Simplify tensor product statement

In summary, the conversation discusses a method for showing that ##(Z \otimes Y)^{\dagger} = Z \otimes Y## by using the properties of the tensor product and the properties of quantum gates. One approach is to use the fact that ##Z^{\dagger} \otimes Y^{\dagger} = Z \otimes Y## because ##Z^{\dagger} = Z## and ##Y^{\dagger} = Y## are self-adjoint. Another approach is to use tensor indexing notation and apply the rules. The conversation also suggests searching for a proof to see how others have done it.
  • #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?
 
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  • #2
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
 

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