On localized and coherent states on some new fuzzy spheres

Abstract

We construct various systems of coherent states (SCS) on the O(D)-equivariant fuzzy spheres $S^d_\Lambda$ ($d=1,2$, $D=d+1$) constructed in Fiore and Pisacane (J Geom Phys 132:423–451, 2018) and study their localizations in configuration space as well as angular momentum space. These localizations are best expressed through the O(D)-invariant square space and angular momentum uncertainties $(\Delta \varvec{x})^2,(\Delta \varvec{L})^2$ in the ambient Euclidean space $\mathbb {R}^D$. We also determine general bounds (e.g., uncertainty relations from commutation relations) for $(\Delta \varvec{x})^2,(\Delta \varvec{L})^2$, and partly investigate which SCS may saturate these bounds. In particular, we determine O(D)-equivariant systems of optimally localized coherent states, which are the closest quantum states to the classical states (i.e., points) of $S^d$. We compare the results with their analogs on commutative $S^d$. We also show that on $S^2_\Lambda$ our optimally localized states are better localized than those on the Madore–Hoppe fuzzy sphere with the same cutoff $\Lambda$.

Appendix

1.1 Proof of Theorems 2.1 and 4.1

Consider a generic Hilbert space ${{\mathcal {H}}}$ carrying a unitary representation of O(3). For any vector $\psi \in {{\mathcal {H}}}$, let $g\in O(3)$ be a $3\times 3$ matrix such that the expectation values of $L_j$ on $\psi$ fulfill

$$\begin{aligned} g_{ij}\langle L_j\rangle =\delta _i^3|\langle \varvec{L}\rangle |. \end{aligned}$$

(68)

The expectation values of the $L_j,\varvec{L}^2$ on the states $\psi$, $\psi ':=U(g)\psi$ fulfill $\langle L_1\rangle '=\langle L_2\rangle '=0$, $\langle L_3\rangle '=|\langle \varvec{L}\rangle '|=|\langle \varvec{L}\rangle |\ge 0$, $\langle \varvec{L}^2\rangle '=\langle \varvec{L}^2\rangle$ (the second equalities hold because U(g) is unitary). Hence, $\psi$ fulfills/saturates (17) iff $\psi '$, respectively, fulfills/saturates

$$\begin{aligned} \langle \varvec{L}^2\rangle ' - \langle L_3\rangle '\left( \langle L_3\rangle '+1\right) \ge 0. \end{aligned}$$

(69)

If ${{\mathcal {H}}}=V_l$ the first term equals $l(l+1)$, the inequality (69) is fulfilled, and it is saturated by $\psi '=|l,l\rangle$, because Spec$(L_3)=\{-l,1-l,\ldots ,l\}$.

Now assume that ${{\mathcal {H}}}$ can be decomposed as the direct sum ${{\mathcal {H}}}={{\mathcal {H}}}_1\oplus {{\mathcal {H}}}_2$ of orthogonal subspaces ${{\mathcal {H}}}_1,{{\mathcal {H}}}_2$ carrying subrepresentations of O(3) and on which (17) is fulfilled; moreover, let $\Gamma _i\subset {{\mathcal {H}}}_i$ be the subsets of vectors saturating (17). Decomposing $\psi '=a_1\psi _1+a_2\psi _2$ and setting $\alpha :=|a_1|^2$, we find $0\le \alpha \le 1$, $|a_2|^2=1-\alpha$, and

$$\begin{aligned}&\langle \varvec{L}^2\rangle '-\langle L_3\rangle '\left( \langle L_3\rangle '+1\right) \nonumber \\&\quad =\alpha \langle \varvec{L}^2\rangle _1+(1-\alpha )\langle \varvec{L}^2\rangle _2 -\left[ \alpha \langle L_3\rangle _1+(1-\alpha )\langle L_3\rangle _2\right] ^2\nonumber \\&\quad -\left[ \alpha \langle L_3\rangle _1+(1-\alpha )\langle L_3\rangle _2\right] =:f(\alpha ), \end{aligned}$$

(70)

where we have abbreviated $\langle A\rangle _i\equiv \langle A\rangle _{\psi _i}$. The polynomial $f'(\alpha )$ vanishes only at one point $\alpha '\in \mathbb {R}$, which, however, is of maximum for $f(\alpha )$, because $f''(\alpha )=-\left[ \langle L_3\rangle _1-\langle L_3\rangle _2\right] ^2\le 0$. Hence, the minimum point of $f(\alpha )$ in the interval [0, 1] is either 0 or 1. But, by our assumptions,

$$\begin{aligned} f(1)= & {} \langle \varvec{L}^2\rangle _1-|\langle \varvec{L}\rangle _1|\left( |\langle \varvec{L}\rangle _1|+1\right) \ge 0,\\ f(0)= & {} \langle \varvec{L}^2\rangle _2-|\langle \varvec{L}\rangle _2|\left( |\langle \varvec{L}\rangle _2|+1\right) \ge 0, \end{aligned}$$

proving that (17) is fulfilled on ${{\mathcal {H}}}$. Moreover, the set of states of ${{\mathcal {H}}}$ saturating the inequality is clearly $\Gamma =\Gamma _1\cup \Gamma _2$.

Choosing first ${{\mathcal {H}}}_1=V_0$ and ${{\mathcal {H}}}_2=V_1$, then ${{\mathcal {H}}}_1=V_0\oplus V_1$ and ${{\mathcal {H}}}_2=V_2$, and so on, one thus iteratively proves the statements of Theorems 2.1 and 4.1 for pure states.

Similarly, we show that also mixed states (i.e., density operators) $\rho$ fulfill (17), but cannot saturate it: abbreviating $\langle A\rangle \equiv \langle A\rangle _\rho :=\text{ tr }(\rho A)$, let $g\in O(3)$ be a $3\times 3$ matrix such that the expectation values of $L_j$ on $\rho$ fulfill (68). Then, the expectation values of $L_j,\varvec{L}^2$ on the state $\rho '=U(g)\rho U^{-1}(g)$ fulfill $\langle L_1\rangle '=\langle L_2\rangle '=0$, $\langle L_3\rangle '=|\langle \varvec{L}\rangle '|=|\langle \varvec{L}\rangle |\ge 0$, $\langle \varvec{L}^2\rangle '=\langle \varvec{L}^2\rangle$, and $\rho$ fulfills/saturates (17) iff $\rho '$ fulfills/saturates (69). If $\rho '=\alpha \rho _1+(1-\alpha )\rho _2$, the left-hand side of (69) again takes the form (70). Hence, reasoning as before, we find that $\rho$ fulfills (17) and that there are no mixed states saturating this inequality.

1.2 Some useful summations

From $h(h+1)(h+2)\ldots (h+j+1)-(h-1)h(h+1)\ldots (h+j)=(j+2)h(h+1)\ldots (h+j)$ (with $j\in \mathbb {N}_0$) it follows

$$\begin{aligned} \sum \limits _{h=1}^n h(h+1)\ldots (h+j)=\frac{1}{j+2} n(n+1)(n+2)\ldots (n+j+1); \end{aligned}$$

(71)

this implies, in particular,

$$\begin{aligned} \sum \limits _{h=1}^n h^2= & {} \sum \limits _{h=1}^n [h(h+1)-h]=\frac{ n(n+1)(n+2)}{3}- \frac{ n(n+1)}{2}\nonumber \\= & {} \frac{ n(n+1)(2n+1)}{6},\quad \end{aligned}$$

(72)

and in the following lines we will also use

$$\begin{aligned} \sum _{h=1}^{n}h^3= & {} \frac{n^2(n+1)^2}{4},\quad \quad \nonumber \\ \sum _{h=1}^{n}h(2h+1)= & {} \frac{4n^3+9n^2+5n}{6}, \end{aligned}$$

(73)

$$\begin{aligned} \sum _{h=1}^{n}h(h+1)(2h+1)= & {} \frac{1}{2}n(n+1)^2(n+2), \end{aligned}$$

(74)

$$\begin{aligned} \sum _{h=1}^{n}\left[ h(h+1)+1\right] (2h+1)= & {} \frac{(n+1)^2(n^2+2n+2)}{2},\quad \quad \nonumber \\ \sum _{h=1}^{n}h\left( 1-\frac{1}{2h}\right)= & {} \frac{n^2}{2}. \end{aligned}$$

(75)

Using the inequalities $1+x/2\ge \sqrt{1+x}\ge 1+x/2-x^2/8$ (the first one is valid for $x\ge -1$, the second for $x\le 8$), we find

$$\begin{aligned}&1+\frac{m(m-1)}{2k}\ge b_m\ge 1+\frac{m(m-1)}{2k} -\frac{m^2(m-1)^2}{(2k)^2}\quad \end{aligned}$$

(76)

$$\begin{aligned}&\quad \Rightarrow \quad n+\frac{(n-1)n(n+1)}{6k}\ge \sum \limits _{m=1}^{n} b_m\ge n+\frac{(n-1)n(n+1)}{6k}\nonumber \\&\quad -\frac{(n-1)n(n+1)(3n^2-2)}{60k^2}. \end{aligned}$$

(77)

Using trigonometric formulae, it is straightforward to show that

$$\begin{aligned} \sum \limits _{m=2}^{n}\cos \left[ \frac{\pi (2m -1)}{2n+2}\right] =0 \end{aligned}$$

(78)

(the terms cancel pairwise: the terms with $m=2,n$ cancel each other, the terms with $m=3,n-1$ cancel each other, etc.), and

$$\begin{aligned} 2\sin \left[ \frac{\pi (n+1+m)}{2n+2}\right] \sin \left[ \frac{\pi (n+m)}{2n+2} \right]= & {} \cos \left[ \frac{\pi }{2n+2}\right] - \cos \left[ \frac{\pi (2n+1 +2m)}{2n+2}\right] \nonumber \\= & {} \cos \left[ \frac{\pi }{2n+2}\right] +\cos \left[ \frac{\pi (2m -1)}{2n+2}\right] . \end{aligned}$$

(79)

1.3 Proofs of some results regarding $S^1_\Lambda$

On a vector $\varvec{\chi }=\sum _{m=-\Lambda }^{\Lambda }{\chi _m\psi _m}$ we find $x_+\varvec{\chi }=\sum _{m=-\Lambda }^{\Lambda -1}\chi _mb_{m+1}\psi _{m+1}$, and

$$\begin{aligned} \langle x_+\rangle _{\varvec{\chi }}= & {} \sum \limits _{m=1-\Lambda }^{\Lambda } \overline{\chi _m}\chi _{m-1}b_m; \end{aligned}$$

(80)

$$\begin{aligned} \langle \varvec{x}^2\rangle _{\varvec{\chi }}= & {} \sum \limits _{m=1-\Lambda }^{\Lambda -1}\left( 1+\frac{m^2}{k}\right) |\chi _m|^2+ \frac{1}{2}\left[ 1+\frac{\Lambda (\Lambda -1)}{k}\right] \, (|\chi _{\Lambda }|^2+|\chi _{-\Lambda }|^2) \nonumber \\= & {} \langle \varvec{\chi },\varvec{\chi }\rangle + \sum \limits _{m=1-\Lambda }^{\Lambda -1}\frac{m^2}{k} |\chi _m|^2\nonumber \\&+ \frac{1}{2}\left[ \frac{\Lambda (\Lambda -1)}{k}-1\right] \, (|\chi _{\Lambda }|^2+|\chi _{-\Lambda }|^2) . \end{aligned}$$

(81)

We first prove (32),

$$\begin{aligned} \left\langle L\right\rangle _{\varvec{\phi }_{\alpha }^{\beta }}= & {} \frac{1}{2\Lambda +1} \sum _{m=-\Lambda }^{\Lambda }m=0,\quad \left\langle L^2\right\rangle _{\varvec{\phi }_{\alpha }^{\beta }}=\frac{1}{2\Lambda +1} \sum _{m=-\Lambda }^{\Lambda }m^2\\= & {} \frac{2}{2\Lambda +1} \sum _{m=1}^{\Lambda }m^2\overset{(72)}{=}\frac{\Lambda (\Lambda +1)}{3}. \end{aligned}$$

Now we prove (33). By (25), (72), (80–81)

$$\begin{aligned} \langle \varvec{x}^2\rangle _{\varvec{\phi }_\alpha ^\beta }= & {} \langle \varvec{\phi }_\alpha ^\beta ,\varvec{x}^2\varvec{\phi }_\alpha ^\beta \rangle = 1+ \frac{2}{2\Lambda +1}\sum \limits _{m=1}^{\Lambda }\frac{m^2}{k}- \frac{1}{2\Lambda +1} \left[ \frac{\Lambda (\Lambda +1)}{k}+1\right] \\= & {} 1+ \frac{\Lambda (\Lambda +1)}{3k}- \frac{1}{2\Lambda +1} \left[ \frac{\Lambda (\Lambda +1)}{k}+1\right] =\frac{2\Lambda }{2\Lambda +1}+ \frac{2(\Lambda -1) \Lambda (\Lambda +1)}{3(2\Lambda +1)k}, \end{aligned}$$

as claimed. Now we are able to prove (34):

$$\begin{aligned} \left( \Delta \varvec{x}\right) ^2_{\varvec{\phi }_\alpha ^\beta }&=\langle \varvec{x}^2\rangle _{\varvec{\phi }_\alpha ^\beta }-|\langle x_+\rangle _{\varvec{\phi }_\alpha ^\beta }|^2 =\frac{2\Lambda }{2\Lambda +1}+ \frac{2(\Lambda ^2-1)}{3(2\Lambda +1)k}-\frac{4}{(2\Lambda +1)^2}\left[ \sum \limits _{m=1}^{\Lambda } b_m\right] ^2\\&\overset{1\le b_m}{\le }\frac{2\Lambda }{2\Lambda +1}+ \frac{2(\Lambda ^2-1)\Lambda }{3(2\Lambda +1)k}-\frac{4\Lambda ^2}{(2\Lambda +1)^2}\\&\overset{(5)}{\le }\frac{2\Lambda }{2\Lambda +1}-\frac{4\Lambda ^2}{(2\Lambda +1)^2}+ \frac{2(\Lambda -1)}{3(2\Lambda +1)\Lambda (\Lambda +1)}< \frac{2\Lambda }{(2\Lambda +1)^2}+\frac{1}{3\Lambda (\Lambda +1)}\\&< \frac{2\Lambda }{4\Lambda (\Lambda +1)}+\frac{1}{3\Lambda (\Lambda +1)}= \frac{1}{\Lambda +1} \left( \frac{1}{2}+\frac{1}{3\Lambda }\right) \overset{\Lambda \ge 2}{\le }\frac{2}{3(\Lambda +1)}. \end{aligned}$$

We now prove (36). On a generic normalized $\varvec{\chi }$ (80–81) with $\Lambda =1$ gives

$$\begin{aligned} \langle \varvec{x}^2\rangle _{\varvec{\chi }}= & {} \frac{1}{2}\left[ 1+|\chi _0|^2\right] =\frac{1}{2}\left[ 1+s\right] ,\quad \langle x_+\rangle _{\varvec{\chi }} =\overline{\chi _0}\chi _{-1}+\overline{\chi _1}\chi _0,\nonumber \\ |\langle x_+\rangle _{\varvec{\chi }}|^2= & {} |\chi _0|^2\left( |\chi _1|^2 +|\chi _{-1}|^2\right) +(\chi _0^2\overline{\chi _1} \overline{\chi _{-1}}+\overline{\chi _0}^2\chi _1\chi _{-1})=s(1-s) +2st\cos \alpha ,\nonumber \\ (\Delta \varvec{x})^2_{\varvec{\chi }}= & {} \langle \varvec{x}^2\rangle _{\varvec{\chi }}-|\langle x_+\rangle _{\varvec{\chi }}|^2 =\frac{1}{2}\left[ 1-s\right] +s^2-2st\cos \alpha \end{aligned}$$

(82)

where $s:=|\chi _0|^2\le 1$, $t:=|\chi _1\chi _{-1}|$, and $\alpha$ is the phase of $\chi _0^2\overline{\chi _1}\overline{\chi _{-1}}$; by the Cauchy–Schwarz inequality $t\le \left( |\chi _1|^2+|\chi _{-1}|^2\right) /2=(1-s)/2$. For fixed s, (82) is minimized by $\alpha =0$ and $t=(1-s)/2$ (namely $|\chi _1|=|\chi _{-1}|=\sqrt{t}=\sqrt{(1-s)/2}$), what then yields

$$\begin{aligned} (\Delta \varvec{x})^2_{\varvec{\chi }} =\frac{1}{2}(1-s)+s^2-s(1-s)=2s^2-\frac{3}{2} s+\frac{1}{2}. \end{aligned}$$

This is minimized by $s=3/8$, and the minimum value is $(\Delta \varvec{x})^2_{\mathrm{{min}}} =7/32$, as claimed. The corresponding minimizing vectors are $\underline{\varvec{\chi }}= \frac{\sqrt{5}}{4}\left[ \mathrm{e}^{i\beta }\psi _{-1}+\mathrm{e}^{i\gamma }\psi _1\right] +\frac{\sqrt{3}}{\sqrt{8}}\mathrm{e}^{i(\beta +\gamma )/2}\psi _0$; the one in (36) is chosen so that $\langle x_+\rangle \in \mathbb {R}$.

Next, we prove (37). Up to normalization, the components of the eigenvector $\varvec{\chi }$ of the Toeplitz matrix $X_0$ with the maximal eigenvalue ($\lambda _M=\cos \left[ \pi /(2\Lambda +2)\right]$) are [see (20)]

$$\begin{aligned} \chi _m=\sin \left[ \frac{\pi (\Lambda +1 +m)}{2\Lambda +2}\right] = \cos \left[ \frac{\pi m}{2\Lambda +2}\right] ; \end{aligned}$$

(83)

then $\langle \varvec{\chi },\varvec{\chi }\rangle =\Lambda +1$,

$$\begin{aligned} \begin{aligned} \langle \varvec{\chi },\varvec{x}^2\varvec{\chi }\rangle&{\mathop {=}\limits ^{(81)}} \langle \varvec{\chi },\varvec{\chi }\rangle + 2\sum \limits _{m=1}^{\Lambda -1}\frac{m^2}{k}\chi _m^2+ \left[ \frac{\Lambda (\Lambda -1)}{k}-1\right] \, \chi _{\Lambda }^2 \\&= \Lambda +1+ 2\sum \limits _{m=1}^{\Lambda -1}\frac{m^2}{k}\chi _m^2+ \left[ \frac{\Lambda (\Lambda -1)}{k}-1\right] \, \chi _{\Lambda }^2\\&\overset{\chi _m^2\le 1}{\le }\Lambda +1+ 2\sum \limits _{m=1}^{\Lambda -1}\frac{m^2}{k}+ \left[ \frac{\Lambda (\Lambda -1)}{k}-1\right] \, \chi _{\Lambda }^2\\&\overset{(5)}{\le }\Lambda +1+ 2\sum \limits _{m=1}^{\Lambda -1}\frac{m^2}{k} {\mathop {=}\limits ^{(72)}} \Lambda +1+\frac{\Lambda (\Lambda -1)(2\Lambda -1)}{3k}, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned} \langle \varvec{x}^2\rangle _{\varvec{\chi }}&= \frac{\langle \varvec{\chi },\varvec{x}^2\varvec{\chi }\rangle }{\langle \varvec{\chi },\varvec{\chi }\rangle }\le 1+ \frac{\Lambda (\Lambda -1)(2\Lambda -1)}{3k(\Lambda +1)} \overset{(5)}{\le }1\\&\quad +\, \frac{\Lambda (\Lambda -1)(2\Lambda -1)}{3\Lambda ^2(\Lambda +1)^3}\le 1+ \frac{1}{(\Lambda +1)^2}. \end{aligned} \end{aligned}$$

(84)

Moreover, due to (76), (77), $\chi _{-m}=\chi _m\in \mathbb {R}$, it is $\langle x_1\rangle _{\varvec{\chi }}=\langle x_+\rangle _{\varvec{\chi }}$ because the latter is real, whence

$$\begin{aligned} \langle \varvec{\chi },x_1\varvec{\chi }\rangle&{\mathop {=}\limits ^{(80)}} 2 \sum \limits _{m=1}^{\Lambda } b_m\sin \left[ \frac{\pi (\Lambda +1+m)}{2\Lambda +2}\right] \sin \left[ \frac{\pi (\Lambda +m)}{2\Lambda +2}\right] \nonumber \\&\overset{b_m\ge 1}{\ge }2 \sum \limits _{m=1}^{\Lambda } \sin \left[ \frac{\pi (\Lambda +1 +m)}{2\Lambda +2}\right] \sin \left[ \frac{\pi (\Lambda +m)}{2\Lambda +2}\right] \nonumber \\&{\mathop {=}\limits ^{(79)}} \sum \limits _{m=1}^{\Lambda } \left\{ \cos \left[ \frac{\pi }{2\Lambda +2}\right] +\cos \left[ \frac{\pi (2m-1)}{2\Lambda +2}\right] \right\} \nonumber \\&\overset{(78)}{=}(\Lambda +1) \cos \left[ \frac{\pi }{2\Lambda +2}\right] \quad \Longrightarrow \nonumber \\ \langle x_1\rangle _{\varvec{\chi }}^2&= \left( \frac{\langle \varvec{\chi },x_1\varvec{\chi }\rangle }{\langle \varvec{\chi },\varvec{\chi }\rangle }\right) ^2 \ge \cos ^2\left[ \frac{\pi }{2\Lambda +2}\right] =1{-}\sin ^2 \left[ \frac{\pi }{2\Lambda +2}\right] {\ge } 1-\left( \frac{\pi }{2\Lambda +2}\right) ^2, \end{aligned}$$

(85)

$$ \begin{aligned} \left( \Delta {\varvec{x}}\right) ^2_{\varvec{\chi }}&=\langle \varvec{x}^2\rangle _{\varvec{\chi }}- \langle x_1\rangle _{\varvec{\chi }}^2 \nonumber \\&\overset{(84) \& (85)}{\le }1+\frac{1}{(\Lambda +1)^2}-1+\left( \frac{\pi }{2\Lambda +2}\right) ^2 =\frac{1+\frac{\pi ^2}{4}}{(\Lambda +1)^2}<\frac{3.5}{(\Lambda +1)^2}. \end{aligned}$$

(86)

1.4 States saturating the Heisenberg UR (13) on $S^1,S^1_\Lambda$

For any $\mu \in \mathbb {R}$, $i=1,2$ let $a^\mu _i:=L-i\mu x_i$, $z_i:=\langle L\rangle - i\mu \langle x_i\rangle$, $A^\mu _i:= a^\mu _i-z_i$. The inequality $0\le \langle A^\mu _i{}^\dagger A^\mu _i\rangle =\left( \Delta L\right) ^2+\mu ^2(\Delta x_i)^2+\mu \epsilon ^{ij}\langle x_j\rangle$ (here $\epsilon ^{11}=\epsilon ^{22}=0$, $\epsilon ^{12}=-\epsilon ^{21}=1$, and a sum over $j=1,2$ is understood) is saturated on the states annihilated by $A^\mu _i$, which are the eigenvectors $\varvec{\chi }=\sum _{n}\chi _n\psi _n$ of $a^\mu _i$; here, the sum runs over $n\in \mathbb {Z}$ for $S^1$ [where by $\psi _n$ we mean $(x_+)^n=\mathrm{e}^{in\varphi }$], over $n\in I_\Lambda :=\{-\Lambda , 1-\Lambda ,\ldots ,\Lambda \}$ for $S^1_\Lambda$. We can just stick to $i=1$; the UR will be thus saturated on the eigenvectors of $a^\mu _1$. The results for $a^\mu _2$ can be obtained by a rotation of $\pi /2$, by the O(2)-equivariance.

One easily checks that $a^\mu _1\varvec{\chi }=z{}\varvec{\chi }$ in ${{\mathcal {H}}}=\mathcal{L}^2(S^1)$ amounts to the equations

$$\begin{aligned} 2\chi _n(n-z{})-i\mu (\chi _{n+1}+\chi _{n-1})=0, \quad n\in \mathbb {Z}. \end{aligned}$$

(87)

One way to fulfill them (with a non trivial $\varvec{\chi }$) is with $\mu =0$; this implies $\chi _n=0$ for all n but one, i.e., $\varvec{\chi }\propto \psi _m$ for some $m\in \mathbb {Z}$, and $z{}=\langle L\rangle =m$. This is actually the only way: if $\mu \ne 0$, then the equations can be used as recurrence relations to determine all the $\chi _n$ as combinations of two, e.g., $\chi _0,\chi _1$; if the latter vanish so do all $\chi _n$, otherwise the resulting sequence does not lead to a $\varvec{\chi }\in {{\mathcal {H}}}$ because $\sum _n|\chi _n|^2=\infty$. In fact, rewriting (87) in the form $\chi _{n+1}=-\chi _{n-1}+C_n\chi _n$, with $C_n:=\frac{2}{i\mu }(n-z{})$ it is easy to iteratively prove the relation

$$\begin{aligned} \chi _{n+1}=\chi _{n}Q_n -\frac{\chi _0}{Q_1Q_2\ldots .Q_{n-1}},\quad Q_1:=C_1,\quad Q_n:=C_n-\frac{1}{Q_{n-1}}. \end{aligned}$$

This implies that as $n\rightarrow \infty$$|C_n|\rightarrow \infty$, $|Q_n|\simeq |C_n|\rightarrow \infty$, $|\chi _{n+1}/\chi _{n}|^2\simeq |Q_n|^2\rightarrow \infty$, whence by the D’Alembert criterion the series $\sum _{n=0}^\infty |\chi _{n}|^2$ diverges. The $\psi _m$ are also eigenvectors of $a^2_{\mu =0}$ and therefore saturate not only (13)$_1$, but also (13)$_2$, and therefore all of (13).

One easily checks that the eigenvalue equation $a^\mu _1\varvec{\chi }=z{}\varvec{\chi }$ in ${{\mathcal {H}}}_\Lambda$ (i.e., on $S^1_\Lambda$) amounts to the equations

$$\begin{aligned} \begin{array}{ll} 2\chi _{-\Lambda }(\Lambda +z{})+i\mu \, b_{1-\Lambda }\chi _{1-\Lambda }=0,&{}\\ 2\chi _n(n-z{})-i\mu (b_{n+1}\chi _{n+1}+b_n\chi _{n-1})=0, \quad &{} n=1-\Lambda ,2-\Lambda ,\ldots ,\Lambda -1,\\ 2\chi _{\Lambda }(\Lambda -z{})-i\mu \, b_{\Lambda }\chi _{\Lambda -1}=0 &{} \end{array} \end{aligned}$$

(88)

(actually the second equations include also the first, third, because for $n=\pm \Lambda$, $b_{-\Lambda }=b_{\Lambda +1}=0$). One way to fulfill (88) is with $\mu =0$; this implies $\chi _n=0$ for all n but one, i.e., $\varvec{\chi }\propto \psi _m$ for some $m\in I_\Lambda$, and $z{}=\langle L\rangle =m$. But nontrivial solutions exist also with nonzero $\mu \ne 0$. In fact, equations (88) can be used as recurrence relations to determine all the $\chi _n$ in terms of one. If we use them in the order to express first $\chi _{1-\Lambda }$ as $\chi _{-\Lambda }$ times a factor, then $\chi _{2-\Lambda }$ as $\chi _{-\Lambda }$ times another factor, etc., then the last equation amounts to the eigenvalue equation, a polynomial equation in z of degree $(2\Lambda +1)$. Note that if z is an eigenvalue and $\varvec{\chi }$ the corresponding eigenvector, then also $z'=-z$ is an eigenvalue with corresponding eigenvector characterized by components $\chi _n'=(-1)^n\chi _{-n}$. Since $a^\mu _2=\mathrm{e}^{-i\pi L/2}a^\mu _1 \mathrm{e}^{i\pi L/2}$, to each eigenvector $\varvec{\chi }$ of $a^\mu _1$ there corresponds the one $\varvec{\chi }'=\mathrm{e}^{-i\pi L/2}\varvec{\chi }$ of $a^\mu _2$ with the same eigenvalue z and components related by $\chi '_n=\chi _n(-i)^n$. Hence, $\varvec{\chi }$ cannot be a simultaneous eigenvector of $a^\mu _1,a^\mu _2$ and therefore again cannot saturate all of (13), but only one of the first two inequalities, unless $\mu =0$, namely unless it is an eigenvector of L; hence, again the $\psi _m$ are the only states saturating all of (13).

We determine the eigenvectors of $a_1^\mu$ for $\Lambda =1$. The eigenvalue equation amounts to $z(z^2-1+\mu ^2/2)=0$. We easily find that (88) admits the following solutions:

$$\begin{aligned} z{}=0, \, \pm \sqrt{1-\frac{\mu ^2}{2}} ,\quad \varvec{\chi }=\chi _{-1} \left\{ \psi _{-1}+ \frac{2i}{\mu }(1+z{})\psi _0- \left[ 1+\frac{4z{}}{\mu ^2}(1+z{})\right] \psi _1\right\} .\nonumber \\ \end{aligned}$$

(89)

$\Vert \varvec{\chi }\Vert ^2=1$ amounts to $\frac{|\chi _{-1}|^2}{\mu ^4}\left\{ \mu ^4+4\mu ^2|1+z{}|^2+ \left| \mu ^2+4z{}(1+z{})\right| ^2\right\} =1$. This leads to

$$\begin{aligned}&z{}=0\quad \Rightarrow \quad |\chi _{-1}|^2=\frac{\mu ^2}{2\mu ^2+4} \\&z{}=\pm \sqrt{1-\frac{\mu ^2}{2}}\Rightarrow z\in \left\{ \begin{array}{l}\mathbb {R},\\ i\mathbb {R},\end{array}\right. \quad |\chi _{-1}|^2 =\left\{ \begin{array}{ll}\frac{\mu ^4}{32(1+z)-8\mu ^2} &{}\quad \text{ if } \mu ^2\le 2,\\ \frac{1}{4}&{}\quad \text{ if } \mu ^2\ge 2.\end{array}\right. \end{aligned}$$

In the $\mu \rightarrow 0$ limit, we recover the eigenvectors $\psi _1,\psi _0,\psi _{-1}$ of L with eigenvalues $-1,0,1$, whereas in the $\mu \rightarrow \infty$ limit we recover the eigenvectors $\varphi _-,\varphi _0,\varphi _+$ of $x_1$ with eigenvalues $-\sqrt{2}/2,0,\sqrt{2}/2$ (we obtain them in the reverse order $\varphi _+,\varphi _0,\varphi _-$ in the limit $\mu \rightarrow -\infty$). On the other hand, if $\mu ^2=2$, then all eigenvalues coincide with the zero eigenvalue, which remains with geometric multiplicity 1; in other words, in this case (only) there is no basis of ${{\mathcal {H}}}_\Lambda$ consisting of eigenvectors of $a_1^\mu$. Moreover, recalling that $z=\langle L\rangle - i\mu \langle x_1\rangle$ we find that if $\mu ^2\le 2$, then $\langle x_1\rangle =0$ on all eigenvectors (because z is real), whereas if $\mu ^2\ge 2$, then $\langle L\rangle =0$ on all eigenvectors (because z is purely imaginary). One easily checks that

$$\begin{aligned} \langle x_1\rangle + i\langle x_2\rangle =\langle x_+\rangle =\frac{2i}{\mu }|\chi _{-1}|^2\left[ 2+z+\bar{z}+\frac{4z}{\mu ^2}|1+z{}|^2\right] , \end{aligned}$$

leading to

$$\begin{aligned} z{}= & {} 0,\quad \mu ^2\le 2 \Rightarrow \langle x_1\rangle =0,\quad \langle x_2\rangle =\frac{2\mu }{\mu ^2+2},\langle \varvec{x}\rangle ^2= \frac{4\mu ^2}{(\mu ^2+2)^2}, \end{aligned}$$

(90)

$$\begin{aligned} z{}= & {} 0,\quad \mu ^2\ge 2\quad \Rightarrow \quad \langle x_1\rangle =0,\quad \langle x_2\rangle =\frac{1}{\mu },\quad \langle \varvec{x}\rangle ^2=\frac{1}{\mu ^2}, \end{aligned}$$

(91)

$$\begin{aligned} z{}= & {} \pm \sqrt{1-\frac{\mu ^2}{2}},\quad \mu ^2\le 2\quad \Rightarrow \quad \langle x_1\rangle =0, \quad \langle x_2\rangle =\frac{\mu }{2},\langle \varvec{x}\rangle ^2=\frac{\mu ^2}{4}, \end{aligned}$$

(92)

$$\begin{aligned} z{}= & {} \pm i\sqrt{\frac{\mu ^2}{2}- 1},\quad \mu ^2\ge 2\quad \Rightarrow \quad \langle x_1\rangle =\frac{\mp 1}{\mu }\sqrt{\frac{\mu ^2}{2}- 1}, \nonumber \\ \langle x_2\rangle= & {} \frac{1}{\mu },\quad \langle \varvec{x}\rangle ^2=\frac{1}{2}. \end{aligned}$$

(93)

As on ${{\mathcal {H}}}_1$ it is $\varvec{x}^2=1-({\tilde{P}}_1+{\tilde{P}}_{-1})/2$ we find

$$\begin{aligned} \langle \varvec{x}^2\rangle =1-\frac{|\chi _{-1}|^2}{2}\left\{ 1+\left| 1+\frac{4z}{\mu ^2}(1+z)\right| ^2\right\} \end{aligned}$$

leading to

$$\begin{aligned} z{}= & {} 0,\quad \mu ^2\le 2 \Rightarrow \langle \varvec{x}^2\rangle =\frac{\mu ^2+4}{2(\mu ^2+2)},\quad (\Delta \varvec{x})^2=\frac{1}{2}+\frac{2-3\mu ^2}{(\mu ^2+2)^2}, \end{aligned}$$

(94)

$$\begin{aligned} z{}= & {} 0,\quad \mu ^2\ge 2\quad \Rightarrow \quad \langle \varvec{x}^2\rangle =\frac{\mu ^2+4}{2(\mu ^2+2)},\quad (\Delta \varvec{x})^2=\frac{\mu ^4+2\mu ^2-4}{2\mu ^2(\mu ^2+2)}, \end{aligned}$$

(95)

$$\begin{aligned} z{}= & {} \pm \sqrt{1-\frac{\mu ^2}{2}},\quad \mu ^2\le 2\quad \Rightarrow \quad \langle \varvec{x}^2\rangle =\frac{1}{2}+\frac{\mu ^2}{8}, \quad (\Delta \varvec{x})^2=\frac{1}{2}-\frac{\mu ^2}{8}, \end{aligned}$$

(96)

$$\begin{aligned} z{}= & {} \pm i\sqrt{\frac{\mu ^2}{2}- 1},\quad \mu ^2\ge 2\quad \Rightarrow \quad \langle \varvec{x}^2\rangle =\frac{3}{4}, \quad (\Delta \varvec{x})^2=\frac{1}{4}. \quad \end{aligned}$$

(97)

We also find

$$\begin{aligned} z{}=0,\quad\Rightarrow & {} \quad \left( \Delta L\right) ^2=\langle L^2\rangle =\frac{\mu ^2}{\mu ^2+2}, \end{aligned}$$

(98)

$$\begin{aligned} z{}=\pm \sqrt{1-\frac{\mu ^2}{2}},\quad \mu ^2\le 2\quad\Rightarrow & {} \quad \left( \Delta L\right) ^2=1- \frac{\mu ^2}{4}- \left[ 1-\frac{\mu ^2(1+z)}{4(1+z)-\mu ^2} \right] ^2, \end{aligned}$$

(99)

$$\begin{aligned} z{}=\pm i\sqrt{\frac{\mu ^2}{2}- 1},\quad \mu ^2\ge 2\quad\Rightarrow & {} \quad \left( \Delta L\right) ^2=\langle L^2\rangle =\frac{1}{2}. \quad \end{aligned}$$

(100)

For all $\mu$$\varvec{\chi }_\alpha :=\mathrm{e}^{i\alpha L}\varvec{\chi }$ is characterized by the same $(\Delta \varvec{L})^2,(\Delta \varvec{x})^2$ as $\varvec{\chi }$. For all $\mu \ne 0$ and any of the eigenvectors $\varvec{\chi }$ of $a^\mu _1$ the system $X:=\{\varvec{\chi }_\alpha \}_{\alpha \in [0,2\pi [}$ is complete (actually overcomplete), but the resolution of the identity $\int _0^{2\pi } \mathrm{d}\alpha \,\varvec{\chi }_\alpha \langle \varvec{\chi }_\alpha ,\cdot \rangle =c I$ does not hold.

1.5 Proof of Theorem 4.2

This is based on the following two lemmas:

Lemma 6.1

Let $P^h=\sum _{l=|h|}^{\Lambda } {\varvec{\psi }}_l^h\langle {\varvec{\psi }}_l^h,\cdot \rangle$ be the projector on the $L_3=h$ eigenspace. Then,

$$\begin{aligned} \int ^{2\pi }_0\mathrm{d}\alpha \, \mathrm{e}^{i\alpha (L_3-h)}=2\pi P^h . \end{aligned}$$

(101)

This can be proved applying both sides to the basis vectors ${\varvec{\psi }}_l^m$. In Sect. 6.6, we prove

Lemma 6.2

If $\vert h\vert ,\vert n\vert \le l,j$ then

$$\begin{aligned} \int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^n\rangle =\frac{2}{2l+1}\delta _{lj}. \end{aligned}$$

(102)

Now let $B:=\int _{SO(3)} \mathrm{d}\mu (g) \,P_g^\beta$, with a generic $\varvec{\omega }=\sum \nolimits _{l=0}^{\Lambda }\sum \nolimits _{h=-l}^{l} \omega _l^h{\varvec{\psi }}_l^h$; we compute $B{\varvec{\psi }}_l^n$ ($|n|\le l$):

$$\begin{aligned} B{\varvec{\psi }}_l^n&=\int ^{2\pi }_0 \mathrm{d}\varphi \int ^{\pi }_0\mathrm{d}\theta \sin \theta \int ^{2\pi }_0\mathrm{d}\alpha \, \mathrm{e}^{i\varphi L_3}\mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega }\, \langle \mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega },\mathrm{e}^{-i\varphi L_3}{\varvec{\psi }}_l^n\rangle \\&{\mathop {=}\limits ^{(39)}}\int ^{2\pi }_0\mathrm{d}\varphi \,\mathrm{e}^{i\varphi (L_3-n)}\int ^{\pi }_0\mathrm{d}\theta \sin \theta \int ^{2\pi }_0\mathrm{d}\alpha \, \mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega }\, \langle \mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega },{\varvec{\psi }}_l^n\rangle \\&{\mathop {=}\limits ^{(101)}}2\pi \sum _{j=|n|}^{\Lambda } {\varvec{\psi }}_{j}^n\int ^{\pi }_0\mathrm{d}\theta \sin \theta \int ^{2\pi }_0\mathrm{d}\alpha \, \langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega }\rangle \, \langle \mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}\varvec{\omega },{\varvec{\psi }}_l^n\rangle \\&\,\,=2\pi \sum _{j=|n|}^{\Lambda } {\varvec{\psi }}_{j}^n\int ^{\pi }_0\mathrm{d}\theta \sin \theta \int ^{2\pi }_0\mathrm{d}\alpha \, \sum _{h=-l}^{l}\overline{\omega _l^h}\sum _{m=-j}^{j} \omega _j^m\langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}{\varvec{\psi }}_{j}^{m}\rangle \, \\&\quad \langle \mathrm{e}^{i\theta L_2}\mathrm{e}^{i\alpha L_3}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^n\rangle \\&\,\,=2\pi \sum _{j=|n|}^{\Lambda } {\varvec{\psi }}_{j}^n\sum _{h=-l}^{l}\overline{\omega _l^h}\sum _{m=-j}^{j}\omega _j^m \int ^{\pi }_0\mathrm{d}\theta \sin \theta \int ^{2\pi }_0\mathrm{d}\alpha \, \mathrm{e}^{i\alpha (m-h)} \langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{m}\rangle \, \\&\quad \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^n\rangle \\&{\mathop {=}\limits ^{(101)}}(2\pi )^2\sum _{j=|n|}^{\Lambda } {\varvec{\psi }}_{j}^n\sum _{h=-m_{jl}}^{m_{jl}}\overline{\omega _l^h}\omega _j^h \int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^n\rangle \\ \end{aligned}$$

where $m_{jl}:=\min \{j,l\}$. By (102) this becomes $B{\varvec{\psi }}_l^n={\varvec{\psi }}_{l}^n\sum \nolimits _{h=-l}^{l} |\omega _l^h|^2 8\pi ^2/(2l+1)$. In order that this equals $C{\varvec{\psi }}_l^n$, i.e., that $B=CI$ with some constant $C>0$, it must be $\sum \nolimits _{h=-l}^{l}|\omega _l^h|^2 =C(2l+1)/8\pi ^2$ for all $l=0,\ldots ,\Lambda$. Summing over l and imposing that ${\varvec{\omega }}$ be normalized, we find

$$\begin{aligned} 1= & {} \Vert \varvec{\omega }\Vert ^2=\sum _{l=0}^{\Lambda }\sum \limits _{h=-l}^{l}|\omega _l^h|^2 =\sum _{l=0}^{\Lambda }\frac{2l+1}{8\pi ^2}C\nonumber \\= & {} \frac{(\Lambda +1)^2}{8\pi ^2}C \quad \Rightarrow \quad C=\frac{8\pi ^2}{(\Lambda +1)^2}, \end{aligned}$$

(103)

as claimed. The strong SCS $\{{\varvec{\omega }}_g\}_{g\in SO(3)}$ is fully O(3)-equivariant if $\omega _l^h=\omega _l^{-h}$ , because then it is mapped into itself also by the unitary transformation ${\varvec{\psi }}_l^h\mapsto {\varvec{\psi }}_l^{-h}$ that corresponds to the transformation of the coordinates (with determinant -1) $(x_1,x_2,x_3)\mapsto (x_1,-x_2,x_3)$.

1.6 Proof of Lemma 6.2

First we recall that, denoting as F(a, b; c; z) the Gauss hypergeometric function and as $(z)_n$ the Pochhammer’s symbol, then, by definition,

$$\begin{aligned} (z)_n:=\frac{\Gamma (z+n)}{\Gamma (z)}\quad \text{ and }\quad F(-n,b;c;z):=\sum _{m=0}^n{n\atopwithdelims ()m} \frac{(-1)^m z^m (b)_m}{(c)_m}. \end{aligned}$$

(104)

According to [44] p. 561 eq 15.4.6, one has

$$\begin{aligned} F(-n,\alpha +1+\beta +n;\alpha +1;x)=\frac{n!}{(\alpha +1)_n}P_n^{(\alpha ,\beta )}(1-2x), \end{aligned}$$

(105)

where $P_n^{(\alpha ,\beta )}$ is the Jacobi polynomial. From p. 556 Eq. 15.1.1 one has

$$\begin{aligned} F(a,b;c;z)=F(b,a;c;z), \end{aligned}$$

(106)

p. 559 Eq. 15.3.3

$$\begin{aligned} F(a,b;c;z)=(1-z)^{c-a-b}F(c-a,c-b;c;z) \end{aligned}$$

(107)

and from p. 774

$$\begin{aligned}&\int _{-1}^1(1-x)^{\alpha }(1+x)^{\beta }P_n^{(\alpha ,\beta )}(x)P_m^{(\alpha ,\beta )}(x)\mathrm{d}x\nonumber \\&\quad =\frac{2^{\alpha +\beta +1}}{2n+\alpha +\beta +1}\frac{\Gamma (n+\alpha +1)\Gamma (n+\beta +1)}{n!\Gamma (n+\alpha +\beta +1)}\delta _{nm}. \end{aligned}$$

(108)

In addition, we need the following

Proposition 6.1

Let $l\ge s\ge h\ge -l$ and

$$\begin{aligned} f(l,h,s):=\left\{ \begin{array}{ll} \prod _{j=h}^{s-1}\left[ l(l+1)-j(j+1)\right] &{}\quad \text{ if }\,\,h<s,\\ 1&{}\quad \text{ if } \,\,h=s; \end{array}\right. \end{aligned}$$

(109)

then

$$\begin{aligned} f(l,h,s)=\frac{(l-h)!(l+s)!}{(l+h)!(l-s)!}. \end{aligned}$$

(110)

Proof

When $h=s$,

$$\begin{aligned} f(l,h,h)=1=\frac{(l-h)!(l+h)!}{(l+h)!(l-h)!}; \end{aligned}$$

assume that $h<s$ and (induction hypothesis)

$$\begin{aligned} f(l,h,s-1)=\frac{(l-h)!(l+s-1)!}{(l+h)!(l-s+1)!}, \end{aligned}$$

so

$$\begin{aligned} f(l,h,s)= & {} f(l,h,s-1)\left[ l(l+1)-(s-1)s\right] \\= & {} \frac{(l-h)!(l+s-1)!}{(l+h)!(l-s+1)!}(l+s)(l-s+1)=\frac{(l-h)!(l+s)!}{(l+h)!(l-s)!}. \end{aligned}$$

$\square$

In the same way, one can prove that when $l\ge s\ge h\ge -l$, and setting

$$\begin{aligned} g(l,h,s):=\left\{ \begin{array}{ll} \prod _{j=h+1}^{s}\left[ l(l+1)-j(j-1)\right] &{}\quad \text{ if }\,\,h<s,\\ 1&{}\quad \text{ if }\,\, h=s; \end{array}\right. \end{aligned}$$

(111)

then

$$\begin{aligned} g(l,h,s)=\frac{(l-h)!(l+s)!}{(l+h)!(l-s)!}; \end{aligned}$$

(112)

so, when $l\ge s\ge h\ge -l$,

$$\begin{aligned} f(l,h,h)= & {} 1=g(l,-h,-h)\quad \text{ and }\nonumber \\ f(l,h,s)= & {} \prod _{j=h}^{s-1}\left[ l(l+1)-j(j+1)\right] \nonumber \\= & {} \prod _{j=-s+1}^{-h}\left[ l(l+1)-j(j-1)\right] =g(l,-s,-h). \end{aligned}$$

(113)

We need to point out that when $0\le n\le h\le l$,

$$\begin{aligned} A&:= \left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^h, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^n\right\rangle \\ \overset{(109)}{=}&\left\langle \sum _{s=h}^l\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{s-h}}{(s-h)!}\sqrt{f(l,h,s)}{\varvec{\psi }}_l^s,\right. \\&\quad \left. \sum _{r=n}^l \frac{\left( -1\right) ^{r-n}\left( \tan {\frac{\theta }{2}}\right) ^{r-n}}{(r-n)!}\sqrt{f(l,n,r)}{\varvec{\psi }}_l^r \right\rangle \\ \overset{n\le h}{=}&\left( -1\right) ^{h-n} \sum _{s=h}^l (-1)^{s-h}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-h}}{(s-h)!}\sqrt{f(l,h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}\\&= \left( -1\right) ^{h-n}\sum _{s=h}^l (-1)^{s-h}\frac{1}{(s-h)!}\sqrt{f(l,h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\\&\quad \left( \sin {\frac{\theta }{2}}\right) ^{2s-n-h} \frac{1}{(s-n)!}\sqrt{f(l,n,s)}\\ \overset{(110)}{=}&\left( -1\right) ^{h-n}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\left( \sin {\frac{\theta }{2}}\right) ^{h-n}\\&\quad \sqrt{\frac{(l-n)!(l-h)!}{(l+n)!(l+h)!}}\sum _{s=h}^{l}(-1)^{s-h}\frac{(l+s)!}{(l-s)!(s-n)!(s-h)!}\left( \sin {\frac{\theta }{2}}\right) ^{2(s-h)}\\ \overset{j=s-h}{=}&\left( -1\right) ^{h-n}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\left( \sin {\frac{\theta }{2}}\right) ^{h-n}\sqrt{\frac{(l-n)!(l-h)!}{(l+n)!(l+h)!}}\sum _{j=0}^{l-h}(-1)^{j}\\&\quad \frac{(l+h+j)!}{(l-h-j)!(h-n+j)!(j)!}\left( \sin {\frac{\theta }{2}}\right) ^{2j},\\ \end{aligned}$$

$$\begin{aligned} A&\overset{(104)}{=} \left( -1\right) ^{h-n}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\left( \sin {\frac{\theta }{2}}\right) ^{h-n}\sqrt{\frac{(l-n)!(l-h)!}{(l+n)!(l+h)!}}\nonumber \\&\qquad \qquad \quad \cdot \frac{(l+h)!}{(l-h)!(h-n)!}F\left( -(l-h),l+h+1;h-n+1;\left( \sin {\frac{\theta }{2}}\right) ^{2} \right) \nonumber \\&\overset{(105)}{=} \left( -1\right) ^{h-n}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\left( \sin {\frac{\theta }{2}}\right) ^{h-n}\nonumber \\&\quad \sqrt{\frac{(l-n)!(l+h)!}{(l+n)!(l-h)!}}\, \frac{(l-h)!(h-n)!}{(h-n)!(l-n)!} \,P_{l-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) \nonumber \\&\quad = \left( -1\right) ^{h-n}\left( \cos {\frac{\theta }{2}}\right) ^{n+h}\left( \sin {\frac{\theta }{2}}\right) ^{h-n}\nonumber \\&\quad \sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}} \, P_{l-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) ,\end{aligned}$$

(114)

$$\begin{aligned}&\left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-n}\right\rangle \nonumber \\&\quad \overset{(111)}{=}\left\langle \sum _{s=-h}^{-l}\left( \cos {\frac{\theta }{2}}\right) ^{-2s} \frac{\left( -1\right) ^{-s-h}\left( \tan {\frac{\theta }{2}}\right) ^{-s-h}}{(-s-h)!}\sqrt{g(l,s,-h)}{\varvec{\psi }}_l^s,\right. \nonumber \\&\quad \left. \sum _{r=-n}^{-l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{-r-n}}{(-r-n)!}\sqrt{g(l,r,-n)}{\varvec{\psi }}_l^r \right\rangle \nonumber \\&\quad \overset{(113)}{=} \left\langle \sum _{s=-h}^{-l}\left( \cos {\frac{\theta }{2}}\right) ^{-2s} \frac{\left( -1\right) ^{-s-h}\left( \tan {\frac{\theta }{2}}\right) ^{-s-h}}{(-s-h)!}\sqrt{f(l,h,-s)}{\varvec{\psi }}_l^s,\right. \nonumber \\&\quad \left. \sum _{r=-n}^{-l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{-r-n}}{(-r-n)!}\sqrt{f(l,n,-r)}{\varvec{\psi }}_l^r \right\rangle \nonumber \\&\quad \overset{r, s\rightarrow -r,-s}{=}\left\langle \sum _{s=h}^{l}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left( -1\right) ^{s-h}\left( \tan {\frac{\theta }{2}}\right) ^{s-h}}{(s-h)!}\sqrt{f(l,h,s)}{\varvec{\psi }}_l^{-s},\right. \nonumber \\&\quad \left. \sum _{r=n}^{l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{r-n}}{(r-n)!}\sqrt{f(l,n,r)}{\varvec{\psi }}_l^{-r} \right\rangle \nonumber \\&\quad \overset{n\le h}{=}\sum _{s=h}^l (-1)^{s-h}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-h}}{(s-h)!}\sqrt{f(l,h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}\nonumber \\&\quad =\sum _{s=h}^l (-1)^{s-h}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-h}}{(s-h)!}\sqrt{f(l,h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}\nonumber \\&\quad \overset{(114)}{=}\left( \cos {\frac{\theta }{2}}\right) ^{n+h} \left( \sin {\frac{\theta }{2}}\right) ^{h-n}\sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}} P_{l-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) , \end{aligned}$$

(115)

$$\begin{aligned} B&:= \left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^{-h}, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^n\right\rangle \nonumber \\&\overset{(109)}{=}\left\langle \sum _{s=-h}^l\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{s+h}}{(s+h)!}\sqrt{f(l,-h,s)}{\varvec{\psi }}_l^s,\right. \nonumber \\&\quad \left. \sum _{r=n}^l \frac{\left( -1\right) ^{r-n}\left( \tan {\frac{\theta }{2}}\right) ^{r-n}}{(r-n)!}\sqrt{f(l,n,r)}{\varvec{\psi }}_l^r \right\rangle \nonumber \\&\overset{-h\le n}{=}\sum _{s=n}^l (-1)^{s-n}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s+h}}{(s+h)!}\sqrt{f(l,-h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}\nonumber \\&=\sum _{s=n}^l (-1)^{s-n}\frac{1}{(s+h)!}\sqrt{f(l,-h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{n-h}\left( \sin {\frac{\theta }{2}} \right) ^{2s-n+h} \nonumber \\&\quad \frac{1}{(s-n)!}\sqrt{f(l,n,s)},\nonumber \\ B&\overset{(110)}{=}\left( \cos {\frac{\theta }{2}}\right) ^{n-h}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l-n)!(l+h)!}{(l+n)!(l-h)!}}\sum _{s=n}^{l}(-1)^{s-n}\nonumber \\&\quad \frac{(l+s)!}{(l-s)!(s-n)!(s+h)!}\left( \sin {\frac{\theta }{2}}\right) ^{2(s-n)}\nonumber \\&\overset{j=s-n}{=}\left( \cos {\frac{\theta }{2}}\right) ^{n-h}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l-n)!(l+h)!}{(l+n)!(l-h)!}}\nonumber \\&\quad \cdot \sum _{j=0}^{l-n}(-1)^{j}\frac{(l+n+j)!}{(l-n-j)!(j)!(h+n+j)!}\left( \sin {\frac{\theta }{2}}\right) ^{2j}\nonumber \\&\overset{(104)}{=}\left( \cos {\frac{\theta }{2}}\right) ^{n-h}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l-n)!(l+h)!}{(l+n)!(l-h)!}}\nonumber \\&\quad \cdot \frac{(l+n)!}{(l-n)!(h+n)!}F\left( -(l-n),l+n+1;h+n+1;\left( \sin {\frac{\theta }{2}}\right) ^{2} \right) \nonumber \\&\overset{(107)}{=}\left( \cos {\frac{\theta }{2}}\right) ^{n-h}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l+n)!(l+h)!}{(l-n)!(l-h)!}}\frac{1}{(h+n)!}\nonumber \\&\quad \cdot \left[ 1-\left( \sin {\frac{\theta }{2}}\right) ^{2}\right] ^{h-n}F\left( l+h+1,-(l-h);h+n+1;\left( \sin {\frac{\theta }{2}}\right) ^{2} \right) \nonumber \\&\overset{(105)}{=}\left( \cos {\frac{\theta }{2}}\right) ^{h-n}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l+n)!(l+h)!}{(l-n)!(l-h)!}}\frac{1}{(h+n)!}\nonumber \\&\quad \cdot \frac{(l-h)!(h+n)!}{(l+n)!} P_{l-h}^{(h+n,h-n)}\left( 1-2\sin ^2{\frac{\theta }{2}} \right) \nonumber \\&=\left( \cos {\frac{\theta }{2}}\right) ^{h-n}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}} P_{l-h}^{(h+n,h-n)}\left( 1-2\sin ^2{\frac{\theta }{2}} \right) \end{aligned}$$

(116)

and

$$\begin{aligned} \begin{aligned} D&:= \left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-n}\right\rangle \\&\overset{(111)}{=}\left\langle \sum _{s=h}^{-l} \left( \cos {\frac{\theta }{2}}\right) ^{-2s} \frac{\left( -1\right) ^{h-s} \left( \tan {\frac{\theta }{2}}\right) ^{h-s}}{(h-s)!}\sqrt{g(l,s,h)}{\varvec{\psi }}_l^s,\right. \\&\quad \left. \sum _{s=-n}^{-l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{-s-n}}{(-s-n)!} \sqrt{g(l,s,-n)}{\varvec{\psi }}_l^s \right\rangle \\&\overset{(113)}{=} \left\langle \sum _{s=h}^{-l}\left( \cos {\frac{\theta }{2}}\right) ^{-2s} \frac{\left( -1\right) ^{h-s}\left( \tan {\frac{\theta }{2}}\right) ^{h-s}}{(h-s)!} \sqrt{f(l,-h,-s)}{\varvec{\psi }}_l^s,\right. \\&\quad \left. \sum _{s=-n}^{-l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{-s-n}}{(-s-n)!}\sqrt{f(l,n,-s)} {\varvec{\psi }}_l^s \right\rangle \\&\overset{s\rightarrow -s}{=}\left\langle \sum _{s=-h}^{l} \left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left( -1\right) ^{s+h} \left( \tan {\frac{\theta }{2}}\right) ^{s+h}}{(s+h)!}\sqrt{f(l,-h,s)}{\varvec{\psi }}_l^{-s},\right. \\&\quad \left. \sum _{s=n}^{l} \frac{\left( \tan {\frac{\theta }{2}}\right) ^{s-n}}{(s-n)!} \sqrt{f(l,n,s)}{\varvec{\psi }}_l^{-s} \right\rangle \\&\overset{-h\le n}{=}\sum _{s=n}^l (-1)^{s+h}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s+h}}{(s+h)!}\sqrt{f(l,-h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s} \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}, \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} D&=(-1)^{h+n}\sum _{s=n}^l (-1)^{s-n}\frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s+h}}{(s+h)!}\sqrt{f(l,-h,s)}\left( \cos {\frac{\theta }{2}}\right) ^{2s}\\&\quad \frac{\left[ \tan {\frac{\theta }{2}} \right] ^{s-n}}{(s-n)!}\sqrt{f(l,n,s)}\\&\overset{(116)}{=} (-1)^{h+n}\left( \cos {\frac{\theta }{2}}\right) ^{h-n}\left( \sin {\frac{\theta }{2}}\right) ^{h+n}\sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}} P_{l-h}^{(h+n,h-n)}\\&\quad \left( 1-2\sin ^2{\frac{\theta }{2}} \right) . \end{aligned} \end{aligned}$$

(117)

Finally, when $l\ge h\ge n\ge 0$, one has

$$\begin{aligned}&\begin{aligned} \frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}\frac{1}{2^{2h}} \cdot \frac{2^{(h+n)+(h-n)+1}}{2(l-h)+(h+n)+(h-n)+1}&\\ \cdot \frac{\Gamma ((l-h)+(h+n)+1)\Gamma ((l-h)+(h-n)+1)}{(l-h)! \Gamma ((l-h)+(h+n)+(h-n)+1)}&=\frac{2}{2l+1}, \end{aligned}\end{aligned}$$

(118)

$$\begin{aligned}&\begin{aligned} \frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}\frac{1}{2^{2h}} \cdot \frac{2^{(h-n)+(h+n)+1}}{2(l-h)+(h-n)+(h+n)+1}&\\ \cdot \frac{\Gamma ((l-h)+(h-n)+1)\Gamma ((l-h)+(h+n)+1)}{(l-h)!\Gamma ((l-h)+(h-n)+(h+n)+1)}&=\frac{2}{2l+1}. \end{aligned} \end{aligned}$$

(119)

We are now ready to prove the aforementioned lemma.

Assume that $0\le n\le h\le l$; by means of the Gauss decomposition, $\mathrm{e}^{i\theta L_2}$ can be written in the “antinormal form” (see, e.g., Eq. (4.3.14) in [35])

$$\begin{aligned} \mathrm{e}^{i\theta L_2} =\mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}\mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}; \end{aligned}$$

(120)

hence

$$ \begin{aligned}&\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^n,\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^n\rangle \\&\quad =\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \overline{\left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_j^h, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_j^n\right\rangle }\\&\quad \left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^h, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+} {\varvec{\psi }}_l^n\right\rangle \\&\quad \overset{(114)}{=}2 (-1)^{2(h-n)} \int _0^{\pi }\mathrm{d}\theta \left( \cos {\frac{\theta }{2}}\right) ^{2(n+h)+1} \left( \sin {\frac{\theta }{2}}\right) ^{2(h-n)+1}\\&\quad \sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}}\sqrt{\frac{(j-h)!(j+h)!}{(j+n)!(j-n)!}} \\&\quad \cdot P_{l-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) P_{j-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) \\&\quad \overset{x=1-2\sin ^2\frac{\theta }{2}}{=}\sqrt{\frac{(l -h)!(l+h)!}{(l+n)!(l-n)!}} \sqrt{\frac{(j-h)!(j+h)!}{(j +n)!(j-n)!}}\\&\quad \int \limits _{-1}^1 \frac{\mathrm{d}x}{2^{2h}}\, (1-x)^{h-n} (1+x)^{h+n}P_{l-h}^{(h-n,h+n)}\left( x\right) P_{j-h}^{(h-n,h+n)}\left( x\right) \\&\quad \overset{(108) \& (118)}{=}\frac{2}{2l+1}\delta _{lj}. \end{aligned}$$

On the other hand, in order to calculate $\displaystyle \int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^{-n},\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{-h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{-h},{\varvec{\psi }}_l^{-n}\rangle ,$ we can use now the “normal form” of the Gauss decomposition (see, e.g., Eq. (4.3.12) in [35])

$$\begin{aligned} \mathrm{e}^{i\theta L_2} =\mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}\mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}, \end{aligned}$$

(121)

and we obtain

$$ \begin{aligned} \begin{aligned}&\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^{-n},\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{-h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{-h},{\varvec{\psi }}_l^{-n}\rangle \\&\quad =\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \overline{\left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_j^{-h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_j^{-n}\right\rangle }\\&\quad \left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-n}\right\rangle \\&\quad \overset{(115)}{=}2 \int _0^{\pi }\mathrm{d}\theta \left( \cos {\frac{\theta }{2}}\right) ^{2(n+h)+1} \left( \sin {\frac{\theta }{2}}\right) ^{2(h-n)+1}\\&\quad \sqrt{\frac{(l-h)! (l+h)!}{(l+n)!(l-n)!}}\sqrt{\frac{(j-h)!(j+h)!}{(j+n)!(j-n)!}} \\&\quad \cdot P_{l-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) P_{j-h}^{(h-n,h+n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) \\&\quad \overset{x=1-2\sin ^2\frac{\theta }{2}}{=}\sqrt{\frac{(l -h)!(l+h)!}{(l+n)!(l-n)!}} \sqrt{\frac{(j-h)!(j+h)!}{(j+n)! (j-n)!}}\frac{1}{2^{2h}}\\&\quad \int \limits _{-1}^1 \mathrm{d}x \,(1-x)^{h-n} (1+x)^{h+n}P_{l-h}^{(h-n,h+n)}\left( x\right) P_{j-h}^{(h-n,h+n)}\left( x\right) \\&\quad \overset{(108) \& (118)}{=}\frac{2}{2l+1}\delta _{lj}. \end{aligned} \end{aligned}$$

Furthermore

$$ \begin{aligned}&\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^{n},\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{-h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{-h},{\varvec{\psi }}_l^{n}\rangle \\&\quad \overset{(120)}{=}\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \overline{\left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_j^{-h}, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_j^n\right\rangle }\\&\quad \left\langle \mathrm{e}^{2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^{-h}, \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_+}{\varvec{\psi }}_l^n\right\rangle \\&\quad \overset{(116)}{=}2 \int _0^{\pi }\mathrm{d}\theta \left( \cos {\frac{\theta }{2}}\right) ^{2(h-n)+1} \left( \sin {\frac{\theta }{2}}\right) ^{2(h+n)+1}\\&\quad \sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}}\sqrt{\frac{(j-h)!(j+h)!}{(j+n)!(j-n)!}} \\&\quad \cdot P_{l-h}^{(h+n,h-n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) P_{j-h}^{(h+n,h-n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) \\&\quad \overset{x=1-2\sin ^2\frac{\theta }{2}}{=}\sqrt{\frac{(l-h)!(l+h)!}{(l+n)! (l-n)!}}\sqrt{\frac{(j-h)! (j+h)!}{(j+n)!(j-n)!}}\frac{1}{2^{2h}}\\&\quad \int \limits _{-1}^1 \mathrm{d}x\, (1-x)^{h+n} (1+x)^{h-n}P_{l-h}^{(h+n,h-n)}\left( x\right) P_{j-h}^{(h+n,h-n)}\left( x\right) \\&\quad \overset{(108) \& (119)}{=}\frac{2}{2l+1}\delta _{lj} \end{aligned}$$

and, finally, as claimed,

$$ \begin{aligned}&E:= \int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \langle {\varvec{\psi }}_{j}^{-n},\mathrm{e}^{i\theta L_2}{\varvec{\psi }}_{j}^{h}\rangle \, \langle \mathrm{e}^{i\theta L_2}{\varvec{\psi }}_l^{h},{\varvec{\psi }}_l^{-n}\rangle \\&\quad \overset{(121)}{=}\int ^{\pi }_0\mathrm{d}\theta \sin \theta \, \overline{\left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_j^{h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_j^{-n}\right\rangle }\\&\qquad \left\langle \mathrm{e}^{-2\log {\left( \cos {\frac{\theta }{2}} \right) }L_0} \mathrm{e}^{-{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{h}, \mathrm{e}^{{\tan {\frac{\theta }{2}}}L_-}{\varvec{\psi }}_l^{-n}\right\rangle \\&\quad \overset{(117)}{=} 2 (-1)^{2(h+n)} \int _0^{\pi }\mathrm{d}\theta \left( \cos {\frac{\theta }{2}}\right) ^{2(h-n)+1}\left( \sin {\frac{\theta }{2}}\right) ^{2(h+n)+1}\\&\qquad \sqrt{\frac{(l-h)!(l+h)!}{(l+n)!(l-n)!}}\sqrt{\frac{(j-h)!(j+h)!}{(j+n)!(j-n)!}} \\&\qquad \cdot P_{l-h}^{(h+n,h-n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) P_{j-h}^{(h+n,h-n)}\left( 1-2\sin ^2\frac{\theta }{2}\right) ,\\&E\overset{x=1-2\sin ^2\frac{\theta }{2}}{=}\sqrt{\frac{(l-h)!(l+h)!}{(l+n)! (l-n)!}}\sqrt{\frac{(j-h)!(j+h)!}{(j+n)!(j-n)!}}\frac{1}{2^{2h}} \\&\qquad \int \limits _{-1}^1 \mathrm{d}x\, (1-x)^{h+n} (1+x)^{h-n}P_{l-h}^{(h+n,h-n)}\left( x\right) P_{j-h}^{(h+n,h-n)}\left( x\right) \\&\qquad \overset{(108) \& (119)}{=}\frac{2}{2l+1}\delta _{lj}. \end{aligned}$$

1.7 Proofs of some results regarding $S^2_\Lambda$

Proof of (57)

$L_+{\varvec{\omega }}^\beta =0$, $L_-{\varvec{\omega }}^\beta$ is a combination of ${\varvec{\psi }}_l^{l-1}$, therefore is orthogonal to ${\varvec{\omega }}^\beta$. Hence,

$$\begin{aligned} \langle L_\pm \rangle _{{\varvec{\omega }}^\beta }=0,\quad \Rightarrow \quad |\langle \varvec{L}\rangle _{{\varvec{\omega }}^\beta }|=\langle L_0\rangle _{{\varvec{\omega }}^\beta }=\sum _{l=0}^{\Lambda }\frac{l(2l +1)}{(\Lambda +1)^2}\overset{(73)_2}{=}\frac{\Lambda (4\Lambda +5)}{6(\Lambda +1)}; \end{aligned}$$

while

$$\begin{aligned} \left\langle \varvec{L}^2\right\rangle _{\varvec{\omega }^{\beta }}=\sum _{l=0}^{\Lambda }\frac{l(l+1)(2l+1)}{(\Lambda +1)^2} \overset{(74)}{=}\frac{\frac{1}{2}\Lambda (\Lambda +1)^2(\Lambda +2)}{(\Lambda +1)^2}=\frac{\Lambda (\Lambda +2)}{2}. \end{aligned}$$

Replacing these results in $(\Delta \varvec{L})^2_{{\varvec{\omega }}^\beta } =\langle \varvec{L}^2\rangle _{{\varvec{\omega }}^\beta }-\langle \varvec{L}\rangle _{{\varvec{\omega }}^\beta }^2$, we find

$$\begin{aligned} (\Delta \varvec{L})^2_{{\varvec{\omega }}^{\beta }}=\frac{\Lambda (\Lambda +2)}{2}-\left( \frac{\Lambda (4\Lambda +5)}{6(\Lambda +1)}\right) ^2=\frac{\Lambda (2\Lambda ^3+32\Lambda ^2+65\Lambda +36)}{36(\Lambda +1)^2}. \end{aligned}$$

On the other hand, $x_0{\varvec{\omega }}^\beta$ is a combination of ${\varvec{\psi }}_{l-1}^{l}, {\varvec{\psi }}_{l+1}^{l}$, therefore is orthogonal to ${\varvec{\omega }}^\beta$, and $\langle x_0\rangle =0$. Hence,

$$\begin{aligned} \left\langle \varvec{x}\right\rangle ^2= & {} \left\langle x_1\right\rangle ^2+ \left\langle x_2\right\rangle ^2+ \left\langle x_3\right\rangle ^2=\frac{\left\langle x_+ +x_- \right\rangle ^2}{4}-\frac{\left\langle x_+-x_-\right\rangle ^2}{4}+\left\langle x_0\right\rangle ^2\\= & {} \left\langle x_+\right\rangle \left\langle x_-\right\rangle =\left| \left\langle x_+\right\rangle \right| ^2,\\ \left( \Delta \varvec{x}\right) ^2= & {} \left\langle \varvec{x}^2\right\rangle -\left| \left\langle x_+\right\rangle \right| ^2. \end{aligned}$$

But

$$\begin{aligned} \left\langle \varvec{x}^2\right\rangle _{\varvec{\omega }^{\beta }}&\overset{(44)_2}{=} 1+\sum _{l=0}^{\Lambda }\frac{l(l+1)+1}{k}\frac{2l+1}{(\Lambda +1)^2} -\left[ 1+\frac{(\Lambda +1)^2}{k}\right] \frac{1}{\Lambda +1}\nonumber \\&=\frac{\Lambda }{\Lambda +1}-\frac{\Lambda +1}{k} +\frac{1}{k(\Lambda +1)^2}\left[ 2\sum _{l=0}^{\Lambda }l(l+1)(l+2)\right. \nonumber \\&\quad \left. -3 \sum _{l=0}^{\Lambda }l(l+1)+\sum _{l=0}^{\Lambda }(2l+1)\right] \nonumber \\&\overset{(75)_1}{=}\frac{\Lambda }{\Lambda +1}-\frac{\Lambda +1}{k}+\frac{1}{k(\Lambda +1)^2} \left[ \frac{\Lambda }{2} (\Lambda +1)(\Lambda +2)(\Lambda +3)\right. \nonumber \\&\quad \left. -\Lambda (\Lambda +1)(\Lambda +2)+(\Lambda +1)^2\right] \nonumber \\&=\frac{\Lambda }{\Lambda +1}-\frac{\Lambda +1}{k} +\frac{1+(\Lambda +1)^2}{2k} =\frac{\Lambda }{\Lambda +1}+\frac{\Lambda ^2}{2k}\nonumber \\&\overset{(5)}{\le }\frac{\Lambda }{\Lambda +1}+\frac{1}{2(\Lambda +1)^2}, \quad \end{aligned}$$

(122)

while

$$\begin{aligned} x_+\varvec{\omega }^{\beta }&\overset{(40)}{=}&\sum _{l=0}^{\Lambda -1}\mathrm{e}^{i\beta _l}\frac{\sqrt{2l+1}}{\Lambda +1}c_{l+1}B_l^{+,l}{\varvec{\psi }}_{l+1}^{l+1}\\&\overset{(41)}{=}&\sum _{l=0}^{\Lambda -1}\mathrm{e}^{i\beta _l}\frac{\sqrt{2l+1}}{\Lambda +1}c_{l+1}\left( -\sqrt{\frac{2l+2}{2l+3}} \right) {\varvec{\psi }}_{l+1}^{l+1}\\= & {} -\sum _{l=1}^{\Lambda }\mathrm{e}^{i\beta _{l-1}}\sqrt{\frac{(2l)(2l-1)}{2l+1}}\frac{c_l}{\Lambda +1}{\varvec{\psi }}_l^l \quad \Longrightarrow \\ \left\langle x_+\right\rangle _{\varvec{\omega }^{\beta }}= & {} -\sum _{l=1}^{\Lambda }\mathrm{e}^{i\left( \beta _{l-1}-\beta _l \right) }\frac{c_l\sqrt{(2l)(2l-1)}}{(\Lambda +1)^2}, \end{aligned}$$

so

$$\begin{aligned} \langle \varvec{x}\rangle _{\varvec{\omega }^{\beta }}^2=\left| \left\langle x_+\right\rangle _{\varvec{\omega }^{\beta }}\right| ^2= \left| \sum _{l=1}^{\Lambda }\frac{c_l2l\sqrt{1-\frac{1}{2l}}}{(\Lambda +1)^2} \mathrm{e}^{i(\beta _{l-1}-\beta _l)}\right| ^2. \end{aligned}$$

Since all $\frac{c_l\sqrt{(2l)(2l-1)}}{(\Lambda +1)^2} > 0$, to maximize $\left| \langle x_+\rangle _{{\varvec{\omega }}^\beta }\right|$, and thus minimize $(\Delta \varvec{x})^2_{{\varvec{\omega }}^\beta }$, we need to take all the $\beta _l$ equal (mod. $2\pi$), in particular $\beta _l=0$. $\square$

In this case, if we use $\sqrt{1-\frac{1}{2l}}\ge 1-\frac{1}{2l}$$\forall l\in \mathbb {N}$ and $c_l\ge 1$, we get (here and below $\varvec{\omega }\equiv \varvec{\omega }^0$)

$$\begin{aligned} \begin{aligned} \langle \varvec{x}\rangle _{\varvec{\omega }}^2&\ge \left[ \frac{2}{(\Lambda +1)^2}\sum _{l=1}^{\Lambda }l \left( 1-\frac{1}{2l}\right) \right] ^2\overset{(75)_2}{=} \left\{ \frac{2}{(\Lambda +1)^2}\left[ \frac{\Lambda ^2}{2}\right] \right\} ^2=\frac{\Lambda ^4}{(\Lambda +1)^4}. \end{aligned} \end{aligned}$$

Finally, we find

$$\begin{aligned} \left( \Delta \varvec{x}\right) ^2_{\varvec{\omega }}= & {} \left\langle \varvec{x}^2\right\rangle _{\varvec{\omega }}-\left\langle \varvec{x}\right\rangle _{\varvec{\omega }}^2\le \frac{\Lambda }{\Lambda +1}+\frac{1}{2(\Lambda +1)^2}-\frac{\Lambda ^4}{(\Lambda +1)^4} \\= & {} \frac{2\Lambda (\Lambda +1)^3+(\Lambda +1)^2-2\Lambda ^4}{2(\Lambda +1)^4}\\= & {} \frac{6\Lambda ^3+7\Lambda ^2+4\Lambda +1}{2(\Lambda +1)^4}<\frac{3\Lambda ^3+9\Lambda ^2+9\Lambda +3}{(\Lambda +1)^4}=\frac{3}{\Lambda +1}. \end{aligned}$$

Proof of (59)

$L_0{\varvec{\phi }}^\beta =0$, while $L_\pm {\varvec{\phi }}^\beta$ are combinations of ${\varvec{\psi }}_l^{\pm 1}$, therefore are orthogonal to ${\varvec{\phi }}^\beta$; similarly, $x_{\pm }{\varvec{\phi }}^\beta$ are combinations of ${\varvec{\psi }}_{l-1}^{\pm 1}, {\varvec{\psi }}_{l+1}^{\pm 1}$, therefore are orthogonal to ${\varvec{\phi }}^\beta$. Hence

$$\begin{aligned} \langle L_a\rangle _{{\varvec{\phi }}^\beta }=0,\,\,\,\langle x_\pm \rangle _{{\varvec{\phi }}^\beta }=0\quad \Rightarrow \quad \langle \varvec{L}\rangle _{{\varvec{\phi }}^\beta }=0,\quad |\langle \varvec{x}\rangle _{{\varvec{\phi }}^\beta }|=|\langle x_0\rangle _{{\varvec{\phi }}^\beta }|. \end{aligned}$$

Replacing these results in $(\Delta \varvec{L})^2=\langle \varvec{L}^2\rangle _{{\varvec{\phi }}^\beta }-\langle \varvec{L}\rangle _{{\varvec{\phi }}^\beta }^2$ and using (58), we find, as claimed

$$\begin{aligned} (\Delta \varvec{L})^2=\langle \varvec{L}^2\rangle _{{\varvec{\phi }}^\beta }= \langle {\varvec{\phi }}^\beta , \varvec{L}^2{\varvec{\phi }}^\beta \rangle =\sum _{l=1}^{\Lambda }\frac{l(l+1)(2l+1)}{(\Lambda +1)^2} {\mathop {=}\limits ^{(74)}} \frac{\Lambda (\Lambda +2)}{2}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \begin{aligned} x^0\varvec{\phi }^{\beta }\overset{(40)}{=}&\sum _{l=0}^{\Lambda }\mathrm{e}^{i\beta _l}\frac{\sqrt{2l+1}}{\Lambda +1}\left( c_lA_l^{0,0}{\varvec{\psi }}_{l-1}^0+c_{l+1}B_l^{0,0}{\varvec{\psi }}_{l+1}^0\right) \\ \overset{(41)}{=}&\sum _{l=0}^{\Lambda }\frac{\mathrm{e}^{i\beta _l}}{\Lambda +1}\left( c_l\frac{l}{\sqrt{2l-1}}{\varvec{\psi }}_{l-1}^0+c_{l+1}\frac{l+1}{\sqrt{2l+3}}{\varvec{\psi }}_{l+1}^0 \right) \\ =&\sum _{l=1}^{\Lambda }\frac{\mathrm{e}^{i\beta _l}c_l}{\Lambda +1}\frac{l}{\sqrt{2l-1}}{\varvec{\psi }}_{l-1}^0+\sum _{l=0}^{\Lambda -1}\frac{\mathrm{e}^{i\beta _l}c_{l+1}}{\Lambda +1}\frac{l+1}{\sqrt{2l+3}}{\varvec{\psi }}_{l+1}^0\\ =&\sum _{l=0}^{\Lambda -1}\frac{\mathrm{e}^{i\beta _{l+1}}c_{l+1}}{\Lambda +1}\frac{l+1}{\sqrt{2l+1}}{\varvec{\psi }}_l^0+\sum _{l=1}^{\Lambda }\frac{\mathrm{e}^{i\beta _{l-1}}c_{l}}{\Lambda +1}\frac{l}{\sqrt{2l+1}}{\varvec{\psi }}_l^0,\quad \Longrightarrow \\ \left\langle x^0\right\rangle _{\varvec{\phi }^{\beta }}=&\sum _{l=0}^{\Lambda -1}\mathrm{e}^{i(\beta _{l+1}-\beta _l)}c_{l+1}\frac{l+1}{(\Lambda +1)^2}+\sum _{l=1}^{\Lambda }\mathrm{e}^{i(\beta _{l-1}-\beta _l)}c_l\frac{l}{(\Lambda +1)^2}\\ =&\sum _{l=1}^{\Lambda }\frac{2lc_l}{(\Lambda +1)^2}\cos {(\beta _{l-1}-\beta _l)}, \end{aligned} \end{aligned}$$

this means that $\left\langle x^0\right\rangle _{\varvec{\phi }^{\beta }}^2\equiv \left\langle \varvec{x}\right\rangle _{\varvec{\phi }^{\beta }}^2$ is maximal when $\beta \equiv 0$, and in this case one has (here and on $\varvec{\phi }\equiv \varvec{\phi }^0$)

$$\begin{aligned} \left\langle \varvec{x}\right\rangle _{\varvec{\phi }}^2\overset{c_l \ge 1}{\ge }\left[ \sum _{l=1}^{\Lambda }\frac{2l}{(\Lambda +1)^2}\right] ^2\overset{(71)}{=}\frac{\Lambda ^2}{(\Lambda +1)^2}. \end{aligned}$$

One easily checks that $\langle \varvec{x}^2\rangle _{{\varvec{\phi }}^\beta }=\langle \varvec{x}^2\rangle _{{\varvec{\omega }}^\beta }$; hence, using (122), on ${\varvec{\phi }}$ it follows, as claimed

$$\begin{aligned} (\Delta \varvec{x})^2_{\varvec{\phi }}= & {} \langle \varvec{x}^2\rangle _{{\varvec{\phi }}}-\langle \varvec{x}\rangle _{{\varvec{\phi }}}^2\le \frac{\Lambda }{\Lambda +1}+\frac{1}{2(\Lambda +1)^2}-\frac{\Lambda ^2}{(\Lambda +1)^2}\\= & {} \frac{2\Lambda (\Lambda +1)+1-2\Lambda ^2}{2(\Lambda +1)^2} =\frac{2\Lambda +1}{2(\Lambda +1)^2}<\frac{1}{\Lambda +1}. \end{aligned}$$

$\square$

Proof of (66)

$$\begin{aligned} \begin{aligned} \left\langle \varvec{x}^2\right\rangle _{\widetilde{\varvec{\chi }}}&\overset{(44)_2}{=} \sum _{l=0}^{\Lambda }\left| \widetilde{\chi }^l \right| ^2 +\frac{\left[ \sum _{l=0}^{\Lambda }l(l+1)\left| \widetilde{\chi }^l \right| ^2\right] +1}{k(\Lambda )}\\&\qquad -\left[ 1+\frac{(\Lambda +1)^2}{k(\Lambda )}\right] \frac{\Lambda +1}{2\Lambda +1}\left| \widetilde{\chi }^{\Lambda } \right| ^2 \\&\overset{\left\| \widetilde{\varvec{\chi }}\right\| _2=1}{=}1+\frac{\left[ \sum _{l=0}^{\Lambda }l (l+1)\left| \widetilde{\chi }^l \right| ^2\right] +1}{k(\Lambda )}\\&\qquad -\left[ 1+\frac{(\Lambda +1)^2}{k(\Lambda )} \right] \frac{\Lambda +1}{2\Lambda +1}\left| \widetilde{\chi }^{\Lambda } \right| ^2 \\&\le 1+\frac{\frac{2}{\Lambda +2}\left[ \sum _{l=1}^{\Lambda }l(l+1)\right] +1}{k(\Lambda )} \overset{(71)}{\le }1+\frac{\frac{2}{3} \Lambda (\Lambda +1)+1}{k(\Lambda )} \overset{(5)}{\le }1\\&\qquad +\frac{\frac{2}{3} \Lambda (\Lambda +1)+1}{\Lambda ^2(\Lambda +1)^2}, \end{aligned} \end{aligned}$$

(123)

so, putting together (65) and (123), we obtain, as claimed,

$$\begin{aligned} (\Delta \varvec{x})^2_{\widetilde{\varvec{\chi }}}= & {} \left\langle \varvec{x}^2\right\rangle _{\widetilde{\varvec{\chi }}}-\left\langle \varvec{x}\right\rangle _{\widetilde{\varvec{\chi }}}^2 =\left\langle \varvec{x}^2\right\rangle _{\widetilde{\varvec{\chi }}}-\left\langle x_0\right\rangle _{\widetilde{\varvec{\chi }}}^2<1-\cos ^2\left( \frac{\pi }{\Lambda +2}\right) \\&+\frac{\frac{2}{3} \Lambda (\Lambda +1)+1}{\Lambda ^2(\Lambda +1)^2}\\= & {} \sin ^2\left( \frac{\pi }{\Lambda +2}\right) +\frac{\frac{2}{3} +\frac{2}{3\Lambda }+\frac{1}{\Lambda ^2}}{(\Lambda +1)^2} \overset{\Lambda \ge 3}{<}\frac{\pi ^2}{(\Lambda +2)^2}\\&+\frac{1}{(\Lambda +1)^2}<\frac{11}{(\Lambda +1)^2}. \end{aligned}$$

$\square$