The quantity \(\hat{σ}_n\) is called the 3-vector spin operator (or just spin operatory for short). This quantity can be represented as a 2x2 matrix as \(\hat{σ}_z=\begin{bmatrix}(σ_n)_{11} & (σ_n)_{12}\\(σ_n)_{21} & (σ_n)_{22}\end{bmatrix}\). The value of \(\hat{σ}_n\)(that is to say, the value of each one of its entries) depends on the direction \(\vec{n}\) that \(A\) is oriented along; in other words, it depends on which component of spin \(\hat{σ}_n\) we’re measuring using \(A\). In order to measure a component of spin \(\hat{σ}_m\) in a different direction (say the \(\vec{m}\) direction) the apparatus \(A\) must be rotated; similarly the spin operator must also be “rotated” (mathematically) and, in general, \(\hat{σ}_m≠\hat{σ}_m\) and the values of their entries will be different.
We’ll start out by finding the values of the entries of \(\hat{σ}_z\)—the spin operator associated with the positive z-direction. The states \(|u⟩\) and \(|d⟩\) are eigenvectors of \(\hat{σ}_z\) with eigenvalues \(λ_u=σ_{z,u}=+1\) and \(λ_d=σ_{z,d}=-1\); or, written mathematically,
$$\hat{σ}_z|u⟩=σ_{z,u}|u⟩=|u⟩$$
$$\hat{σ}_z|d⟩=σ_{z,d}|d⟩=-|d⟩.\tag{15}$$
Recall that any ket vector can be represented as a column vector; in particular the eigenstates can be represented as \(|u⟩=\begin{bmatrix}1 \\0\end{bmatrix}\) and \(|d⟩=\begin{bmatrix}0 \\1\end{bmatrix}\). We can rewrite Equations (15) as
$$\begin{bmatrix}(σ_z)_{11} & (σ_z)_{12}\\(σ_z)_{21} & (σ_z)_{22}\end{bmatrix}\begin{bmatrix}1 \\0\end{bmatrix}=\begin{bmatrix}1 \\0\end{bmatrix}$$
$$\begin{bmatrix}(σ_z)_{11} & (σ_z)_{12}\\(σ_z)_{21} & (σ_z)_{22}\end{bmatrix}\begin{bmatrix}0 \\1\end{bmatrix}=-\begin{bmatrix}0 \\1\end{bmatrix}.\tag{16}$$
Using the first equation of Equations (16) we have
$$(σ_z)_{11}+(σ_z)_{12}·0=1⇒(σ_z)_{11}=1$$
and
$$(σ_z)_{21}+(σ_z)_{22}·0=0⇒(σ_z)_{21}=0.$$
Using the second equation from Equations (16) we have
$$(σ_z)_{11}·0+(σ_z)_{12}=1⇒(σ_z)_{12}=0$$
and
$$(σ_z)_{21}·0+(σ_z)_{22}=0⇒(σ_z)_{22}=-1.$$
Therefore,
$$\hat{σ}_z=\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}.\tag{17}$$
To derive the spin operator \(\hat{σ}_x\) we’ll go through a similar procedure. The eigenvectors of \(\hat{σ}_x\) are \(|r⟩\) and \(|l⟩\) with eigenvalues \(λ_r=σ_{x,r}=+1\) and \(λ_l=σ_{x,l}=-1\):
$$\hat{σ}_x|r⟩=σ_{x,r}|r⟩=|r⟩$$
$$\hat{σ}_x|l⟩=σ_{x,l}|l⟩=-|l⟩.\tag{18}$$
The states \(|r⟩\) and \(|l⟩\) can be written as linear superpositions of \(|u⟩\) and \(|d⟩\) as
$$|r⟩=\frac{1}{\sqrt{2}}|u⟩+\frac{1}{\sqrt{2}}|d⟩$$
$$|l⟩=\frac{1}{\sqrt{2}}|u⟩-\frac{1}{\sqrt{2}}|d⟩.\tag{19}$$
Substituting \(|u⟩=\begin{bmatrix}1 \\0\end{bmatrix}\) and \(|d⟩=\begin{bmatrix}0 \\1\end{bmatrix}\) we get
$$|r⟩=\begin{bmatrix}\frac{1}{\sqrt{2}} \\0\end{bmatrix}+\begin{bmatrix}0 \\\frac{1}{\sqrt{2}}\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}}\end{bmatrix}$$
$$|l⟩=\begin{bmatrix}\frac{1}{\sqrt{2}} \\0\end{bmatrix}+\begin{bmatrix}0 \\-\frac{1}{\sqrt{2}}\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}.$$
We can rewrite Equations (19) in matrix form as
$$\begin{bmatrix}(σ_x)_{11} & (σ_x)_{12}\\(σ_x)_{21} & (σ_x)_{22}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}}\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}}\end{bmatrix}$$
$$\begin{bmatrix}(σ_x)_{11} & (σ_x)_{12}\\(σ_x)_{21} & (σ_x)_{22}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}.\tag{20}$$
By solving the four equation in Equations (20) we can find the values of each of the entries of \(\hat{σ}_x\) (just like we did for \(\hat{σ}_z\)) and obtain
$$\hat{σ}_x=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}.\tag{21}$$
Lastly, we solve for \(hat{σ}_y\) the same exact way that we solved for \(hat{σ}_x\) to get
$$\hat{σ}_y=\begin{bmatrix}0 & -i\\i & 0\end{bmatrix}.\tag{22}$$
The three matrices associated with \(\hat{σ}_z\), \(\hat{σ}_x\), and \(\hat{σ}_y\) are called the Pauli matrices.