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).
Best Answer
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.