Three Commutator Identities for a Constant Commutator

These three identities show up all the time in quantum mechanics: in coherent states, in the Baker–Campbell–Hausdorff formula, in the time evolution of observables and in Weyl quantization. The proofs are short, but there is one trick in them (turning the identity into an ODE) that is worth knowing on its own, so I write them out carefully here.

Setup#

Let $A$ and $B$ be operators satisfying

[A, B] = c

where $c$ is a c-number (a constant, not an operator). We want to prove:

$e^A B e^{-A} = B + c$
$e^A B^n e^{-A} = (B + c)^n$
$e^A e^B e^{-A} e^{-B} = e^c$

The trick: turn it into an ODE#

The standard approach for identities of this type is to embed the LHS into a one-parameter family, differentiate, and solve the resulting differential equation.

Define

\mathcal{F}(x) = e^{xA}\, B\, e^{-xA}

so that $\mathcal{F}(0) = B$ (the initial condition) and $\mathcal{F}(1)$ is exactly what we want to prove equals $B + c$ .

Differentiating an operator exponential#

First, a small but important fact. For any operator $A$ ,

\frac{d}{dx} e^{xA} = \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^n}{n!} A^n = \sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!} A^n = A\, e^{xA} = e^{xA} A

So $A$ commutes with $e^{xA}$ , and the operator exponential differentiates exactly like a scalar one (it doesn’t matter whether $A$ ends up on the left or the right).

Solving for $\mathcal{F}(x)$ #

Differentiating $\mathcal{F}(x)$ :

\frac{d\mathcal{F}}{dx} = e^{xA} A B\, e^{-xA} - e^{xA} B A\, e^{-xA} = e^{xA} [A, B]\, e^{-xA}

Since $[A, B] = c$ is a constant, it commutes with everything:

\frac{d\mathcal{F}}{dx} = e^{xA}\, c\, e^{-xA} = c

Integrating both sides from $0$ to $x$ :

\mathcal{F}(x) = \mathcal{F}(0) + cx = B + cx

Setting $x = 1$ :

\boxed{e^A B e^{-A} = B + c}

Powers of $B$ : $e^A B^n e^{-A} = (B+c)^n$ #

Insert $e^{-A}e^A = \mathbf{1}$ between each factor of $B$ :

e^A B^n e^{-A} = e^A B \underbrace{e^{-A} e^A}_{1} B \underbrace{e^{-A} e^A}_{1} \cdots B\, e^{-A}

Each adjacent pair $e^A B e^{-A}$ is replaced by $(B + c)$ from the first identity, giving $n$ such factors:

\boxed{e^A B^n e^{-A} = (B+c)^n}

Exponentials: $e^A e^B e^{-A} e^{-B} = e^c$ #

Expand $e^B$ as a power series and sandwich it:

e^A e^B e^{-A} = e^A \left(\sum_{n=0}^{\infty} \frac{B^n}{n!}\right) e^{-A} = \sum_{n=0}^{\infty} \frac{e^A B^n e^{-A}}{n!}

Applying the second identity:

= \sum_{n=0}^{\infty} \frac{(B+c)^n}{n!} = e^{B+c}

So:

e^A e^B e^{-A} e^{-B} = e^{B+c} e^{-B} = e^{B+c-B} = e^c

where the last step uses the fact that $B+c$ and $B$ commute (their commutator is zero), so the exponentials combine freely.

\boxed{e^A e^B e^{-A} e^{-B} = e^c}

Where these come up#

The first identity is the infinitesimal generator version of a similarity transformation. If $A = -i\theta \hat{n} \cdot \vec{J}/\hbar$ is a rotation generator, $e^A B e^{-A}$ gives the rotated observable, and the deviation from $B$ is exactly the commutator structure.

The third identity says that the group commutator $e^A e^B e^{-A} e^{-B}$ equals $e^c$ . For general $A, B$ that’s only the leading-order term of the Baker–Campbell–Hausdorff formula; here it’s exact, because $c$ commutes with everything.

Both show up in coherent states. For the harmonic oscillator with $[a, a^\dagger] = 1$ , take the displacement operator $D(\alpha) = e^{\alpha a^\dagger - \alpha^* a}$ . The first identity with $A = -(\alpha a^\dagger - \alpha^* a)$ and $B = a$ (so $[A, B] = \alpha$ ) gives $D^\dagger a D = a + \alpha$ . So $D(\alpha)|0\rangle$ is an eigenstate of $a$ with eigenvalue $\alpha$ , which is exactly a coherent state. The third one then gives the Weyl relation $D(\alpha)D(\beta)D(\alpha)^{-1}D(\beta)^{-1} = e^{\alpha\beta^* - \alpha^*\beta}$ : displacements commute only up to a phase.