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intermediate · Physics · Spin-½ Systems & Pauli Algebra

Spin Operators Sᵢ = (ħ/2)σᵢ

The Pauli matrices are dimensionless. To turn them into physical observables — operators whose eigenvalues carry units of angular momentum — we attach the natural quantum of action \hbar. The spin operators are defined as

Sx=2σx,Sy=2σy,Sz=2σz.S_x = \frac{\hbar}{2}\,\sigma_x, \qquad S_y = \frac{\hbar}{2}\,\sigma_y, \qquad S_z = \frac{\hbar}{2}\,\sigma_z.

The factor of 12\tfrac{1}{2} is exactly what makes this a spin-½ particle.

Eigenvalues are ±ħ/2

Because σz\sigma_z has eigenvalues ±1\pm 1, the operator Sz=2σzS_z = \tfrac{\hbar}{2}\sigma_z has eigenvalues

Sz0=+20,Sz1=21.S_z\,|0\rangle = +\frac{\hbar}{2}\,|0\rangle, \qquad S_z\,|1\rangle = -\frac{\hbar}{2}\,|1\rangle.

A measurement of the zz-component of spin can only return +2+\tfrac{\hbar}{2} ("spin up") or 2-\tfrac{\hbar}{2} ("spin down"). The same is true along any axis, because SxS_x and SyS_y are likewise 2\tfrac{\hbar}{2} times an involution.

The angular-momentum commutators

Scaling the Pauli commutators by (/2)2(\hbar/2)^2 and using [σj,σk]=2iεjklσl[\sigma_j,\sigma_k] = 2i\varepsilon_{jkl}\sigma_l gives the canonical angular-momentum algebra

[Sx,Sy]=iSz,[Sy,Sz]=iSx,[Sz,Sx]=iSy.[S_x, S_y] = i\hbar\,S_z, \qquad [S_y, S_z] = i\hbar\,S_x, \qquad [S_z, S_x] = i\hbar\,S_y.

This is the same algebra obeyed by orbital angular momentum L=r×p\mathbf{L} = \mathbf{r}\times\mathbf{p}. Spin is genuine angular momentum, even though no particle is literally spinning.

The total spin and its magnitude

The squared total spin operator is

S2=Sx2+Sy2+Sz2=3(2)2I=342I,S^2 = S_x^2 + S_y^2 + S_z^2 = 3\left(\frac{\hbar}{2}\right)^2 I = \frac{3}{4}\hbar^2\,I,

using σk2=I\sigma_k^2 = I. This matches the general angular-momentum formula S2=s(s+1)2S^2 = s(s+1)\hbar^2 with s=12s = \tfrac{1}{2}:

s(s+1)2=12322=342.s(s+1)\hbar^2 = \tfrac{1}{2}\cdot\tfrac{3}{2}\,\hbar^2 = \tfrac{3}{4}\hbar^2.

Every spin-½ state is an eigenstate of S2S^2 with the same eigenvalue 342\tfrac{3}{4}\hbar^2 — the magnitude of the spin is fixed, only its direction is quantum.

Try it

Working in units where =1\hbar = 1, compute the spin-up eigenvalue of SzS_z (the larger of the two). It should equal 2=0.5\tfrac{\hbar}{2} = 0.5.

Run your code to see the quantum state.

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