Eigenvalue for the creation operator for a coherent state For a coherent state 
$$
|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}|n\rangle
$$
I can't solve the eigenvalue problem for $\hat{a}^{\dagger}|\alpha\rangle$ where $\hat{a}^{\dagger}$ is the creation operator. I can only get this far
$$
\begin{align}
\hat{a}^{\dagger}|\alpha\rangle&=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\hat{a}^{\dagger}|n\rangle\\
&=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n}\frac{\alpha^{n}}{\sqrt{n!}}\sqrt{n+1}|n+1\rangle
\end{align}
$$
Ultimately, I want to calculate $\langle \alpha |a\hat{a}^{\dagger}|\alpha\rangle$, but I don't know $\hat{a}^{\dagger}|\alpha\rangle$.
 A: Using the definition of the creation operator, $a^\dagger = c(m\omega \hat x - i\hat p)$ where $c$ is a constant, and $\hat p = -i\hbar\partial_x$, you can write the eigenvalue problem in the position representation as $$(m\omega x - \hbar\partial_x)\psi = \alpha\psi.$$
You can solve this differential equation to find $$\psi = C\exp(m\omega x^2/\hbar - \alpha x/\hbar)$$
which is clearly not normalizable. Hence the creation operator has no normalizable eigenstates.
A: To add to Innisfree's correct answer, I'd like to emphasize something that the OP does not seem to know and that is that the creation operator has no eigenvectors (nor, therefore, eigenvalues). It is easy to see this: write a general state as a row vector $(\psi_0,\,\psi_1,\,\cdots)$ of superposition weights for the number states $|0\rangle,\,|1\rangle,\,\cdots$ and in this notation, our eigenvalue equation (in $\lambda$) for $a^\dagger$ is: 
$$a^\dagger (\psi_0,\,\psi_1,\,\cdots) =(0,\,\psi_0,\,\sqrt{2} \psi_1,\,\sqrt{3} \psi_2,\,\cdots) =  \lambda (\psi_0,\,\psi_1,\,\cdots)$$
whence we get $\lambda \,\psi_n=\sqrt{n}\psi_{n-1}$ and $\lambda\,\psi_0=0$. If $\lambda = 0$ it follows straight away that $\psi_n=0\,\forall\,n\in\mathbb{N}$. If $\lambda\neq 0$, then $\psi_0=0$, whence (by induction through $\psi_n = \sqrt{n}\psi_{n-1}/\lambda$) $\psi_n=0\,\forall\,n\in\mathbb{N}$. There is therefore no normalizable superposition of number states that is an eigenvector for $a^\dagger$. It's therefore not surprising that the OP was having difficulty!
A: A coherent state is, amongst other interesting things, an eigenstate of the annihilation operator. It is not an eigenstate of the creation operator; hence, I'm not sure this "eigenvalue problem" makes much sense.
This is easy to realize. You can quickly see that $\langle0|a^\dagger|\alpha\rangle=0$, whereas $\langle0|\alpha\rangle\neq0$.
If you really want to find $\langle \alpha | a a^\dagger|\alpha\rangle
$ in e.g. 
$$
\langle x^2 \rangle \propto \langle \alpha | (a+a^\dagger)(a+a^\dagger)|\alpha\rangle
$$
you can commute the operators $a$ and $a^\dagger$ with the rule $[a,a^\dagger] = 1$, such that
$$
\langle \alpha | a a^\dagger|\alpha\rangle = \langle \alpha | a^\dagger a|\alpha\rangle + 1 = 1 + |\alpha|^2
$$
You can also verify this the long way around by acting the the operators.
