Spin-1/2 System

This spin is for 2-dimensional Hilbert Space. Electrons, protons, neutrons and quarks have spin-1/2. For spin-1/2 systems,

Sx≜ℏ2σxSy≜ℏ2σySz≜ℏ2σzS_x\triangleq\frac{\hbar}{2}\sigma_x\quad S_y\triangleq\frac{\hbar}{2}\sigma_y\quad S_z\triangleq\frac{\hbar}{2}\sigma_z

i.e.,

Sx≜ℏ2(0110)Sy≜ℏ2(0−ii0)Sz≜ℏ2(100−1)S_x\triangleq\frac{\hbar}{2}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\quad S_y\triangleq\frac{\hbar}{2}\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}\quad S_z\triangleq\frac{\hbar}{2}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}

Note that

∣z+⟩=∣↑⟩=∣0⟩=(10)∣z−⟩=∣↓⟩=∣1⟩=(01)\left\lvert z+ \right\rangle=\left\lvert \uparrow \right\rangle = \left\lvert 0 \right\rangle=\begin{pmatrix}1 \\ 0\end{pmatrix}\quad \left\lvert z- \right\rangle=\left\lvert \downarrow \right\rangle = \left\lvert 1 \right\rangle=\begin{pmatrix}0 \\ 1\end{pmatrix}

And

∣x+⟩=∣+⟩=12(∣0⟩+∣1⟩)=12(11)∣x−⟩=∣−⟩=12(∣0⟩−∣1⟩)=12(1−1)\left\lvert x+ \right\rangle=\left\lvert + \right\rangle=\frac{1}{\sqrt{2}}(\left\lvert 0 \right\rangle+\left\lvert 1 \right\rangle)=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}\quad \left\lvert x- \right\rangle=\left\lvert - \right\rangle=\frac{1}{\sqrt{2}}(\left\lvert 0 \right\rangle-\left\lvert 1 \right\rangle)=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix} ∣y+⟩=12(∣z+⟩+i∣z−⟩)=12(1i),∣y−⟩=12(∣z+⟩−i∣z−⟩)=12(1−i)|y+\rangle = \frac{1}{\sqrt{2}}\big(|z+\rangle + i|z-\rangle\big) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}, \qquad |y-\rangle = \frac{1}{\sqrt{2}}\big(|z+\rangle - i|z-\rangle\big) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}