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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] https://arxiv.org/pdf/quant-ph/0608197 page 6

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  • $\begingroup$ 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? $\endgroup$ – Norbert Schuch Nov 3 '18 at 15:28
  • $\begingroup$ @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 $\lvert\phi_{L}\rangle=\Lambda$. $\endgroup$ – Alessandro Nov 3 '18 at 18:20
  • $\begingroup$ RG doesn't make much sense without a notion of translational invariance. Note that e.g. Eq. (3) in said paper is tinv. $\endgroup$ – Norbert Schuch Nov 3 '18 at 18:22
  • $\begingroup$ @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 $\Lambda^{[k]}$. $\endgroup$ – Alessandro Nov 3 '18 at 18:45
  • $\begingroup$ All right. And where are they talking about the eigenvectors of the non-tinv E? $\endgroup$ – Norbert Schuch Nov 3 '18 at 18:54
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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.

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