What is the density matrix of a GHZ state when onle a qubit is in a decoherence channel?

Question

Suppose Alice, Bob and Rob share a GHZ state. Now consider Rob's qubit is in a bit-flip channel. How to obtain the density matrix in this senario? Also i would be glad to get some articles adrresing these kind of topics.

Answer 1

Suppose Rob's channel applies a bit flip with probability $p$. It means that with probability $1-p$, the state is: $$\rho_0=\frac{|000\rangle\!\langle000|+|000\rangle\!\langle111|+|111\rangle\!\langle000|+|111\rangle\!\langle111|}{2}$$ and with probability $1-p$ this state is $$\rho_1=\frac{|001\rangle\!\langle001|+|001\rangle\!\langle110|+|110\rangle\!\langle001|+|110\rangle\!\langle110|}{2}.$$ Overall, the state is $(1-p)\rho_0+p\rho_1$, since the convex combination represents the classical uncertainty about the state.

More generally, what you're looking for are probably Kraus operators. Every CPTP map $\Phi$ can be written as: $$\Phi=\rho\mapsto\sum_{i=1}^nK_i\rho K_i^\dagger$$ with $\sum\limits_{i=1}^nK_i^\dagger K_i=I$, with $I$ being the identity.

In particular, in your case, the quantum channel you're describing is represented by the following Kraus operators: $$K_0=\sqrt{p}I_2\otimes I_2\otimes X$$ and $$K_1=\sqrt{1-p}I_8.$$ Note that we do have $K_0^\dagger K_0+K_1^\dagger K_1=I_8$. You can then compute $$K_0\sigma K_0^\dagger+K_1\sigma K_1^\dagger$$ with $\sigma$ being the density matrix of a GHZ state to recover the expression above.