Commutators and Operator Exponentials

Commutators measure the failure of operators to commute, and nested commutators measure how that failure propagates through products, conjugations, and exponentials. They control three basic operations: $e^A B e^{-A}$ is generated by repeated commutation with $A$; the logarithm of $e^Ae^B$ is organized by the Baker–Campbell–Hausdorff series; and finite symmetry transformations are exponentials of infinitesimal generators. Ordering corrections vanish when the relevant commutators vanish; nested expansions truncate when sufficiently deep commutators vanish.

Required background. Vector Spaces, Duals, Linear Maps, and Bases supplies composition, endomorphisms, and basis-independent map notation.

The matrix statements on this page are finite dimensional, where every exponential series converges. The canonical-commutation examples involve unbounded operators or smeared fields and are stated on a common invariant domain. This distinction is essential: exact canonical commutation relations have no nontrivial finite-dimensional matrix representation.

The commutator as a derivation

For endomorphisms $A$ and $B$ of one vector space, define

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

This page uses the ordinary commutator. For homogeneous graded operators, the graded bracket and its Koszul signs are treated on Exterior, Graded, and Grassmann Algebra.

The commutator is bilinear and antisymmetric,

$$[A,B]=-[B,A],$$

and it satisfies the Jacobi identity,

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

For fixed $A$, the map

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

acts as a derivation:

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

Induction gives

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

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

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

Consequently, for a power series $f$ convergent on the relevant matrix,

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

under the same commuting hypothesis. Without it, the unsimplified sum is the correct ordering formula.

The trace provides an immediate finite-dimensional check:

$$\operatorname{tr}[A,B]=0.$$

This identity is one reason canonical commutation relations cannot be represented by finite matrices.

Matrix exponentials

For a finite matrix or finite-dimensional endomorphism,

$$e^A = \sum_{n=0}^{\infty}\frac{A^n}{n!}$$

converges absolutely in every matrix norm. It is always invertible:

$$(e^A)^{-1}=e^{-A}.$$

Similarity commutes with exponentiation,

$$e^{SAS^{-1}}=Se^AS^{-1},$$

and

$$\det(e^A)=e^{\operatorname{tr}A}.$$

If $A$ and $B$ commute, their exponentials behave like scalar exponentials:

$$e^{A+B}=e^Ae^B=e^Be^A.$$

The converse need not hold globally because exponentials are not injective.

For a differentiable matrix $A(t)$, the scalar-looking rule

$$\frac{d}{dt}e^{A(t)} = \dot A(t)e^{A(t)}$$

is valid when $[A(t),\dot A(t)]=0$. In general the correct formula is

$$\frac{d}{dt}e^{A(t)} = \int_0^1 e^{(1-s)A(t)} \dot A(t) e^{sA(t)} \,ds.$$

It retains the ordering information between $A$ and $\dot A$.

Conjugation and the Hadamard lemma

Define

$$F(t)=e^{tA}Be^{-tA}.$$

Differentiation gives

$$\frac{dF}{dt} = [A,F(t)], \qquad F(0)=B.$$

Solving this linear equation on the vector space of matrices yields the Hadamard lemma:

$$\begin{aligned} e^ABe^{-A} &= e^{\operatorname{ad}_A}B\\ &= B+[A,B] +\frac1{2!}[A,[A,B]] +\frac1{3!}[A,[A,[A,B]]] +\cdots. \end{aligned}$$

The series converges for finite matrices. It terminates whenever some nested commutator

$$\operatorname{ad}_A^N(B)=0.$$

Two common special cases are

$$[A,B]=c\mathbb 1 \quad\Longrightarrow\quad e^ABe^{-A}=B+c\mathbb 1,$$

and

$$[A,B]=\lambda B \quad\Longrightarrow\quad e^ABe^{-A}=e^\lambda B.$$

Conjugation preserves products and commutators:

$$e^A(BC)e^{-A} = (e^ABe^{-A})(e^ACe^{-A}),$$ $$e^A[B,C]e^{-A} = [e^ABe^{-A},e^ACe^{-A}].$$

Thus an exponential implements an automorphism of the operator algebra.

Composing exponentials

Near $A=B=0$, one can write

$$e^Ae^B=e^Z$$

with the Baker–Campbell–Hausdorff expansion

$$\begin{aligned} Z ={}& A+B+\frac12[A,B]\\ &+ \frac1{12}[A,[A,B]] +\frac1{12}[B,[B,A]] +O\!\left((A,B)^4\right). \end{aligned}$$

Here $O((A,B)^4)$ denotes Lie words containing at least four occurrences of $A$ and $B$. The BCH series is always meaningful as a formal power series. As an analytic equality for matrices, it is local: convergence and the chosen branch of the matrix logarithm must be controlled. The existence of $e^Ae^B$ does not make a globally single-valued logarithm automatic. Hall 2015, Chapters 2, 3, and 5 develops the exponential, adjoint action, and local Baker–Campbell–Hausdorff statement.

If $[A,B]$ commutes with both $A$ and $B$, every higher nested commutator vanishes and BCH becomes exact:

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

Equivalently,

$$e^Ae^B = e^Be^A e^{[A,B]}.$$

This central-commutator case is the algebraic source of the phases in Weyl relations and displacement operators.

For finite matrices, $[A,B]=c\mathbb 1$ forces $c=0$ by the trace identity. The nonzero scalar-central case is a formal Lie-algebra identity or an unbounded-operator or Weyl-representation statement, where exponentiation needs additional hypotheses.

The Lie–Trotter product formula gives another composition principle:

$$e^{t(A+B)} = \lim_{n\to\infty} \left( e^{tA/n}e^{tB/n} \right)^n$$

for finite matrices. BCH shows why the leading commutator error per step is of order $n^{-2}$ and why the accumulated error tends to zero. For unbounded operators, a Trotter formula needs domain and semigroup hypotheses.

Time-dependent generators

For a matrix-valued generator $K(t)$, the evolution problem

$$\frac{d}{dt}U(t,t_0) = K(t)U(t,t_0), \qquad U(t_0,t_0)=\mathbb 1$$

has the time-ordered solution

$$U(t,t_0) = \mathrm T \exp\left( \int_{t_0}^{t}K(s)\,ds \right).$$

A simple sufficient condition for reducing it to the ordinary exponential of the integral is $[K(t_1),K(t_2)]=0$ for all relevant times. Without such a condition, time ordering cannot generally be dropped. The first terms of the Dyson series are

$$\begin{aligned} U(t,t_0) ={}& \mathbb 1 +\int_{t_0}^{t}dt_1\,K(t_1)\\ &+ \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\, K(t_1)K(t_2) +\cdots. \end{aligned}$$

The Magnus expansion rewrites the same evolution as one exponential,

$$U(t,t_0)=e^{\Omega(t,t_0)},$$

beginning with

$$\Omega_1 = \int_{t_0}^{t}dt_1\,K(t_1),$$ $$\Omega_2 = \frac12 \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\, [K(t_1),K(t_2)].$$

For quantum evolution $K=-iH$ with Hermitian $H$, every finite Magnus truncation is anti-Hermitian, so its exponential is unitary. Whether the truncations converge to the exact evolution is a separate question. Blanes et al. 2009, §§ 2–3 gives the Magnus construction and its convergence qualifications.

QFT-facing examples

The unitary-action and free-field conventions used in these examples are cross-checked against MIT OpenCourseWare 2017, Lectures 3, 6, 8, and 18 and Tong 2006–2007, § 2.

A finite symmetry action

Let $Q^\dagger=Q$, $\alpha\in\mathbb R$, and

$$U(\alpha)=e^{-i\alpha Q}.$$

Then $U(\alpha)$ is unitary. With the operator convention

$$\mathcal O(\alpha) = U(\alpha)^\dagger \mathcal O U(\alpha),$$

the infinitesimal transformation is

$$\left. \frac{d\mathcal O(\alpha)}{d\alpha} \right|_{\alpha=0} = i[Q,\mathcal O],$$

and the finite transformation is

$$\mathcal O(\alpha) = e^{i\alpha\operatorname{ad}_Q}\mathcal O.$$

Choosing $U\mathcal O U^\dagger$ instead reverses the commutator sign. The side on which $U$ acts must therefore be stated.

For the oscillator number operator $N=a^\dagger a$, the algebraic relations

$$[a,a^\dagger]=\mathbb 1, \qquad [N,a]=-a, \qquad [N,a^\dagger]=a^\dagger$$

give

$$e^{i\alpha N}a e^{-i\alpha N} = e^{-i\alpha}a,$$ $$e^{i\alpha N}a^\dagger e^{-i\alpha N} = e^{+i\alpha}a^\dagger.$$

These are identities on a common invariant oscillator domain, not identities between finite matrices.

Canonical translations and the trace obstruction

In units with $\hbar=1$, assume $p$ is self-adjoint, $T(a)=e^{-iap}$, and $T(a)$ preserves a common core $\mathcal D\subset\operatorname{Dom}q$ on which $a\mapsto T(a)^\dagger qT(a)\psi$ is differentiable and

$$[q,p]\psi=i\psi, \qquad \psi\in\mathcal D.$$

Then the Hadamard argument, or direct differentiation in $a$, gives on $\mathcal D$

$$T(a)^\dagger q\,T(a) = e^{iap}q e^{-iap} = q+a\mathbb 1.$$

The result has the expected translation sign because the convention uses $T^\dagger qT$. If $q$ and $p$ were $n\times n$ matrices, taking traces of the canonical relation would give

$$0 = \operatorname{tr}[q,p] = in,$$

which is impossible for $n>0$. Canonical commutation relations therefore require an infinite-dimensional or algebraic setting.

For fields, the equal-time relation is distributional:

$$[\phi(t,\mathbf x),\pi(t,\mathbf y)] = i\delta^{(d-1)}(\mathbf x-\mathbf y)\mathbb 1.$$

It becomes an operator statement only after smearing with test functions and specifying a common domain or an exponentiated algebra. The physical distinction among the canonical algebra, its representation, and its state is developed on Canonical Quantization: Algebra, Representation, and State.

Common pitfalls

Splitting $e^{A+B}$ without checking a commutator. Commutativity guarantees $e^{A+B}=e^Ae^B$. Without it, do not assume the factorization; accidental global equalities can occur because the exponential is not injective.

Differentiating a noncommuting exponential as if it were scalar. When $[A,\dot A]\neq0$, the derivative contains an integral with $\dot A$ inserted between exponentials.

Using BCH as a global logarithm formula. The formal series and the local analytic series are useful, but a matrix logarithm has branch and convergence issues. A large product of exponentials need not be represented by the displayed local branch.

Dropping the side of conjugation. The transformations $U^\dagger\mathcal O U$ and $U\mathcal O U^\dagger$ have opposite infinitesimal commutator signs.

Ignoring domains for unbounded products. For unbounded $A$ and $B$, the expressions $AB$, $BA$, and $[A,B]$ may be defined on different domains. Formal commutator manipulation does not prove that the resulting operator identity holds.

Exercises

1. With the Pauli matrices, use $[\sigma_z,\sigma_x]=2i\sigma_y$ and $[\sigma_z,\sigma_y]=-2i\sigma_x$ to show that

   $$e^{-i\theta\sigma_z/2} \sigma_x e^{+i\theta\sigma_z/2} = \cos\theta\,\sigma_x +\sin\theta\,\sigma_y.$$
   Solution

   Successive commutators alternate between $\sigma_x$ and $\sigma_y$. Applying the Hadamard series with $A=-i\theta\sigma_z/2$ gives the even powers as $\cos\theta\,\sigma_x$ and the odd powers as $\sin\theta\,\sigma_y$.

2. In a formal associative algebra with $[A,B]=c\mathbb 1$ central, derive both

   $$e^Ae^B = e^{A+B+\frac12c\mathbb 1}$$

   and

   $$e^Ae^B=e^Be^Ae^{c}.$$
   Solution

   The central commutator makes all higher BCH terms vanish, giving the first formula. Interchanging $A$ and $B$ gives $e^Be^A=e^{A+B-\frac12c\mathbb 1}$. Comparing the two expressions yields $e^Ae^B=e^Be^Ae^c$.

3. Show directly that no finite-dimensional matrices satisfy $[q,p]=i\mathbb 1$.

   Solution

   Cyclicity of the finite-dimensional trace gives

   $$\operatorname{tr}[q,p] = \operatorname{tr}(qp)-\operatorname{tr}(pq) = 0.$$

   The proposed right-hand side has trace $i\operatorname{tr}\mathbb 1=in$, which is nonzero for dimension $n>0$.

4. Suppose $[K(t_1),K(t_2)]=0$ for every pair of times. Show that the Dyson series sums to

   $$U(t,t_0) = \exp\left(\int_{t_0}^{t}K(s)\,ds\right).$$
   Solution

   Pairwise commutativity makes each ordered product symmetric in its time arguments. The ordered $n$-simplex integral is therefore $1/n!$ times the integral over the full $n$-cube:

   $$\int_{t_0<t_n<\cdots<t_1<t} K(t_1)\cdots K(t_n) \,\mathrm dt_1\cdots\mathrm dt_n = \frac1{n!} \left(\int_{t_0}^{t}K(s)\,ds\right)^n.$$

   Summing over $n$ gives the ordinary exponential series.

References

• Sergio Blanes, Fernando Casas, José A. Oteo, and José Ros, “The Magnus Expansion and Some of Its Applications”, Physics Reports 470 (2009), 151–238, for time-ordered evolution, nested commutators, and convergence qualifications.
• Brian C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Graduate Texts in Mathematics 222, Springer, 2015, Chapters 2, 3, and 5, for matrix exponentials, adjoint actions, and the Baker–Campbell–Hausdorff formula.
• MIT OpenCourseWare, Quantum Theory I: Lecture Notes, 8.321, Fall 2017, Lectures 3, 6, 8, and 18, Massachusetts Institute of Technology, for unitary transformations, time-dependent quantum evolution, and continuous symmetries.
• David Tong, Lectures on Quantum Field Theory, Cambridge Part III lecture notes, University of Cambridge, 2006–2007, § 2, for the harmonic oscillator and free-field commutators.