# Commutator of Pauli Matrices in two qubit system

For two qubits 1 and 2 with Pauli Matrices $\sigma_{x_1}, \sigma_{x_2}, \sigma_{z_1}, \sigma_{z_2}$, prove that: $$[\sigma_{x_1} \sigma_{x_2}, \sigma_{z_1} \sigma_{z_2}] = 0$$

I am very confused about what the question means. I thought the Pauli Matrices would be the same regardless of which qubit they belong to, which would mean that $\sigma_{x_1} \sigma_{x_2} = I$, and the same for z. This would in turn give me the correct commutator relation, but I feel that is not what the question is getting at (especially since the next part asks to prove that eigenvalues are $\pm 1$).

I should be able to get the commutator relation myself once I understand what is happening here. Is it the case that the Pauli Matrices for this system are a linear combination of the standard Pauli Matrices?

• Could someone explain the downvotes? – infinitylord Jan 30 '17 at 19:20
• ok that should do it. – ZeroTheHero Jan 30 '17 at 19:35

Start with $\sigma_x=\left(\begin{array}{cc}0&1 \\ 1&0\end{array}\right)$ and $\sigma_z=\left(\begin{array}{cc}1&0 \\ 0 & -1\end{array}\right)$. The next step is to construct $\sigma_{x1}\sigma_{x2}$ which is the tensor product $$\sigma_{x_1}\otimes\sigma_{x_2} = \left(\begin{array}{cc} 0\times \sigma_{x}&1\times \sigma_{x} \\ 1\times \sigma_{x}& 0\times \sigma_{x}\end{array}\right)$$ i.e. $A\otimes B$ means you take all the entries of the first matrix $A$ and multiply these by the matrix $B$. (The tensor product is also known as Kronecker product.) The result is a $4\times 4$ matrix $$\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right)$$ Doing the same for $\sigma_{z_1}\sigma_{z_2}=\sigma_{z_1}\otimes \sigma_{z_2}$ you will obtain $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)\, .$$ You can then verify the resulting $4\times 4$ matrices actually commute. This is not elegant as there are shortcuts available for the more complicated cases but this should be enough for now.