Finding a basis of $\mathbb{C}^4$ in which both $\sigma_1\otimes\sigma_1$ and $\sigma_2\otimes\sigma_2$ are diagonal Let $(\sigma_i)_{i=0,...,3}$ be the Pauli matrices. I'm looking for a way to find a basis of $\mathbb{C}^4$ in which both $\sigma_1\otimes\sigma_1$ and $\sigma_2\otimes\sigma_2$ are diagonal. I thought about finding eigenvectors for them, but this wouldn't be enough it seems. Any hint would be deeply appreciated.
 A: Let $T_1=\sigma_1\otimes\sigma_1$ and $T_2=\sigma_2\otimes\sigma_2$. As $\sigma_1\sigma_2=-\sigma_2\sigma_1$ we see that $T_1$ and $T_2$ commute. We can call upon the fact that commuting diagonalizable matrices are simultaneously diagonalizable, or we can just calculate.
Let us first consider $T_1$. The eigenvectors of $\sigma_1$ are $\binom 1{\pm1}$ belonging to eigenvalues $\lambda_{\pm}=\pm1$. Therefore the eigenspace $V_+$ of $T_1$ belonging to eigenvalue $+1$ is spanned by $v_1:=\binom 11\otimes \binom 11$ and
$v_2:=\binom 1{-1}\otimes \binom 1{-1}$.
Because $T_1$ and $T_2$ commute, the space $V^+$ must be stable under $T_2$. We see that
$\sigma_2\binom 11=\binom{-i}i=-i\binom1{-1}$. Therefore
$$
T_2v_1=\binom{-i}i\otimes \binom{-i}i=(-i)^2\binom 1{-1}\otimes \binom1{-1}=-v_2.
$$
Similarly $\sigma_2\binom1{-1}=\binom i i=i\binom 11$, and thus
$$
T_2v_2=\binom i i\otimes \binom i i = i^2\binom11\otimes\binom11=-v_1.
$$
It follows that

*

*$v_1+v_2$ is an eigenvector of $T_1$ belonging to $+1$ and an eigenvector of $T_2$ belonging to $-1$, and

*$v_1-v_2$ is an eigenvector of $T_1$ belonging to $+1$ and an eigenvector of $T_2$ belonging to $+1$.

Similarly, the span $V_-$ of $v_3:=\binom 11\otimes \binom 1{-1}$ and
$v_4:=\binom 1{-1}\otimes \binom 11$ is the 2-dimensional eigenspace of $T_1$
belonging to the eigenvalue $-1$. It, too, must be stable under $T_2$. A similar calculation (verify this, I did not doublecheck) shows that
$$
T_2v_3=v_4\quad\text{and}\quad T_2v_4=v_3.
$$
This implies that

*

*$v_3+v_4$ is an eigenvector of $T_1$ belonging to $-1$ and an eigenvector of $T_2$ belonging to $+1$, and

*$v_3-v_4$ is an eigenvector of $T_1$ belonging to $-1$ and an eigenvector of $T_2$ belonging to $-1$.

There you have it. The joint four eigenvectors are $v_1\pm v_2$ and $v_3\pm v_4$. They form a basis of $\Bbb{C}^2\otimes \Bbb{C}^2$. You can identify this space with $\Bbb{C}^4$ by your chosen way.
