Annihilation operator in the exponent? My 1D harmonic oscillator of mass $m$ is in a potential $V(x) = \frac{m\omega^2x^2}{2}$. Here's its state: $\left|\psi\right> = Ne^{\lambda a^\dagger}\left|0\right>$, where 
$\lambda$ is given, $\lambda \in \mathcal{C}$, and $a^\dagger$ is a ladder operator, $a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i}{m\omega}\hat{p}\right)$.
I'm asked to normalize it (find $N$), but I don't understand what's the effect of having an $a^\dagger$ in the exponent, would that mean that my state decays in phase space? Or is just a mathematic exercise with no practical physical meaning?
UPD: I don't question the existence of such operator, or mathematical ways to compute $N$. This question is about physical picture of such operator.
 A: The exponential of the creation operator has a lot of physical meaning, the state is very important.
There are several classes of wisdom about it, including


*

*how it's calculated

*what algebraic properties the exponential or its action on a state obeys

*what is the physical importance or interpretation of these things


The exponential of an operator may be calculated e.g. using Taylor series
$$ \exp(X) = \sum_{n=0}^\infty \frac{X^n}{n!}$$
This holds even if $X$ is an operator, e.g. a creation operator. The power $X^n$ is calculated from the usual matrix multiplication (with itself). An alternative calculation is
$$ \exp(X) = \lim_{M\to\infty} (1+X/M)^M $$
which resembles the compound interest in banking.
More general functions of an operator may be found by diagonalizing the operator as 
$$ X = U D U^{-1}$$
and then
$$ f(X) = U f(D) U^{-1}$$
which holds because $U^{-1}$ and $U$ cancel in the power $X^N$, and $f(D)$, a function of a diagonal matrix, is calculated by applying the function $f$ on each diagonal matrix entry.
At any rate, the operator $\exp(\lambda a^\dagger)$ perfectly exists, is calculable, and is a well-defined linear operator on the same Hilbert space.
The normalization $N$ may be calculated if you evaluate the squared norm
Insert e.g. the "limit" form of the exponential for the exponential above. You will have 
$$\langle 0 | (1+\lambda a/N)^N (1+\lambda a^\dagger)^N |0 \rangle$$
in the limit $N\to\infty$. How much is the matrix element above? If you move all the factors $(1+\lambda a/N)^N$ to the right, next to the ket $|0\rangle$, their action on the ket will be simply $1$ and you may forget about those. Similarly with the action of the Hermitian conjugate factors on the bra-vector.
To permute them, you need to exchange the order of the $(1+\lambda a/N)$ and $(1+\lambda a^\dagger/N)$ factors. There are $N^2$ pairs of such factors that you need to exchange and you use
$$ (1+\lambda a/N) (1+\lambda a^\dagger/N) = (1+\lambda a^\dagger/N) (1+\lambda a/N) (1+\lambda^2 [a,a^\dagger]/N^2) $$
where the last multiplicative correction just adjusts the terms of order $1/N^2$, in a $1/N$ expansion. Also, $[a,a^\dagger]=1$. When $N^2$ such permutations are made, the $(1+\lambda^2 [a,a^\dagger]/N^2)$ factors multiplicatively combine to
$$\exp(\lambda^2) $$
by the same limit-based definition of the exponential – the limit $N^2\to \infty$ is the same as $N\to\infty$ and the $1/N^2$ factor cancels.
So the norm is $\exp(\lambda^2)$ and to guarantee the normalization to one, as you need it, you need
$$ N_{\rm yours} = \exp(-|\lambda|^2/2)$$
I added the absolute value because $\lambda$ may be complex. In that case, one-half of the $\lambda$'s above, all the bra-style ones, need to be complex conjugated and we end up with the absolute value.
The exponential of $\lambda a^\dagger$ acting on the vacuum has the effect of moving the Gaussian wave function in the $x$ and $p$ direction of the phase space – according to the real and imaginary part of $\lambda$. The state is known as a coherent state and you may prove that it's an eigenstate of $a$. Just like $a$ itself annihilates $|0\rangle$, its action on the coherent state is still simple, the latter is an eigenstate, and the eigenvalue is $\lambda$ or so (or perhaps its complex conjugate, or it over two etc., find it yourself).
There are lots of calculations to be made here and they're important and may be done in various ways. But I want to stress the more general point. The exponentials of all similar operators are meaningful and very important in quantum mechanics and lots of important laws and states and objects are indeed written in this way. It is not just some silly mathematics where objects are unnaturally substituted where they don't belong. They certainly do belong there and many important if not elementary objects have a similar mathematical form.
A: I did a quick calculation and (using the earlier mentioned relations) I found the following (beware, I may have been sloppy)
$|\psi\rangle=N\sum_n\frac{\sqrt{(n+1)}\lambda^n}{\sqrt{n!}}|n+1\rangle$. Inserting the eigenfunctions of the Hamiltonian
$$
 \psi_n(x) = N_n \cdot e^{
- \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right)
$$
where the normalization constants are 
$$
 \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}
$$
Inserting this in the above sum we find
$$
\begin{align}
 |\psi\rangle=&N\sum_n\frac{\lambda^n}{2^{n/2} n!} \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{
- \frac{m\omega x^2}{2 \hbar}} \cdot H_{n+1}\left(\sqrt{\frac{m\omega}{\hbar}} x \right) \\
=& N e^{\lambda/\sqrt{2}} \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{
- \frac{m\omega x^2}{2 \hbar}} \cdot H_{n+1}\left(\sqrt{\frac{m\omega}{\hbar}} x \right)
\end{align}
$$
So your normalization constant seems to be is
$$
 N = e^{-\lambda/\sqrt{2}} \left(\frac{m\omega}{\pi \hbar}\right)^{-1/4}
$$
