_d^{[i_2]} \end{pmatrix}...\begin{pmatrix} A_u^{[i_L]} & 0 \\ 0 & A_d^{[i_L]} \end{pmatrix}] |i_1 i_2,...,i_L\rangle$$

Let's call the MPS in the symmetric cat state $$A_c^{[i]} = \begin{pmatrix} A_u^{[i]} & 0 \\ 0 & A_d^{[i]} \end{pmatrix}$$

Note that $$A_c^{[+]} = \frac{1}{\sqrt{2}} \left( \begin{pmatrix} A_u^{[0]} & 0 \\ 0 & A_d^{[0]} \end{pmatrix} + \begin{pmatrix} A_u^{[1]} & 0 \\ 0 & A_d^{[1]} \end{pmatrix} \right) = \frac{1}{\sqrt{2}} \begin{pmatrix} A_u^{[0]}+A_u^{[1]} & 0 \\ 0 & A_d^{[0]}+A_d^{[1]} \end{pmatrix}$$ and note that we can use the relationship between $A_u$ and $A_d$ to rewrite $A_d^{[0]}+A_d^{[1]}$ as $A_u^{[0]}+A_u^{[1]}$, giving

$$A_c^{[+]} = \frac{1}{\sqrt{2}} \begin{pmatrix} A_u^{[0]}+A_u^{[1]} & 0 \\ 0 & A_u^{[0]}+A_u^{[1]} \end{pmatrix} = I \otimes \frac{A_u^{[0]}+A_u^{[1]}}{\sqrt{2}}. $$

A similar computation shows that

$$A_c^{[-]} = \sigma^z \otimes \frac{A_u^{[0]}-A_u^{[1]}}{\sqrt{2}}.$$

These are exactly of the postulated form, confirming the guess.