Coherent States and ladder operators 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?
 A: You might, or might not,
go for 
$$
|z\rangle= e^{-|z|^2/2} e^{za^\dagger}|0\rangle\qquad \Longrightarrow  \\
a^\dagger |z\rangle= e^{-|z|^2/2} \frac{\partial}{\partial  z} \left( e^{|z|^2/2} |z\rangle \right)\ .
$$
A: 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:$\hat a^\dagger \vert z \rangle = (\frac{\partial}{\partial z}+\frac{z^*}{2})\vert z \rangle$.
