Commutation and common eigenvectors in two dimensions

Write $M=\begin{pmatrix}p&q\\r&s\end{pmatrix}$. We calculate

$$\begin{aligned} \sigma_zM=\begin{pmatrix}p&q\\-r&-s\end{pmatrix},\qquad M\sigma_z=\begin{pmatrix}p&-q\\r&-s\end{pmatrix}. \end{aligned}$$

Equality requires $q=-q$ and $r=-r$, hence $q=r=0$: the matrices that commute with $\sigma_z$ are the diagonal matrices. Similarly,

$$\begin{aligned} \sigma_xM=\begin{pmatrix}r&s\\p&q\end{pmatrix},\qquad M\sigma_x=\begin{pmatrix}q&p\\s&r\end{pmatrix}, \end{aligned}$$

and equality requires $r=q$ and $s=p$: the matrices that commute with $\sigma_x$ have the form $\begin{pmatrix}p&q\\q&p\end{pmatrix}=p\,\mathbf{1}+q\,\sigma_x$. In both cases, they are the linear combinations of the identity and the matrix itself.

Apply $\hat A$ to the vector $\hat B\ket{a_i}$, using $\hat A\hat B=\hat B\hat A$:

$$\hat A\bigl(\hat B\ket{a_i}\bigr)=\hat B\hat A\ket{a_i}=a_i\,\hat B\ket{a_i}.$$

The vector $\hat B\ket{a_i}$ is therefore either zero or an eigenvector of $\hat A$ with eigenvalue $a_i$. In either case, it belongs to the eigenspace associated with $a_i$, which is the line spanned by $\ket{a_i}$ because the eigenvalue is simple. There is therefore a number $b_i$ such that $\hat B\ket{a_i}=b_i\ket{a_i}$: each $\ket{a_i}$ is also an eigenvector of $\hat B$, and the matrix of $\hat B$ in this basis is diagonal. This is what we found in the first question: $\sigma_z$ has two distinct eigenvalues, and the matrices that commute with it are diagonal in its eigenbasis.

We have $\sigma_x\sigma_z=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and $\sigma_z\sigma_x=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, hence $[\sigma_x,\sigma_z]=\begin{pmatrix}0&-2\\2&0\end{pmatrix}=-2i\sigma_y\neq0$. The eigenvectors of $\sigma_z$ are, up to a factor, the basis vectors $(1,0)$ and $(0,1)$, since its two eigenvalues are distinct. But $\sigma_x(1,0)=(0,1)$ and $\sigma_x(0,1)=(1,0)$: neither is an eigenvector of $\sigma_x$. There is therefore no common eigenvector. Physically, there is no state in which both observables simultaneously have a definite value: after a measurement of $\sigma_z$, the state is an eigenvector of $\sigma_z$, and a measurement of $\sigma_x$ then gives either result with probability $1/2$. These are incompatible observables, like the $z$ and $x$ components of spin in the Stern—Gerlach experiment.

No. The identity commutes with every matrix, in particular with $\sigma_x$. Every vector is an eigenvector of $\mathbf{1}$, and the canonical basis $(1,0)$, $(0,1)$ is therefore an eigenbasis of $\hat A$; but $\sigma_x$ is not diagonal in this basis. When $\hat A$ has a degenerate eigenvalue, the argument of the second question shows only that $\hat B$ leaves each eigenspace of $\hat A$ invariant, without necessarily being diagonal in an arbitrary eigenbasis of $\hat A$. There is nevertheless an eigenbasis of $\hat A$ in which $\hat B$ is diagonal: here, the basis $\frac{1}{\sqrt2}(1,\pm1)$ of eigenvectors of $\sigma_x$. This is the simultaneous diagonalisation theorem, studied in the next exercise.