2-D covariant affine integral quantization(s)

Abstract

Covariant affine integral quantization is studied and applied to the motion of a particle in a punctured plane ${\mathbb {R}}_{*}^2{:}{=}{\mathbb {R}}^2{\setminus }\{0\}$, for which the phase space is ${\mathbb {R}}_{*}^2 \times {\mathbb {R}}^2$. We examine the consequences of different quantizer operators built from weight functions on ${\mathbb {R}}_{*}^2 \times {\mathbb {R}}^2$. To illustrate the procedure, we examine two examples of weights. The first one corresponds to 2-D coherent state families, while the second one corresponds to the affine inversion in the punctured plane. The later yields the usual canonical quantization and a quasi-probability distribution (2-D affine Wigner function) which is real, marginal in both position $\mathbf{q}$ and momentum $\mathbf{p}$.

Appendices

A A 2D-Dirac formula

We start by calculating the following Fourier transform in the distribution sense. On one hand we have

$$\begin{aligned} \int _{{\mathbb {R}}^2} \text {d}^2 \mathbf{p}\, e^{\mathsf {i}\mathbf{p}\cdot \left( \frac{ \mathbf{q}}{ {\mathbf{x}}}- {\mathbf{x}}\right) }= 4\pi ^2 \delta \left( \frac{ \mathbf{q}}{ {\mathbf{x}}}- {\mathbf{x}}\right) . \end{aligned}$$

(A.1)

Alternatively, we first change the argument of the exponential as

$$\begin{aligned} \mathbf{p}\cdot \left( \frac{ \mathbf{q}}{ {\mathbf{x}}}- {\mathbf{x}}\right) = \mathbf{p}\cdot \frac{ \mathbf{q}}{ {\mathbf{x}}}- \mathbf{p}\cdot {\mathbf{x}}= \frac{ \mathbf{p}}{ {\mathbf{x}}^{*}}\cdot \mathbf{q}- \mathbf{p}\cdot {\mathbf{x}}= \frac{\text {R}(\theta ) \mathbf{p}}{x}\cdot \mathbf{q}- \mathbf{p}\cdot {\mathbf{x}}, \end{aligned}$$

(A.2)

with $x=(x,\theta )$. Now we change

$$\begin{aligned} \mathbf{p}^{\prime }= \frac{\text {R}(\theta ) \mathbf{p}}{x}\, ,\quad \text {d}^2\mathbf{p}^{\prime }= \frac{\text {d}^2 \mathbf{p}}{x^2}, \end{aligned}$$

(A.3)

to obtain finally

$$\begin{aligned} \int _{{\mathbb {R}}^2} \text {d}^2 \mathbf{p}\, e^{\mathsf {i}\mathbf{p}\cdot \left( \frac{ \mathbf{q}}{ {\mathbf{x}}}- {\mathbf{x}}\right) }= x^2\int _{{\mathbb {R}}^2} \text {d}^2 \mathbf{p}^{\prime }\, e^{\mathsf {i}\mathbf{p}^{\prime }\cdot \left( \mathbf{q}- {\mathbf{x}}^2\right) }=4\pi ^2x^2\delta \left( \mathbf{q}- {\mathbf{x}}^2\right) . \end{aligned}$$

(A.4)

Thus

$$\begin{aligned} \delta \left( \frac{ \mathbf{q}}{ {\mathbf{x}}}- {\mathbf{x}}\right) =x^2\delta \left( {\mathbf{x}}^2- \mathbf{q}\right) . \end{aligned}$$

(A.5)

Now, we must be cautious with square roots of $\mathbf{q}$ as they appear above as roots of ${\mathbf{x}}^2- \mathbf{q}=0$. We should not forget that ${\mathbf{x}}$ and $\mathbf{q}$ are confined to the punctured plane ${\mathbb {R}}_*^2 \sim {\mathbb {C}}_*$. So, let us proceed with the change of variables

$$\begin{aligned} {\mathbf{x}}= (x,\theta ) \mapsto \mathbf{y}= (y,\omega ) = {\mathbf{x}}^2= (x^2,2 \theta )\, , \quad \theta \in [0,2\pi ) \Rightarrow \omega \in [0,4\pi ). \end{aligned}$$

(A.6)

Thus we deal here with the two sheets of the Riemann surface for the variable $\mathbf{y}$:

$$\begin{aligned} {\mathcal {R}}_1{:}{=} \{ \mathbf{y}= (y,\omega ), \, 0\le \omega< 2\pi \}\, , \quad {\mathcal {R}}_2{:}{=} \{ \mathbf{y}= (y,\omega ), \, 2\pi \le \omega < 4\pi \}, \end{aligned}$$

(A.7)

with the cut $\{\mathbf{y}=-y, \, y\in {\mathbb {R}}^{+}\}$. From

$$\begin{aligned} \int _{{\mathbb {R}}^{2}_{*}} \text {d}^2{\mathbf{x}}= \int _0^{+\infty } x\text {d}x\int _0^{2\pi }\text {d}\theta = \frac{1}{2}\int _0^{+\infty }\text {d}y\int _0^{4\pi }\text {d}\omega , \end{aligned}$$

we derive the decomposition formula

$$\begin{aligned} \int _{{\mathbb {R}}^{2}_{*}} \text {d}^2{\mathbf{x}}= \frac{1}{2}\left[ \int _{{\mathcal {R}}_1} \frac{\text {d}\mathbf{y}}{y} +\int _{{\mathcal {R}}_2} \frac{\text {d}\mathbf{y}}{y}\right] . \end{aligned}$$

(A.8)

This formula gives a precise sense to the Dirac distribution $\delta \left( {\mathbf{x}}^2- \mathbf{q}\right)$:

$$\begin{aligned} \int _{{\mathbb {R}}^{2}_{*}} \text {d}^2{\mathbf{x}}\, \delta \left( {\mathbf{x}}^2- \mathbf{q}\right) \,\varphi ({\mathbf{x}})= \frac{1}{2q}\left[ \varphi (\sqrt{\mathbf{q}}) + \varphi (-\sqrt{\mathbf{q}})\right] , \end{aligned}$$

(A.9)

for all test function in some suitable space, e.g. $C^{\infty }$ functions on ${\mathbb {R}}^{2}_{*}$ which rapidly decrease at 0 and $\infty$, and where $\pm \sqrt{\mathbf{q}}$ are defined for $\mathbf{q}=(q,\theta )$, $\theta \in [0, 2\pi )$, as

$$\begin{aligned} \sqrt{\mathbf{q}}\,{:=} \,\left. \mathbf{q}^{1/2}\right| _{\mathbf{q}\in {\mathcal {R}}_1}= (\sqrt{q}, \theta /2)\, , \quad -\sqrt{\mathbf{q}}\,{:=}\, \left. \mathbf{q}^{1/2}\right| _{\mathbf{q}\in {\mathcal {R}}_2}. \end{aligned}$$

(A.10)

Eventually, we get from (A.5) a formula relevant to the present paper:

$$\begin{aligned} \int _{{\mathbb {R}}^{2}_{*}} \text {d}^2{\mathbf{x}}\, \delta \left( {\mathbf{x}}- \frac{\mathbf{q}}{{\mathbf{x}}} \right) \,\varphi ({\mathbf{x}})= \frac{1}{2}\left[ \varphi (\sqrt{\mathbf{q}}) + \varphi (-\sqrt{\mathbf{q}})\right] , \end{aligned}$$

(A.11)

B Quantization with 2-D affine coherent states (ACS) : a summary

In this appendix, which completes the technical content of [18], we particularize the general method of 2-D affine covariant integral quantisation to the case when the quantiser operator is the projector $|\psi \rangle \langle \psi |$ on an admissible fiducial vector, defined by the function (5.1), as it was described in Sect. 5.1.

1.1 B.1 Quantization with ACS

Let us implement the integral quantization scheme described in this paper by restricting the method to the specific case of the rank-one density operator or projector $\mathsf {M}^{\varpi _{\psi }}=|\psi \rangle \left\langle \psi \right|$ where $\psi$ is a unit-norm admissible state, or fiducial vector, or wavelet, i.e., is in $L^{2}({{\mathbb {R}}_{*}^2},\text {d}^2{\mathbf{x}}) \cap L^{2}\left( {{\mathbb {R}}_{*}^2},\dfrac{{\text {d}}^2{\mathbf{x}}}{x^2}\right)$. We have from (4.2) and (5.2):

$$\begin{aligned} \varOmega ( \mathbf{u})(\equiv \varOmega _{0}( \mathbf{u}))= & {} \frac{2\pi }{u^2} \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}\, , \quad \varOmega _{\beta }(1)\nonumber \\= & {} 2\pi \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^{2+\beta }}\,\vert \psi ( \mathbf{q})\vert ^2. \end{aligned}$$

(B.1)

Proposition 11

With implicit assumptions on the existence of derivatives of $\psi$ and of their (square) integrability, we have

$$\begin{aligned} (\pmb {\nabla } \varOmega )(1)= & {} -2\varOmega (1)-i2\pi \left\langle \frac{ \mathbf{P}}{ \mathbf{Q}}\psi |\psi \right\rangle , \end{aligned}$$

(B.2)

$$\begin{aligned} (\varDelta \varOmega )(1))= & {} 4\varOmega (1)+8\mathsf {i}\pi \left\langle \frac{ \mathbf{P}}{ \mathbf{Q}}\psi |\psi \right\rangle -2\pi \left\langle \mathbf{P}^{2}\psi |\psi \right\rangle , \end{aligned}$$

(B.3)

which implies for $\psi$ real:

$$\begin{aligned} 2 +\frac{\pmb {\varOmega }^{(1)}(1)}{\varOmega (1)}=\mathbf 0 , \end{aligned}$$

(B.4)

and,

$$\begin{aligned} 4+4\frac{\varOmega _1^{(1)}(1)}{\varOmega (1)}+\frac{\varOmega ^{(2)}(1)}{\varOmega (1)}= -2\pi \langle \mathbf{P}^{2}\psi |\psi \rangle . \end{aligned}$$

(B.5)

Proof

For (B.2),

$$\begin{aligned}&\pmb {\nabla }_{ \mathbf{u}}\left[ \frac{2\pi }{u^2} \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}\right] _{ \mathbf{u}=1}= \pmb {\nabla }_{ \mathbf{u}}\left[ \frac{2\pi }{u^2} \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}\right] _{ \mathbf{u}=1}\\&\quad = -2\varOmega (1)-2\pi \int _{{\mathbb {R}}^2_{*}}\text {d}^2 \mathbf{q}\, \mathbf{q}^{-1}\,\pmb {\nabla }_{ \mathbf{q}}\{{\overline{\psi ( \mathbf{q})}}\}\psi ( \mathbf{q})\\&\quad = -2\varOmega (1)-\mathsf {i}2\pi \left\langle \frac{ \mathbf{P}}{ \mathbf{Q}}\psi |\psi \right\rangle . \end{aligned}$$

For (B.3),

$$\begin{aligned}&\pmb {\nabla }\cdot \pmb {\nabla }_{ \mathbf{u}}\left[ \frac{2\pi }{u^2} \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}\right] _{ \mathbf{u}=1}\nonumber \\&\quad = \left[ \left( \frac{4}{u^2}\right) 2\pi \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}+2\left( \frac{-2 \mathbf{u}}{u^4}\right) \right. \nonumber \\&\qquad \cdot 2\pi \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, \frac{- \mathbf{q}^{\,*}}{ \mathbf{u}^{2*}}\pmb {\nabla }_{\frac{ \mathbf{q}}{ \mathbf{u}}}\left( {\overline{\psi \left( \frac{ \mathbf{q}}{ \mathbf{u}}\right) }}\right) \nonumber \\&\qquad +\left. \frac{2\pi }{u^2} \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, \frac{q^2}{u^6}\pmb {\nabla }_{\frac{ \mathbf{q}}{ \mathbf{u}}}\cdot \pmb {\nabla }_{\frac{ \mathbf{q}}{ \mathbf{u}}}{\overline{\psi (\frac{ \mathbf{q}}{ \mathbf{u}}}})\right] _{ \mathbf{u}= 1}\nonumber \\&\quad = 4\times 2\pi \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}}{q^2}\, \psi ( \mathbf{q})\, {\overline{\psi \left( \mathbf{q}\right) }}+4\times 2\pi \int _{{\mathbb {R}}_{*}^2}\text {d}^2 \mathbf{q}\, \psi ( \mathbf{q})\, \frac{ \mathbf{q}^{\,*}}{q^2}\pmb {\nabla }_{ \mathbf{q}}{\overline{\psi \left( \mathbf{q}\right) }} \nonumber \\&\qquad +2\pi \int _{{\mathbb {R}}_{*}^2}\text {d}^2 \mathbf{q}\, \psi ( \mathbf{q})\, \pmb {\nabla }_{ \mathbf{q}}\cdot \pmb {\nabla }_{ \mathbf{q}}{\overline{\psi \left( \mathbf{q}\right) }}\nonumber \\&\quad =4\varOmega (1)+8\mathsf {i}\pi \left\langle \frac{ \mathbf{P}}{ \mathbf{Q}}\psi |\psi \right\rangle -2\pi \left\langle \mathbf{P}^{2}\psi |\psi \right\rangle . \end{aligned}$$

(B.6)

Note that $\left\langle \frac{ \mathbf{P}}{ \mathbf{Q}}\psi |\psi \right\rangle$ is purely imaginary and cancels for real $\psi$. $\square$

Therefore, by applying the general formalism, we recover a set of results already given in previous works, e.g. in [9] in the 1-D affine case, and in [18] in the 2-D case. As was pointed out after stating the orthogonality relations (3.17), the action of the UIR operators $U( \mathbf{q}, \mathbf{p})$ on $\psi$ produces all affine coherent states, i.e. wavelets, defined as $| \mathbf{q}, \mathbf{p}\rangle \ =U( \mathbf{q}, \mathbf{p})|\psi \rangle$. Given a certain function $f( \mathbf{q}, \mathbf{p})$ on the phase space, the corresponding affine coherent state (ACS) quantization reads as

$$\begin{aligned} f\ \mapsto \ A_{f}=\int _{\varGamma }f( \mathbf{q}, \mathbf{p})| \mathbf{q}, \mathbf{p}\rangle \langle \mathbf{q}, \mathbf{p}|\dfrac{\text {d}^2 \mathbf{q}\text {d}^2\, \mathbf{p}}{(2\pi )^2 c_{0}}, \end{aligned}$$

(B.7)

which arises from the resolution of the identity

$$\begin{aligned} \int _{\varGamma }| \mathbf{q}, \mathbf{p}\rangle \langle \mathbf{q}, \mathbf{p}|\,\dfrac{\text {d}^2 \mathbf{q}\text {d}^2 \mathbf{p}}{(2\pi )^2 c_{0}}=\mathbb {1}, \end{aligned}$$

(B.8)

where we adopt for convenience the simplified notations of [9],

$$\begin{aligned} c_{\beta }:= & {} \int _{{\mathbb {R}}_{*}^2}|\psi ( {\mathbf{x}})|^{2}\,\frac{\text {d}^2 {\mathbf{x}}}{x^{2+\beta }} = \frac{\varOmega _{\beta }(1)}{2\pi }. \end{aligned}$$

(B.9)

$$\begin{aligned} c_{\beta \nu _1\nu _2}:= & {} \int _{{\mathbb {R}}_{*}^2}\,\frac{\text {d}^2 {\mathbf{x}}}{x^{2+\beta }}|\psi ( {\mathbf{x}})|^{2}{x_1}^{\nu _1}{x_2}^{\nu _2} \nonumber \\= & {} \, \frac{\varOmega _{(\beta ,\nu _1,\nu _2)}(1)}{2\pi }, \quad c_{\beta \,0\,0}\equiv c_{\beta }. \end{aligned}$$

(B.10)

Thus, a necessary condition to have (B.8) true is that $c_{0} < \infty$, which implies $\psi (\mathbf 0 ) = 0$, a well-known requirement in wavelet analysis.

Choosing $\psi$ real, and using equations (B.4) and (B.5) in the quantization formulas established in Sect. 6.1, one gets easily:

$$\begin{aligned} A_{ \mathbf{p}}= \mathbf{P}, \quad A_{q^{\beta }}=\frac{c_{\beta }}{c_{0}} \,Q^{\beta }, A_{ \mathbf{q}}=\frac{\;c_{210}}{c_{0}} \mathbf{Q}. \end{aligned}$$

(B.11)

whereas $\mathbf{Q}$ is essentially self-adjoint, we recall that the operator $\mathbf{P}$ is symmetric but has no self-adjoint extension.

The quantization of the kinetic energy gives

$$\begin{aligned} A_{p^{2}}=P^{2}+KQ^{-2}, \quad K=2\pi \langle P^{2}\psi |\psi \rangle , \end{aligned}$$

(B.12)

We see that a repulsive potential $\propto \, 1/Q^2$ is obtained with strength $K>0$. In a semi-classical interpretation, this extra centrifugal term prevents the particle to reach the singular point, whatever the value of its angular momentum [18]. With a suitable choice of $\psi$, we can make this coefficient large enough, namely $K \ge 1$ [5, 27, 31], to make (B.12) an essentially self-adjoint kinetic operator, i.e. no boundary condition is needed, and then quantum dynamics of the free motion on the punctured plane is unique. Whilst canonical quantization, based on Weyl–Heisenberg symmetry which is unnatural in the present case, introduces ambiguity on the quantum level, ACS quantization with suitable fiducial vector removes this ambiguity.

The quantum states and their dynamics have semi-classical phase space representations through symbols. For the state $|\phi \rangle$ the corresponding symbol reads

$$\begin{aligned} \varPhi ( \mathbf{q}, \mathbf{p})=\frac{\langle \mathbf{q}, \mathbf{p}|\phi \rangle }{(2\pi )^2}, \end{aligned}$$

(B.13)

with the associated probability distribution on phase space given by

$$\begin{aligned} \varUpsilon _{\phi }( \mathbf{q}, \mathbf{p})=\dfrac{1}{2\pi c_{0}}|\langle \mathbf{q}, \mathbf{p}|\phi \rangle |^{2}. \end{aligned}$$

(B.14)

Having the (energy) eigenstates of some quantum Hamiltonian $\mathsf {H}$ at our disposal, the most natural one being in this context the quantized $A_h$ of a classical Hamiltonian $h( \mathbf{q}, \mathbf{p})$, we can compute the time evolution

$$\begin{aligned} \varUpsilon _{\phi }( \mathbf{q}, \mathbf{p},t){:}{=}\dfrac{1}{2\pi c_{0}}|\langle \mathbf{q}, \mathbf{p}|e^{-\mathsf {i}\mathsf {H}t}|\phi \rangle |^{2}, \end{aligned}$$

(B.15)

for any state $\phi$.

The map (2.8), yielding lower symbols from classical f reads in the present case:

$$\begin{aligned} \check{f}( \mathbf{q}, \mathbf{p})= & {} \frac{(2\pi )^2}{c_0}\int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2 \mathbf{q}^{\prime }}{q^2\;{q^{\prime }}^2}\,\int _{{\mathbb {R}}_{*}^2}\text {d}^2 {\mathbf{x}}\,\int _{{\mathbb {R}}_{*}^2}\text {d}^2 \mathbf{y}\,e^{\mathsf {i}\mathbf{p}(\mathbf{y}- {\mathbf{x}})}{{\hat{f}}}_p( \mathbf{q}^{\prime },\mathbf{y}- {\mathbf{x}}) \nonumber \\&\times \psi \left( \frac{ {\mathbf{x}}}{ \mathbf{q}}\right) \, \psi \left( \frac{ {\mathbf{x}}}{ \mathbf{q}^{\prime }}\right) \,\psi \left( \frac{y}{ \mathbf{q}}\right) \, \psi \left( \frac{\mathbf{y}}{ \mathbf{q}^{\prime }}\right) , \end{aligned}$$

(B.16)

where ${{\hat{f}}}_p$ stands for the partial inverse Fourier transform introduced in (4.3), and with implicit hypotheses on f and $\psi$ allowing derivations, Fourier transform and permutation of integrals allowed by the Fubini theorem.

For functions f depending on $\mathbf{q}$ only, expression (B.16) simplifies to a lower symbol depending on $\mathbf{q}$ only:

$$\begin{aligned} \check{f}( \mathbf{q})= \frac{(2\pi )^2}{c_0}\int _{{\mathbb {R}}_{*}^2} \frac{\text {d}^2 \mathbf{q}^{\prime }}{q^2\;{q^{\prime }}^2}\, f( \mathbf{q}^{\prime }) \int _{{R_{*}^2}}\text {d}^2 {\mathbf{x}}\,\psi ^2\left( \frac{ {\mathbf{x}}}{ \mathbf{q}}\right) \,\psi ^2\left( \frac{ {\mathbf{x}}}{ \mathbf{q}^{\prime }}\right) . \end{aligned}$$

(B.17)

For instance, any power of $q= \sqrt{q_1^2 + q_2^2}$ is transformed into the same power up to a constant factor

$$\begin{aligned} q^{\beta } \mapsto \check{q^{\beta }}=\frac{c_{\beta }c_{-\beta -2}}{c_{0}} \, q^{\beta }. \end{aligned}$$

(B.18)

For $f( \mathbf{q}, \mathbf{p})= \mathbf{p}$, we prove that:

$$\begin{aligned} \check{ \mathbf{p}}= \mathbf{p}. \end{aligned}$$

(B.19)

For $f( \mathbf{q}, \mathbf{p})= p^2$, we have:

$$\begin{aligned} \check{p^2}=p^2+\frac{\gamma ^2}{q^2}\, , \quad \gamma ^{2}=2\left\langle P^2\psi |\psi \right\rangle . \end{aligned}$$

(B.20)

Another interesting formula in the semi-classical context concerns the Fubini-Study metric derived from the symbol of total differential $\text {d}$ with respect to parameters $\mathbf{q}$ and $\mathbf{p}$ affine coherent states,

$$\begin{aligned} \text {d}\sigma ( \mathbf{q}, \mathbf{p})^2=\Vert \text {d}| \mathbf{q}, \mathbf{p}\rangle \Vert ^2-\vert \langle \mathbf{q}, \mathbf{p}|\text {d}| \mathbf{q}, \mathbf{p}\rangle \vert ^2 \end{aligned}$$

(B.21)

Choosing a $\psi$ such that $c_{-2 0 1}=0$ gives:

$$\begin{aligned} \vert \langle \mathbf{q}, \mathbf{p}|\text {d}| \mathbf{q}, \mathbf{p}\rangle \vert ^2=c_{-2 1 0}^2[q_1^2 \text {d}p_1^2+q_2^2 \text {d}p_2^2+2q_1 q_2 \text {d}p_1 \text {d}p_2]. \end{aligned}$$

(B.22)

For the squared norm of $\text {d}| \mathbf{q}, \mathbf{p}\rangle$,

$$\begin{aligned} \text {d}\sigma ( \mathbf{q}, \mathbf{p})^2&= q^{-4}(A^2 q_1^2+B^2 q_2^2) \text {d}q_1^2+q^{-4}(B^2q_1^2+A^2 q_2^2) \text {d}q_2^2 +\nonumber \\&\quad + 2q^{-4}((A^2-B^2)q_1 q_2+C(q_1^2-q_2^2) )\text {d}q_1 \text {d}q_2 +\nonumber \\&\quad +((E^2 -c_{-201}^2) q_1^2+F^2 q_2^2) \text {d}p_1^2\nonumber \\&\quad +(F^2q_1^2+(E^2-c_{-201}^2) q_2^2) \text {d}p_2^2 +\nonumber \\&\quad + 2((E^2-F^2-c_{-201}^2)q_1 q_2+G(q_1^2-q_2^2) )\text {d}p_1 \text {d}p_2, \end{aligned}$$

(B.23)

where:

$$\begin{aligned}&A= \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2\mathbf{y}}{y^2}\,\left( \mathbf{y}\cdot \pmb {\nabla }_{\mathbf{y}}[y\psi (\mathbf{y})]\right) ^2 \, , \nonumber \\&B= \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2\mathbf{y}}{y^2}\,\left( \mathbf{y}\times \pmb {\nabla }_{\mathbf{y}}[y\psi (\mathbf{y})]\right) ^2\, ,\nonumber \\&C= \int _{{\mathbb {R}}_{*}^2}\frac{\text {d}^2\mathbf{y}}{y^2}\,\left( \mathbf{y}\cdot \pmb {\nabla }_{\mathbf{y}}[y\psi (\mathbf{y})]\right) \,\left( \mathbf{y}\times \pmb {\nabla }_{\mathbf{y}}[y\psi (\mathbf{y})]\right) \, ,\nonumber \\&D= \int _{{\mathbb {R}}_{*}^2} \text {d}^2 \mathbf{y}(\psi (\mathbf{y}))^2 y_1^2\, ,\nonumber \\&E= \int _{{\mathbb {R}}_{*}^2} \text {d}^2 \mathbf{y}(\psi (\mathbf{y}))^2 y_2^2\, ,\nonumber \\&F= \int _{{\mathbb {R}}_{*}^2} \text {d}^2 \mathbf{y}(\psi (\mathbf{y}))^2 y_1 y_2. \end{aligned}$$

(B.24)