part a) is fine. part b) is not.
A commutator is defined as, for operators $A$ and $B$, $[A,B]=AB-BA$.
part a) is fine. part b) is not.
A commutator is defined as, for operators $A$ and $B$, $[A,B]=AB-BA$.
I get that,
$H(\lambda)=e^{-\lambda D}Ce^{\lambda D}$,
$H'(\lambda)=-De^{-\lambda D}Ce^{\lambda D}+e^{-\lambda D}CDe^{\lambda D}$,
$H''(\lambda)=D^2e^{-\lambda D}Ce^{\lambda D}-2De^{-\lambda D}CDe^{\lambda D}+e^{-\lambda D}CD^2e^{\lambda D}$.
Now, we know that $D$ and $e^{-\lambda D}$ commute and we have that $H(\lambda) = \sum_{n=0}^\infty \frac{H^{(n)}(0)}{n!} \lambda^n$
$H(0)=C$,
$H'(0)=-DC+CD=[C,D]$,
$H''(0)=D^2C-2DCD+CD^2=DDC-DCD-DCD+CDD=D[D,C]-[D,C]D=[D,[D,C]]=0$
So $H(\lambda)=C+\lambda[C,D]$