Given a qubit state $|\psi\rangle \in \mathcal{H}$, and two bipartite general mixed states $\rho$ and $\sigma$, such that, $$\langle \psi|\otimes \langle \psi|\rho - \sigma |\psi\rangle \otimes |\psi \rangle \ \leqslant \epsilon$$ Now suppose the reduced state of $\rho, \sigma$ be such that, $$ \rho_r = Tr_1(\rho) = Tr_2(\rho), \hspace{5mm} \sigma_r = Tr_1(\sigma) = Tr_2(\sigma)$$ Then can we say something about the closeness of the reduced state in terms of epsilon? In other words, $$\langle \psi| \rho_r - \sigma_r|\psi\rangle \leqslant ? $$

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    $\begingroup$ Can you explain (otherwise edit) the notation $|\psi\rangle \in SU(2)$ since it is not commonplace? $\endgroup$ Jul 13 '20 at 9:07
  • $\begingroup$ I think you could use a distance measure to compare how close the states are considering distance measures are used to check for distinguishability. In Nielsen & Chuang, there's some discussion about how the distance measures decrease when a system is partially traced over i.e. the distinguishability further decreases or stays the same, $\endgroup$ Jul 13 '20 at 19:50
  • $\begingroup$ Hi Purva, the closeness notion that I was looking for with respect to a fixed state $|\psi\rangle$. This is a weak notion of distinguishability. And as DaftWullie wrote in the answer, it is not possible to give non-trivial measure of the closeness of the reduced state with respect to the state $|\psi\rangle$. If, alternatively, I had the notion of distinguishability over trace 1 norm (which is the maximum overlap of $\rho - \sigma$ with $|\psi\rangle$, where the choice of state $|\psi\rangle$ is over entire bloch sphere, then I could be able to make statements about the reduced density matrix. $\endgroup$ Jul 14 '20 at 4:38

No, there's not a lot you can say. Consider these two cases, both with $\epsilon=0$.

First, the obvious one, $\rho=\sigma=|\psi\rangle\langle\psi|\otimes |\psi\rangle\langle\psi|$. Clearly $\rho_r-\sigma_r=0$.

Second, let $|\psi^\perp\rangle$ be orthogonal to $|\psi\rangle$. You can have $$\rho=(|\psi\rangle\langle\psi|\otimes |\psi^\perp\rangle\langle\psi^\perp|+|\psi^\perp\rangle\langle\psi^\perp|\otimes |\psi\rangle\langle\psi|)/2$$ and $$\sigma=|\psi^\perp\rangle\langle\psi^\perp|\otimes |\psi^\perp\rangle\langle\psi^\perp|.$$ Now you have $$ \langle\psi|\rho_r-\sigma_r|\psi\rangle=\frac12, $$ which is more or less as far away as you can get.

  • $\begingroup$ Hi DaftWullie, in your proof, you have not considered the property that the reduced state of $\rho$ (or $\sigma$) obtained by tracing either the first subsystem or the second subsystem is the same. $\endgroup$ Jul 13 '20 at 7:07
  • $\begingroup$ Ah, sorry. Missed that aspect. $\endgroup$
    – DaftWullie
    Jul 13 '20 at 7:13
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    $\begingroup$ Does that fix it? $\endgroup$
    – DaftWullie
    Jul 13 '20 at 7:17
  • $\begingroup$ Thank you DaftWullie. This indeed shows that if some more information is known about $\rho$ and $\sigma$, then one can hope for something better. $\endgroup$ Jul 13 '20 at 7:27

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