- #1
Albert01
- 12
- 0
Hi,
if I wanted to show ##(Z \otimes Y)^{\dagger} = Z \otimes Y##, then I could simply multiply out the matrices belonging to the operators of quantum gates ##Z## and ##Y##.
But my question is whether this is also solvable via the properties of the tensor product and the properties of the gates.
My approach would be the following:
##(Z \otimes Y)^{\dagger} = Z^{\dagger} \otimes Y^{\dagger}##
holds because this is a property of the tensor product. Continue with
##Z^{\dagger} \otimes Y^{\dagger} = Z \otimes Y##
which holds because ##Z^{\dagger} = Z## and ##Y^{\dagger} = Y## are self-adjoint.
My question, is it possible to do it this way now or did I miss something?
if I wanted to show ##(Z \otimes Y)^{\dagger} = Z \otimes Y##, then I could simply multiply out the matrices belonging to the operators of quantum gates ##Z## and ##Y##.
But my question is whether this is also solvable via the properties of the tensor product and the properties of the gates.
My approach would be the following:
##(Z \otimes Y)^{\dagger} = Z^{\dagger} \otimes Y^{\dagger}##
holds because this is a property of the tensor product. Continue with
##Z^{\dagger} \otimes Y^{\dagger} = Z \otimes Y##
which holds because ##Z^{\dagger} = Z## and ##Y^{\dagger} = Y## are self-adjoint.
My question, is it possible to do it this way now or did I miss something?