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Above is a circuit I am provided. I am asked to write down the matrix representation of a $4 \times 4$ matrix representation of a unitary matrix for this circuit.

Using small endian notation (left to right):

The initial state is $|\psi\rangle = |x\rangle |y \rangle$

Then, $(H \otimes I) |x \otimes y \rangle = H|x\rangle \otimes I|y\rangle$ for the post - gate state.

The matrix representation for the unitary matrix $(H \otimes I) $ is

$\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix}$

Am I correct?

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  • $\begingroup$ Whether this is correct depends on the endian-ness convention you are using. Is H on the top qubit and I on the bottom qubit $H \otimes I$ or is it $I \otimes H$. You've correctly computed $I \otimes H$. I think that's the less common convention, but it is used e.g. in Quirk. $\endgroup$
    – Craig Gidney
    Oct 7 at 16:18

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No, you are incorrect. The tensor product is defined as

kronecker product

With this in mind, the answer is

$$ H \otimes I = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\times I & 1 \times I \\ 1 \times I & -1 \times I \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix}. $$

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