Pauli Spin Matrices

Equivalent forms

\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\ \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}

\{\sigma_i,\sigma_j\}=2\delta_{ij}I

[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k

The smallest non-trivial quantum system — a single qubit — is completely described by three anticommuting matrices.

Symbol	Name	SI unit	Dimension	Range
$Sz$	Spin-z eigenvalueoutput Measured value of S_z, either +hbar/2 or -hbar/2	J*s	M L^2 T^-1	-1 – 1
$theta$	Polar angle Angle of the spin state on the Bloch sphere	rad	dimensionless	0 – 3.14159
$Pup$	Spin-up probabilityoutput Probability of measuring spin up along z	1	dimensionless	0 – 1

Unit systems

Symbol	SI	CGS	Imperial
$Sz$	J*s	erg*s	J*s
$theta$	rad	rad	rad
$Pup$	1	1	1

Where it holds

Exact for any spin-1/2 or two-level system; higher spins use larger (2s+1)-dimensional spin matrices with the same algebra.

Dimensional analysis

S_i has units J s (angular momentum)

(hbar/2) sigma_i: hbar in J s, sigma dimensionless

sigma_i is a pure number matrix, so S_i inherits hbar's units.

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Discovery

Wolfgang Pauli · 1927

In 1927 Pauli introduced these matrices to describe the electron's newly discovered spin, adding a two-component 'spinor' to the wavefunction. A year later Dirac's relativistic equation revealed spin as an inevitable consequence of relativity.

Try this

How do you write down the spin of an electron with just three 2x2 matrices?

Spin-1/2 has no classical analogue, yet its entire algebra fits in three 2x2 matrices. The Pauli matrices generate every rotation of a qubit and are the building blocks of spin operators, quantum gates, and the Dirac equation.

Research status: stable

Real-world applications

Single-qubit gates (X, Y, Z, Hadamard) in every quantum computer
Spin dynamics and pulse sequences in NMR/MRI
The spin structure of the Dirac equation and particle physics

Common misconceptions

sigma_y contains i but is still Hermitian (i sits off the diagonal with its conjugate)
The factor is hbar/2, not hbar — spin-1/2 eigenvalues are +/- hbar/2
A 2pi rotation gives a minus sign on the spinor; you need 4pi to return to the start

Experimental verification

Stern-Gerlach experiments show exactly two spin outcomes, the +/- hbar/2 eigenvalues of S_z. Electron-spin-resonance and single-qubit gate tomography reproduce the Pauli algebra to high precision.

Derivation

Spin operators must satisfy [S_i,

S_j] = i

hbar epsilon_{ijk} S_k with eigenvalues +/- hbar/2.

Writing

S_i =

(hbar/2) sigma_i, the sigma_i are the unique (up to basis) 2x2 Hermitian, traceless, unit-determinant matrices obeying

\sigma _i \sigma _j = \delta _{ij} I + i \varepsilon _{ijk} \sigma _k

.

This single relation packs both the anticommutator {sigma_i,

\sigma _j} = 2 \delta _{ij} I

and the commutator [sigma_i,

\sigma _j] = 2i \varepsilon _{ijk} \sigma _k

.

Exponentiating gives rotations: exp(-i theta sigma_n/2) rotates a spinor by theta about axis n — the single-qubit gates of quantum computing.

Limiting cases

\theta = 0

⟶ state |up>,

P_up = 1

Aligned with z: measuring S_z gives +hbar/2 with certainty.

\theta = pi/2

⟶ equal superpositionIn-plane spin gives 50/50 along z.

\sigma _i^2 = I

⟶ eigenvalues +/-1Each Pauli matrix squares to the identity — spin components read +/- hbar/2.

What if…

What if you rotate a spinor by 2pi?

It picks up a factor of -1, not +1 — a hallmark of spin-1/2 confirmed by neutron interferometry. A full 4pi restores it.

What if the system had spin 1 instead?

You would use 3x3 spin matrices with eigenvalues -hbar, 0, +hbar and three measurement outcomes.

1

sigma_x eigenstates

Given ·

matrix:: sigma_x

Find · eigenvalues and states

Steps

$det(\sigma _x - \lambda I) = \lambda ^2 - 1$
$\lambda = +/-1$
eigenvectors (1,+/-1)/sqrt(2)

Result · Eigenvalues +/-1 with eigenstates (|up> +/- |down>)/sqrt(2) — spin pointing along +/-x.

2

Probability from Bloch angle

Given ·

theta:: 60 deg

Find · P_up

Steps

state $= \cos (\theta /2)$ |up> + sin(theta/2)|down>
$P_up = \cos ^2(\theta /2)$
$= \cos ^2(30) = 0.75$

Result ·

P_up = \cos ^2(\theta /2) = \cos ^2(30\,\mathrm{deg}) = 0.75

.

Spin-1/2 Operators→Angular Momentum Operators→Stern–Gerlach Deflection→Rabi Oscillations→