Spin

1925: Goudsmit and Uhlenbeck

Every electron has an intrinsic angular momentum (spin) of $ℏ / 2$ which corresponds to a magnetic moment of one Bohr magneton, $μ_{B} = \frac{q ℏ}{2 m}$ .

Stern-Gerlach (1922) Used an atom in 1s state $(L = 0)$ $H = H_{0} + H_{1} + H_{2}$ , $H_{0} = \frac{p^{2}}{2 m_{e}} - \frac{e^{2}}{r}$ , Paramagnetic term: $H_{1} = - \vec{M} \cdot \vec{B}$ , Diamagnetic term: $H_{2} = \frac{q^{2} B^{2}}{8 m_{e}} (x^{2} + y^{2})$ , $M = \frac{q}{2 m_{e}} \vec{L}$ . Diamagnetic term is typically ignored unless paramagnetic term is zero. The force is then expected to be: $\vec{F} = - \nabla (- \vec{M} \cdot \vec{B}) \approx M_{z} (\nabla B)_{z}$ This then raised the suspicion that $\vec{M} = \frac{q}{2 m_{e}} \vec{J} = \vec{M_{S}} + \vec{M_{L}}$ $S_{z} = \pm \frac{ℏ}{2}$
Anomalous Zeeman Effect
A given particle is characterized by a unique value of $s$
$[{\vec{L}}^{2}, {\vec{S}}^{2}] = 0$ (They act on different spaces, different quantum numbers)
General properties:
- $e^{-} \Rightarrow s = \frac{1}{2} \Rightarrow 2 s + 1 = 2$
- $| + ⟩ = | \frac{1}{2} \frac{1}{2} ⟩$
- $| - ⟩ = | \frac{1}{2} \frac{- 1}{2} ⟩$
- $⟨ \pm | \mp ⟩ = 0$
- $⟨ \pm | \pm ⟩ = 1$
- $P_{+} + | + ⟩ ⟨ + | + | - ⟩ ⟨ - | = I$
- ${\vec{S}}^{2} | \pm ⟩ = ℏ^{2} \frac{3}{4} ℏ^{2} | \pm ⟩$
- $S_{z} | \pm ⟩ = \pm \frac{ℏ}{2} | \pm ⟩$
- $S_{\pm} = S_{x} \pm i S_{y}$
- $S_{\pm} | \pm ⟩ = 0$
- $S_{+} | - ⟩ = c_{-} | + ⟩$
- $S_{-} | + ⟩ = c_{+} | - ⟩$
- $J_{\pm} | j m ⟩ = ℏ \sqrt{j (j + 1) - m (m \pm 1)} | j (m \pm 1) ⟩$

$[S_{i}, S_{j}] = i ℏ ε_{i j k} S_{k}$ , ${\vec{S}}^{2} | s m_{s} ⟩ = ℏ^{2} s (s + 1) | s m_{s} ⟩$ , $S_{z} | s m_{s} ⟩ = ℏ m_{s} | s m_{s} ⟩$ , $| m_{s} | \leq s$

Matrix Representation

$| + ⟩ = (\begin{matrix} 1 \\ 0 \end{matrix})$
$| - ⟩ = (\begin{matrix} 0 \\ 1 \end{matrix})$
$| α ⟩ = ⟨ + | α ⟩ | + ⟩ + ⟨ - | α ⟩ | - ⟩ = (\begin{matrix} ⟨ + | α ⟩ \\ ⟨ - | α ⟩ \end{matrix})$
$S_{z} = (\begin{matrix} \frac{ℏ}{2} & 0 \\ 0 & - \frac{ℏ}{2} \end{matrix})$ .
$σ_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$
$σ_{y} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix})$
$σ_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$
They all have negative 1 determinants (parity operators?, not quite since they reside in a half integer space)
$[σ_{i}, σ_{j}] = 2 i σ_{k} ε_{i j k}$
$σ_{i}$ Unitary
$σ_{i}$ Hermitian
Zero trace
${σ_{i}, σ_{j}} = 0$

Examples

Inverses

$(2 I + σ_{x})^{- 1} = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) = \frac{1}{3} (\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix})$

$(2 I + σ_{x})^{- 1} = \frac{1}{2} {(I + \frac{σ_{x}}{2})}^{- 1} \approx \frac{1}{2} (I - \frac{σ_{x}}{2} + {(\frac{σ_{x}}{2})}^{2} + \dots) = \frac{1}{2} (I (1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + \dots)) - \frac{σ_{x}}{2} (1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + \dots) = \frac{2}{3} I - \frac{1}{3} σ_{x}$ . $\sum_{n = 0}^{\infty} q^{n} = \frac{1}{1 - q}$ . $σ_{x}^{2} = I$ .

Tensor Products

$| Ψ ⟩ \in E_{x} \leftrightarrow | Ψ ⟩ \in E_{\vec{r}}$

$Ψ (\vec{r}) = Ψ_{x} (x) Ψ_{y} (y) Ψ_{z} (z) \to | Ψ ⟩ = | Ψ_{x} ⟩ \otimes | Ψ_{y} ⟩ \otimes | Ψ_{z} ⟩$ $E_{\vec{r}} = E_{x} \otimes E_{y} \otimes E_{z}$

$\otimes =$ tensor (direct) product.

The vector space $E$ is the tensor product of $E_{1}$ and $E_{2}$ if there is a vector $| ϕ_{1} ⟩ \otimes | χ_{2} ⟩ \in E$ associated with $| ϕ_{1} ⟩ \in E_{1}$ and $| χ_{2} \in E_{2}$ which satisfies:

linear with respect to multiplication by complex numbers $λ | ϕ_{1} ⟩ \otimes | χ_{2} ⟩ = λ (| ϕ_{1} ⟩ \otimes | χ_{2} ⟩)$
Distributive with respect to vector addition $| ϕ_{1} ⟩ \otimes (| χ_{2} ⟩ + | κ_{2} ⟩) = | ϕ_{1} ⟩ \otimes | χ_{2} ⟩ + | ϕ_{1} ⟩ \otimes | κ_{2} ⟩$
Basis: ${| u_{i, 1} ⟩} \in E_{1}$ , ${| v_{i, 2} ⟩} \in E_{2}$ gives an overall basis, ${| u_{i, 1} ⟩ \otimes | v_{j, 2} ⟩} = {| u_{i, 1}} \times {| v_{j, 2} ⟩}$ . Behaves like a Cartesian product. Thus, if $E_{i}$ has dimension $N_{i}$ then the overall space has dimension $N = N_{1} N_{2}$ .
${\vec{L}}^{2} | n ℓ m ⟩ \otimes | s m_{s} ⟩$ operates on $E_{\vec{r}}$ hence acts on $| n ℓ m ⟩$ An operator that acts on the entire space would be ${\vec{L}}^{2} \otimes I_{s} \in E$ . Or for the reverse case, $I_{\vec{r}} \otimes {\vec{S}}^{2}$ . Note this ordering is assumed by writing ${\vec{L}}^{2}$ to be implicitly ${\vec{L}}^{2} \otimes I_{s}$ or whatever identities are needed.
Operators acting on different spaces commute $[A_{1} \otimes I_{2}, I_{1} \otimes B_{2}] | Ψ ⟩ = 0, Ψ = u_{1} \otimes u_{2}$
. $| u_{1} ⟩ = (1 0), | u_{2} ⟩ = (0 1)$ $| v_{1} ⟩ = (1 0 0), | v_{2} ⟩ = (0 1 0), | v_{3} ⟩ = (0 0 1)$ Multiply top number by each of the other basis’s elements, then go down one number in the first and repeat $| u_{i} ⟩ \otimes | v_{j} ⟩ = (\begin{matrix} u_{i, 1} v_{j, 1} \\ u_{i, 1} v_{j, 2} \\ u_{i, 1} v_{j, 3} \\ u_{i, 2} v_{j, 1} \\ u_{i, 2} v_{j, 2} \\ u_{i, 2} v_{j, 3} \end{matrix})$ $| u_{1} ⟩ \otimes | v_{1} ⟩ = (\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix})$ $| u_{2} ⟩ \otimes | v_{1} ⟩ = (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{matrix})$ $| u_{2} ⟩ \otimes | v_{3} ⟩ = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix})$