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

Question

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

Qmechanic · Accepted Answer

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.)

Jorge Lavín · Answer

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

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 $i=x,y,z$

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

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