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A pure quantum state $$\tag{1}|\Psi\rangle=\sum_{j_1,\ldots,j_N=1}^{d}\psi_{j_1j_2\ldots j_N} |j_1, \dots, j_N\rangle\,,$$ can be represented exactly in the MPS form \begin{equation}\tag{2} |\Psi\rangle = \sum_{j_1,\ldots,j_N=1}^{d} A^{[j_1]} A^{[j_2]} \ldots A^{[j_N]} |j_1, \dots, j_N\rangle\,. \end{equation} For the left-canonical form, it holds true that $\sum_{j_i} A^{\dagger[j_i]}A^{[j_i]}=I$. Why is this the case?

I understand that at each SVD $U^{\dagger}U=I$, but I'm not sure about the sum over $j_i$. Or more concretely, in this publication - Eq. 38, I get that

\begin{equation}\tag{3} \left(U^{\dagger}\right)_{a_{\ell},\left(a_{\ell-1} \sigma_{\ell}\right)} U_{\left(a_{\ell-1} \sigma_{\ell}\right), a_{\ell}} = I. \end{equation} But I don't understand the sum over $a_{\ell-1} \sigma_{\ell}$.

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    $\begingroup$ What is your question? $\endgroup$ Mar 23 at 23:06
  • $\begingroup$ @NorbertSchuch My question is $\sum_{j_i} A^{\dagger[j_i]}A^{[j_i]}=I$? I don't follow the derivation in the publication (Eq. 38) $\endgroup$
    – jayjay
    Mar 24 at 11:39
  • $\begingroup$ Yes, it is. Does this answer your question? $\endgroup$ Mar 24 at 15:13
  • $\begingroup$ @NorbertSchuch no my question is why. Sorry the question was miswritten. $\endgroup$
    – jayjay
    Mar 24 at 17:50
  • $\begingroup$ I'm puzzled that you say you understand that the svd gives $U^\dagger U =I$, but then you say you don't understand the other formula. It is just the svd, rewritten. Matrix multiplication means summing over some index. Now, this index consists of two indices. One sum is written explicitly, one is implicit in the matrix multiplication. --- Note that your (3) does not have any index being summed over, so it cannot be anything like the identity matrix. (This is also not what the reference says, there is a $\delta_{xy}$.) $\endgroup$ Mar 24 at 19:04

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Iterative SVDs of the rank-$N$ coefficient tensor $\psi$ (see Eq. (1)) result in the representation of the quantum state as an MPS of the form written in Eq. (2). For a left-canonical MPS, let $A^{[j_l]}_{a_{l-1}, a_l}=U_{(a_{l-1},j_l), a_l }$, since it holds that $U^{\dagger}U=I$ at each SVD, the tensors obey the following relationship

\begin{equation} \begin{aligned} \delta_{a_{\ell}, a_{\ell}^{\prime}} &=\sum_{a_{\ell-1} j_{\ell}}\left(U^{\dagger}\right)_{a_{\ell},\left(a_{\ell-1} j_{\ell}\right)} U_{\left(a_{\ell-1} j_{\ell}\right), a_{\ell}^{\prime}} \\ &=\sum_{a_{\ell-1} [j_{\ell}]}\left(A^{\dagger[j_{\ell}] }\right)_{a_{\ell}, a_{\ell-1}} A_{a_{\ell-1}, a_{\ell}^{\prime}}^{[j_{\ell}]} = \sum_{[j_{\ell}]} \sum_{a_{\ell-1}} \left(A^{\dagger[j_{\ell}] }\right)_{a_{\ell}, a_{\ell-1}} A_{a_{\ell-1}, a_{\ell}^{\prime}}^{[j_{\ell}]}\\ &=\sum_{j_{\ell}}\left(A^{\dagger[j_{\ell}] } A^{[j_{\ell}]}\right)_{a_{\ell}, a_{\ell}^{\prime}}, \end{aligned} \end{equation} or more succintly, $\sum_{j_{\ell}}\left(A^{\dagger[j_{\ell}] } A^{[j_{\ell}]}\right) = I$ (this implies $a_{\ell} = a_{\ell}^{\prime}$).

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