The spin operators are established analogously to those for orbital angular momentum.

$\hat{l}_x$ $\hat{l}_y$ $\hat{l}_z$ $\hat{l}^2$
$\hat{s}_x$ $\hat{s}_y$ $\hat{s}_z$ $\hat{s}^2$

Where, $\hat{s}^2$, is the square of the spin angular momentum and $\hat{s}_z$, the z component of the spin angular momentum.

Spin angular momentum has the same commutation relations as orbital angular momentum:

\begin{equation} [\hat{l}_x,\hat{l}_y]=i\hbar \hat{l}_z\; \;\rightarrow [\hat{s}_x,\hat{s}_y]=i\hbar \hat{s}_z \end{equation}

\begin{equation} [\hat{l}_y,\hat{ l}_z]=i\hbar \hat{l}_x\;\;\rightarrow [\hat{s}_y,\hat{s}_z]=i\hbar \hat{s}_x \end{equation}

\begin{equation} [\hat{l}_z,\hat{l}_x]=i\hbar \hat{l}_y\;\;\rightarrow [\hat{s}_z,\hat{s}_x ]=i\hbar \hat{s}_y \end{equation}

The same commutation relations also hold for the square of angular momentum and its components.

\begin{equation} [\hat{l}^2,\hat{l}_x]=[\hat{l}^2,\hat{l}_y]=[\hat{l}^2,\hat{l}_z]=0 \end{equation}

\begin{equation} [\hat{s}^2,\hat{s}_x]=[\hat{s}^2,\hat{s}_y] =[\hat{s}^2,\hat{s}_z]=0 \end{equation}

These last commutation relations indicate that the angular momentum squared of spin commutes with its components, there being a complete set of functions owned by both operators. That is, if we consider the operators $\hat{s}^2$ and $\hat{s}_z$, there is a complete set of functions common to both of them. This complete set of eigenfunctions consists of two members, named $\alpha$ and $\beta$. Any quantum spin state is given by these functions or by linear combinations of them. We write the eigenvalue equations for $\hat{s}^2$ and $\hat{s}_z$ by analogy with those for $\hat{l}^2$ and $\hat{l}_z$

\begin{equation} \hat{l}^2\psi=\hbar^2l(l+1)\psi \end{equation} where, l=0,1,2,3....

\begin{equation} \hat{s}^2\alpha=\hbar^2s(s+1)\alpha \end{equation}

where, s=0,1/2,1,3/2,2......

\begin{equation} \hat{l}_z\psi=m\hbar\psi \end{equation}

where, m=-l,.....,0,.....+l

\begin{equation} \hat{s}_z\alpha=m\hbar\alpha \end{equation}

where, s=-s,......+s Experience shows that the electron has $s=1/2$, which gives us $m_s=\pm\frac{1}{2}$, leaving the previous equations as follows:

\begin{eqnarray} \hat{s}^2\alpha & = & 3/4\hbar^2\ alpha\\ \hat{s}^2\beta & = & 3/4\hbar^2\beta\\ \hat{s}_z\alpha & = & \frac{1}{2}\hbar\alpha\\ \hat{s}_z\beta & = & \frac{-1}{2}\hbar\beta \end{eqnarray}

$\alpha$ and $\beta$ are considered to depend exclusively on $m_s$ The condition normalization of a spatial wave function whose variables oscillate continuously between $-\infty$ and $\infty$ is:

\begin{equation} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \psi^{\ast}\psi dxdydz=1 \end{equation}

In the case of spin functions the variable $m_s$ only takes values $\pm\frac{1}{2}$ and the normalization condition It is expressed by an extended summation of these values.

\begin{equation} \sum_{m_s=-1/2}^{1/2}\alpha^{\ast}(m_s)\alpha(m_s)=1 \end{equation}

This last equation implies that $\alpha(+1/2)=1$ and $\alpha(-1/2)=0$. The normalization condition for $\beta$ can be written analogously.