In my lectures, the professor discussed that for exponential linear operators it is $$ \exp(\lambda A + \lambda B) \neq \exp(\lambda A)\exp(\lambda B) $$ for $AB\neq BA$.
Now I know that the exponential operator has a similar definition as the exponential function. So there must be also the notion of a Taylor series for operators.
Let us assume that $\lambda \ll 1$ such that we can neglect $O(\lambda^3)$ but we have to keep to second order in the real number $\lambda$. If the exponent of the exponential operator consists of only one small operator with the form $$ \exp(A+\lambda B) $$ can we then expand to second order in $\lambda$ such that $$ \exp(A+\lambda B)\approx \exp(A) + \exp(A)\lambda B + \frac{1}{2}\exp(A) \lambda^2 B^2 + .....? $$ I believe that this is not the case because we also have $$ \exp(A)\exp(\lambda B)\approx \exp(A) (I+\lambda B + \frac{1}{2} \lambda^2 B^2 + .....) $$ which is the same as above and that contradicts $AB\neq BA$.
One can also apply the definition of the exponential operator directly by $$ \exp(A+\lambda B) = \sum_{n=0}^\infty \frac{(A+\lambda B)^n}{n!} $$ but I am not sure how to decompose the "$A$" part (assuming we can neglect $O(\lambda^3)$) and also how the Taylor expansion would work applied to such an operator without bumping into that contradiction.
Thanks in advance!