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I understand the basics of how to evaluate tensor products. For example: $$\sigma_z \otimes I = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{pmatrix}$$ where $\sigma_z$ is the pauli z matrix and $I$ is the 2x2 identity.

My question arises when applying this math to two qubits. There are four operations to consider: $$1)\hspace{1cm}\sigma_z^{(1)} \otimes I^{(2)}$$ $$2)\hspace{1cm}I^{(2)} \otimes\sigma_z^{(1)}$$ $$3)\hspace{1cm}\sigma_z^{(2)} \otimes I^{(1)}$$ $$4)\hspace{1cm}I^{(1)} \otimes \sigma_z^{(2)}$$

Where the superscript denotes the qubit that the operation (gate) is applied to. So are $1)$ and $2)$ the same $4\times 4$ matrix? Or are $1)$ and $3)$ the same matrix. I guess my question can be summarized is it the order or the superscript/qubit that matters when evaluating out these tensor products into matrix form?

Mauricio
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  • Can you write the four matrices? It will be clearer to point out where you are getting it wrong that way. – Mauricio Apr 14 '23 at 16:39
  • It is to my understanding that if you add the qubit labels, the tensor product in your notation is commutative. – Mauricio Apr 14 '23 at 16:41
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    Based on @Mauricio's comment, where it is true that the tensor product is abstractly commutative if you label the operators with a qubit label, then 1 and 2 are the same operator, and 3 and 4 are the same operator. When you write operators in a basis and make a matrix representation, then the order absolutely matters, and you have to choose one of $\sigma_z^{(1)}\otimes I^{(2)}$ or $I^{(2)}\otimes \sigma_z^{(1)}$ to perform the matrix tensor product. – march Apr 14 '23 at 18:49

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