# Left and Right Eigenvectors of Transfer Matrix in Matrix Product States (MPS)

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

• I'm not sure why you want to find the eigenvector. Are you assuming translational invariance? Can you give a reference for the last formula you give? – Norbert Schuch Nov 3 '18 at 15:28
• @NorbertSchuch I'm not assuming translational invariance. I want to find the right and left eigenvectors corresponding to the eigenvalue 1 (assumed to be non-degenerate) because under RG transformation the tensor $E$ flows to the fixed point $E^{\infty}=\lim_{ n\to\infty} E^{n}$ which is given by the tensor product of these two eigenvectors. Here arxiv.org/pdf/quant-ph/0410227, at page 3 , $E^{\infty}=\lvert\phi_{R}\rangle\langle\phi_{L}\rvert$ where $\lvert\phi_{R}\rangle=I$ and $\lvert\phi_{L}\rangle=\Lambda$. – Alessandro Nov 3 '18 at 18:20
• RG doesn't make much sense without a notion of translational invariance. Note that e.g. Eq. (3) in said paper is tinv. – Norbert Schuch Nov 3 '18 at 18:22
• @NorbertSchuch Yes, you're right, I considered the wrong example. Here arxiv.org/pdf/1008.3745.pdf at page 5 they consider a non translation invariant mps, but they're using a slight different expression for $A_{i_k}^{[k]}$, and so the eingenvectors are $\delta^{[k]}_{\alpha\gamma}\lambda^{[k]}_{\alpha}$ and $\delta^{[k+1]}_{\beta\chi}\lambda^{[k+1]}_{\beta}$ where here ${(\lambda^{[k]}_{\alpha})}^2$ are the eigenvalues of $\Lambda^{[k]}$. – Alessandro Nov 3 '18 at 18:45
• All right. And where are they talking about the eigenvectors of the non-tinv E? – Norbert Schuch Nov 3 '18 at 18:54

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.