New answers tagged

2 votes

Measuring one register of the state $\frac{1}{2^{m}} \sum_{x}\sum_{k} (-1) ^ {x\cdot k} |k\rangle |f(x)\rangle$

Your state before measurement is $$ |\psi\rangle=\frac{1}{2^m}|0\rangle\sum_x|f(x)\rangle+\frac{1}{2^m}\sum_x\sum_{k\neq 0}(-1)^{x\cdot k}|k\rangle|f(x)\rangle. $$ How do we calculate the probability ...
DaftWullie's user avatar
  • 56.9k
1 vote

How to prove that the mutual information is subadditive?

Unlike entropy, mutual information can be either subadditive or superadditive. If $A$, $B$, $C$ are bits with equal probabilities and $A=B=C$, then $$ I(A,B:C) = 1,\\ I(A:C) + I(B:C) = 2, $$ ...
Adam's user avatar
  • 11

Top 50 recent answers are included