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

Question

$A,B$ are quantum mechanical operators.

• $[A,B]\neq 0$ that is given.

• $e^{A}=\sum_{n=1}^{\infty} \frac{A^n}{n!} $

Is the following correct?

$$[e^{A},e^{B}]=e^{A}e^{B}-e^{B}e^{A}=e^{A+B}-e^{B+A}=0 $$

Answer 1

An even simpler example: \begin{align} s_z=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)\, ,& \qquad s_y=\left( \begin{array}{cc} 0 & i \\ -i & 0 \\ \end{array} \right)\\ e^{-i\alpha s_z}=\left( \begin{array}{cc} e^{-i \alpha } & 0 \\ 0 & e^{i \alpha } \\ \end{array} \right)&\qquad e^{-i\beta s_y}=\left( \begin{array}{cc} \cos (\beta ) & \sin (\beta ) \\ -\sin (\beta ) & \cos (\beta ) \\ \end{array} \right) \end{align} and clearly $[e^{-i\alpha s_z} ,e^{-i\beta s_y}]\ne 0$.

Note that, in connection to the comments on the answer of @user124864, $e^{-i\alpha s_z}e^{-i\beta s_y}\ne e^{-i\alpha s_z-i\beta s_y-\frac{1}{2}\alpha\beta[s_z,s_y]}$ either.

You can easily verify using the first few terms of the explicit expansion that, in general $$ e^A e^B= \sum_n \frac{A^n}{n!}\,\sum_m \frac{B^m}{m!} \ne e^{A+B}=\sum_{p}\frac{(A+B)^p}{p!}\, . $$ If anything: \begin{align} e^Ae^B &=\left(1+A+\frac{A^2}{2!}+\ldots \right) \left(1+B+\frac{B^2}{2!}+\ldots \right)\\ &= 1+ (A+B)+ \frac{1}{2!} (A^2+2AB+B^2)+\ldots \tag{1} \end{align} but $$ (A^2+2AB+B^2)\ne (A+B)^2= A^2+AB+BA+B^2 \tag{2} $$ unless $AB=BA$, i.e. unless $[A,B]=0$.

Answer 2

No this is not correct. The problem is that you have used $e^Ae^B = a^{A+B}$. This holds when $A$ and $B$ are numbers, so you might expect that it would also holds for matrices or even general operators. This is not in general the case, as the other answers show explicitly.