Particle on the sphere:
group-theoretic quantization
in the presence of a magnetic monopole

A magnetic field 2-form $B$ on $S^{2}$ admits an infinite dimensional symmetry group that acts transitively on $S^{2}$, provided that $B$ is nowhere vanishing. This is the group of “area” preserving diffeomorphisms, where $B$ defines the area element. We are interested in quantizing an $SO(3)$ subgroup of this group, which not only is the smallest group that can act transitively on $S^{2}$ but is also the group of isometries of a round metric. There are infinitely many such subgroups, which all lead to equivalent phase space quantizations. We write the magnetic field as

$$B=g\epsilon\,,$$

(3.1)

where the 2-form $\epsilon$ is scaled so that $\int_{S^{2}}\epsilon=4\pi$, and $g$ is a dimensionful coefficient. The total flux through the sphere is simply $4\pi g$, so $g$ can be interpreted as the magnetic charge of a monopole “inside”⁹⁹9If the magnetic field were not nowhere vanishing, we could still separate it as $B=g\epsilon+dA$, where $A$ is a globally defined potential 1-form that could be included in the Hamiltonian, while the $g\epsilon$ term could be included in the symplectic form.. The kinetic energy term in the Hamiltonian for a charged particle on $S^{2}$ involves a particular metric on the $S^{2}$. In order to naturally quantize this Hamiltonian, we will ultimately base the quantization on the $SO(3)$ group of isometries of this metric, however that choice plays no role until we come to quantizing the Hamiltonian.

We denote by $l_{R}(x)$ the action of a rotation $R\in SO(3)$ on a point $x$ on the sphere. As with any action on the configuration space, there is a natural lifted action to the cotangent bundle, defined by

$$L_{R}(p)=l_{R}^{-1*}p\,,$$

(3.2)

which maps the fiber over $x$ to that over $l_{R}(x)$, i.e. it satisfies $\pi\circ L_{R}=l_{R}\circ\pi$. The canonical potential 1-form $\theta$ is invariant under the lift of any point transformation (diffeomorphism of the configuration space).¹⁰¹⁰10Let $\phi$ be a diffeomorphism of the configuration space, and let $\Phi$ be its lift to the phase space, i.e., $\Phi(p):=\phi^{-1*}p$. If $V\in T_{p}\mathcal{P}$, then $\Phi^{*}\theta(V)=\theta(\Phi_{*}V)=(\Phi(p))(\pi_{*}\Phi_{*}V)=(\phi^{-1*}p)(\pi_{*}\Phi_{*}V)=p(\phi_{*}^{-1}\pi_{*}\Phi_{*}V)=p(\pi_{*}V)=\theta(V)$. Alternatively, using coordinates, $p^{\prime}_{i}dx^{\prime i}=p_{j}\frac{\partial x^{j}}{\partial x^{\prime i}}\frac{\partial x^{\prime i}}{\partial x^{k}}dx^{k}=p_{k}dx^{k}$. In particular, we have $L_{R}^{*}\theta=\theta$, and therefore $L_{R}^{*}d\theta=dL_{R}^{*}\theta=d\theta$. Moreover $\pi^{*}B$ is invariant under rotations: $L_{R}^{*}(\pi^{*}B)=(\pi\circ L_{R})^{*}B=(l_{R}\circ\pi)^{*}B=\pi^{*}B$. Therefore the symplectic form (1.1) is invariant,

$$L_{R}^{*}\omega=\omega\,.$$

(3.3)

That is, for all $R$, $L_{R}$ is a symmetry of the symplectic form.

The quantizing group should be larger than just $SO(3)$, since the rotations act only “horizontally” on the phase space. For the quantizing group to act transitively, it should include elements that move points along the fibers of the cotangent bundle. The simplest “vertical” action is a translation of momentum,

$$F_{\alpha}(p)=p-\alpha\,,$$

(3.4)

where $\alpha$ is a 1-form field on $S^{2}$. (For notational simplicity we leave implicit the point $\pi(p)$ at which $\alpha$ is evaluated.) This acts on the symplectic form (1.1) as

$$F^{*}_{\alpha}\omega=d(F^{*}_{\alpha}\theta)+e(\pi\circ F_{\alpha})^{*}B\,.$$

(3.5)

The term $\pi^{*}B$ is invariant since $\pi\circ F_{\alpha}=\pi$. The symplectic potential, however, transforms non-trivially. For any $V\in T_{p}\mathcal{P}$, we have

	$\displaystyle F^{*}_{\alpha}\theta(V)=\theta(F_{\alpha*}V)=(F_{\alpha}p)(\pi_{*}F_{\alpha*}V)$
	$\displaystyle=(p-\alpha)(\pi_{*}V)=(\theta-\pi^{*}\alpha)(V)\,,$		(3.6)

so that

$$F^{*}_{\alpha}\omega=\omega-\pi^{*}d\alpha\,.$$

(3.7)

In order for $F_{\alpha}$ to be a symplectic symmetry, we must require $\alpha$ to be closed. We will restrict further to exact 1-forms, $\alpha=df$, with $f$ globally defined on $S^{2}$ to ensure that the associated charges will be globally defined.

The (infinite-dimensional) space of all exact 1-form fields is unnecessarily large, so we look for a “minimal” set of momentum translations that act transitively along the fibers and are consistent with the spherical symmetry, in the sense that they extend the chosen $SO(3)$ into a larger group. As observed by Isham (in a more general setting), a suitable set can be generated by realizing the configuration space $S^{2}$ as an orbit of a representation of $SO(3)$ in a vector space, and defining the momentum translations as the pullback to the orbit of the “constant” 1-forms on that vector space. In particular, we can choose the fundamental representation on $\mathbb{R}^{3}$, and identify the $S^{2}$ with the orbit passing through $u=(0,0,1)$ in $\mathbb{R}^{3}$, that is, with the set of unit vectors $x\in\mathbb{R}^{3}$ such that $x=Ru$ for some $R\in SO(3)$.¹¹¹¹11This identification of the configuration space with the unit sphere in the abstract $\mathbb{R}^{3}$ should not be confused with the physical sphere, which may have its own geometry. The orbit is a “unit sphere” with respect to the inner product $\langle v,u\rangle=\sum_{i=1}^{3}v_{i}u_{i}$ on the abstract $\mathbb{R}^{3}$. (The notation “$x$” for these vectors coincides with that which we used already to label the points in $S^{2}$.) Any dual vector $\alpha\in\mathbb{R}^{3*}$, can naturally be seen as a 1-form field on $\mathbb{R}^{3}$. Moreover, this 1-form field is exact, for it can be written as $df_{\alpha}$, where the function $f_{\alpha}:\mathbb{R}^{3}\rightarrow\mathbb{R}$ is defined by

$$f_{\alpha}(x):=\alpha(x)\,.$$

(3.8)

These 1-form fields can be pulled-back to $S^{2}$ to define the corresponding action along the fibers of $\mathcal{P}$.

Combining the $L_{R}$ and $F_{\alpha}$ transformations, we get a transitive group of symplectic symmetries of the phase space: the semidirect product $G=\mathbb{R}^{3*}\rtimes SO(3)$, acting on $\mathcal{P}$ as

$$\Lambda_{(\alpha,R)}(p)=l_{R}^{-1*}p-\alpha\,,$$

(3.9)

which satisfies the product rule

$$(\alpha,R)(\alpha^{\prime},R^{\prime})=(\alpha+l_{R}^{-1*}\alpha^{\prime},RR^{\prime})\,.$$

(3.10)

Since $\mathbb{R}^{3*}\sim\mathbb{R}^{3}$ and the co-representation of $SO(3)$ in $\mathbb{R}^{3*}$ is equivalent to its representation in $\mathbb{R}^{3}$,¹²¹²12In a matrix realization, $R^{-1*}\alpha=(R^{-1})^{T}\alpha=R\alpha$. this group is isomorphic to the Euclidean group, $E_{3}=\mathbb{R}^{3}\rtimes SO(3)$.

We take this group, $G$, to be the quantizing group. Note that it is independent of the magnetic term in the symplectic form. Since its algebra does not admit any non-trivial central extension by 2-cocycles $z(\xi,\eta)$,¹³¹³13Since we found no explicit demonstration of this statement in the literature, we include one in Appendix A for completeness. This formal proof is not really necessary for our purposes, however, as in the next section we explicitly compute the Poisson algebra and show that no central charges arise. no (non-trivial) central charge can appear in the associated Poisson algebra. The quantizing algebra is thus the same as in the uncharged case. However, as we shall see in Sec. 4.2, the magnetic term makes itself felt through the value of a Casimir invariant of the corresponding Poisson bracket algebra, which carries over to the quantum theory.

We next compute the classical charges associated with the quantizing group $G$, beginning with the $SO(3)$ generators. Let $n$ be an element of the algebra $\mathfrak{so}(3)$, and denote its exponential by $R_{n}=\exp(n)$. (Here $\exp:\mathfrak{g}\rightarrow G$ is the usual Lie group exponential map.) Let $\overline{X}_{n}$ be the vector field (on $S^{2}$) induced by $n$ through the action of $SO(3)$ on $S^{2}$, and let $X_{n}$ be the vector field (on $\mathcal{P}$) induced by $n$ through the lifted action of $SO(3)$ on $\mathcal{P}$ defined in (3.2). Since the group action on $\mathcal{P}=T^{*}S^{2}$ maps fibers into fibers, we note that $\overline{X}_{n}$ is just the projection of $X_{n}$ to the sphere, i.e., $\overline{X}_{n}=\pi_{*}X_{n}$. The corresponding charge $P_{n}$ is defined by

	$\displaystyle dP_{n}$	$\displaystyle=-\imath_{X_{n}}\omega$
		$\displaystyle=-\imath_{X_{n}}(d\theta+eg\,\pi^{*}\epsilon)$
		$\displaystyle=d[\theta(X_{n})]-eg\,\pi^{*}\imath_{\overline{X}_{n}}\epsilon\,.$		(3.11)

The first term in the last line follows from $0=\pounds_{X_{n}}\theta=\imath_{X_{n}}d\theta+d\imath_{X_{n}}\theta$, and $\theta$ is invariant under any point transformation, as discussed in the paragraph leading to (3.3). Like the first term, the second term is also an exact 1-form: it is closed since $d(\pi^{*}\imath_{\overline{X}_{n}}\epsilon)=\pi^{*}d\imath_{\overline{X}_{n}}\epsilon=\pi^{*}{\cal L}_{\overline{X}_{n}}\epsilon=0$, and since $S^{2}$ is simply connected it is therefore also exact, i.e., $\imath_{\overline{X}_{n}}\epsilon=d\Gamma_{n}$, for some $\Gamma_{n}:S^{2}\rightarrow\mathbb{R}$. The charge $P_{n}$ is thus given by

$$P_{n}=p(\overline{X}_{n})-eg\,\Gamma_{n}\circ\pi$$

(3.12)

up to an additive constant. The use of the symbol “$P$” is motivated by the fact that the charges associated with spatial transformations are the analogue of momentum coordinates. The term $p(\overline{X}_{n})$ alone is the usual orbital angular momentum associated with the rotation Killing vector field $\overline{X}_{n}$, while $P_{n}$ is the canonical angular momentum, i.e., the charge that generates rotations on the phase space with symplectic form $\omega$ (1.1). On a phase space with the “usual” symplectic form $d\theta$, the canonical angular momentum would have been simply $p(\overline{X}_{n})$.

Next we compute the charges associated with the $\mathbb{R}^{3*}$ part of the group. Since the group $\mathbb{R}^{3*}$ is a vector space, it can be naturally identified with its Lie algebra. Let $Y_{\alpha}$ be the momentum translation vector field on $\mathcal{P}$ induced by an element $\alpha$ of the Lie algebra of $\mathbb{R}^{3*}$ . The corresponding charge $Q_{\alpha}$ is defined by

	$\displaystyle dQ_{\alpha}$	$\displaystyle=-\imath_{Y_{\alpha}}\omega$
		$\displaystyle=-\imath_{Y_{\alpha}}d\theta$
		$\displaystyle=-\pounds_{Y_{\alpha}}\theta$
		$\displaystyle=-\left.\frac{d}{dt}F_{t\alpha}^{*}\theta\right|_{t=0}$
		$\displaystyle=\pi^{*}\alpha$
		$\displaystyle=\pi^{*}df_{\alpha}$
		$\displaystyle=d(f_{\alpha}\circ\pi)\,,$		(3.13)

where in the second line we used that $\imath_{Y_{\alpha}}\epsilon=0$; in the third line that $\imath_{Y_{\alpha}}\theta=0$; in the fourth line that the flow induced by $\alpha$ is $p\mapsto F_{t\alpha}(p)$; in the fifth line we used (3.6); and in the sixth line we used that $df_{\alpha}=\alpha$, as defined in (3.8).¹⁴¹⁴14Alternatively, in local coordinates adapted to the bundle structure of $T^{*}S^{2}$, the flow generated by $\alpha=\alpha_{i}dq^{i}$ is given by $F_{t\alpha}(q^{i},p_{i})=(q^{i},p_{i}-t\alpha_{i})$, so $Y_{\alpha}=-\alpha_{i}\frac{\partial}{\partial p_{i}}$, and the charge differential is $dQ_{\alpha}=-i_{Y_{\alpha}}(dp_{i}\wedge dq^{i}+eg\,\epsilon_{ij}dq^{i}\wedge dq^{j})=\alpha_{i}dq^{i}=\pi^{*}\alpha$. In the last step the $\pi^{*}$ appears because, in this equation, $q^{i}$ are coordinates on $T^{*}S^{2}$, while in the definition of $\alpha$ they are coordinates on $S^{2}$. The charge associated with $\alpha$ is therefore

$$Q_{\alpha}=f_{\alpha}\circ\pi\,$$

(3.14)

up to an additive constant. The use of the symbol “$Q$” here is motivated by the fact that the charges associated with momentum translations are the analogue of position coordinates.

To be more concrete, it is convenient to use the realization of $S^{2}$ as the unit sphere in the abstract $\mathbb{R}^{3}$, which was introduced in the previous subsection. We identify $n\in\mathfrak{so}(3)$ with the vector in $\mathbb{R}^{3}$ whose direction is the corresponding axis of rotation and whose magnitude $|n|$ gives the angle of rotation of $\exp(n)$, according to the right-hand rule. Then, using adapted spherical coordinates in which $n$ is aligned with $\theta=0$, we have $\overline{X}_{n}=|n|\partial_{\phi}$, hence $\imath_{\overline{X}_{n}}\epsilon=|n|\,\imath_{\partial_{\phi}}\!\sin\theta\,d\theta\wedge d\phi=|n|\,d(\cos\theta)=d(n\cdot x)$, so we can choose $\Gamma_{n}=n\cdot x$. Therefore, the canonical charges are given, up to an additive constant, by

	$\displaystyle P_{n}=p(\overline{X}_{n})-eg\,n\cdot x$		(3.15)
	$\displaystyle Q_{\alpha}=\alpha(x)\,.$		(3.16)

The notation is somewhat abbreviated here. Strictly speaking, $P_{n}$ and $Q_{\alpha}$ are functions on the phase space ${\cal P}$, which is specified above by giving their values at a point $p\in{\cal P}$. The vector field $\overline{X}_{n}$ is implicitly evaluated at $\pi(p)$, and $x$ is the unit vector representative of $\pi(p)$ in the embedded realization of $S^{2}\subset\mathbb{R}^{3}$.

We next consider the Poisson bracket algebra of the charges. By construction, this algebra matches the Lie algebra of the canonical group that defined the charges, up to a possible central extension. If a central extension appears in such an algebra, in general it may or may not be removable using the freedom to shift the charges by addition of constants. As mentioned above, the Euclidean algebra in itself (i.e. apart from any canonical realization) does not admit any non-trivial central extension (by 2-cocycles), so that it must be possible to choose the additive constants such that the Poisson algebra matches the Lie algebra. In fact, the choices we have made in (3.15) and (3.16) satisfy this criterion, and the Poisson algebra takes the form

	$\displaystyle\{P_{n},P_{n^{\prime}}\}=P_{[n,n^{\prime}]}$
	$\displaystyle\{Q_{\alpha},P_{n}\}=Q_{{\cal L}_{X_{n}}\alpha}$
	$\displaystyle\{Q_{\alpha},Q_{\alpha^{\prime}}\}=0\,,$		(3.17)

which matches the semi-direct product structure of the algebra $\mathfrak{g}=\mathbb{R}^{3*}\oplus_{S}\mathfrak{so}(3)$ of $G$ without central charges. That is, denoting elements of $\mathfrak{g}$ by $(\alpha,n)\in\mathbb{R}^{3*}\oplus_{S}\mathfrak{so}(3)$, the product rule reads

	$\displaystyle[(0,n),(0,n^{\prime})]=(0,[n,n^{\prime}])$
	$\displaystyle[(\alpha,0),(0,n)]=(\mathcal{L}_{X_{n}}\alpha,0)$
	$\displaystyle[(\alpha,0),(\alpha^{\prime},0)]=0\,,$		(3.18)

revealing how the linear association $(\alpha,n)\mapsto P_{n}+Q_{\alpha}$ is a (true) homomorphism.

To verify that the choices (3.15) and (3.16) lead to no central charges, and for later purposes, it is convenient to introduce a basis for $\mathfrak{g}$. Using the identification $\mathfrak{so}(3)\sim\mathbb{R}^{3}$, choose an orthonormal basis $\{e_{i}\}$ ($i=1,2,3$) in $\mathbb{R}^{3}$. (Note that $\exp(e_{i})$ implements a right-handed rotation by the angle $1$ around the $i$-axis.) It is straightforward to check that $[e_{i},e_{j}]=\varepsilon_{ijk}e_{k}$, where $\varepsilon_{ijk}$ is the Levi-Civita symbol.¹⁵¹⁵15 As $SO(3)$ acts on $\mathbb{R}^{3}$ from the left, the algebra element $n\in\mathbb{R}^{3}$ induces the vector field $\left.\overline{X}_{n}\right|_{x}=n\times x$, where $x\in\mathbb{R}^{3}$. The Lie bracket of two such vector fields is given by $[\overline{X}_{n},\overline{X}_{n^{\prime}}]=-\overline{X}_{n\times n^{\prime}}$. Together with $[\overline{X}_{\xi},\overline{X}_{\eta}]=\overline{X}_{[\eta,\xi]}$ (see third paragraph of section 2) this yields $[n,n^{\prime}]=n\times n^{\prime}$. Let $\{e^{i}\}$ denote the dual basis, satisfying $e^{i}(e_{j})=\delta^{i}_{\,\,j}$. We define

		$\displaystyle J_{i}:=P_{e_{i}}$
		$\displaystyle N_{i}:=Q_{e^{i}}\,,$		(3.19)

which satisfy the algebra

	$\displaystyle\{J_{i},J_{j}\}=\varepsilon_{ijk}J_{k}$
	$\displaystyle\{J_{i},N_{j}\}=\varepsilon_{ijk}N_{k}$
	$\displaystyle\{N_{i},N_{j}\}=0\,.$		(3.20)

This is the algebra of the Euclidean group, presented in terms of a basis of generators of rotation and translation.

We can express (3.15) and (3.16) in this basis. If $x\in S^{2}\subset\mathbb{R}^{3}$, we can write $\overline{X}_{e_{i}}=e_{i}\times x$. Also, if $p$ is a co-vector on $S^{2}$ at $x$, we can (abusing the notation) associate it with a vector $p$ in $\mathbb{R}^{3}$, tangent to $S^{2}$ at $x$, such that $p\cdot v=p(v)$, where $v$ is any vector on $\mathbb{R}^{3}$ tangent to $S^{2}$ at $x$. In this way, we have $p(\overline{X}_{e_{i}})=p\cdot(e_{i}\times x)=e_{i}\cdot(x\times p)$, which is the familiar orbital angular momentum about the axis $e_{i}$ in $\mathbb{R}^{3}$. Thus, $J_{i}=e_{i}\cdot(x\times p-eg\,x)$. Also, $N_{i}$, evaluated at any point in the fiber over $x$, can be written as $N_{i}=e^{i}(x)=e_{i}\cdot x$. In a 3-vector notation,

	$\displaystyle J$	$\displaystyle=x\times p-eg\,x$		(3.21)
	$\displaystyle N$	$\displaystyle=x\,,$		(3.22)

so we have $J_{i}=e_{i}\cdot J$ and $N_{i}=e_{i}\cdot N$.

To establish (3.20), i.e., to verify that indeed there are no missing central terms, we may evaluate the brackets at points in the phase space where both sides of the equation vanish. For example, recall that $\{J_{1},J_{2}\}=-\omega(X_{e_{1}},X_{e_{2}})$. The vector field $X_{e_{1}}$ vanishes at the points in phase space with zero momentum and located at the rotation axis $e_{1}$ on the $S^{2}$ (i.e., the two points in the intersection of the zero section with the fibers over $x=\pm e_{1}$). Hence $\{J_{1},J_{2}\}$ vanishes there. On the other hand, according to (3.21), at the same point the function $J_{3}$ is equal to $-eg\,e_{3}\cdot e_{1}=0$. Any constant added to $J_{3}$ would spoil this agreement. This argument works for all of the $J_{i}$ brackets, so we conclude that no central term need be added on the right hand side of the first bracket in (3.20). The argument just given also implies that $\{J_{1},N_{2}\}=-\omega(X_{e_{1}},Y_{e^{2}})$ vanishes at the axis point $e_{1}$, while $N_{3}$ equals $e^{3}(e_{1})=0$ at that same point. Hence no central term appears in the second bracket either. As for the last brackets in (3.20), since the right hand side vanishes, it is unaffected by addition of a constant to any charge. In conclusion, as we claimed, the charges defined in (3.15) and (3.16) provide a realization of the quantizing algebra without central charges.

The quantum version of the classical canonical observables $J_{i}$ and $N_{i}$ are the self-adjoint generators of the corresponding unitary transformations in some Hilbert space $\mathcal{H}$. That is, given some unitary irreducible representation $U$ of $G$ (or its universal cover), and denoting elements of $\mathfrak{g}$ by $(\alpha,n)$, we define the operators $\widehat{J}_{i}$ and $\widehat{N}_{i}$ by

	$\displaystyle U[\exp(0,\lambda e_{i})]=:e^{-i\lambda\widehat{J}_{i}/\hbar}$
	$\displaystyle U[\exp(\lambda e^{i},0)]=:e^{-i\lambda\widehat{N}_{i}/\hbar}\,,$		(4.1)

where $\lambda$ is an arbitrary real parameter. It follows from the group structure that the quantized algebra satisfies

	$\displaystyle[\widehat{J}_{i},\widehat{J}_{j}]=i\hbar\,\varepsilon_{ijk}\widehat{J}_{k}$
	$\displaystyle[\widehat{J}_{i},\widehat{N}_{j}]=i\hbar\,\varepsilon_{ijk}\widehat{N}_{k}$
	$\displaystyle[\widehat{N}_{i},\widehat{N}_{j}]=0\,.$		(4.2)

The quantization map is $J_{i}\rightarrow\widehat{J}_{i}$ and $N_{i}\rightarrow\widehat{N}_{i}$, and (4.2) is the quantization of the Poisson algebra (3.20). For notational simplicity we shall henceforth omit the “hat” symbol over quantum operators, since it should be clear from the context whether we are referring to the classical or the quantum observables.

Casimir operators, by definition, commute with all elements of the algebra, and their eigenvalues can therefore be used to label its irreducible representations¹⁷¹⁷17If an operator commutes with all other operators in an irreducible representation, then according to Schur’s lemma it must be proportional to the identity operator.. There are two independent Casimir operators associated with the algebra $\mathfrak{g}$,

	$\displaystyle N^{2}=\sum_{i}(N_{i})^{2}$
	$\displaystyle N\cdot J=\sum_{i}N_{i}J_{i}\,.$		(4.3)

Their classical correspondents Poisson-commute with everything in the classical algebra and, using (3.22) and (3.21), we have

	$\displaystyle N^{2}=x^{2}=1$
	$\displaystyle N\cdot J=-eg\,x^{2}=-eg\,,$		(4.4)

revealing that these two quantities are constant classical observables. Since no operator ordering ambiguities arise in quantizing $N^{2}$ and $N\cdot J$ ($=J\cdot N$), we take as part of the quantization prescription that the quantum theory must carry the representation for which the values of the two Casimirs (4.3) are given by precisely the corresponding classical values (4.4). Note that the value of $N\cdot J$ is the only way the presence of the magnetic monopole is felt in the quantum theory.

While the representation theory of the Euclidean group is well known from Mackey’s theory of induced representations carried by a space of wavefunctions [30, 31], we will present it here using a rather simpler abstract ladder-operator approach.¹⁸¹⁸18After completing this work we found that a similar realization of the representation appears in [12], although no explicit derivation is presented there. The method is reminiscent of the usual derivation of the unitary irreducible representations of $SU(2)$. Some of the details are left to Appendix B. In Appendix C we discuss an alternative derivation based on Mackey theory, which provides a construction of the Hilbert space based on wave functions on $S^{2}$.

Note first that the Euclidean algebra (4.2) is invariant under rescaling of $N$. Thus, without loss of generality, we can particularize to the representation with $N^{2}=1$. The value of the other Casimir is left arbitrary,

$$N\cdot J=s\hbar\,,$$

(4.5)

with $s$ some real parameter.

Let us start with a basis of simultaneous eigenvectors of $J^{2}$ and $J_{3}$, denoted as $|j,m\rangle$, defined by

	$\displaystyle J^{2}|j,m\rangle=j(j+1)\hbar^{2}|j,m\rangle$
	$\displaystyle J_{3}|j,m\rangle=m\hbar|j,m\rangle\,.$		(4.6)

At this point it is not clear which values of $j$ are allowed, for a given $s$, nor if there is more than one state with a given value of $j$ and $m$. Note that $J_{\pm}=J_{1}\pm iJ_{2}$ act as raising and lowering operators for $m$ and, from the standard analysis of the angular momentum algebra, we know that, for a given $j$, $m$ varies from $-j$ to $j$ in integer steps. Also, $j$ can only be a non-negative integer or half-integer (i.e., $j\in\frac{1}{2}\mathbb{Z}^{0+}$), but it may be that only a subset of that is included.

Before systematically deriving the properties of the irreducible representations, it is enlightening to guess, by a simple but non-rigorous reasoning, which values of $s$ and $j$ are included. Supposing that there is (in a limiting sense) a state localized at the north pole of the sphere, the operator $N\cdot J$ acts on such a state as $J_{3}$ (see (4.18) for details). This implies that $2s$ must be integer valued in order for the representation to be non-trivial. Since it has the $J_{3}$ eigenvalue $s\hbar$, such a state must be constructed from states with $j\geq s$. By virtue of rotational symmetry, the same can be said about states localized at any other point on the $S^{2}$, and one would expect an irreducible representation to be constructed from the span of these states with $j\geq s$. Moreover, since $N$ is a vector operator (in the way it transforms under commutation with the $J_{i}$), its action can change the value of $j$ by plus or minus unity, which suggests that repeated action of $N$ will both raise the $j$ values without bound and lower them until they reach a floor, which presumably lies at $j=s$, since $j\geq s$ is the only apparent constraint. That is, the representation must include all values $j=s+n$, for non-negative integers $n$. Indeed this is the correct spectrum, as we now show by explicit construction of the representation.¹⁹¹⁹19The earliest mention of this spectrum that we have found appears in [32], although no derivation was given there.

We now present the rigorous derivation of the irreducible representations, analyzing how certain operators in the algebra act as “ladder operators” to shift the values $j$ and $m$. From the Wigner-Eckart theorem we know that when the vector operator $N_{i}$ acts on a state, it can only change the value of $j$ by $-1$, $0$ or $1$. And, since acting on a state with $N_{\pm}=N_{1}\pm iN_{2}$ changes the $m$ value by $\pm 1$, it follows that $N_{+}|j,j\rangle\propto|j+1,j+1\rangle$. Thus $N_{+}$ acts as a raising operator for edge states $|j,j\rangle$, i.e. states with the maximal $m=j$ for a given $j$. For now, let us assume that if $|j,j\rangle$ is in the Hilbert space then so is $N_{+}|j,j\rangle$, i.e., that its norm is positive. This assumption will be justified later.

Now let $|j_{0},j_{0}\rangle$ be the ground state, in the sense that there are no states with $j<j_{0}$ in the representation being constructed. (We know that there must be such a lowest $j$ state, since $j$ is non-negative.) Since we are constructing an irreducible representation, the whole Hilbert space $\mathcal{H}$ must be generated by acting with all elements of the algebra on any given state, in particular the ground state. Consider then the set of states

$$|j,m\rangle:=(J_{-})^{j-m}(N_{+})^{j-j_{0}}|j_{0},j_{0}\rangle\,,$$

(4.7)

where $j-j_{0}\in\mathbb{Z}^{0+}$ and $-j\leq m\leq j$. Since these states have distinct eigenvalues for the the self-adjoint operators $J^{2}$ and/or $J_{3}$, they are necessarily orthogonal (although not normalized). We will show that a representation exists only if $s\in\frac{1}{2}\mathbb{Z}$, in which case there is a unique irreducible representation, spanned by the states (4.7) with $j_{0}=|s|$. To establish this, we first prove that these states are closed under the action of the entire algebra, and then we show that that they all have positive norm, provided the $s$ quantization condition holds and $j_{0}$ has the required value.

It is clear how to define the action of $J_{i}$ on this basis, using just the angular momentum algebra. That is, $J_{i}|j,m\rangle$ can be written as a linear combination of $|j,m-1\rangle$, $|j,m\rangle$ and $|j,m+1\rangle$ with coefficients determined by the algebra. So let us focus on defining the action of the $N$’s. It is easy to see that, since the commutator of an $N$ with a $J$ gives an $N$, the action of $N_{i}$ on any state is well-defined provided that the $N$’s have a well-defined action on the edge states $|j,j\rangle$. Similarly, if the edge states have positive norm then so do the rest of the states.

By the definition of the basis states (4.7) we have

$$N_{+}|j,j\rangle=|j+1,j+1\rangle\,.$$

(4.8)

Next, using only the algebra and the Casimir invariants, we show in Appendix B that

$$N_{3}|j,j\rangle=\frac{s}{(j+1)}|j,j\rangle-\frac{1}{2(j+1)}|j+1,j\rangle$$

(4.9)

and, for $j>j_{0}$,

	$\displaystyle N_{-}|j,j\rangle=$	$\displaystyle\frac{2j}{2j+1}\left(1-\frac{s^{2}}{j^{2}}\right)|j-1,j-1\rangle+$
		$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +s\frac{|j,j-1\rangle}{j(j+1)}-\frac{|j+1,j-1\rangle}{2(2j+1)(j+1)}\,.$		(4.10)

(The action of the algebra on the edge state $|j,j\rangle$ is depicted on the left in Fig. 2.) It then remains to determine $N_{-}|j_{0},j_{0}\rangle$, to establish that the states $|j,j\rangle$ with $j\geq j_{0}$ have positive norm, and to show that the algebra is consistent with the assumption that $j$ cannot be lowered below $j_{0}$.

Observe that the form of (4.10) indicates that the operator

$$L^{(j)}=N_{-}-\frac{s}{j(j+1)}J_{-}+\frac{1}{2(2j+1)(j+1)}(J_{-})^{2}N_{+}$$

(4.11)

acts as a $j$-lowering operator for $|j,j\rangle$, i.e. , $L^{(j)}|j,j\rangle\propto|j-1,j-1\rangle$, modulo the assumption that $j>j_{0}$. In fact, using only the algebra it can be established for any $j$, including $j_{0}$, that

$$[J^{2},L^{(j)}]|j,j\rangle=-2j\hbar^{2}L^{(j)}|j,j\rangle\,,$$

(4.12)

hence $J^{2}L^{(j)}|j,j\rangle=j(j-1)\hbar^{2}L^{(j)}|j,j\rangle$.