3
$\begingroup$

Consider two bipartite quantum states $|\phi\rangle^{AB}$ and $|\psi\rangle^{AB}$ (in a finite dimensional Hilbert space $\mathcal H_A\otimes \mathcal H_B$), such that $$\| |\phi\rangle\langle\phi|^{AB} - |\psi\rangle\langle\psi|^{AB} \|_p \leq \varepsilon\ .$$ Does this imply that there exist Schmidt decompositions \begin{align} |\phi\rangle^{AB}&=\sum_i\sqrt{p_i}|e_i\rangle^A|\tilde{e}_i\rangle^B\ ,\\ |\psi\rangle^{AB}&=\sum_i\sqrt{q_i}|f_i\rangle^A|\tilde{f}_i\rangle^B \end{align} which are also close to each other, i.e., for which \begin{align} 1-|\langle e_i|f_i\rangle|^2 & \leq g(\varepsilon)\ ,\\ 1-|\langle \tilde e_i|\tilde f_i\rangle|^2 & \leq g(\varepsilon)\ ,\\ |p_i-q_i|&\le h(\varepsilon)\ , \end{align} where $g(\varepsilon),h(\varepsilon)\to0$ as $\varepsilon\to0$?


Cross-posted on QC.SE

$\endgroup$
2
  • $\begingroup$ Hint: First, prove a relation between p-norm distance and overlap. Second, use this to show a relation between the Schmidt decompositions. $\endgroup$ Sep 18, 2023 at 7:45
  • 1
    $\begingroup$ In the end, this boils down to understand whether closeby matrices have closeby SVDs, cf. math.stackexchange.com/questions/3389899/… $\endgroup$ Sep 19, 2023 at 16:43

1 Answer 1

3
$\begingroup$

No.

To this end, consider $$ \lvert\phi\rangle = a\lvert0\rangle\lvert0\rangle + b \lvert1\rangle\lvert1\rangle\ , $$ and $$ \lvert\psi\rangle = a\lvert+\rangle\lvert+\rangle + b \lvert-\rangle\lvert-\rangle\ , $$ where $a=\sqrt{\tfrac12-\varepsilon}$, $b=\sqrt{\tfrac12+\varepsilon}$ [and with $\lvert \pm\rangle = \tfrac12(\lvert0\rangle\pm\lvert1\rangle)$].

$\lvert\phi\rangle$ and $\lvert\psi\rangle$ are in their Schmidt decomposition, and it is unique (as long as $\varepsilon\ne 0$). Moreover, $$ \|\lvert\phi\rangle\langle\phi\rvert-\lvert\phi\rangle\langle\phi\rvert\|_p \to 0 $$ as $\varepsilon\to 0$.

Yet, their Schmidt vectors do not become close to each other; in fact, they are completely independent of $\varepsilon$.


Thus, the only way in which this can be made to work is if you insist that you are sufficiently far (as comapred to $\varepsilon$) from a state with degenerate Schmidt coefficients.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.