2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/92152In this comment, we want to emphasize that when the state of the two qubit is $|0>_{B}|1>_{A}$ or $|1>_{A}|0>_{B}$, Bob's Bell basis measurement outcome can not be $|ϕ^{\pm}>$. Also, when the product state is $|0> |0>$ or $|1> |1>$, Bob's measurement outcome can not be $|ψ^{\pm}>$.PACS:03.67.HkQuantum Physicstext