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I want to know that there is a relation between the distance of two vectors and the corresponding elements of the Schmidt bases.

We assume that two bipartite vectors $|\phi\rangle^{AB}$ and $|\psi\rangle^{AB}$ can be decomposed such that

$$|\phi\rangle^{AB}=\sum_i\sqrt{p_i}|e_i\rangle^A|\tilde{e}_i\rangle^B\ \qquad(p_1\geq p_2 \geq ...), $$ $$|\psi\rangle^{AB}=\sum_i\sqrt{q_i}|f_i\rangle^A|\tilde{f}_i\rangle^B\ \qquad (q_1 \geq q_2 \geq ...).$$

These Schomidt coefficients are arranged in decreasing order, and $|e_i\rangle^{A}|\tilde{e}_i\rangle^{B}$ and $|f_i\rangle^A|\tilde{f}_i\rangle^B$ are Schmidt basis corresponding to coefficient $\sqrt{p_i}$, $\sqrt{q_i}$ respectively.

If $$\| |\phi\rangle\langle\phi|^{AB} - |\psi\rangle\langle\psi|^{AB} \|_p \leq \varepsilon$$

then is there relation that $$|\langle e_i|^A\langle \tilde{e}_i|^B|f_i\rangle^A|\tilde{f}_i\rangle^B| \geq 1- g(\varepsilon)\ ,$$ where $g(\varepsilon)$ is some kind of function of $\varepsilon$ ?

Certainly, when it comes to degenerate cases, Schmidt bases are not uniquely determined. The question arises whether it's possible to pair up "Schmidt bases which satisfy that relation for any $i$."

Does a relation like this exist or not?


Cross-posted on P.SE

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  • $\begingroup$ Start from an arbitrary basis. Schmidt decomposition is the SVD of the matrix of coefficients in this basis. You can Taylor-expand the SVD vectors to get the first order approximation (but this may be quite a bit of work) $\endgroup$
    – EvgeniyZh
    Sep 19, 2023 at 8:46

2 Answers 2

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No.

Here is an example without small Schmidt coefficients.

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. Note that this is also the issue with the other example by @AdamZalcman: There is one zero Schmidt coefficient and one very small one, which are thus almost degenerate.

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  • $\begingroup$ Reproduced from the same question on pse. $\endgroup$ Sep 19, 2023 at 16:58
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    $\begingroup$ +1 This is a better answer since it works for qubits, too. $\endgroup$ Sep 20, 2023 at 1:15
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    $\begingroup$ @AdamZalcman The basic insight is that it depends on degeneracies. As far as I can tell the SVD is continuous (or rather can be chosen continuously) except for that. The third level is basically necessary since you have two small (or zero) singular values. $\endgroup$ Sep 20, 2023 at 10:20
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TL;DR: No such relation exists, because the upper bound on the norm fails to impose any constraints whatsoever on the basis elements corresponding to very small Schmidt coefficients $\sqrt{p_i}$ and $\sqrt{q_i}$.

Let's construct an explicit counterexample. Define the following states of two qutrits $A$ and $B$ $$ \begin{align} |\phi\rangle^{AB}:=\sqrt{1-\delta^2}|0\rangle|0\rangle+\delta|1\rangle|1\rangle\tag1\\ |\psi\rangle^{AB}:=\sqrt{1-\delta^2}|0\rangle|0\rangle+\delta|2\rangle|2\rangle\tag2 \end{align} $$ where $\delta:=\varepsilon/2$. Note that the matrix of $D:=|\phi\rangle\langle\phi|^{AB} - |\psi\rangle\langle\psi|^{AB}$ in the computational basis has six non-zero elements: two diagonal ones equal to $\pm\delta^2$ and four off-diagonal equal to $\pm\delta\sqrt{1-\delta^2}$. Therefore, the Frobenius$^1$ norm of $D$ is $$ \begin{align} \|D\|_2&=\sqrt{2\delta^4+4\delta^2(1-\delta^2)}\tag3\\ &=\sqrt{4\delta^2-2\delta^4}\tag4\\ &\le 2\delta=\varepsilon.\tag5 \end{align} $$ If $\delta$ is so small that $\sqrt{1-\delta^2}\gt\delta$, then $$ \langle e_2|^A\langle \tilde{e}_2|^B|f_2\rangle^A|\tilde{f}_2\rangle^B=\langle 1|^A\langle 1|^B|2\rangle^A|2\rangle^B=0\tag6 $$ so the only possible $g(\varepsilon)$ is $g(\varepsilon)=1$ which yields the trivial bound $|\langle e_i|^A\langle \tilde{e}_i|^B|f_i\rangle^A|\tilde{f}_i\rangle^B| \geq 0$.


$^1$ I assume that $\|.\|_p$ in the question denotes the Schatten $p$-norm, so we get Frobenius norm for $p=2$. Norm equivalence will provide tight bounds on other norms, too. For example, for the trace norm we have $\|D\|_1\leq\varepsilon\sqrt{3}$.

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