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, $$\langle \psi| \rho_r - \sigma_r|\psi\rangle \leqslant ? $$

  • 1
    $\begingroup$ Can you explain (otherwise edit) the notation $|\psi\rangle \in SU(2)$ since it is not commonplace? $\endgroup$ – keisuke.akira 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, $\endgroup$ – Purva Thakre 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. $\endgroup$ – Niraj Kumar 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. $\endgroup$ – Niraj Kumar Jul 13 '20 at 7:07
  • $\begingroup$ Ah, sorry. Missed that aspect. $\endgroup$ – DaftWullie Jul 13 '20 at 7:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Niraj Kumar Jul 13 '20 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.