Quantum Mechanics: Creation and Annihilation operators Is an eigenvector/eigenstate of the creation operator an eigenvector/eigenstate of the annihilation operator too? Why? 
 A: Here's a simple argument showing that $a$ and $a^\dagger$ cannot have a common eigenvector using only the commutation relation between them.  
Suppose,by way of contradiction, there existed a vector that were a common eigenvector of both, namely a nonzero vector $|\psi\rangle$ such that
\begin{align}
  a|\psi\rangle &= \alpha|\psi\rangle, \\
  a^\dagger|\psi\rangle &= \beta|\psi\rangle
\end{align}
for some complex numbers $\alpha$ and $\beta$.  Then we would have
\begin{align}
  [a,a^\dagger]|\psi\rangle = (aa^\dagger-a^\dagger a)|\psi\rangle &= (\beta\alpha-\alpha\beta)|\psi\rangle=0.
\end{align}
On the other hand, recall that
\begin{align}
  [a,a^\dagger] = I,
\end{align}
so that
\begin{align}
  [a,a^\dagger]|\psi\rangle = |\psi\rangle.
\end{align}
Putting these facts together gives
\begin{align}
  |\psi\rangle = 0,
\end{align}
a contradiction.
A: An eigenfunction of the creation operator would be a state that satisfies $a^\dagger |\psi\rangle \propto |\psi\rangle$. Think about the consequences of this for a creation operator.
EDIT:
Since this got downvoted, let me be more specific.
Suppose $|\psi\rangle = \sum_{n=0}^\infty c_n |n\rangle$, and we require that
$a^\dagger |\psi\rangle =\lambda |\psi\rangle$. Then
$a^\dagger |\psi\rangle = \sum_{n=0}^\infty c_n \sqrt{n+1}|n+1\rangle$.
But the RHS contains no $|0\rangle$. Do you see the problem?
EDIT 2:
What I wrote above and in the comments is not quite complete. The annihilation operator can have an eigenstate. The state 
$|\alpha\rangle = e^{-\frac{|\alpha|}{2}^2} e^{\alpha a^{\dagger}}  \left|0\right\rangle$ is a coherent state with eigenvalue $\alpha$.
