Three Commutator Identities for a Constant Commutator

These three identities show up constantly in quantum mechanics — in the derivation of coherent states, the Baker–Campbell–Hausdorff formula, time evolution of observables, and Weyl quantization. The proofs are elegant enough to be worth writing down carefully.

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 key 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 the operator exponential commutes with its own derivative in the exponent — it behaves just like a scalar exponential in this regard.

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 $x = 1$ :

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

Part 2: $e^A B^n e^{-A} = (B+c)^n$ #

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 Part 1, giving $n$ such factors:

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

Part 3: $e^A e^B e^{-A} e^{-B} = e^c$ #

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 Part 2:

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

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.

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

Why do these come up?#

Part 1 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 $B$ — and the deviation from $B$ is exactly the commutator structure.

Part 3 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 $D(\alpha) = e^{\alpha a^\dagger - \alpha^* a}$ . Part 1 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. Part 3 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.