Commutator Identities

Definition

This site uses

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

Products act from right to left:

$$AB\lvert\psi\rangle = A \left( B\lvert\psi\rangle \right).$$

The commutator measures the difference between the two orderings. It is an operator, not a scalar magnitude of noncommutativity.

At a glance

Identity	Formula
Antisymmetry	$[A,B]=-[B,A]$
Bilinearity	$[A,\alpha B+\beta C]=\alpha[A,B]+\beta[A,C]$
Right product rule	$[A,BC]=[A,B]C+B[A,C]$
Left product rule	$[AB,C]=A[B,C]+[A,C]B$
Inverse rule	$[A,B^{-1}]=-B^{-1}[A,B]B^{-1}$
Positive power	$[A,B^n]=\sum_{r=0}^{n-1}B^r[A,B]B^{n-1-r}$
Jacobi identity	$[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$
Adjoint	$[A,B]^\dagger=-[A^\dagger,B^\dagger]$
Similarity covariance	$S[A,B]S^{-1}=[SAS^{-1},SBS^{-1}]$
Finite trace	$\operatorname{Tr}[A,B]=0$
Tensor factors	$[A\otimes I,I\otimes B]=0$

Every row is exact for finite matrices. For unbounded operators, it is an identity only on vectors for which all products in that row are defined, or on a stated common invariant core.

Basic algebra

The commutator is linear in each slot:

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

It is antisymmetric:

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

Scalar multiples of the identity commute with every operator:

$$[A,\lambda I]=0.$$

The reverse implication is representation dependent. In an irreducible finite-dimensional representation, an operator commuting with the entire full matrix algebra is a scalar multiple of $I$. Commuting with one selected operator is much weaker.

Product rules

Associativity gives

$$\begin{aligned} [A,BC] &= ABC-BCA\\ &= (AB-BA)C + B(AC-CA)\\ &= [A,B]C+B[A,C]. \end{aligned}$$

Similarly,

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

Operator order must be preserved. The commutator acts as a derivation, not as a multiplicative map.

For an ordered product of $n$ factors,

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

An empty product on either side of $[A,B_r]$ is interpreted as $I$.

Likewise,

$$\begin{aligned} &[A_1A_2\cdots A_n,B]\\ &\quad= \sum_{r=1}^n A_1\cdots A_{r-1} [A_r,B] A_{r+1}\cdots A_n. \end{aligned}$$

Inverse rules

Assume $B$ is invertible and all products are defined. Since

$$[A,BB^{-1}] = [A,I] = 0,$$

the product rule gives

$$[A,B]B^{-1} + B[A,B^{-1}] = 0.$$

Multiplying on the left by $B^{-1}$ yields

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

Similarly,

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

These formulas preserve operator order. The shortcut

$$[A,B^{-1}] \neq - [A,B]B^{-2}$$

in general. It reduces to that form only if $[A,B]$ commutes with $B$.

For an invertible product,

$$(BC)^{-1} = C^{-1}B^{-1}.$$

Apply the product and inverse rules in the displayed reversed order.

Powers

For a positive integer $n$,

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

Likewise,

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

If

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

so that $[A,B]$ commutes with $B$, then

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

Without that condition, the $n$ ordered terms cannot be collapsed.

For negative powers, combine the inverse and positive-power rules. If $[A,B]$ commutes with $B$ and $B$ is invertible, then for any integer $n$,

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

For noncommuting $[A,B]$, keep the ordered sums and inverse factors explicit.

Functions and resolvents

If $f$ is represented by a convergent power series on the relevant spectrum and $[A,B]$ commutes with $B$, then

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

Equivalent ordering is allowed because the two factors commute. This derivative rule is not valid in general.

If $[A,B]=0$, then

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

for polynomial, continuous, or holomorphic functional calculi under their usual hypotheses.

For the resolvent

$$R_B(z) = \left( zI-B \right)^{-1},$$

the inverse rule gives

$$[A,R_B(z)] = R_B(z) [A,B] R_B(z).$$

This identity remains ordered and is useful even when $[A,B]$ does not commute with $B$.

For a holomorphic function written as a contour integral,

$$f(B) = \frac{1}{2\pi i} \oint_\Gamma f(z) \left( zI-B \right)^{-1} \,dz,$$

one obtains

$$[A,f(B)] = \frac{1}{2\pi i} \oint_\Gamma f(z) R_B(z) [A,B] R_B(z) \,dz,$$

under the boundedness and contour assumptions needed to exchange the commutator and integral.

Jacobi identity

The Jacobi identity is

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

An equivalent rearrangement is

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

This says that the map

$$\operatorname{ad}_A(B) \equiv [A,B]$$

acts as a derivation of the commutator bracket:

$$\operatorname{ad}_A([B,C]) = [ \operatorname{ad}_A(B), C ] + [ B, \operatorname{ad}_A(C) ].$$

Jacobi consistency is what makes commutators into a Lie bracket and underlies operator algebras for rotations, translations, and other continuous symmetries.

Adjoint and Hermiticity

Taking an adjoint reverses product order:

$$(AB)^\dagger = B^\dagger A^\dagger.$$

Therefore

$$\begin{aligned} [A,B]^\dagger &= B^\dagger A^\dagger - A^\dagger B^\dagger\\ &= - [A^\dagger,B^\dagger]. \end{aligned}$$

If $A$ and $B$ are Hermitian,

$$[A,B]^\dagger = - [A,B].$$

Their commutator is anti-Hermitian, while

$$\frac1i[A,B]$$

is Hermitian. Consequently,

$$\langle[A,B]\rangle$$

is purely imaginary in every state for which the expectation is defined.

Similarity transformations

For invertible $S$,

$$\begin{aligned} [SAS^{-1},SBS^{-1}] &= SABS^{-1} - SBAS^{-1}\\ &= S[A,B]S^{-1}. \end{aligned}$$

Commutation relations are covariant under similarity transformations. For a unitary basis change $S=U$, Hermiticity and inner products are preserved as well.

If

$$[A,B]=cI,$$

then the same central commutator holds after similarity transformation.

Nested commutators and conjugation

Define

$$\operatorname{ad}_A^0(B) = B,$$ $$\operatorname{ad}_A^{n+1}(B) = [ A, \operatorname{ad}_A^n(B) ].$$

The Hadamard lemma is

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

whenever the series is formal or convergent in the relevant sense.

If the nested chain terminates, the result is exact after finitely many terms. For example, if

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

then

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

The exponential product

$$e^Ae^B$$

and logarithm of that product belong to the Baker–Campbell–Hausdorff card.

Traces and finite-dimensional constraints

For finite matrices,

$$\operatorname{Tr}[A,B] = \operatorname{Tr}(AB) - \operatorname{Tr}(BA) = 0.$$

The same identity holds when the products are trace class and cyclicity is justified. It cannot be applied blindly to divergent or conditionally defined traces.

There are no finite-dimensional matrices satisfying

$$[A,B]=cI$$

with $c\neq0$. Taking the trace would give

$$0 = c\,d,$$

a contradiction in dimension $d$. Canonical commutation relations therefore require infinite-dimensional representations or finite approximations that modify the algebra.

Tensor products

Operators on distinct tensor factors commute:

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

More generally,

$$\begin{aligned} &[A\otimes B,C\otimes D]\\ &\quad= \frac12 \left( [A,C]\otimes\left\lbrace B,D\right\rbrace + \left\lbrace A,C\right\rbrace\otimes[B,D] \right). \end{aligned}$$

This form is often faster than expanding many-body product operators.

For a local sum

$$G = \sum_jG_j,$$

linearity gives

$$[G,O] = \sum_j[G_j,O].$$

Terms acting on tensor factors disjoint from the support of $O$ vanish.

Canonical-variable shortcuts

With

$$[x,p] = i\hbar I,$$

the central commutator condition holds. Therefore

$$[x,p^n] = i\hbar n p^{n-1},$$ $$[x^n,p] = i\hbar n x^{n-1},$$

and, for suitable functions,

$$[x,g(p)] = i\hbar g'(p),$$ $$[f(x),p] = i\hbar f'(x).$$

Reversing the order changes the sign:

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

For several Cartesian degrees of freedom,

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

The tempting rule

$$[f(x),g(p)] \stackrel{\mathrm{generally}}{\neq} i\hbar f'(x)g'(p)$$

is not exact in general. Higher ordering corrections appear.

Ladder-operator shortcuts

For

$$[a,a^\dagger]=I, \qquad N=a^\dagger a,$$

the product rule gives

$$[N,a]=-a, \qquad [N,a^\dagger]=a^\dagger.$$

Repeated use yields

$$[N,(a^\dagger)^r] = r(a^\dagger)^r,$$ $$[N,a^s] = -sa^s,$$

and

$$\left[ N, (a^\dagger)^r a^s \right] = (r-s) (a^\dagger)^r a^s.$$

These formulas encode how a monomial changes occupation number.

Domain conditions

For bounded operators, every product is defined on the full Hilbert space. For unbounded $A$ and $B$,

$$\mathcal D([A,B]) = \mathcal D(AB) \cap \mathcal D(BA).$$

A formal identity should be stated on a common invariant domain $\mathcal C$ such that every displayed product maps vectors in $\mathcal C$ where needed. Typical choices include smooth compactly supported functions or the Schwartz space, depending on the problem.

Additional cautions:

• an inverse may be unbounded or fail to exist at zero in the spectrum;
• a function $f(B)$ requires a specified functional calculus;
• equality on a dense core does not automatically settle equality of closed extensions;
• vanishing algebraic commutator on a core need not imply strong commutation of spectral measures;
• integration by parts can introduce boundary terms that invalidate a formal differential-operator calculation.

Assumptions

• Operator multiplication is associative.
• Scalars commute with operators.
• Every displayed product is defined on the stated vectors or common domain.
• Inverse formulas require an actual inverse on the relevant space.
• Power-series and contour formulas satisfy convergence and spectral hypotheses.
• Trace cyclicity is used only when the relevant products are trace class.
• Canonical derivative shortcuts use the stated commutation relation and suitable function domains.

Calculation checks

• Reversing a commutator must change its sign.
• Setting two entries equal must give zero.
• Product expansions must retain the original factor order.
• The inverse rule must give zero when $[A,B]=0$.
• If $A$ and $B$ are Hermitian, the result must be anti-Hermitian.
• Similarity-transforming every operator must transform the commutator covariantly.
• Finite-dimensional commutators must have zero trace.
• Operators on disjoint tensor factors must commute.
• Dimensions of $[A,B]$ must equal the product of dimensions of $A$ and $B$.
• Test a proposed identity on small noncommuting matrices before using a scalar-looking simplification.

Minimal worked uses

Quadratic momentum

From

$$[x,p]=i\hbar I,$$

the product rule gives

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

Resolvent

For

$$R(z) = \left( zI-H \right)^{-1},$$ $$[A,R(z)] = R(z)[A,H]R(z).$$

If $A$ commutes with $H$, it commutes with the resolvent and with suitable functions of $H$.

Derivation and canonical home

Commutators and Anticommutators is the canonical algebra catalog. Commutators owns the physical interpretation, compatibility, transformations, dynamics, and domain discussion.

Commutator Table is the dense lookup companion. Canonical Commutation Relations owns the position–momentum algebra and its representations.

Worked examples

• Matrix-Mechanics Examples
• Operator Identities
• Position and Momentum Representations
• Ladder-Operator Solution: First Encounter

Common mistakes

• Writing $[A,BC]=[A,B][A,C]$.
• Reordering factors after applying a product rule.
• Collapsing the power sum without checking $[B,[A,B]]=0$.
• Writing $[A,B^{-1}]=-[A,B]B^{-2}$ without a commutation condition.
• Using $[A,f(B)]=[A,B]f'(B)$ for arbitrary noncommuting $[A,B]$.
• Treating $\langle[A,B]\rangle=0$ in one state as the operator identity $[A,B]=0$.
• Assuming a formal commutator of unbounded operators is defined everywhere.
• Using trace cyclicity for divergent products.
• Claiming finite matrices realize $[A,B]=i\hbar I$ exactly.
• Forgetting that adjoints reverse order.
• Applying ordinary commutators where a graded commutator is required for fermionic parity.

Related formulas

• Canonical Commutation Relations
• Baker–Campbell–Hausdorff
• Uncertainty Relations
• Heisenberg Equation
• Angular-Momentum Algebra
• Matrix Functions and Exponentials

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, chs. 4, 7, and 9.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chs. 5 and 14.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, rev. ed., Academic Press, 1980.

Exercises

Exercise 1: derive the inverse rule

Assume $B$ is invertible. Derive $[A,B^{-1}]$ from $[A,BB^{-1}]=0$.

Solution

The product rule gives

$$0 = [A,BB^{-1}] = [A,B]B^{-1} + B[A,B^{-1}].$$

Multiplying on the left by $B^{-1}$,

$$0 = B^{-1}[A,B]B^{-1} + [A,B^{-1}].$$

Therefore

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

Exercise 2: inverse momentum

On a domain where $p^{-1}$ exists, use $[x,p]=i\hbar I$ to compute $[x,p^{-1}]$.

Solution

The inverse rule gives

$$\begin{aligned} [x,p^{-1}] &= - p^{-1}[x,p]p^{-1}\\ &= - i\hbar p^{-2}. \end{aligned}$$

The result is also the derivative shortcut

$$[x,g(p)] = i\hbar g'(p)$$

with $g(p)=p^{-1}$. The domain must exclude or otherwise control zero momentum.

Exercise 3: occupation-number change

Given

$$[N,a]=-a, \qquad [N,a^\dagger]=a^\dagger,$$

compute

$$\left[ N, (a^\dagger)^3a^2 \right].$$

Solution

The product and power rules give

$$[N,(a^\dagger)^3] = 3(a^\dagger)^3,$$ $$[N,a^2] = - 2a^2.$$

Therefore

$$\begin{aligned} \left[ N, (a^\dagger)^3a^2 \right] &= [N,(a^\dagger)^3]a^2 + (a^\dagger)^3[N,a^2]\\ &= (3-2) (a^\dagger)^3a^2\\ &= (a^\dagger)^3a^2. \end{aligned}$$

The monomial raises occupation by one.