Commutators and Anticommutators

The commutator and anticommutator are the antisymmetric and symmetric parts of an operator product. They measure different kinds of algebraic structure: order-dependence for commutators, and symmetric pairing for anticommutators.

The Core Formalism interpretations are Commutators and Anticommutators. This page collects the reusable finite-dimensional algebra.

Definitions

For two linear operators or square matrices $A$ and $B$ on the same vector space, the commutator is

$$[A,B] = AB-BA.$$

The anticommutator is

$$\{A,B\} = AB+BA.$$

Together they split the two possible orderings:

$$AB = \frac12 \left( \{A,B\}+[A,B] \right), \qquad BA = \frac12 \left( \{A,B\}-[A,B] \right).$$

Thus $[A,B]=0$ says the orderings agree, while $\{A,B\}=0$ says the two orderings cancel.

Basic Commutator Identities

Commutators are bilinear:

$$[A,\alpha B+\beta C] = \alpha[A,B]+\beta[A,C],$$

and similarly in the first slot.

They are antisymmetric:

$$[A,B]=-[B,A], \qquad [A,A]=0.$$

They satisfy product rules:

$$[A,BC] = [A,B]C+B[A,C],$$

and

$$[AB,C] = A[B,C]+[A,C]B.$$

The Jacobi identity is

$$[A,[B,C]] +[B,[C,A]] +[C,[A,B]] = 0.$$

This identity is what makes the commutator into a Lie bracket.

Powers and Functions

Repeated use of the product rule gives

$$[A,B^n] = \sum_{r=0}^{n-1} B^r[A,B]B^{n-1-r}.$$

If $[A,B]$ also commutes with $B$, this reduces to

$$[A,B^n] = n[A,B]B^{n-1}.$$

If $[A,B]=0$, then $A$ commutes with every polynomial in $B$. Under the usual convergence hypotheses, it also commutes with functions of $B$ defined by power series:

$$[A,f(B)]=0.$$

This is the algebra behind many simultaneous-diagonalization and time-evolution simplifications. The function side is developed in Matrix Functions and Exponentials.

Anticommutator Identities

Anticommutators are bilinear and symmetric:

$$\{A,B\}=\{B,A\}.$$

They do not obey the same derivation rule as commutators. A useful mixed identity is

$$[A,\{B,C\}] = \{[A,B],C\} +\{B,[A,C]\}.$$

For any operator,

$$\{A,A\} = 2A^2.$$

For Hermitian observables, anticommutators appear in covariance formulas because they capture the symmetric part of the product. If

$$\Delta A=A-\langle A\rangle I, \qquad \Delta B=B-\langle B\rangle I,$$

then the symmetric covariance term is built from

$$\frac12 \langle\{\Delta A,\Delta B\}\rangle.$$

The commutator supplies the antisymmetric uncertainty term, while the anticommutator supplies the covariance refinement.

Commuting Observables

In finite-dimensional quantum mechanics, Hermitian operators $A$ and $B$ can be simultaneously diagonalized exactly when

$$[A,B]=0.$$

This is the algebraic core of compatibility. It means the operators can be represented as diagonal matrices in one common orthonormal basis, though degeneracies may require care in choosing that basis.

The physics statement is Compatible Observables. In infinite-dimensional settings, unbounded operators require domain and spectral-projector care; formal commutator algebra alone may not settle compatibility.

Lie Algebra Preview

The commutator turns suitable spaces of matrices into Lie algebras. For example, angular momentum operators satisfy

$$[J_i,J_j] = i\hbar \sum_k \epsilon_{ijk}J_k.$$

The Jacobi identity guarantees consistency of nested commutators. This is why commutators are the natural language for generators of rotations, translations, and other continuous symmetries.

For the classical bracket that plays the analogous role in Hamiltonian mechanics, see Poisson Brackets. For the symmetry interpretation, see Commutators and Conservation Laws and Angular Momentum Algebra.

Pauli Matrix Example

The Pauli matrices satisfy

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

Taking the antisymmetric and symmetric parts gives

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

and

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

Thus the commutator encodes the spin algebra, while the anticommutator encodes the Clifford-algebra relation. See Pauli Matrices.

Fermionic Preview

Anticommutators become structural in fermionic many-body theory. Fermionic creation and annihilation operators satisfy canonical anticommutation relations such as

$$\{c_i,c_j^\dagger\} = \delta_{ij}, \qquad \{c_i,c_j\}=0.$$

This page does not develop Fock space or identical-particle physics. The point here is only the algebraic contrast: bosonic oscillator variables are organized by commutators, while fermionic variables are organized by anticommutators.

Worked Example

Use the product rule to compute $[A,B^2]$:

$$[A,B^2] = [A,B]B+B[A,B].$$

If $[A,B]$ commutes with $B$, then

$$[A,B^2] = 2[A,B]B.$$

Without that additional commutation, the two terms must stay in the displayed order.

Common Mistakes

• Replacing $AB$ by $BA$ after expanding an expression.
• Using $[A,BC]=[A,B][A,C]$, which is false.
• Assuming $[A,B]=0$ just because one expectation value $\langle[A,B]\rangle$ vanishes.
• Forgetting that anticommutators are not commutators with a sign changed.
• Treating Pauli, bosonic, and fermionic algebra conventions as interchangeable.
• Applying finite-dimensional commutator identities to unbounded operators without checking domains.
• Assuming $\exp(A+B)=\exp(A)\exp(B)$ when $[A,B]\ne0$.

Cross-Links

• Commutators
• Poisson Brackets
• Compatible Observables
• General Uncertainty Relations
• Commutator Table
• Canonical Commutation Relations
• Matrix Functions and Exponentials
• Pauli Matrices
• Angular Momentum Algebra
• Commutators and Conservation Laws

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

Exercises

1. Prove the product rule

$$[A,BC] = [A,B]C+B[A,C].$$

Solution

Expand the right-hand side:

$$[A,B]C+B[A,C] = (AB-BA)C+B(AC-CA) = ABC-BAC+BAC-BCA = ABC-BCA.$$

This is exactly

$$[A,BC].$$

2. Compute $[\sigma_x,\sigma_y]$ and $\{\sigma_x,\sigma_y\}$.

Solution

The Pauli algebra gives

$$\sigma_x\sigma_y=i\sigma_z, \qquad \sigma_y\sigma_x=-i\sigma_z.$$

Therefore

$$[\sigma_x,\sigma_y] = 2i\sigma_z, \qquad \{\sigma_x,\sigma_y\} = 0.$$

3. Suppose $[A,B]=0$. Show that $[A,B^3]=0$.

Solution

Use the powers identity:

$$[A,B^3] = [A,B]B^2+B[A,B]B+B^2[A,B].$$

Every term vanishes because $[A,B]=0$.

4. Why does $\{A,B\}=0$ not mean that $A$ and $B$ commute?

Solution

The equation $\{A,B\}=0$ says $AB=-BA$. Commutation would require $AB=BA$. Both can hold together only when $AB=0$. In general, anticommuting operators are maximally order-sensitive, not commuting.