Exercise 2.29

Suppose \(A\) is a \(m\times m\) unitary operator and \(B\) is a \(n\times n\) unitary operator, we have \(A^{\dagger}A = AA^{\dagger} = I_{A}\) and \(B^{\dagger}B = BB^{\dagger} = I_B\), where \(I_{A}\) is \(m\times m\) identity matrix and \(I_B\) is \(n\times n\) identity matrix. We could also obtain tensor product between \(A\) and \(B\) as \(A\otimes B\), and the Hermitian conjugate of \(A\otimes B\) as

(1)\[ (A\otimes B)^{\dagger} = A^{\dagger}\otimes B^{\dagger} \]

Then we could compute \((A\otimes B)^{\dagger}(A\otimes B)\) as

(2)\[ (A\otimes B)^{\dagger}(A\otimes B) = (A^{\dagger}\otimes B^{\dagger})(A\otimes B) = A^{\dagger}A\otimes B^{\dagger}B = I_{A}\otimes I_{B} = I_{A\otimes B} \]

where \(I_{A\otimes B}\) is \(mn\times mn\) identity matrix. Similarly,

(3)\[ (A\otimes B)(A\otimes B)^{\dagger} = (A\otimes B)(A^{\dagger}\otimes B^{\dagger}) = AA^{\dagger}\otimes BB^{\dagger} = I_{A}\otimes I_{B} = I_{A\otimes B} \]

From eq. (2) and eq. (3), we conclude that the tensor product of two unitary operators is unitary.