Baker–Campbell–Hausdorff

Purpose

The Baker–Campbell–Hausdorff formula, abbreviated BCH, replaces a product of noncommuting exponentials by one exponential:

$$e^A e^B=e^Z, \qquad Z=\log(e^Ae^B).$$

The exponent $Z$ is built from $A$, $B$, and their nested commutators. BCH is therefore the right tool when the order of two finite transformations matters. The closely related Hadamard lemma evaluates a conjugation $e^A B e^{-A}$.

This page is a formula and diagnostic reference. The extended physical development is at Baker–Campbell–Hausdorff Formula.

At a glance

Define the adjoint action

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

The principal formulas are:

Case	Result
General BCH series	$\log(e^Ae^B)=A+B+\frac12[A,B]+\cdots$
Commuting operators	$[A,B]=0\ \Longrightarrow\ e^Ae^B=e^{A+B}$
Central commutator	$[A,[A,B]]=[B,[A,B]]=0\ \Longrightarrow\ e^Ae^B=e^{A+B+\frac12[A,B]}$
Central reordering	$e^Ae^B=e^{[A,B]}e^Be^A$
Hadamard lemma	$e^A B e^{-A}=e^{\operatorname{ad}_A}B$
Leading splitting correction	$e^{tA}e^{tB}=\exp\!\left(t(A+B)+\frac{t^2}{2}[A,B]+O(t^3)\right)$

The central-commutator rows require $[A,B]$ to commute with both $A$ and $B$. They do not follow merely from $[A,B]$ being easy to calculate.

BCH series

Through commutator degree four,

$$\begin{aligned} Z &= A+B+\frac12[A,B] \\ &\quad +\frac1{12}[A,[A,B]] +\frac1{12}[B,[B,A]] \\ &\quad -\frac1{24}[B,[A,[A,B]]] +\cdots . \end{aligned}$$

Here degree counts occurrences of $A$ and $B$, not the number of bracket symbols. Thus $[A,B]$ has degree two and $[A,[A,B]]$ has degree three. Equivalent-looking expressions can be related by antisymmetry and the Jacobi identity. Fix one ordering convention before comparing signs.

A reliable way to organize approximations is to introduce a bookkeeping parameter:

$$\begin{aligned} \log(e^{tA}e^{tB}) &= t(A+B) +\frac{t^2}{2}[A,B] \\ &\quad +\frac{t^3}{12} \left( [A,[A,B]] +[B,[B,A]] \right) \\ &\quad -\frac{t^4}{24} [B,[A,[A,B]]] +O(t^5). \end{aligned}$$

This form makes the order of a discarded term explicit. An expansion in $t$ is controlled only when the operator scales and the relevant time or step size make the neglected terms small.

Why commutators appear

For scalars, the power series for $e^Ae^B$ can be freely reordered and collected into $e^{A+B}$. Operator products cannot generally be reordered. The first mismatch is

$$\begin{aligned} e^Ae^B &= I+(A+B) \\ &\quad +\frac12(A^2+B^2)+AB+\cdots , \end{aligned}$$

whereas

$$\begin{aligned} e^{A+B} &= I+(A+B) \\ &\quad +\frac12(A^2+AB+BA+B^2)+\cdots . \end{aligned}$$

Their quadratic difference is

$$\frac12(AB-BA)=\frac12[A,B].$$

BCH recursively packages every higher ordering mismatch into nested commutators. If $A$ and $B$ belong to a Lie algebra closed under commutators, then $Z$ remains in that Lie algebra wherever the logarithm and series are well defined.

Important truncations

Commuting case

If

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

all nested commutators vanish and

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

Pairwise commutation is essential when extending this statement to more than two exponentials.

Central-commutator case

Suppose

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

Equivalently, $[A,B]$ is central within the algebra generated by $A$ and $B$. Then the BCH series stops after its first commutator:

$$e^Ae^B = \exp\!\left( A+B+\frac12[A,B] \right).$$

Reversing $A$ and $B$ changes the commutator sign:

$$e^Be^A = \exp\!\left( A+B-\frac12[A,B] \right).$$

Because the commutator is central,

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

This identity is the source of the phase or scalar factor that appears when Weyl operators and oscillator displacement factors are reordered.

Nilpotent commutator chains

More generally, BCH terminates if sufficiently long nested commutators vanish. For example, if every commutator of degree three vanishes, only $A+B+\frac12[A,B]$ remains. Closure on a finite-dimensional operator space does not by itself imply termination: the infinite series may instead sum to a finite combination of basis operators, as it does for rotations.

Hadamard lemma

Conjugation by an exponential is generated by repeated commutation:

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

The opposite ordering changes the sign:

$$e^{-A}Be^A = e^{-\operatorname{ad}_A}B = B-[A,B] +\frac1{2!}[A,[A,B]] -\cdots .$$

To derive the series, define

$$F(s)=e^{sA}Be^{-sA}.$$

Differentiation gives

$$\frac{dF}{ds} = [A,F(s)] = \operatorname{ad}_A F(s), \qquad F(0)=B.$$

The formal solution is $F(s)=e^{s\operatorname{ad}_A}B$; setting $s=1$ produces the formula. For unbounded operators, this differentiation assumes a domain on which the products and derivatives are legitimate.

Early termination test

Calculate nested commutators in order:

$$C_1=[A,B], \qquad C_2=[A,C_1], \qquad C_3=[A,C_2], \quad\ldots$$

If $C_{n+1}=0$, Hadamard’s series stops at $C_n/n!$. If instead the $C_n$ close on a small operator basis, convert the commutator action into a matrix on that basis and exponentiate that smaller matrix.

Oscillator disentangling

Let $[a,a^\dagger]=I$ and define

$$D(\alpha) = \exp\!\left( \alpha a^\dagger-\alpha^*a \right).$$

Take

$$A=\alpha a^\dagger, \qquad B=-\alpha^*a.$$

Their commutator is central:

$$[A,B] = |\alpha|^2 I.$$

The central BCH formula gives

$$e^{\alpha a^\dagger}e^{-\alpha^*a} = e^{|\alpha|^2/2}D(\alpha),$$

or the normally ordered factorization

$$D(\alpha) = e^{-|\alpha|^2/2} e^{\alpha a^\dagger} e^{-\alpha^*a}.$$

Reversing the factors yields

$$D(\alpha) = e^{|\alpha|^2/2} e^{-\alpha^*a} e^{\alpha a^\dagger}.$$

The scalar prefactor is not optional. Dropping it changes normalization and composition phases.

Hadamard’s lemma also gives

$$D^\dagger(\alpha)aD(\alpha) = a+\alpha I,$$

because the first commutator is $\alpha I$ and all further nested commutators vanish.

Translation and canonical variables

For the canonical pair $[X,P]=i\hbar I$, the translation operator

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

acts by conjugation as

$$T^\dagger(a)XT(a)=X+aI.$$

Indeed, with $A=iaP/\hbar$,

$$[A,X]=aI,$$

and the Hadamard series truncates immediately. The result is a useful sign check: $T(a)$ translates states to the right, while $T^\dagger X T$ shifts the position observable by $+a$ in the transformed state.

For

$$U(a)=e^{-iaP/\hbar}, \qquad V(b)=e^{ibX/\hbar},$$

the exponent commutator is central, and reordering gives the Weyl relation

$$U(a)V(b) = e^{-iab/\hbar}V(b)U(a).$$

This exponentiated relation is often better behaved than manipulating the unbounded operators $X$ and $P$ directly.

Product formulas and error estimates

For a small parameter $t$,

$$e^{tA}e^{tB} = \exp\!\left( t(A+B)+\frac{t^2}{2}[A,B]+O(t^3) \right).$$

Thus replacing $e^{t(A+B)}$ by $e^{tA}e^{tB}$ has a local discrepancy whose leading exponent is quadratic in $t$ and proportional to $[A,B]$. For a Hamiltonian split $H=H_A+H_B$, set

$$A=-\frac{i}{\hbar}H_A, \qquad B=-\frac{i}{\hbar}H_B.$$

Then

$$\frac{t^2}{2}[A,B] = -\frac{t^2}{2\hbar^2}[H_A,H_B].$$

The symmetric product

$$e^{tA/2}e^{tB}e^{tA/2}$$

cancels the degree-two commutator correction; its exponent differs from $t(A+B)$ first at order $t^3$. BCH diagnoses these local error structures. Norm convergence over many steps belongs to the Trotter Product Formula.

Pauli-matrix sign check

Let

$$A=i\alpha\sigma_x, \qquad B=i\beta\sigma_y,$$

with real $\alpha$ and $\beta$. Since

$$[\sigma_x,\sigma_y]=2i\sigma_z,$$

the first commutator is

$$[A,B] = -2i\alpha\beta\sigma_z.$$

Therefore, for small angles,

$$\log(e^{i\alpha\sigma_x}e^{i\beta\sigma_y}) = i\alpha\sigma_x +i\beta\sigma_y -i\alpha\beta\sigma_z +O(\alpha^2\beta,\alpha\beta^2).$$

The generated $\sigma_z$ term is a direct signature of noncommuting rotations. Its sign reverses when the two exponentials are reversed.

Interpretation and scope

BCH has three common mathematical interpretations:

1. Formal series. Treat $A$ and $B$ as noncommuting symbols and organize terms by degree. This is enough for perturbative algebra but makes no analytic convergence claim.
2. Local matrix or bounded-operator identity. Near the identity, $e^Ae^B$ has a consistent logarithm branch and the convergent BCH series represents it.
3. Lie-theoretic coordinate formula. Locally, multiplying two Lie-group elements corresponds to a nonlinear composition law in the Lie algebra.

Even for finite matrices, a global logarithm is branch dependent, so the symbol $\log(e^Ae^B)$ must be interpreted locally or with a specified branch. For unbounded quantum operators, $A+B$, $[A,B]$, and their nested products may have different domains. The existence of the unitary exponentials does not by itself validate termwise BCH manipulation on the full Hilbert space.

When $A$ and $B$ are anti-Hermitian finite matrices, $e^A$ and $e^B$ are unitary. Every displayed BCH term is then anti-Hermitian, so any consistent finite-order truncation preserves the expected generator type, although the truncation need not reproduce the exact product.

Calculation workflow

1. State the ordering to be combined: $e^Ae^B$ is not interchangeable with $e^Be^A$.
2. Compute $[A,B]$ and then the nested commutators needed at the desired order.
3. Test whether $[A,B]$ is central or whether a nested-commutator chain terminates.
4. Introduce a small parameter when using a truncation, and count total degree consistently.
5. Check Hermiticity or anti-Hermiticity, dimensions, and the commuting limit.
6. For unbounded operators, identify a common invariant domain or use a rigorous exponentiated relation.

For conjugation, use Hadamard’s lemma directly. Computing $\log(e^Ae^B)$ first is usually unnecessary.

Common mistakes

• Writing $e^{A+B}=e^Ae^B$ without verifying $[A,B]=0$.
• Using the central formula after checking only that $[A,B]\ne0$, rather than checking both second nested commutators.
• Forgetting that reversing $A$ and $B$ changes the sign of the first commutator correction.
• Confusing $e^ABe^{-A}$ with $e^{-A}Be^A$.
• Mixing commutator degree with the number of nested brackets.
• Dropping the scalar or phase factor when displacement or Weyl operators are reordered.
• Treating an asymptotic or formal truncation as an exact identity.
• Assuming a finite-dimensional closed commutator algebra makes the BCH series terminate.
• Confusing BCH with time ordering. BCH combines discrete exponentials; time ordering organizes evolution generated at continuously varying times.
• Ignoring logarithm branches for finite matrices or domains for unbounded operators.

Cross-links

• Canonical Derivation and Applications
• Commutator Identities
• Canonical Commutation Relations
• Matrix Functions and Exponentials
• Lie Algebras
• Trotter Product Formula
• Time Ordering
• Displacement Operator
• Commutator Table

References

• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015, Chapters 2–3.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, Sections 14.2–14.3.
• N. J. Higham, Functions of Matrices: Theory and Computation, SIAM, 2008, Chapters 10–11.
• M. W. Reinsch, “A simple expression for the terms in the Baker–Campbell–Hausdorff series,” Journal of Mathematical Physics 41, 2434–2442 (2000).
• A. Van-Brunt and M. Visser, “Special-case closed form of the Baker–Campbell–Hausdorff formula,” Journal of Physics A: Mathematical and Theoretical 48, 225207 (2015).
• W. Magnus, “On the exponential solution of differential equations for a linear operator,” Communications on Pure and Applied Mathematics 7, 649–673 (1954).
• M. Suzuki, “Generalized Trotter’s formula and systematic approximants of exponential operators and inner derivations with applications to many-body problems,” Communications in Mathematical Physics 51, 183–190 (1976).
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Suppose $[A,B]=cI$. Express $e^Ae^B$ as a single exponential and then reorder it as a multiple of $e^Be^A$.

Solution

Because $cI$ commutes with both $A$ and $B$, all nested commutators vanish:

$$e^Ae^B = \exp\!\left( A+B+\frac{c}{2}I \right).$$

Reversing the factors gives

$$e^Be^A = \exp\!\left( A+B-\frac{c}{2}I \right).$$

The exponents differ by $cI$, which commutes with everything, so

$$e^Ae^B=e^c e^Be^A.$$

2. Derive the normally ordered factorization of $D(\alpha)$ from $[a,a^\dagger]=I$.

Solution

Set $A=\alpha a^\dagger$ and $B=-\alpha^*a$. Then

$$\begin{aligned} [A,B] &= -|\alpha|^2[a^\dagger,a] \\ &= |\alpha|^2 I. \end{aligned}$$

This commutator is central, so

$$e^Ae^B = \exp\!\left( A+B+\frac{|\alpha|^2}{2}I \right) = e^{|\alpha|^2/2}D(\alpha).$$

Therefore

$$D(\alpha) = e^{-|\alpha|^2/2} e^{\alpha a^\dagger} e^{-\alpha^*a}.$$

3. Let $H=H_A+H_B$. Find the leading correction in the exponent of $e^{-itH_A/\hbar}e^{-itH_B/\hbar}$ relative to $e^{-it(H_A+H_B)/\hbar}$.

Solution

Choose

$$A=-\frac{i}{\hbar}H_A, \qquad B=-\frac{i}{\hbar}H_B.$$

BCH gives

$$\log(e^{tA}e^{tB}) = t(A+B)+\frac{t^2}{2}[A,B]+O(t^3).$$

Since

$$[A,B] = -\frac{1}{\hbar^2}[H_A,H_B],$$

the leading correction is

$$-\frac{t^2}{2\hbar^2}[H_A,H_B].$$

It vanishes when the two Hamiltonian pieces commute.

4. Use Hadamard’s lemma to evaluate $e^{i\lambda P/\hbar}Xe^{-i\lambda P/\hbar}$ when $[X,P]=i\hbar I$.

Solution

Take $A=i\lambda P/\hbar$. Then

$$\begin{aligned} [A,X] &= \frac{i\lambda}{\hbar}[P,X] \\ &= \frac{i\lambda}{\hbar}(-i\hbar I) = \lambda I. \end{aligned}$$

All further nested commutators vanish because $[A,I]=0$. Hence

$$e^{i\lambda P/\hbar} X e^{-i\lambda P/\hbar} = X+\lambda I.$$