Pauli Matrix Identity Index

This index collects the Pauli matrix identities that recur in spin-$1/2$ systems, two-level Hamiltonians, Bloch-vector calculations, and qubit notation. The identities use a fixed right-handed basis so signs can be compared directly.

For the conceptual spin home, use Pauli Matrices. For a compact table, use Pauli Matrices. For the linear-algebra basis viewpoint, use Pauli Matrices.

Pauli Matrices and Spin Components

The standard $S_z$ basis is

$$\lvert\uparrow\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \qquad \lvert\downarrow\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}.$$

In that basis,

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

Spin operators are

$$S_i=\frac{\hbar}{2}\sigma_i.$$

The matrices $\sigma_i$ are dimensionless. The physical spin components $S_i$ carry the factor $\hbar/2$.

Product and Commutator Identities

The most used identity is the Pauli product rule:

$$\sigma_i\sigma_j = \delta_{ij}I + i\sum_k\epsilon_{ijk}\sigma_k.$$

It immediately gives

$$[\sigma_i,\sigma_j] = 2i\sum_k\epsilon_{ijk}\sigma_k, \qquad \{\sigma_i,\sigma_j\} = 2\delta_{ij}I.$$

For spin operators this becomes the angular-momentum commutator

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

Need	Identity	Derivation or table
Multiply two Pauli matrices	$\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$	Pauli Matrix Table
Compute spin commutators	$[S_i,S_j]=i\hbar\epsilon_{ijk}S_k$	Angular Momentum Algebra
Use anticommutation	$\{\sigma_i,\sigma_j\}=2\delta_{ij}I$	Spin-Half Matrices
Reduce a Pauli word	$XY=iZ$, $YZ=iX$, $ZX=iY$	Quantum Gates

The cyclic identities above assume the standard convention $\epsilon_{xyz}=+1$. Reversing the sign convention for $\sigma_y$ reverses several signs, so compare tables only after checking conventions.

Vector Identities

For commuting scalar-component vectors $\mathbf a$ and $\mathbf b$, real or complex, use the bilinear dot product $\mathbf a\cdot\mathbf b=\sum_i a_i b_i$ without complex conjugation. Then

$$(\mathbf a\cdot\boldsymbol\sigma) (\mathbf b\cdot\boldsymbol\sigma) = (\mathbf a\cdot\mathbf b)I + i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.$$

Useful consequences include

$$(\mathbf a\cdot\boldsymbol\sigma)^2 = (\mathbf a\cdot\mathbf a)I,$$

and, for a real unit vector $\hat{\mathbf n}$,

$$\sigma_{\hat n} = \hat{\mathbf n}\cdot\boldsymbol\sigma, \qquad \sigma_{\hat n}^2=I.$$

Therefore $\sigma_{\hat n}$ has eigenvalues $\pm1$, and the spin component $S_{\hat n}$ has eigenvalues $\pm\hbar/2$.

Use Spin Rotations for the exponential consequences and Bloch Sphere for the geometric interpretation.

Trace and Basis Identities

The Pauli matrices, together with $I$, form an orthogonal basis for two-by-two matrices with respect to the trace inner product:

$$\operatorname{tr}I=2, \qquad \operatorname{tr}\sigma_i=0, \qquad \operatorname{tr}(\sigma_i\sigma_j)=2\delta_{ij}.$$

Any two-by-two matrix $A$ can be expanded as

$$A = \frac12\operatorname{tr}(A)I + \frac12 \sum_i \operatorname{tr}(A\sigma_i)\sigma_i.$$

For a Hermitian matrix, the coefficients in this expansion are real. This is the algebraic reason every two-level Hermitian Hamiltonian can be written as

$$H=c_0I+\mathbf b\cdot\boldsymbol\sigma.$$

The canonical two-level-system use is Two-State Hamiltonians.

Projectors and Measurements

The projectors onto spin up and spin down along a real unit vector $\hat{\mathbf n}$ are

$$P_\pm(\hat{\mathbf n}) = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

They satisfy

$$P_\pm^2=P_\pm, \qquad P_+P_-=0, \qquad P_++P_-=I.$$

For a pure state with Bloch vector $\hat{\mathbf r}$,

$$\rho = \frac12 \left( I+\hat{\mathbf r}\cdot\boldsymbol\sigma \right),$$

the probability for the $+$ outcome along $\hat{\mathbf n}$ is

$$p_+(\hat{\mathbf n}) = \operatorname{tr}(\rho P_+) = \frac12 \left( 1+\hat{\mathbf r}\cdot\hat{\mathbf n} \right).$$

Use Stern–Gerlach Revisited for the measurement interpretation and Bloch Sphere for the state geometry.

Rotations and Exponentials

Since $(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I$, the spinor rotation exponential reduces exactly:

$$\exp \left( -\frac{i\theta}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma \right) = \cos\frac{\theta}{2}I -i\sin\frac{\theta}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

The half-angle belongs to spinors. The corresponding Bloch vector rotates by the ordinary angle $\theta$.

For time evolution under a time-independent Hermitian Hamiltonian with real $c_0$ and $\mathbf b$,

$$H=c_0I+\mathbf b\cdot\boldsymbol\sigma,$$

the scalar term contributes an overall phase, while the traceless part generates precession about $\mathbf b$:

$$\exp\left(-\frac{iHt}{\hbar}\right) = e^{-ic_0t/\hbar} \left[ \cos\frac{\lvert\mathbf b\rvert t}{\hbar}I -i\sin\frac{\lvert\mathbf b\rvert t}{\hbar} \hat{\mathbf b}\cdot\boldsymbol\sigma \right],$$

Here $\hat{\mathbf b}=\mathbf b/|\mathbf b|$ for $\mathbf b\ne0$. At $\mathbf b=0$, take the continuous limit $U=e^{-ic_0t/\hbar}I$.

Use Spin Rotations, Spin in Magnetic Fields, and Larmor Precession for the physics of these formulas.

Density Matrices and Bloch Vectors

Every qubit density matrix has the form

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right), \qquad \lvert\mathbf r\rvert\leq1.$$

Its purity is

$$\operatorname{tr}(\rho^2) = \frac12 \left( 1+\lvert\mathbf r\rvert^2 \right).$$

Pure states have $\lvert\mathbf r\rvert=1$; the maximally mixed state has $\mathbf r=0$. The same Pauli coordinates appear in the density-operator treatment of the Bloch Sphere.

Qubit and Pauli-String Identities

In the computational basis, the single-qubit Pauli gates are

$$X=\sigma_x, \qquad Y=\sigma_y, \qquad Z=\sigma_z.$$

They obey

$$X^2=Y^2=Z^2=I, \qquad XY=iZ, \qquad YZ=iX, \qquad ZX=iY.$$

Reversing the order gives the minus sign:

$$YX=-iZ, \qquad ZY=-iX, \qquad XZ=-iY.$$

For tensor-product Pauli strings, two strings commute if the number of tensor positions where their nonidentity factors anticommute is even; they anticommute if that number is odd. Stabilizer calculations use this rule constantly, and their subspaces are developed in Stabilizer Identities.

Time Reversal Warning

For a spin-$1/2$ particle, the time-reversal operator is antiunitary. In a common convention,

$$\Theta=-i\sigma_yK,$$

where $K$ is complex conjugation in the $S_z$ basis. It reverses spin:

$$\Theta\sigma_i\Theta^{-1}=-\sigma_i.$$

The complex conjugation is not optional. Treating $\Theta$ as merely $-i\sigma_y$ gives wrong transformation laws and misses Kramers degeneracy. Use Time Reversal for Spin-1/2 Particles for the canonical discussion.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Quick Checks

1. Use the vector identity to compute $(\sigma_x+\sigma_z)^2$.

Solution

Here $\mathbf a=(1,0,1)$, so

$$\sigma_x+\sigma_z = \mathbf a\cdot\boldsymbol\sigma.$$

Therefore

$$(\sigma_x+\sigma_z)^2 = (\mathbf a\cdot\mathbf a)I = 2I.$$

2. What are the projectors onto spin up and spin down along $\hat z$?

Solution

Set $\hat{\mathbf n}=\hat z$. Then

$$P_\pm(\hat z) = \frac12(I\pm\sigma_z).$$

Using the displayed $\sigma_z$ convention,

$$P_+(\hat z) = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad P_-(\hat z) = \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}.$$

3. Do $X\otimes Z$ and $Z\otimes X$ commute?

Solution

At the first tensor position, $X$ and $Z$ anticommute. At the second tensor position, $Z$ and $X$ also anticommute. There are two anticommuting positions, an even number, so the full Pauli strings commute.