# Tag Info

## New answers tagged fidelity

2

First, take the case without encoding. Let there be an $X$ error with probability $p$, so the output state is $$\rho=(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X.$$ Normally, I would just talk about the probability of an error (it lets you avoid some of the averaging I'm about to do), but we can talk about fidelity if you want.  F=\langle\...

0

I believe the answer can be proven to be yes as follows: Suppose $|\phi^{AB}\rangle$ is a 2-qubit state $a_0|00\rangle + a_1|11\rangle$. Then $F(\phi^{AB},(\Phi^A\otimes I_B)(\phi^{AB}))\\ = |a_0|^2F(|0\rangle\langle 0|,\Phi^A(|0\rangle\langle 0|))+|a_1|^2F(|1\rangle\langle 1|, \Phi^A(|1\rangle\langle1|))$ Thus if \$F(\psi, \Phi_1^A(\psi)) \ge F(\psi, \phi_2^...

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