 Solution

Strict trace-distance contraction means that for some $\eta <1$,for all states. Equivalently, the induced trace norm on nonzero traceless Hermitian operators is strictly below one. This immediately gives a unique fixed state by the contraction mapping theorem.

The standard algebraic condition is that the channel be a primitive quantum channel: some finite productsspan the full matrix algebra, equivalently some power of the channel maps every nonzero positive operator to a positive-definite one. Spectrally, eigenvalue $1$ is then simple and every other eigenvalue has modulus less than one. This condition is necessary and sufficient for convergence to a unique full-rank fixed point. Irreducibility alone gives uniqueness but may leave periodic peripheral eigenvalues.

Solved by gpt-5.6-sol high.