Given the following pair of operators $a$ and $a^{\dagger}$ that satisfy the usual bosonic CCR:
$$[a,a]=[a^{\dagger},a^{\dagger}] = 0;\ [a,a^{\dagger}] = 1$$
For what values of $\alpha \in\mathbb C$ are the following expressions well defined?
$$a^{\alpha} \ \text{and}\ (a^{\dagger})^{\alpha}$$
For integer $\alpha$, I know that these expressions are well defined, but I am interested in knowing what kind of constraints are imposed on $\alpha$, if such an operator must be appropriately definable. If the above expressions can be defined in precise manner, then please also provide the appropriate construction of these operators.

Some meandering background:
The actual problem I'm trying to solve has nothing to do with quantum mechanics! It is instead a reformulation of the dynamical equations of a polymer (hydrodynamically interacting with a solvent) using the rouse modes as "bosons" [Ref]. So there are no nice hamiltonians and unitary operators to work with. The dynamical equation is of the following form:
$$\dfrac{\partial{\left|\rho\right\rangle}}{\partial t}+\mathcal{L}\left|\rho\right\rangle=0$$
Now the operator $\mathcal{L}$ has a representation in the boson operators as
$$\mathcal{L}=g(a^{\dagger}a+\zeta a^{\dagger}a^{\dagger})$$
where $g$ and $\kappa$ are real constants. The kets live in a standard bosonic Fock space, and a complete basis of nuber states is present constructed in a fashion equivalent to the case of the harmonic oscillator. On trying to solve the eigenvalue equation for $\mathcal{L}$, one can write down the eigenstates, indexed by the eigenvalue as
$$\left|\psi;\lambda\right\rangle=exp\left(-\frac{\zeta}{2}a^{\dagger}a^{\dagger}\right)(a^{\dagger})^{\lambda}\left|0\right\rangle$$
such that $\mathcal{L}\left|\psi;\lambda\right\rangle=\lambda\left|\psi;\lambda\right\rangle$. This is where my original question comes into the forefront - Deciding on how these operators behave on being raised to arbitrary powers, I can fix the possible eigenvalues of $\mathcal{L}$. If it is possible to arrive at a conclusion on the eigenvalues of the given $\mathcal{L}$ operator in any other fashion, without resorting to this approach, then please go ahead and tell me.