Quantcast
Channel: Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases? - Quantum Computing Stack Exchange
Viewing all articles
Browse latest Browse all 3

Answer by Norbert Schuch for Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?

$
0
0

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.


Viewing all articles
Browse latest Browse all 3

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>