PX262 - M1 - introduction to spin

experimental evidence points to an electron possessing an intrinsic spin
there is an associated magnetic moment

\hat{\vec{μ}} = - \frac{g e}{2 m_{e}} \hat{\vec{s}}

where, $g = 2.00232 \dots$

{\hat{H}}_{i n t} = - \hat{\vec{μ}} \cdot {\vec{B}}^{e x t} = - {\hat{μ}}_{z} B^{e x t}

orbital motion also generates a moment $\propto \hat{l}$ , with ${\hat{H}}_{i n t} = - e / 2 m_{e} \hat{l} \cdot {\vec{B}}^{e x t}$
${\hat{s}}_{z}$ is associated with spin, $s = 1 / 2$
the operator has two eigenvalues: $\pm ℏ / 2$ ( $- s, - 2 + 1, \dots, + s$ )
choosing $2 \times 2$ matrices:

{\hat{s}}_{z} = \frac{ℏ}{2} (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})

checking:

\begin{matrix} \frac{ℏ}{2} (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = λ (\begin{matrix} a \\ b \end{matrix}) \\ \frac{ℏ}{2} a = λ a \\ - \frac{ℏ}{2} b = λ b \\ λ = \frac{ℏ}{2} ⟹ a = 1, b = 0 \\ λ = - \frac{ℏ}{2} ⟹ a = 0, b = 1 \end{matrix}

for the eigenvalue, $λ = ℏ / 2 :$

(\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix}) = X_{1 / 2, m_{s} = 1 / 2}

for the eigenvalue, $λ = ℏ / 2 :$

(\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} 0 \\ 1 \end{matrix}) = X_{1 / 2, m_{s} = - 1 / 2}

spin is an intrinsic angular momentum, so it is set up with the same structure as regular angular momentum

{\hat{l}}^{2} Y_{l m} (θ, φ) = l (l + 1) ℏ^{2} Y_{l m} (θ, φ)

where, $l \geq 0$

{\hat{l}}_{z} Y_{l m} (θ, φ) = m_{l} ℏ Y_{l m} (θ, φ)

where, $m_{l} = [- l, + l]$

these are commutable, ie:

[{\hat{l}}^{2}, {\hat{l}}_{z}] = 0

but:

\begin{matrix} [{\hat{l}}_{x}, {\hat{l}}_{y}] = i ℏ {\hat{l}}_{z} \\ [{\hat{l}}_{y}, {\hat{l}}_{z}] = i ℏ {\hat{l}}_{x} \end{matrix}

similarly, setting up a structure for spin angular momentum:
so, $s = \frac{1}{2} ⟹ m_{} = 1 / 2, - 1 / 2$

{\hat{s}}_{z} X_{s, m_{s}} = m_{s} ℏ X_{s, m_{s}}

[{\hat{S}}_{x}, {\hat{S}}_{y}] = i ℏ {\hat{S}}_{z}

by ensuring the above pattern, it is found that:

\begin{matrix} {\hat{S}}_{x} = \frac{ℏ}{2} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = \frac{ℏ}{2} {\hat{σ}}_{x} \\ {\hat{S}}_{y} = \frac{ℏ}{2} (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) = \frac{ℏ}{2} {\hat{σ}}_{y} \\ {\hat{S}}_{z} = \frac{ℏ}{2} (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) = \frac{ℏ}{2} {\hat{σ}}_{z} \end{matrix}

\begin{matrix} {\hat{σ}}_{x}^{2} = {\hat{σ}}_{y}^{2} = {\hat{σ}}_{z}^{2} = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) \\ {\hat{S}}^{2} = {\hat{S}}_{x}^{2} + {\hat{S}}_{y}^{2} + {\hat{S}}_{z}^{2} = \frac{3 ℏ^{2}}{4} (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ {\hat{S}}^{2} = \frac{1}{2} (\frac{1}{2} + 1) ℏ^{2} (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ [{\hat{S}}^{2}, {\hat{S}}_{z}] = 0, etc. \end{matrix}

this is the right pattern for an angular momentum with spin $= 1 / 2$
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