Baker–Campbell–Hausdorff Formula

The Baker–Campbell–Hausdorff formula, often abbreviated BCH, describes how exponentials of noncommuting operators combine. In quantum dynamics it answers two recurring questions:

1. What is the effective exponent of a product $e^Ae^B$?
2. How does an operator change under conjugation $e^A B e^{-A}$?

Both answers are controlled by nested commutators. That is why BCH appears in Heisenberg evolution, interaction-picture transformations, coherent-state displacement operators, spin rotations, product-formula errors, and Lie-algebraic approximation schemes.

For the compact formula-card version, see Baker–Campbell–Hausdorff. For the general finite-dimensional construction of operator exponentials, see Matrix Functions and Exponentials.

Why Noncommuting Exponentials Are Hard

If $A$ and $B$ commute, then

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

This identity is safe because the power series can be rearranged like scalar products. If $[A,B]\ne0$, the rearrangement fails. The product $e^Ae^B$ is still meaningful, but its logarithm is not usually $A+B$.

The first correction is already enough to explain many quantum effects:

$$\log(e^Ae^B) = A+B+\frac12[A,B]+\cdots .$$

The ellipsis matters. Unless nested commutators vanish or are controlled, replacing the product by only the first few terms is an approximation.

Product Form

The BCH series begins

$$\begin{aligned} \log(e^Ae^B) &= A+B+\frac12[A,B] \\ &\quad +\frac1{12}[A,[A,B]] +\frac1{12}[B,[B,A]] +\cdots . \end{aligned}$$

Equivalently,

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

This is an identity under appropriate convergence assumptions, and it is also used formally in Lie-algebraic calculations. In quantum mechanics, formal use is common, but it should not hide domain questions for unbounded operators.

Central-Commutator Truncation

A particularly useful special case occurs when

$$[A,B]$$

commutes with both $A$ and $B$. Then all higher nested commutators vanish, and

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

The same assumption gives the reordering identity

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

This central-commutator case is the algebra behind many Weyl, displacement, and canonical-commutator manipulations. It is also a common place to make mistakes: the truncation is not valid merely because $[A,B]$ is simple-looking; it must commute with $A$ and $B$.

Similarity Transformations

The conjugation identity is often the more useful dynamics formula. Define

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

Then

$$e^A B e^{-A} = e^{\operatorname{ad}_A}B,$$

meaning

$$e^A B e^{-A} = B+[A,B] +\frac{1}{2!}[A,[A,B]] +\frac{1}{3!}[A,[A,[A,B]]] +\cdots .$$

This identity is sometimes called the Hadamard lemma. It is the local engine behind operator picture transformations: conjugation by an exponential becomes a series of commutators.

Heisenberg Evolution

For a time-independent Hamiltonian,

$$A_H(t) = e^{iHt/\hbar}A e^{-iHt/\hbar}.$$

Use the similarity formula with

$$A_{\rm gen} = \frac{iHt}{\hbar}.$$

Then

$$\begin{aligned} A_H(t) &= A + \frac{it}{\hbar}[H,A] \\ &\quad + \frac{1}{2!} \left( \frac{it}{\hbar} \right)^2 [H,[H,A]] +\cdots . \end{aligned}$$

Differentiating this expansion at $t=0$ gives the Heisenberg equation

$$\dot A_H = \frac{i}{\hbar}[H,A_H].$$

The nested commutators describe the finite-time orbit of $A$ under Hamiltonian conjugation. If they close on a finite-dimensional operator subspace, the dynamics can often be solved by finite-dimensional linear algebra.

Product-Formula Error

BCH also explains why splitting a Hamiltonian into noncommuting pieces produces finite-step errors. Let

$$X = -\frac{i}{\hbar}H_A\Delta t, \qquad Y = -\frac{i}{\hbar}H_B\Delta t.$$

Then

$$e^Xe^Y = \exp \left[ X+Y+\frac12[X,Y]+O(\Delta t^3) \right].$$

Since

$$[X,Y] = -\frac{\Delta t^2}{\hbar^2}[H_A,H_B],$$

one step contains the correction

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

This is the local source of the first-order Trotter error. The convergence and higher-order splitting discussion belongs to Trotter Product Formula.

Displacement Operators

For one oscillator mode, define

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

Let

$$K = \alpha a^\dagger-\alpha^*a.$$

Using $[a,a^\dagger]=1$,

$$[K,a] = -\alpha, \qquad [K,[K,a]]=0.$$

The similarity formula truncates:

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

and equivalently

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

This is why displacement operators create coherent states from the oscillator ground state. The coherent-state page uses this result as a construction; the operator reference card records common convention warnings in Displacement Operator.

Spin Rotations

Angular-momentum rotations are another closed commutator system. With

$$U_z(\theta) = \exp \left( -\frac{i}{\hbar}\theta S_z \right),$$

and

$$[S_z,S_x]=i\hbar S_y, \qquad [S_z,S_y]=-i\hbar S_x,$$

the nested commutators alternate between $S_x$ and $S_y$. One convention gives

$$U_z(\theta)S_xU_z^\dagger(\theta) = S_x\cos\theta+S_y\sin\theta.$$

The series has summed to an ordinary rotation. The spinor-specific half-angle formula is developed in Spin Rotations; BCH explains the operator-algebra mechanism behind the rotation of spin components.

Time Ordering Is Different

BCH combines a finite product of exponentials into one effective exponential. Time ordering solves a different problem: it organizes a continuous product when the Hamiltonian at different times does not commute with itself.

For a time-dependent Hamiltonian, one cannot generally write

$$U(t,t_0) = \exp \left[ -\frac{i}{\hbar} \int_{t_0}^t H(s)\,ds \right].$$

The correct object is a time-ordered exponential unless

$$[H(t_1),H(t_2)]=0$$

for the times being compared, or unless a special transformation reduces the problem. For that organization, see Time Ordering.

Common Mistakes

• Using $e^{A+B}=e^Ae^B$ for noncommuting operators.
• Applying the central-commutator truncation without checking nested commutators.
• Treating BCH as a convergence theorem for unbounded operators without domain assumptions.
• Confusing BCH with the Trotter product formula or a time-ordered exponential.
• Dropping the phase generated when displacement operators are reordered.
• Forgetting that active and passive rotation conventions change signs in conjugation formulas.
• Keeping only the first commutator term while needing a finite-time transformation.

Cross-Links

• Operator Identities
• Baker–Campbell–Hausdorff Reference Card
• Matrix Functions and Exponentials
• Commutator Dynamics
• Trotter Product Formula
• Time Ordering
• Commutators and Anticommutators
• Lie Algebras
• Coherent States
• Displacement Operator
• Spin Rotations

References

• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• N. J. Higham, Functions of Matrices: Theory and Computation, SIAM, 2008.
• 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.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.

Exercises

1. Suppose $[A,B]$ commutes with both $A$ and $B$. What is $e^Ae^B$ as a single exponential?

Solution

Under this central-commutator assumption, all nested commutators beyond $[A,B]$ vanish in the BCH series. Therefore

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

2. Use the similarity formula to show that if $[A,B]=cI$, then $e^A B e^{-A}=B+cI$.

Solution

The similarity expansion is

$$e^A B e^{-A} = B+[A,B] +\frac{1}{2!}[A,[A,B]] +\cdots .$$

If $[A,B]=cI$, then $[A,[A,B]]=[A,cI]=0$. Thus the series truncates:

$$e^A B e^{-A} = B+cI.$$

3. For $D(\alpha)=\exp(\alpha a^\dagger-\alpha^*a)$, derive $D^\dagger(\alpha)aD(\alpha)=a+\alpha$.

Solution

Let $K=\alpha a^\dagger-\alpha^*a$, so $D=e^K$ and $D^\dagger=e^{-K}$. Since

$$[K,a] = \alpha[a^\dagger,a]-\alpha^*[a,a] = -\alpha,$$

one has

$$[-K,a] = \alpha.$$

All higher nested commutators vanish because $\alpha$ is a scalar. Therefore

$$D^\dagger aD = e^{-K}ae^K = a+\alpha.$$

4. Identify the leading BCH correction in one first-order split step with $X=-iH_A\Delta t/\hbar$ and $Y=-iH_B\Delta t/\hbar$.

Solution

BCH gives

$$e^Xe^Y = \exp \left( X+Y+\frac12[X,Y]+O(\Delta t^3) \right).$$

Now

$$[X,Y] = \left[ -\frac{i}{\hbar}H_A\Delta t, -\frac{i}{\hbar}H_B\Delta t \right] = -\frac{\Delta t^2}{\hbar^2}[H_A,H_B].$$

The leading correction in the exponent is therefore

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