Matrix representations for spin ½ systems

# Base kets

For spin $\frac{1}{2}$ systems, base kets are $|S_z;\pm\rangle$, which we can write as $|\pm\rangle$.

$$ \Lambda_{a'} = |a'\rangle\langle a'|, \qquad \sum_{a'}\Lambda_{a'} = I = 1 $$

In this case the identity can be written as:

$$ \boxed{\sum_{a'}|a'\rangle\langle a'| = |+\rangle\langle+| + |-\rangle\langle-| = 1} $$

since it can only take in two values.

# Writing Sz with outer products

The eigenvalue of the state is $\pm\frac{\hbar}{2}$. $\therefore$ we must be able to say $\hat S_z = \sum_{a'}a'\Lambda_{a'}$ (from the matrix representations notes):

$$ \hat S_z = \frac{\hbar}{2}|+\rangle\langle+| + \left(-\frac{\hbar}{2}\right)|-\rangle\langle-| $$

$$ \implies \hat S_z = \frac{\hbar}{2}\left(|+\rangle\langle+| - |-\rangle\langle-|\right) $$

Applying this to say $|+\rangle$:

$$ \hat S_z|+\rangle = \frac{\hbar}{2}\left(|+\rangle\langle+|+\rangle - |-\rangle\langle-|+\rangle\right) $$

since these are orthogonal: $\langle+|-\rangle = \langle-|+\rangle = 0$ and $\langle+|+\rangle = 1$ (normal)

$$ \boxed{\hat S_z|+\rangle = \frac{\hbar}{2}|+\rangle \quad\implies\quad \hat S_z|\pm\rangle = \pm\frac{\hbar}{2}|\pm\rangle \quad\text{(general)}} $$

This is true only because $|+\rangle$ and $|-\rangle$ are

1. basis vectors
2. orthogonal
3. normal.

# Raising and lowering operators

There are two more operators:

$$ \begin{aligned} S_+ &= \hbar|+\rangle\langle-| \quad \to \text{spin down to up} \\ S_- &= \hbar|-\rangle\langle+| \quad \to \text{up to down} \end{aligned} $$

again due to orthonormality,

$$ \boxed{S_+|+\rangle = 0, \qquad S_+|-\rangle = \hbar|+\rangle} $$

The physical sense of $S_\pm$ is that they raise (or lower) the spin component by $\hbar$; if it cannot be raised (or lowered), we get a null ket. We’ll see that these are just $S_x \pm iS_y$ in the measurements notes.

# The matrices

Coming back to matrices:

$$ |+\rangle \doteq \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |-\rangle \doteq \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

$$ S_z \doteq \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad S_+ \doteq \hbar\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad S_- \doteq \hbar\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$

(This is where the Pauli matrices come from: $S_z = \frac{\hbar}{2}\sigma_z$.)

Why? Using the outer product as a matrix:

Important:

$$ \boxed{ |\alpha\rangle\langle\beta| \doteq \begin{pmatrix} \langle a^{(1)}|\alpha\rangle\langle a^{(1)}|\beta\rangle^{\ast} & \langle a^{(1)}|\alpha\rangle\langle a^{(2)}|\beta\rangle^{\ast} & \cdots \\ \langle a^{(2)}|\alpha\rangle\langle a^{(1)}|\beta\rangle^{\ast} & \langle a^{(2)}|\alpha\rangle\langle a^{(2)}|\beta\rangle^{\ast} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} } $$

$$ \hat S_z = \frac{\hbar}{2}\left(|+\rangle\langle+| - |-\rangle\langle-|\right) $$

$$ \begin{aligned} \hat S_z \doteq \frac{\hbar}{2}\Bigg[ &\begin{pmatrix} \langle+|+\rangle\langle+|+\rangle^{\ast} & \langle+|+\rangle\langle-|+\rangle^{\ast} \\ \langle-|+\rangle\langle+|+\rangle^{\ast} & \langle-|+\rangle\langle-|+\rangle^{\ast} \end{pmatrix} \\ {-} &\begin{pmatrix} \langle+|-\rangle\langle+|-\rangle^{\ast} & \langle+|-\rangle\langle-|-\rangle^{\ast} \\ \langle-|-\rangle\langle+|-\rangle^{\ast} & \langle-|-\rangle\langle-|-\rangle^{\ast} \end{pmatrix} \Bigg] \end{aligned} $$

$$ \hat S_z \doteq \frac{\hbar}{2}\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right] = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$

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