# 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.

• The derivative on the RHS of the target identity (as written) acts on both the ket and the bra. If you don't include that then there's no chance of equating the two sides. Aug 13 '20 at 9:29
• @EmilioPisanty Aha. Ok that would make sense. I will try it out. thanks a lot! Aug 13 '20 at 9:31
• @EmilioPisanty I was able to solve it. Thanks. But the derivative wrt $\alpha$ doesn't really act on the bra, because it only depends on $\alpha^*$. Nevertheless one needs to take the bra into account, otherwise one gets a factor of 1/2 in the first term of the identity. Thanks again. Aug 13 '20 at 10:11
• If you've solved your problem, you should post your solution as an answer. Aug 13 '20 at 10:23
• @EmilioPisanty Ok, I will do that. Give me some time. Aug 13 '20 at 10:26

The missing term in the summation containing the vacuum state is independent of $$\alpha$$, isn't it?
In other words, $$\frac{\partial}{\partial \alpha} \sum_{n=0}^\infty \frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle = \frac{\partial}{\partial \alpha} \left( \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle - |0\rangle \right) = \frac{\partial}{\partial \alpha} \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle$$
• I actually found another way around that problem, which I find more elegant. If one doesn't introduce the derivative one has that prefactor $n+1$. Since this factor vanishes for $n=-1$ one can then change the summation from $-1$ to $\infty$. Then one can relabel the indices and apply the derivative. Works out really nicely. Aug 14 '20 at 9:33
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