The Spin Matrices

A quantum state is given by $|\psi> = \sum c_a |\psi>_a$ where $|\psi>_a$ are the eigenvectors of some operator $\hat{A}$ which represents the observable $A$. This means that $$\hat{A}|\psi>_a = a|\psi>_a$$

$$\hat{S}_z = \frac{1}{2}\begin{bmatrix} 1&0\\0&-1 \end{bmatrix} \hspace{10mm} |\psi>_{z,1/2} = \begin{bmatrix} 1\\0 \end{bmatrix} \hspace{10mm} |\psi>_{z,-1/2} = \begin{bmatrix} 0\\1 \end{bmatrix}$$ $$\hat{S}_x = \frac{1}{2}\begin{bmatrix} 0&1\\1&0 \end{bmatrix} \hspace{10mm} |\psi>_{x,+1/2} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\1 \end{bmatrix} \hspace{10mm} |\psi>_{x,-1/2} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\-1 \end{bmatrix}$$

Spin 1/2 Matrices

$$S_x = \frac{\hbar}{2}\begin{bmatrix} 0&1\\1&0 \end{bmatrix} \hspace{10mm} S_y = \frac{\hbar}{2}\begin{bmatrix} 0&-i\\i&0 \end{bmatrix} \hspace{10mm}S_z = \frac{\hbar}{2}\begin{bmatrix} 1&0\\0&-1 \end{bmatrix} \hspace{10mm} S^2 = S_x^2 + S_y^2 + S_z^2$$

$$S^2=S_x^2+S_y^2+S_z^2$$

$$[S_i, S_j]=i\hbar\epsilon_{ijk}S_k$$

Sya@Sxa

$$\left\{S_i, S_j\right\}=\hbar^2/4\times 2 \delta_{i,j}I=\frac{\hbar^2}{2}\delta_{i,j}I$$

$$Tr(S_i)=0$$

$$Tr(S_iS_j)=\hbar^2/4\times2\delta_{ij}=\frac{\hbar^2}{2}\delta_{ij}$$

Spin 1 Matrices

$$S_x = \frac{\hbar}{\sqrt{2}}\begin{bmatrix} 0&1&0\\1&0&1\\0&1&0 \end{bmatrix} \hspace{10mm} S_y = \frac{\hbar}{\sqrt{2}}\begin{bmatrix} 0&-i&0\\i&0&-i\\0&i&0 \end{bmatrix} \hspace{10mm}S_z = \hbar \begin{bmatrix} 1&0&0\\0&0&0\\0&0&-1 \end{bmatrix} \hspace{10mm} S^2 = S_x^2 + S_y^2 + S_z^2$$

What do these matrices mean?

These matrices are expressed in the basis of $S_z$ eigenvectors. This means that $S_z$ is written in a form such that $\left<1, 0\right>$ and $\left<0,1\right>$ are eigenvectors ($s=1/2$) or $\left<1,0,0\right>$ and $\left<0,1,0\right>$ and $\left<0,0,1\right>$ are eigenvectors ($s=1$)

Comparing Numpy and Sympy for evaluating eigenvalues and eigenvectors of a Matrix

Furthermore, since $S^2$ commutes with $S_z$, they share common eigenvectors

Addition of Spins

Now if we put a spin 1 and spin 1/2 particle together, we need a vector that takes into account all possible states

$$\begin{bmatrix} \left<1,0,0\right> \text{and} \left<1,0\right>\\ \left<0,1,0\right> \text{and} \left<1,0\right>\\ \left<0,0,1\right> \text{and} \left<1,0\right>\\ \left<1,0,0\right> \text{and} \left<0,1\right>\\ \left<0,1,0\right> \text{and} \left<0,1\right>\\ \left<0,0,1\right> \text{and} \left<0,1\right>\\ \end{bmatrix} \hspace{20mm} \text{so} \hspace{5mm} \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ \end{bmatrix} \hspace{5mm} \text{refers to} \hspace{5mm} \left<0,0,1\right> \text{and} \left<1,0\right> $$

What matrix will then give us $S_z$ for the spin-1 particle then? Can check that

$$\begin{bmatrix} S_z& [0]_{3x3}\\ [0]_{3x3} & S_z \end{bmatrix} = \begin{bmatrix} 1&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&-1\\\end{bmatrix}$$

gives the right values

But what about $S_z$ for the spin 1/2 particle? Can check that

$$\frac{1}{2}\begin{bmatrix} 1I_{3x3}&0I_{3x3}\\0I_{3x3}&-1I_{3x3} \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&-1&0&0\\ 0&0&0&0&-1&0\\ 0&0&0&0&0&-1\\\end{bmatrix} $$

gives the right values

In order to get the sum of the z-components, we need to add both matrices

So we can create these matrices for the total $S_x$, $S_y$, and $S_z$

Like before, the total spin $S^2$ and $S_z$ matrices commute, meaning they share a common eigenspace:

So we can get the eigenvectors $\psi_{S^2}$ for total $S^2$. The sympy solver finds these eigenvectors such that also happen to be eigenvectors of $S_z$ as well (this is a coincidence). Furthermore, they will be written in terms of the basis representing the z-spin of each of the two particles:

$$\begin{bmatrix} \left<1,0,0\right> \text{and} \left<1,0\right>\\ \left<0,1,0\right> \text{and} \left<1,0\right>\\ \left<0,0,1\right> \text{and} \left<1,0\right>\\ \left<1,0,0\right> \text{and} \left<0,1\right>\\ \left<0,1,0\right> \text{and} \left<0,1\right>\\ \left<0,0,1\right> \text{and} \left<0,1\right>\\ \end{bmatrix}$$

This is precisely what the Clebsh-Gordan Coefficients give us.

Get the eigenvectors $\psi_{S^2}$ of $S^2$ and $S_z$. Can show that with this particular solver, these are also eigenvectors of $S_z$, so we get a simultaenous eigenbasis.

We can get the eigenvalues seperately, and noting that $S^2 \psi_{S^2} = \hbar^2s(s+1) \psi_{S^2}=\lambda \psi_{S^2}$.

$s = \frac{-1 \pm \sqrt{1+\frac{4\lambda}{\hbar^2}}}{2}$, where $s$ is the total spin eigenvalue and $\lambda=\hbar^2s(s+1)$.

We can check that for this particular solver that all the eigenvectors $\psi_{S^2}$ of $S^2$ are also eigenvectors for $S_z$. Since $S_z \psi_{S^2} = s_z \psi_{S^2}$ and the $\psi$s are normalized we get

$$s_z = \psi_{S^2}^{^\dagger} S_z \psi_{S^2} $$

The vector is

$$\begin{bmatrix} \left<1,0,0\right> \text{and} \left<1,0\right>\\ \left<0,1,0\right> \text{and} \left<1,0\right>\\ \left<0,0,1\right> \text{and} \left<1,0\right>\\ \left<1,0,0\right> \text{and} \left<0,1\right>\\ \left<0,1,0\right> \text{and} \left<0,1\right>\\ \left<0,0,1\right> \text{and} \left<0,1\right>\\ \end{bmatrix}$$

Or in Griffith's notation

$$|1/2,-1/2> = -\sqrt{\frac{2}{3}}|1,-1>|1/2,1/2> + \sqrt{\frac{1}{3}}|1,0>|1/2,-1/2>$$

fig39.png

Relation between Clebsh-Gordon Coefficient and Wigner3j Coefficient

fig40.png