Quantum Information – Relation Between Distance and Schmidt Basis

linear algebraquantum mechanicsquantum-entanglementquantum-informationquantum-states

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

Best Answer

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.

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