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I am working on a problem involving ladder operators and coherent states. I know that $$a|z\rangle=z|z\rangle$$ and $$\langle z|a^\dagger = \langle z|z^*.$$ I am wondering how I could figure out what $a^\dagger|z\rangle$ is?

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  • $\begingroup$ There is no “clean” form: must use the expansion of $\vert \alpha\rangle$ in terms of number states $\vert n\rangle$. $\endgroup$ – ZeroTheHero Jun 26 '19 at 23:17
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You might, or might not, go for $$ |z\rangle= e^{-|z|^2/2} e^{za^\dagger}|0\rangle\qquad \Longrightarrow \\ a^\dagger |z\rangle= e^{-|z|^2/2} \frac{\partial}{\partial z} \left( e^{|z|^2/2} |z\rangle \right)\ . $$

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    $\begingroup$ Very clean answer. $\endgroup$ – gented Jun 27 '19 at 10:02
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Another idea is as follows:

The coherent states are defined to be the eigenstate of annihilation operator : $\hat a \vert z\rangle=z\vert z \rangle$. And ladder operator is $\hat D(z)$ which is defined $\hat D(z)=e^{z\hat a^\dagger - z^* \hat a}$ and it makes $\vert z \rangle=\hat D(z)\vert 0 \rangle$.

Using the representation of the coherent state in the basis of Fock states:From Wiki $$ \vert z \rangle=e^{-\frac{\lvert z \rvert^2}{2}}e^{z\hat a^\dagger}\vert 0 \rangle=e^{-\frac{\lvert z \rvert^2}{2}}\sum_{n=0}^\infty \frac{(z\hat a^\dagger)^n}{n!} \vert 0 \rangle=e^{-\frac{\lvert z \rvert^2}{2}}\sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}} \vert z \rangle$$

So: $$ \hat a^\dagger \vert z\rangle = e^{-\frac{\lvert z \rvert^2}{2}} \sum_{n=0}^\infty \frac{z^n {\hat a^\dagger}^{n+1}}{n!} \vert 0 \rangle = \frac{\partial}{\partial z}(e^{-\frac{\lvert z \rvert^2}{2}} \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle) -\frac{\partial}{\partial z} e^{-\frac{\lvert z \rvert^2}{2}}\cdot \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle $$ Because $$e^{-\frac{\lvert z \rvert^2}{2}} \cdot \frac{\partial}{\partial z} \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle= e^{-\frac{\lvert z \rvert^2}{2}} \cdot \sum_{n=0}^\infty \frac{z^n {\hat a^\dagger}^{n+1}}{n!} \vert 0 \rangle $$ And: $$ \frac{\partial}{\partial z}(e^{-\frac{\lvert z \rvert^2}{2}} \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle) =\frac{\partial}{\partial z}(e^{-\frac{\lvert z \rvert^2}{2}} \sum_{n=-1}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle - e^{-\frac{\lvert z \rvert^2}{2}}\vert 0 \rangle) =\frac{\partial}{\partial z} \vert z\rangle+\frac{z^*}{2} e^{-\frac{\lvert z \rvert^2}{2}}\vert 0 \rangle $$ $$ \frac{\partial}{\partial z} e^{-\frac{\lvert z \rvert^2}{2}}\cdot \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle =-\frac{z^*}{2}e^{-\frac{\lvert z \rvert^2}{2}}\cdot \sum_{n=0}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle =-\frac{z^*}{2}e^{-\frac{\lvert z \rvert^2}{2}}\cdot \sum_{n=-1}^\infty \frac{z^{n+1} {\hat a^\dagger}^{n+1}}{(n+1)!} \vert 0 \rangle+\frac{z^*}{2}e^{-\frac{\lvert z \rvert^2}{2}} \vert 0 \rangle $$

Therefore:$\hat a^\dagger \vert z \rangle = (\frac{\partial}{\partial z}+\frac{z^*}{2})\vert z \rangle$.

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