# Pauli Matrices

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.