Left and right eigenvectors of transfer matrix in matrix product states (MPS)

Question

Let

$$\lvert{\psi}\rangle=\sum_{i_1i_2...i_n}Tr(A^{[1]}_{i_1}A^{[2]}_{i_2}...A^{[n]}_{i_n})\lvert{i_1 i_2...i_n}\rangle$$

be a MPS, where $i_k=1,2...d$ and $A^{[k]}_{i_k}$ are $D\times D$ matrices on site $k$. We know we can construct the "Transfer Matrix" $E^{[k]}$ as:

$$E^{[k]}=\sum_{i_k} A_{i_k}^{[k]}\otimes {A_{i_k}^{*}}^{[k]}.$$

We also have the freedom to choose the $A^{[k]}_{i_k}$ matrices such that [1]:

$$\sum_{i_k} A_{i_k}^{[k]}{A_{i_k}^{\dagger}}^{[k]}=I \tag1$$

$$\sum_{i_k} {A^\dagger}_{i_k}^{[k]}\Lambda^{[k-1]}{A_{i_k}}^{[k]}=\Lambda^{[k]} \tag2$$

where $\Lambda^{[k]}$ is a diagonal matrix with $Tr(\Lambda^{[k]})=1$ containing the eigenvalues of the reduced density matrix $\rho_k=Tr_{k+1,...n}\lvert\psi\rangle\langle\psi\rvert$.

We can think of $E^{[k]}$ as a $D^2\times D^2$ matrix and I need to find the right and left eigenvectors of $E^{[k]}$ corresponding to the eigenvalue 1.

Using $(1)$ it is easy to see that $I$ is a right eigenvector:

$$E^{[k]}(I)=I$$

but from $(2)$:

$${E^{*}}^{[k]}(\Lambda^{[k-1]})=\Lambda^{[k]}\neq\Lambda^{[k-1]}$$

so $\Lambda^{[k-1]}$ should not be a left eigenvector from my understanding but in the literature it is treated as such and $E^{[k]}$ is expressed as:

$$E^{[k]}=\lvert I \rangle\langle\Lambda^{[k-1]}\rvert + \cdots$$

Where am I wrong?

Reference

[1] D. Perez-Garcia, F. Verstraete, M.M. Wolf and J.I. Cirac, "Matrix product state representations", Quantum Inf. Comput. 7, 401 (2007), arXiv:quant-ph/0608197 (page 6).

Answer 1

I think the confusion arises because the notation hints at an inner product, which is nowhere defined. I will maybe make things a bit more complicated than is needed in practice, but i hope it helps.

Denote by $M^{[k]}$ the space of $D_k\times D_k$-matrices, and let $1_k$ be the unit matrix of $M^{[k]}$. Then the transfer operator is a map

$$ E^{[k]} : M^{[k]} \rightarrow M^{[k+1]} \ . $$

One demands that

$$ E^{[k]}(1_k) = 1_{k+1} $$

The above enables us to write

$$ E^{[k]} = 1_{k+1} \otimes \lambda_k + \cdots $$

where $\lambda_k \in (M^{[k]})^*$ is the linear functional dual to $1_k$. In order to go to Dirac bra-ket notation, we have to identify $M^{[k]}$ with its dual space. The most straightforward way to do this is probably to equip $M^{[k]}$ with the Hilbert-Schmidt inner product $(X,Y) \mapsto \langle X , Y \rangle = \text{tr}(X^*Y)$ and then define the anti-linear Riesz isomorphism $R(X) = \langle X, \cdot\rangle$. Then a matrix $X$ is represented by the ket $| X \rangle$, and one may write $R(X) = \langle X |$.

We may thus write

$$E^{[k]} = |1_{k+1}\rangle \langle \Lambda_k | + \cdots $$

If one sets $D_k = D$ a constant, then, of course, one may identify all the $M^{[k]}$ and the $1_k$'s.