Quantization of $c$-number Dirac-Bracket I have a question concerning the quantization of phase-space variables $(q_1, q_2, q_3, p_1, p_2, p_3)$ with the Hamiltonian
$$
H = \frac{3}{2}(p_1^2+p_2^2 +p_3^2)
$$
and the following non-commuting second class constraints:
$$
\Phi_1 = q_1+q_2+q_3=0\\
\Phi_2 = p_1+p_2+p_3=0.
$$
The general method proposed by Dirac in such a case is to compute the Dirac Bracket
$$
[F,G]_D = [F,G] - [F, \Phi_i]c^{ij}[\Phi_j,G] 
$$
where $c^{ij} = [\Phi_i,\Phi_j]^{-1}$ and $[\cdot, \cdot]$ is the usual Poisson bracket. In this case one has to deal with c-number Dirac brackets, i.e brackets which amount to a complex/real number:
$$
[q_i,q_j]_D  = 0= [p_i,p_j]_D \\
[q_i,p_j]_D = \delta_{ij} - \frac{1}{3}.
$$
Quoting Henneaux & Teitelboim (https://press.princeton.edu/books/paperback/9780691037691/quantization-of-gauge-systems), Chapter 13, page 273ff, it should be possible to quantize such a structure. Unfortunately, they never state how exactly one should approach such a task. I experimented with different combination of first-order differential operators and coordinates, similiar to the classical Poisson-bracket, but to no avail. Is there a general recipe how this is done?
Caveat: If feasible, the whole algebra should be quantized. I know it is possible to find new (Darboux) coordinates on the constraint surface, express a new bracket and quantize it. But this is not the primary goal here.
 A: *

*Let us for physical reasons separate out the total momentum $P:=p_1+p_2+p_3$ and CoM coordinate $Q:=\frac{q^1+q^2+q^3}{3}$. Relative to the CoM frame, we define coordinates
$$q^{\prime j}~:=~q^j-Q\quad\text{and}\quad p^{\prime}_k~:=~p_k-P.\tag{A}$$


*The non-zero Dirac brackets are
$$\{q^{\prime j},p^{\prime}_k\}_D~=~\{q^j,p_k\}_D~=~\delta^j_k-\frac{1}{3}.\tag{B}$$


*OP's Hamiltonian becomes weakly diagonal
$$\begin{align}\frac{2}{3}H~=~&\sum_{k=1}^3 p_k^2~\approx~\sum_{k=1}^3 p_k^{\prime 2}~\approx~p_1^{\prime 2}+p_2^{\prime 2}+(p^{\prime}_1+p^{\prime}_2)^2\cr 
~=~&2(p_1^{\prime 2}+p_2^{\prime 2}+p^{\prime}_1p^{\prime}_2)~=~3p_+^{\prime 2}+p_-^{\prime 2}, \end{align}\tag{C} $$
if we define
$$p^{\prime}_{\pm}~:=~\frac{p^{\prime}_1\pm p^{\prime}_2}{\sqrt{2}}. \tag{D}$$


*Finally choose appropriate linear combinations $q^{\prime\pm}$ of $q^{\prime 1}$ and $q^{\prime 2}$ to make $(q^{\prime +},q^{\prime -},p^{\prime}_+,p^{\prime}_-)$ Darboux/canonical coordinates.


*When we use the coordinate system $(q^{\prime +},q^{\prime -}.Q,p^{\prime}_+,p^{\prime}_-,P)$, we see that the Hamiltonian (C) is just 2 independent 1D free particles, which we already know how to quantize.
