In order to prove that the Controlled Z gate can create entanglement, I'm trying to show that using these two arbitrary $2\times 2$ matrices and their tensor product: $$ \begin{bmatrix}a_1&b_1\\c_1&d_1\end{bmatrix}\quad \begin{bmatrix}a_2&b_2\\c_2&d_2\end{bmatrix} $$ that the controlled Z gate creates a contradiction and cannot be represented as the tensor product of two matrices. How would I go about doing this?


1 Answer 1


Showing that the matrix of a gate cannot be written as the tensor product is a tedious approach. It's easier to come up with a product state that the gate entangles. For this, it's helpful to have a simple criterion that detects entanglement in two-qubit quantum states.

Criterion for entanglement in $2$-qubit states

A two-qubit state with amplitudes $[a,b,c,d]^T$ is a product state if and only if it can be written as the Krokecker product $[u,v]^T\otimes[x,y]^T=[ux,uy,vx,vy]^T$. This is the case if and only if $$ \det\begin{bmatrix}a&b\\c&d\end{bmatrix}=0.\tag1 $$ Moreover, the criterion is unaffected by global phase and normalization of the state.

Entangling action of the CZ gate

Now, the result of applying the CZ gate to the product state $[1,1]^T\otimes[1,1]^T=[1,1,1,1]^T$ is $[1,1,1,-1]^T$. But $$ \det\begin{bmatrix}1&1\\1&-1\end{bmatrix}=-2\ne 0\tag2 $$ so the CZ gate can create entanglement.


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