# Commutator with exponential $[A, \exp(B)]$

How can I tell if $A$ and $\exp(B)$ commute?

For $[A, B]$ it's simply $AB-BA$ and for $[\exp(A), \exp(B)]$ I think it'd be $\exp(A)\exp(B) - \exp(B)\exp(A) = \exp(A+B) - \exp(B+A) = 0$. Update: it's not generally true.

Is there a 'simple' way to find $[A, \exp(B)]$?

Or is this one of those problems where, if you encounter them at all, you are probably doing something wrong? The example I am encountering is $[\vec{S}, \exp(S_z)]$).

• Note that it is NOT true in general that for matrices $A$ and $B$, $e^Ae^B=e^{A+B}$. This is true provided $A$ and $B$ commute. I'm not sure if there is any simple formula for [A, e^B] for arbitrary matrices, but I doubt that there is. Commented Jan 23, 2013 at 23:06

If OP wants to evaluate $[A,e^B]$ in terms of $[A,B]$, there is a formula

$$\tag{1} [A,e^B] ~=~\int_0^1 \! ds~ e^{(1-s)B} [A,B] e^{sB}.$$

Proof of eq.(1): The identity (1) follows by setting $t=1$ in the following identity

$$\tag{2} e^{-tB} [A,e^{tB}] ~=~ \int_0^t\!ds~e^{-sB}[A,B]e^{sB} .$$

To prove equation (2), first note that (2) is trivially true for $t=0$. Secondly, note that a differentiation wrt. $t$ on both sides of (2) produces the same expression

$$\tag{3} e^{-tB}[A,B]e^{tB},$$

where we use the fact that

$$\tag{4}\frac{d}{dt}e^{tB}~=~Be^{tB}~=~e^{tB}B.$$

So the two sides of eq.(2) must be equal.

Remark: See also this related Phys.SE post. (It is related because $[A, \cdot]$ acts as a linear derivation.)

$$[A,e^{B}]=[A,\sum_{i=0}^{\infty}\frac{B^{i}}{i!}]=\sum_{i=0}^{\infty}\frac{[A,B^{i}]}{i!}$$

so in order to $A$ and $e^B$ to commute, $A$ should commute with $B$ and hence with any power of $B$. You can apply this to $[\vec{S},e^{S_{z}}]$

$$[S_{i},e^{S_{z}}]=[S_{i},\sum_{j=0}^{\infty}\frac{S_{z}^{j}}{j!}]=\sum_{j=0}^{\infty}\frac{[S_{i},S_{z}^{j}]}{j!}$$

for $i=x,y,z$

• Thanks! Is there an expression for $[A, \exp(B)]$ if $[A, B]$ isn't 0?
– Mark
Commented Jan 24, 2013 at 0:03