 Solution

While the sequence is remembered, the conditional quantum state is the known pure endpoint $C|0⟩$, so its Bloch vector is one of the six Pauli vertices. Partial forgetting leaves a probability distribution $p_j$ over possible endpoints, givingHence the vector lies in the stabilizer octahedron. The mixing describes lost classical information, not a different unitary evolution of the physical qubit.

Under complete unbiased uncertainty over the six endpoints, $p_j=1/6$, the opposite vectors cancel andThis is the maximally mixed state at the center. Mere loss of the particular sequence does not mathematically force uniform probabilities: a known biased distribution of endpoint states would leave its corresponding noncentral mixture. The central conclusion uses the unbiased randomization implicit in complete forgetting in this setting.