Commutators

The commutator

$$[A,B]\equiv AB-BA$$

measures the failure of two ordered products to agree. It acts as a derivation, defines the Lie algebra of symmetry generators, controls Heisenberg evolution, and supplies the algebraic input to uncertainty relations.

This page is a calculation table. The canonical Commutators page develops the physical meaning of compatibility and noncommutativity; the Commutators and Anticommutators page develops the linear-algebra structure.

Scope and Convention

The anticommutator is

$$\lbrace A,B\rbrace\equiv AB+BA.$$

Products are read in the written order, operators act on kets from the right, and the canonical Cartesian convention is

$$[x_i,p_j]=i\hbar\delta_{ij}I, \qquad p_j=-i\hbar\frac{\partial}{\partial x_j}$$

in the position representation.

The purely algebraic identities below hold in any associative algebra wherever the displayed products exist. For bounded operators on a Hilbert space, this causes no domain problem. For unbounded operators, an equality initially means an equality on a stated common invariant domain or core. Closures, self-adjoint extensions, boundary conditions, and strong commutation may require separate analysis.

Basic Algebra

Let $\alpha$ and $\beta$ be scalars.

Identity	Formula	Condition or use
Self-commutator	$[A,A]=0$	Immediate from the definition
Identity	$[A,I]=0$	$I$ is the multiplicative identity
Antisymmetry	$[A,B]=-[B,A]$	Reversing order changes the sign
Bilinearity	$[A,\alpha B+\beta C]=\alpha[A,B]+\beta[A,C]$	Also linear in the first slot
Right product rule	$[A,BC]=[A,B]C+B[A,C]$	Commutator as a derivation
Left product rule	$[AB,C]=A[B,C]+[A,C]B$	Preserve factor order
Jacobi identity	$[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$	Lie-algebra consistency
Adjoint	$[A,B]^\dagger=-[A^\dagger,B^\dagger]$	A commutator of Hermitian operators is anti-Hermitian
Similarity covariance	$S[A,B]S^{-1}=[SAS^{-1},SBS^{-1}]$	For invertible $S$
Finite trace	$\operatorname{tr}[A,B]=0$	Finite matrices, or trace-class products where cyclicity applies

The product rule iterates to

$$[A,B_1B_2\cdots B_n] = \sum_{r=1}^{n} B_1\cdots B_{r-1}[A,B_r]B_{r+1}\cdots B_n.$$

Powers and inverses

For a positive integer $n$,

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

If $[A,B]$ commutes with $B$, the first formula reduces to

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

If $B$ is invertible and all products are defined,

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

More generally, if $[A,B]$ commutes with $B$ and $f$ has an appropriate power series or functional calculus,

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

Do not use this derivative rule when $[A,B]$ fails to commute with $B$.

Tensor products

Operators on different tensor factors commute:

$$[A\otimes I,I\otimes B]=0.$$

A useful general identity is

$$\begin{aligned} [A\otimes B,C\otimes D] =\frac{1}{2}\Big( &[A,C]\otimes\lbrace B,D\rbrace \\ &+\lbrace A,C\rbrace\otimes[B,D] \Big). \end{aligned}$$

It is often faster than expanding two many-body product operators entry by entry.

Canonical Position–Momentum Identities

The following entries use one Cartesian degree of freedom unless indices are shown.

Identity	Result	Assumptions
Canonical relation	$[x,p]=i\hbar I$	Common invariant domain
Cartesian components	$[x_i,p_j]=i\hbar\delta_{ij}I$	Cartesian coordinates
Position components	$[x_i,x_j]=0$	Standard position representation
Momentum components	$[p_i,p_j]=0$	No gauge-covariant momentum substituted
Momentum power	$[x,p^n]=i\hbar n p^{n-1}$	Positive integer $n$
Position power	$[x^n,p]=i\hbar n x^{n-1}$	Positive integer $n$
Position function	$[f(x),p]=i\hbar f'(x)$	Sufficiently regular $f$
Momentum function	$[x,g(p)]=i\hbar g'(p)$	Sufficiently regular $g$
Reversed position function	$[p,f(x)]=-i\hbar f'(x)$	Same convention
Quadratic example	$[x^2,p^2]=2i\hbar(xp+px)$	Common domain for all products

For several Cartesian degrees of freedom,

$$[f(\boldsymbol x),p_j] =i\hbar\,\partial_j f(\boldsymbol x), \qquad [x_j,g(\boldsymbol p)] =i\hbar\,\frac{\partial g}{\partial p_j}.$$

These identities are exact for suitable polynomials and extend to broader classes only with appropriate domain or functional-calculus hypotheses.

Translation check

With the active translation operator

$$T(a)=e^{-iap/\hbar},$$

the canonical relation gives

$$[x,T(a)]=aT(a), \qquad T^\dagger(a)xT(a)=x+aI.$$

This is a useful sign check: $T(a)$ shifts a localized state toward larger position by $a$ under the stated convention.

Dynamics

In the Heisenberg picture,

$$\frac{dA_H}{dt} =\frac{i}{\hbar}[H_H,A_H] +\left(\frac{\partial A}{\partial t}\right)_H.$$

For

$$H=\frac{p^2}{2m}+V(x),$$

the relevant commutators are

Commutator	Result	Heisenberg equation
$[x,H]$	$i\hbar p/m$	$\dot x=p/m$
$[p,H]$	$-i\hbar V'(x)$	Use the reversed order with care
$[H,x]$	$-i\hbar p/m$	$\dot x=(i/\hbar)[H,x]=p/m$
$[H,p]$	$i\hbar V'(x)$	$\dot p=(i/\hbar)[H,p]=-V'(x)$

The two rows for each observable are deliberately redundant: they expose the sign change caused by reversing the commutator.

If $A$ has no explicit time dependence, $[H,A]=0$ implies that its Heisenberg operator and expectation value are constant in time, subject to the domain conditions needed for the evolution.

Rotation and Spin Algebras

Use $\epsilon_{xyz}=+1$.

Objects	Relation	Comment
Angular momentum	$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k$	Generator algebra of rotations
Casimir	$[J^2,J_i]=0$	$J^2=J_x^2+J_y^2+J_z^2$
Ladder with $J_z$	$[J_z,J_\pm]=\pm\hbar J_\pm$	$J_\pm=J_x\pm iJ_y$
Ladder pair	$[J_+,J_-]=2\hbar J_z$	Sign follows the ladder definition
Vector operator	$[J_i,V_j]=i\hbar\epsilon_{ijk}V_k$	Defines Cartesian vector transformation
Pauli matrices	$[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$	Dimensionless spin-$1/2$ matrices
Spin one-half	$[S_i,S_j]=i\hbar\epsilon_{ijk}S_k$	$S_i=\hbar\sigma_i/2$

For orbital angular momentum $\boldsymbol L=\boldsymbol x\times\boldsymbol p$, the canonical commutation relations imply the same angular-momentum algebra. The derivation belongs in the canonical Angular Momentum Algebra page.

Oscillator and Mode Algebras

One harmonic oscillator

Objects	Relation
Ladder pair	$[a,a^\dagger]=I$
Number and lowering	$[N,a]=-a$
Number and raising	$[N,a^\dagger]=a^\dagger$
Number definition	$N=a^\dagger a$
Hamiltonian and lowering	$[H,a]=-\hbar\omega a$
Hamiltonian and raising	$[H,a^\dagger]=\hbar\omega a^\dagger$

Here

$$H=\hbar\omega\left(N+\frac12 I\right).$$

The relations show directly that $a$ and $a^\dagger$ lower and raise energy by $\hbar\omega$, provided the resulting state is nonzero and lies in the relevant domain.

Bosonic and fermionic modes

For independent bosonic modes,

$$[a_r,a_s^\dagger]=\delta_{rs}I, \qquad [a_r,a_s]=[a_r^\dagger,a_s^\dagger]=0.$$

Fermionic modes instead obey canonical anticommutation relations:

$$\lbrace c_r,c_s^\dagger\rbrace=\delta_{rs}I, \qquad \lbrace c_r,c_s\rbrace =\lbrace c_r^\dagger,c_s^\dagger\rbrace=0.$$

Replacing the fermionic anticommutators by commutators changes the algebra and is not a harmless notation choice.

Exponentials and Nested Commutators

Define

$$\operatorname{ad}_A(B)\equiv[A,B], \qquad \operatorname{ad}_A^0(B)\equiv B.$$

The Hadamard lemma is

$$e^A B e^{-A} = \sum_{n=0}^{\infty} \frac{1}{n!}\operatorname{ad}_A^n(B),$$

when the series converges or when it is used as a controlled formal expansion. If the nested commutators terminate, the expression is finite.

An exact integral identity useful when $[A,B]$ is not central is

$$[A,e^B] = \int_0^1 e^{sB}[A,B]e^{(1-s)B}\,ds.$$

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

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

The Baker–Campbell–Hausdorff expansion begins

$$\begin{aligned} \log(e^Ae^B) =\;&A+B+\frac12[A,B] \\ &+\frac{1}{12}[A,[A,B]] +\frac{1}{12}[B,[B,A]] +\cdots. \end{aligned}$$

When $[A,B]$ commutes with both $A$ and $B$, all displayed nested commutators vanish and

$$e^Ae^B =e^{A+B+\frac12[A,B]}.$$

The central-commutator condition is essential. Applying this truncated formula to generic matrices gives a wrong result.

Domains and Strong Commutation

For unbounded operators, $AB\psi$ exists only if

$$\psi\in\mathcal D(B) \quad\text{and}\quad B\psi\in\mathcal D(A).$$

Thus $[A,B]\psi$ requires both $AB\psi$ and $BA\psi$. A formal differential calculation commonly establishes an identity first on a dense invariant core, such as a suitable smooth rapidly decreasing domain. Whether the identity extends to self-adjoint closures is a separate question.

Two useful warnings follow:

1. In finite dimensions, commuting Hermitian matrices admit a simultaneous orthonormal eigenbasis. For unbounded self-adjoint operators, vanishing of a formal commutator on a small domain need not imply that their spectral projections commute. The stronger spectral condition is called strong commutation.
2. No finite-dimensional matrices can satisfy the exact canonical relation. If the dimension is $d$, cyclicity gives

$$\operatorname{tr}[X,P]=0,$$

whereas

$$\operatorname{tr}(i\hbar I)=i\hbar d\ne0.$$

Finite matrix approximations therefore modify the canonical commutator, typically near a truncation boundary.

How to Use This Table

1. Fix the order of the requested commutator before substituting identities.
2. Expand products with the derivation rule while preserving factor order.
3. Use a simplified derivative formula only after checking that the basic commutator is central or commutes with the relevant operator.
4. For unbounded operators, identify a common domain or state the calculation as formal.
5. Check signs by swapping the order, taking an adjoint, or acting in a concrete representation.
6. For symmetry algebras, verify at least one cyclic case and the Jacobi identity.

Common Mistakes

• Writing $[A,BC]=[A,B][A,C]$ instead of using the product rule.
• Reordering factors after expanding a commutator.
• Forgetting the sign change between $[A,B]$ and $[B,A]$.
• Using $[A,f(B)]=[A,B]f'(B)$ without checking the centrality condition.
• Confusing the commutator of Pauli matrices with that of physical spin operators and losing a factor of $\hbar/2$.
• Replacing fermionic anticommutators by commutators.
• Applying the short Baker–Campbell–Hausdorff formula when nested commutators do not vanish.
• Claiming that finite matrices obey $[X,P]=i\hbar I$ exactly.
• Inferring simultaneous spectral measurements from a merely formal commutator calculation.
• Ignoring boundary terms when differential operators act on domains with boundaries.

Verification

Every algebraic row follows from associativity and the defining commutator. Standard quantum rows can be checked independently in the position representation, a finite spin representation, or the number-state basis. For an audit:

1. expand each product identity directly;
2. verify the canonical derivative identities on a smooth test function;
3. derive one angular-momentum cyclic relation;
4. apply oscillator commutators to $\lvert n\rangle$;
5. compare an exponential identity through second order in its parameters.

Last reviewed: 2026-08-19.

Exercises

Exercise 1: Momentum squared

Use only the product rule and $[x,p]=i\hbar I$ to calculate $[x,p^2]$.

Solution

The right product rule gives

$$\begin{aligned} [x,p^2] &=[x,p]p+p[x,p] \\ &=i\hbar p+p\,i\hbar \\ &=2i\hbar p. \end{aligned}$$

The scalar $i\hbar$ commutes with $p$.

Exercise 2: Force and sign order

For $H=p^2/(2m)+V(x)$, derive both $[p,H]$ and $[H,p]$, then recover the Heisenberg equation for $\dot p$.

Solution

Because $p$ commutes with $p^2$,

$$[p,H]=[p,V(x)]=-i\hbar V'(x).$$

Antisymmetry then gives

$$[H,p]=i\hbar V'(x).$$

The Heisenberg equation uses the second ordering:

$$\dot p =\frac{i}{\hbar}[H,p] =\frac{i}{\hbar}\,i\hbar V'(x) =-V'(x).$$

This is the operator form of force equals minus the potential gradient.

Exercise 3: Why truncation changes the canonical relation

Suppose $X$ and $P$ are $d\times d$ matrices. Prove that they cannot satisfy $[X,P]=i\hbar I_d$ exactly. What does this imply for a numerical oscillator basis truncated to finitely many states?

Solution

Cyclicity of the finite-dimensional trace gives

$$\operatorname{tr}[X,P] =\operatorname{tr}(XP)-\operatorname{tr}(PX) =0.$$

If the canonical relation held exactly, the same trace would equal

$$\operatorname{tr}(i\hbar I_d)=i\hbar d,$$

which is nonzero for $\hbar\ne0$ and $d>0$. This contradiction proves the claim. A finite oscillator truncation can reproduce the canonical relation well on low-lying states, but it must contain a compensating boundary correction, usually concentrated near the highest retained basis state.

Canonical Links

• Commutators
• Commutators and Anticommutators
• Canonical Commutation Relations
• Angular Momentum Algebra
• Pauli Matrices
• Harmonic Oscillator Spectrum
• Heisenberg Picture
• Unbounded Operators

References

• 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. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• R. Gilmore, Lie Groups, Physics, and Geometry, Cambridge University Press, 2008.