Timeline for Why does $\max\{S(\rho_A)-\sum_i p_i S(\rho_A^i)\}$ measure classical correlations, and not full (classical + quantum) correlation?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2021 at 17:44 | vote | accept | Gold | ||
| Feb 23, 2019 at 12:02 | history | bounty ended | CommunityBot | ||
| Feb 22, 2019 at 19:24 | history | edited | glS | CC BY-SA 4.0 |
added 40 characters in body
|
| Feb 22, 2019 at 13:56 | comment | added | user1936752 | Let us continue this discussion in chat. | |
| Feb 22, 2019 at 13:52 | comment | added | user1936752 | According to the defintion of $J$, what you do is take a bipartite state. So here we have $I_{AB} = I_A\otimes I_B$. $S(\rho_A) = 1$ since the reduced state is also maximally mixed. According to the defintion, what you do next is measure system $B$ in the computational basis assuming the outcome is $i$, check $S(\rho_A^i)$. This is still maximally mixed since measuring in $B$ can't change the $A$ part. So for the maximally mixed state, $J(\rho) = 0$. If Alice and Bob hold a maximally mixed state, Bob can do whatever he wants with his state and there can be no effect on Alice's side. | |
| Feb 22, 2019 at 13:47 | history | edited | glS | CC BY-SA 4.0 |
edited body
|
| Feb 22, 2019 at 12:58 | comment | added | user1936752 | Sorry, I was wrong. $J(\rho) = 0$ for max mixed state (the POVM is on B!). And this is correct since there are no correlations of any kind in the maximally mixed state. | |
| Feb 22, 2019 at 12:20 | comment | added | glS | @user1936752 well, not really. For the max mixed state, you have a purely classical situation, with $X$ and $Y$ being fully correlated. The correlation is maximal because knowing $Y$ gives you full information about $X$, and thus $J(\rho)=I(X:Y)=H(\rho_A)=H(\vec p)$. The fact that you get the same value for the maximally entangled state is because you are not actually exploiting the entanglement of the state, and therefore you cannot see the "quantumness" at all. This is like $|0\rangle+|1\rangle\sim |0\rangle\!\langle0|+|1\rangle\langle1|$ if you only measure in the computational basis | |
| Feb 22, 2019 at 11:52 | comment | added | user1936752 | I would add that it's a bit weird though that $J(\rho) = 1$ for both the maximally mixed state and the maximally entangled state. Surely, a measure of correlations ought to give you 0, at least for the maximally mixed state? | |
| Feb 22, 2019 at 9:55 | history | answered | glS | CC BY-SA 4.0 |