What if we skip polarization in geometric quantization? In QM, we most frequently work with "position-space" representation of the CCR
$$
\mathcal{H} = L_2(\mathbb{R}, dx), \quad
X = x, \quad
P = - i \hbar \frac{d}{dx}.
$$
Sometimes it's useful to work with the "momentum-space" representation
$$
\mathcal{H} = L_2(\mathbb{R}, dp), \quad
X = i \hbar \frac{d}{dp}, \quad
P = p.
$$
It's a well-known fact that these two representations are unitarily equivalent, the unitary map between the two being the Fourier transform.
In fact, a more general statement is the Stone-von Neumann theorem: any two representations of the CCR are unitarily equivalent (well, actually, of the Weyl algebra, but it's essentially the same thing for the purposes of this question).
I am interested in the following "symmetric" representation of the CCR:
$$
\mathcal{H} = L_2(\mathbb{R}^2, dx dp), \quad
X = x + \frac{i \hbar}{2} \frac{\partial}{\partial p}, \quad
P = p - \frac{i \hbar}{2} \frac{\partial}{\partial x}.
$$
It's straightforward to see that $[X, P] = i \hbar$ is satisfied.
My questions are:

*

*How can I build an isomorphism between this "symmetric" representation and the "position-space" representation the existence of which is guaranteed by the Stone-von Neumann theorem?

*Does this isomorphism have any physical significance that's noteworthy?


What follows is some (optional) background on why this representation is interesting to me.
This formula is inspired by Geometric Quantization, where we build the "pre-quantum Hilbert space" before we choose the polarization to cut down the "excessive size". In this example, polarisation would account for only letting the wavefunction depend on say $x$, and adjusting the operators accordingly.
Since geometric quantization is most usually applied to compact manifolds (sphere, torus, etc.) where the quantum theory acts on a finite dimensional vector space, polarization is an important step to have the "correct" quantum theory.
However, for the case of a plane $(x,p)$, both the pre-quantum and the polarised spaces are isomorphic, because they both are just the abstract Hilbert space. Furthermore, both the prequantum operators and polarised operators give representations of the CCR which must be equivalent due to the Stone-von Neumann theorem.
This seems to suggest that polarisation is somehow not necessary when working with non-compact phase spaces?
 A: Whenever you have two representations $(V_1,\pi_1)$ and $(V_2,\pi_2)$ of a group or algebra and an isomorphism $\phi : V_1 \to V_2$, then for any $g$ in the group or algebra that isomorphism must send eigenvectors of $\pi_1(g)$ to eigenvectors of $\pi_2(g)$ with the same eigenvalues by definition of an isomorphism of representations (this is the "intertwiner" part).
So when the eigenstates of $\pi_i(g)$ are non-degenerate and a basis of $V_i$, this already uniquely fixes the isomorphism because for each eigenvector there is only a single possible target. There is a slight subtlety here because $x$ and $p$ only have generalized eigenstates, but the same idea still works: Any $v_1\in V_1$ has a representation as
$$ v_1 = \int f(x)\lvert x\rangle_1\mathrm{d}x$$
where the $\lvert x\rangle_1$ are eigenstates of $\pi_1(x)$ and this gets sent to
$$ \phi(v_1) = \int f(x)\lvert x\rangle_2\mathrm{d}x,$$
i.e. you're just matching wavefunctions. Another way to say this is that you can always construct an isomorphism to the default representation on $L^2(\mathbb{R})$ where $x$ is multiplication by mapping any state $\lvert \psi\rangle$ to its position wavefunction $\psi(x) = \langle x\vert \psi\rangle$, and then concatenating these isomorphisms gives you isomorphisms between any two representations of the CCR.

Note that the above requires that you already know the isomorphism $\phi$ should exist. Your symmetric representation is a representation of the CCR, but in order for us to know that it should be isomorphic to the standard representation on $L^2(\mathbb{R},\mathrm{d}x)$ via Stone-von Neumann you also need to know that it is irreducible. But this is not the case here: You have four operators $x,\partial_x,p,\partial_p$, and so besides your $X = x+\frac{1}{2}\mathrm{i}\partial_p$ and $P = p - \frac{1}{2}\mathrm{i}\partial_x$, there are also the linearly-independent operators $X' = x-\frac{1}{2}\mathrm{i}\partial_p$ and $P' = p + \frac{1}{2}\mathrm{i}\partial_x$ which also fulfill $[X',P'] = [x-\mathrm{i}\partial_p, p + \mathrm{i}\partial_x] = \mathrm{i}$.
So what you have here is a representation of a two-dimensional CCR algebra, and in particular the generalized eigenstates of $X$ or $X'$ alone are not non-degenerate. For $X,X'$ together, you get non-degenerate eigenstates $\lvert x,x'\rangle$, and then the argument from above gives you maps to wavefunctions $\psi(x,x')$ that allow you to map this to the standard position representation $L^2(\mathbb{R}^2,\mathrm{d}x_1\mathrm{d}x_2)$ of the CCR in two spatial dimensions.
A: *

*Concerning OP's 3 representations$^1$ of the Heisenberg algebra, we can put them on equal footing as follows:
$${\cal A}(x,\partial_x)~\cong~ \{ A\in {\cal A}(x,p,\partial_x,\partial_p)\mid [A,p]~=~0 ~\wedge~ [\partial_p,A]~=~0\}, $$
$$ {\cal A}(p,\partial_p)~\cong~ \{ A\in {\cal A}(x,p,\partial_x,\partial_p)\mid [A,x]~=~0 ~\wedge~ [\partial_x,A]~=~0\}, $$
$$\begin{align} & {\cal A}(x+ \frac{i \hbar}{2}\partial_p,p- \frac{i \hbar}{2}\partial_x)\cr
~\cong~& \{ A\in {\cal A}(x,p,\partial_x,\partial_p)\mid [A,x- \frac{i \hbar}{2}\partial_p]~=~0 ~\wedge~ [A,p+ \frac{i \hbar}{2}\partial_x]~=~0\}. \end{align}$$
These algebras are isomorphic, which can be seen via appropriate change of coordinates.


*Similarly, we can consider Poisson algebras ${\cal A}(x^1,x^2,p_1,p_2)$ with 2 second-class constraints. The $L^2$ Hilbert spaces of sough-for Schrödinger representations/polarizations is in this setting then a matter of choosing$^2$ a Lagrangian subspace, possibly via appropriate change of coordinates.
--
$^1$ Here ${\cal A}(t_1,\ldots,t_n)$ denotes the algebra generated by the elements $t_1,\ldots,t_n$.
$^2$ We should here choose 1 coordinate out of 4.
