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 $[\vec{S}, \exp(S_z)]$).

  • 3
    $\begingroup$ 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. $\endgroup$ – joshphysics Jan 23 '13 at 23:06
  • 1
    $\begingroup$ $[A,BC]=[A,B]C+B [A,C]$, specifically $[A,B^{n+1}]=[A,B B^n]=[A,B]B^n+B [A,B^n]$ and $f(B)=\sum a_n B^n$. $\endgroup$ – Nikolaj-K Jan 23 '13 at 23:25
  • $\begingroup$ @NickKidman: it'd be a good idea to state what you mean by $f(B)$ and $a_n$... $\endgroup$ – Vibert Jan 23 '13 at 23:48

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


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.)

| cite | improve this answer | |


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 $[\vec{S},e^{S_{z}}]$


for $i=x,y,z$

| cite | improve this answer | |
  • $\begingroup$ Thanks! Is there an expression for $[A, \exp(B)]$ if $[A, B]$ isn't 0? $\endgroup$ – Mark Jan 24 '13 at 0:03
  • $\begingroup$ The first equation Burzum wrote is such an expression... You can simplify or rewrite the infinite series once you know [A,B] $\endgroup$ – zakk Jan 24 '13 at 0:05
  • $\begingroup$ Sorry, what I meant was an expression without infinite sum. E.g. with an expontentiation. Is it reducable to something like that? $\endgroup$ – Mark Jan 24 '13 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.