# What can be inferred about the closeness of reduced qubit states from the closeness of the bipartite quantum state?

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

• Can you explain (otherwise edit) the notation $|\psi\rangle \in SU(2)$ since it is not commonplace? – keisuke.akira Jul 13 '20 at 9:07
• 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, – Purva Thakre Jul 13 '20 at 19:50
• 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. – 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.
• 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. – Niraj Kumar Jul 13 '20 at 7:07
• Thank you DaftWullie. This indeed shows that if some more information is known about $\rho$ and $\sigma$, then one can hope for something better. – Niraj Kumar Jul 13 '20 at 7:27