Physics Asked by Viktor Zelezny on July 26, 2020

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.

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 $$q^{prime j}~:=~q^j-Qquadtext{and}quad p^{prime}_k~:=~p_k-P.tag{A}$$

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 $$begin{align}frac{2}{3}H~=~&sum_{k=1}^3 p_k^2~approx~sum_{k=1}^3 p_k^{prime 2}~approx~p_1^{prime 2}+p_2^{prime 2}+(p^{prime}_1+p^{prime}_2)^2cr ~=~&2(p_1^{prime 2}+p_2^{prime 2}+p^{prime}_1p^{prime}_2)~=~3p_+^{prime 2}+p_-^{prime 2}, end{align}tag{C} $$

if we define $$p^{prime}_{pm}~:=~frac{p^{prime}_1pm p^{prime}_2}{sqrt{2}}. tag{D}$$Finally choose appropriate linear combinations $q^{primepm}$ 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.

Answered by Qmechanic on July 26, 2020

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