# Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?

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

• 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) Sep 19, 2023 at 8:46

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.

• Reproduced from the same question on pse. Sep 19, 2023 at 16:58
• +1 This is a better answer since it works for qubits, too. Sep 20, 2023 at 1:15
• @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. Sep 20, 2023 at 10:20

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 $$\|D\|_1\leq\varepsilon\sqrt{3}$$.