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For a coherent state $$|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}(a^{\dagger})^n}{n!}|0\rangle$$ I want to find a simplified expression for $a^{\dagger}|\alpha\rangle.$ I can only get this $$\begin{align} a^{\dagger}|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}(a^{\dagger})^{n+1}}{n!}|0\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}}{\sqrt{n!}}\sqrt{n+1}|n+1\rangle \end{align}$$ or $$a^{\dagger}|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}a^{\dagger}e^{\alpha a^{\dagger}}|0\rangle.$$ Is it possible to get something more "beautiful" and "useful"?(I apologize for the unscientific lexicon.)

Ultimately, I want to find a simplified expression for $N|\alpha\rangle=a^{\dagger}a|\alpha\rangle,$ but I don't know such an expression for $a^{\dagger}|\alpha\rangle.$

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There's no easy expression for $a^\dagger\vert\alpha\rangle$ but you are interested in $\hat N\vert \alpha\rangle$ the easy way is \begin{align} \hat N\vert\alpha\rangle &= \hat N e^{-\vert\alpha\vert^2/2} \sum_n \frac{\alpha^n}{\sqrt{n!}}\hat N\vert n\rangle\, ,\\ &= \hat N e^{-\vert\alpha\vert^2/2} \sum_n \frac{\alpha^n}{\sqrt{n!}}n\vert n\rangle\, . \tag{1} \end{align} What is simple and useful is $$ \langle \alpha\vert a^\dagger =\alpha^*\langle \alpha\vert \tag{2} $$ obtained by taking the transpose conjugate of $a\vert\alpha\rangle=\alpha\vert\alpha\rangle$. The calculation of $\langle N\rangle$ then easily follows.

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The following expression can sometimes be useful: $$ a^\dagger |\alpha\rangle = \left( \partial_\alpha + \frac{\alpha^\ast}{2} \right) |\alpha\rangle . $$

To prove this, just calculate $$ \partial_\alpha |\alpha\rangle = \partial_\alpha \left( \mathrm e^{-|\alpha|^2 / 2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle \right) $$ using the product rule and $\partial_\alpha |\alpha|^2 = \alpha^\ast$.

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