part a) is fine. part b) is not.
A commutator is defined as, for operators $A$ and $B$, $[A,B]=AB-BA$.
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1$\begingroup$ The following may help: $D$ and $e^{-\lambda D}$ commute. This is because the terms in the expansion of $D e^{-d\lambda D}$ consists of a scalar times a power of $D$, so you can factor $D$ on either the right or the left. $\endgroup$ – Hugh Denoncourt May 18 '14 at 15:05
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1$\begingroup$ Also, the expansion in $\lambda$ consists of evaluating $H(0)$, $H'(0)$, etc, and writing $H(\lambda) = \sum_{n=0}^\infty \frac{H^{(n)}(0)}{n!} \lambda^n$. Since the desired answer stops at the $\lambda$ term, we expect (or at least hope!) the second derivative (and hence all remaining derivatives) to be 0. So, you likely need to use $[C, [C,D]] = 0$ or $[D,[C,D]] = 0$ to show that. $\endgroup$ – Hugh Denoncourt May 18 '14 at 15:12
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1$\begingroup$ No, $H'(0)$ isn't $0$, since $e^0$ is the identity operator. By commuting, I mean that $D e^{-\lambda D} = e^{-\lambda D} D$. If you are wondering what I mean by expansion, I mean this: $De^{-\lambda D} = D - \lambda D^2 + \lambda^2 / 2 D^3 +...$ You can factor this as $D(I - \lambda D + ...)$ or as $(I - \lambda D + ...) D$. This is why the two operators commute. $\endgroup$ – Hugh Denoncourt May 18 '14 at 16:48
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1$\begingroup$ It's not that $H'(\lambda) = -DC + CD$; that's not true according to your first calculation. It's that $H'(0) = -DC + CD$. $\endgroup$ – Hugh Denoncourt May 18 '14 at 16:50
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1$\begingroup$ Good point. I suppose we don't need the commuting operator. We just need to evaluate at $\lambda = 0$ to see it. I saw it easier after rearranging and applying commutativity where I could. $\endgroup$ – Hugh Denoncourt May 18 '14 at 16:52
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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 $H(\lambda)=C+\lambda[C,D]$
