Relation between representations of boson operators? I have a simple (I think !) question about the representations of boson operators and how they are related. First of all let's define two conjugate observables $Q$ and $P$ (i.e. $\left[Q,P\right]=i$ and $Q^\dagger=Q$, $P^\dagger=P$).  If we further define:
\begin{equation}
a=\sqrt{\frac{\alpha}{2}}\left(Q+\frac{i}{\alpha}P\right)~~~~~~~~~~~~
a^\dagger=\sqrt{\frac{\alpha}{2}}\left(Q-\frac{i}{\alpha}P\right)~~~~~~~\alpha\in \mathbb{C},
\end{equation}
(as in the harmonic oscillator problem) we have that $\left[a,a^{\dagger}\right]=1$. We can there identify $a^{(\dagger)}$ as boson annihilation (creation) operators. However we can also define:
\begin{equation}
b=\sqrt{Q}e^{iP}~~~~~~~~~~~~~~~~b^\dagger=e^{-iP}\sqrt{Q}
\end{equation}
which will verify $\left[b,b^{\dagger}\right]=1$ (this requires a bit more algebra though).
Question : Is there a relation between these two representations ? These are specific examples, but one could probably think of other representations. Since these representations implement the same commutation relations, does it mean that there are related by some transformation (a unitary transformation in particular) ?
(I give here specific examples for bosonic operators, but I guess one may extend the discussion to any type of operator satisfying some commutation relation).
 A: Your non standard representation does not produce a well-behaved canonical theory. 
The most evident and direct way to face it, barring theoretical remarks based on the absence of rigorous hypotheses sufficient to apply some theorem (Stone von Neumann, Nelson, FS^3, Dixmier...), is the following.
To construct a representation of your bosonic theory (a) you have to build up the orthonormal set  of occupation numbers states $\{|n\rangle\}_{n=0,1,2,\ldots}$ and (b) you have to prove that this set is complete (i.e., maximal)($^*$). 
By definition, where $C_n \neq 0$ is a normalization coefficient:
$$|n\rangle := C_n(b^\dagger)^n|0\rangle \qquad (1)$$
with: 
$$b|0\rangle =0\quad\mbox{and}\quad \langle 0|0\rangle =1\:.\qquad (2)$$
The former equation in (2), making explicit the form of the operator $b$ in the Hilbert space of the theory, $L^2(\mathbb R)$, and writing down the equation using the wavefunction $\psi_0$ of $|0\rangle$ in position representation, reads:
$$\sqrt{x}\psi_0(x+1)=0  \quad \mbox{(almost everywhere)}\:,\qquad (3)$$
where I have exploited the fact that $\{e^{-i\lambda P}\}_{\lambda \in \mathbb R}$ is the unitary representation of the group of $x$-translations.
The only $L^2$ solution of (3) is trivially: $$\psi_0(x) = 0 \quad \mbox{almost everywhere.}$$ Consequently the latter condition in (2) is untenable and all  the construction aborts here.

footnotes
$(^*)$ Technically speaking, these vectors are consequently analytic vectors for all the involved operators and this a guarantee for the validity of several crucial properties like essentially self-adjointness of the new canonical variables.
A: There is only one unitary representation for the algebra of bosonic operators. Given a set of creation and annihilation operators,
$$
[b,b^\dagger] = 1,
$$
you can define a set of canonical position and momentum operators,
$$
[Q,P] = iC,\quad [C,P] = [C,Q] = 0.
$$
which is known as Heisenberg algebra. $C$ is the center of this algebra. There is only one unitary representation for Heisenberg algebra (Stone-von Neumann theorem).
As for the "new representation" you mentioned, those ($P'$ and $Q'$) are just action-angle variables. $Q'$ in the "new representation" is the amplitude of oscillation, whereas $P'$ is roughly the phase angle. More specifically (set $\alpha=1$),
$$
Q = \sqrt{2Q'}\cos(P');\quad P = \sqrt{2Q'}\sin(P').
$$
$P'$ and $Q'$ are the "new" variables. They also form canonical conjugate pair. You can check this for both classical and quantum oscillators.
A: Yes, your b excitations are known: they are tweaked coherent states, based on the Displacement operator of optical phase space. 
For simplicity, take α=2, so that 
$$
Q=\frac{a+a^\dagger}{2}, \qquad iP=a-a^\dagger,
$$
and hence 
$$
b=\sqrt{\frac{a+a^\dagger}{2}} e^{a- a^\dagger}= \sqrt{\frac{a+a^\dagger}{2}} D^\dagger(1),\\  
b^\dagger= e^{a^\dagger -a}   \sqrt{\frac{a+a^\dagger}{2}} =D(1)\sqrt{\frac{a+a^\dagger}{2}} ~,
$$
where the displacement operator is defined as $D(1)= e^{a^\dagger -a}$. 
Then, evidently,
$$[b,b^\dagger]=\frac{a+a^\dagger}{2}   - D(1)\frac{a+a^\dagger}{2}D^\dagger(1)=1  ~. $$
Acting on the Fock vacuum annihilated by a, different α than the above!, the displacement operators 
define the coherent state $D(\alpha=1)|0\rangle=|\alpha=1\rangle$, the eigenstate of the annihilation operator, but I am not sure of the drift of the rest of your question. 
Slight modifications of these maps are popular in deformed oscillator algebras, section 4.g).
