I Simplify tensor product statement

[Albert01]

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?

 

[jedishrfu]

That seems reasonable although is the product reversed to be Y x Z?

Another approach would be to use Tensor indexing notation and apply the rules. That way you can see the details of what you have.

Have you searched for a proof to see how others have done it?

https://en.wikipedia.org/wiki/Hermitian_matrix