Quantum Computing Asked by Niraj Kumar on December 23, 2020
Given a qubit state $|psirangle in mathcal{H}$, and two bipartite general mixed states $rho$ and $sigma$, such that,
$$langle psi|otimes langle psi|rho – sigma |psirangle otimes |psi rangle leqslant epsilon$$
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,
$$langle psi| rho_r – sigma_r|psirangle leqslant ? $$
No, there's not a lot you can say. Consider these two cases, both with $epsilon=0$.
First, the obvious one, $rho=sigma=|psiranglelanglepsi|otimes |psiranglelanglepsi|$. Clearly $rho_r-sigma_r=0$.
Second, let $|psi^perprangle$ be orthogonal to $|psirangle$. You can have $$rho=(|psiranglelanglepsi|otimes |psi^perpranglelanglepsi^perp|+|psi^perpranglelanglepsi^perp|otimes |psiranglelanglepsi|)/2$$ and $$sigma=|psi^perpranglelanglepsi^perp|otimes |psi^perpranglelanglepsi^perp|.$$ Now you have $$ langlepsi|rho_r-sigma_r|psirangle=frac12, $$ which is more or less as far away as you can get.
Correct answer by DaftWullie on December 23, 2020
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