Creation Operator acting on Coherent State I am trying to prove that $$a^\dagger |\alpha\rangle\langle\alpha|=\left(\alpha^*+\frac{\partial}{\partial \alpha}\right)|\alpha\rangle\langle\alpha|.$$
The way I calculate this is, is just by acting with the creation operator on the coherent state.
When I do this I get:
$$e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{(n+1)\alpha^n}{\sqrt{(n+1)!}}|n+1\rangle.$$
Since $\frac{\partial(\alpha^{n+1})}{\partial\alpha}=(n+1)\alpha^n$ I can write this as:
$$e^{-|\alpha|^2/2}\frac{\partial}{\partial\alpha}\sum_{n=0}^{\infty}\frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle.$$
One can then move the derivative outside to the very beginning and apply the product rule and that is how  one would get the first part of the solution, however the rest is not a coherent state, because it doesn't have the vacuum state in the summation.
I hope somebody can help with this.
 A: The missing term in the summation containing the vacuum state is independent of $\alpha$, isn't it?
In other words, $$\frac{\partial}{\partial \alpha} \sum_{n=0}^\infty \frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle = \frac{\partial}{\partial \alpha} \left( \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle - |0\rangle \right) = \frac{\partial}{\partial \alpha} \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle $$
Does that help you get your answer?
A: Starting from the last equation and taking into account the braket one gets:
$$e^{-|\alpha|^2}\frac{\partial}{\partial\alpha}\sum_{n=0}^{\infty}\frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|$$
Relabeling the index according to $l=n+1$:
$$e^{-|\alpha|^2}\frac{\partial}{\partial\alpha}\sum_{l=1}^{\infty}\frac{\alpha^{l}}{\sqrt{l!}}|l\rangle\langle\alpha|\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|$$
the first term looks very close to the coherent state, however the summation doesn't have the vacuum in it. So we can add that term and subsequenty subtract it so that we get:
$$e^{-|\alpha|^2}\frac{\partial}{\partial\alpha}\left((\sum_{l=0}^{\infty}\frac{\alpha^{l}}{\sqrt{l!}}|l\rangle-|0\rangle)\right)\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|$$
Since the vacuum term is independent of $\alpha$ and we are differentiating wrt $\alpha$ we can omit this term, as Philip wrote. Doing that and moving the derivative of the beginning and using the product rule one gets:
$$\frac{\partial}{\partial\alpha}e^{-|\alpha|^2}\left(\sum_{l=0}^{\infty}\frac{\alpha^{l}}{\sqrt{l!}}|l\rangle\right)\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|+\alpha^*e^{-|\alpha|^2}\left(\sum_{l=0}^{\infty}\frac{\alpha^{l}}{\sqrt{l!}}|l\rangle\right)\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|$$ The summations including the exponential fore factor are representations of the coherent sate so one finally arrives at:
$$\left(\alpha^*+\frac{\partial}{\partial\alpha}\right)|\alpha\rangle\langle\alpha|$$
as desired
