Spin-½ States

Electron spin as an intrinsic two-state system

When silver atoms pass through the Stern–Gerlach magnet they split into exactly two beams — not a continuous smear, not three beams, but two. The conclusion is unavoidable: electrons carry an intrinsic angular momentum called spin that can point in only two distinguishable directions relative to any chosen axis. We label them spin-up $|\!\uparrow\rangle$ and spin-down $|\!\downarrow\rangle$ along the $z$ -axis.

Because there are exactly two orthogonal outcomes, spin- $\tfrac{1}{2}$ is the smallest non-trivial quantum system. It lives in a two-dimensional complex Hilbert space — the same space as a qubit.

Connecting spin to the computational basis

By convention we identify:

|\!\uparrow\rangle \;\equiv\; |0\rangle \;=\; \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |\!\downarrow\rangle \;\equiv\; |1\rangle \;=\; \begin{pmatrix} 0 \\ 1 \end{pmatrix}.

This is purely a labeling choice, but it is universal in quantum information. With it, every tool built for qubits — gates, circuits, the Bloch sphere — applies immediately to a single spin- $\tfrac{1}{2}$ particle.

General spin states and the Born rule

The most general state of a spin- $\tfrac{1}{2}$ particle is a superposition

|\psi\rangle = \alpha\,|0\rangle + \beta\,|1\rangle, \qquad |\alpha|^2 + |\beta|^2 = 1,

where $\alpha, \beta \in \mathbb{C}$ . The normalization condition $|\alpha|^2 + |\beta|^2 = 1$ ensures the Born-rule probabilities sum to one. If we measure $S_z$ in this state:

we obtain $+\hbar/2$ (spin-up, outcome $0$ ) with probability $|\alpha|^2$ ,
we obtain $-\hbar/2$ (spin-down, outcome $1$ ) with probability $|\beta|^2$ .

Notice that the two eigenvalues of $S_z$ are $\pm\hbar/2$ , not $\pm 1$ . The qubit labels $0$ and $1$ are just tags for the two orthogonal states; the actual physical eigenvalues carry the factor $\hbar/2 \approx 5.27 \times 10^{-35}\,\text{J\,s}$ .

Visualizing spin states on the Bloch sphere

The Bloch sphere is a unit sphere whose north pole is $|0\rangle$ (spin-up) and south pole is $|1\rangle$ (spin-down). Every pure spin- $\tfrac{1}{2}$ state corresponds to exactly one point on its surface. A point on the equator, for instance, represents an equal-weight superposition with a particular relative phase — meaning the electron has definite spin along some horizontal axis, but is in an equal superposition of up and down along $z$ .

Try it

The simulator represents spin- $\tfrac{1}{2}$ states as single-qubit statevectors. The qubit starts in $|0\rangle$ (spin-up). Apply the gate that flips it to $|1\rangle$ (spin-down) and press Check to verify that your statevector matches $|\!\downarrow\rangle$ .

Run your code to see the quantum state.