**2. Invariance and Degeneracy**

We took into account only the linear operators having a discrete spectrum. The following definitions and results, which are well known in the literature, are helpful to characterize our setup and to establish the notation that is used.

**Definition 1** (**Invariant operator**)**.** *A linear operator* O*, defined on a Hilbert space, is said to be invariant under a linear transformation* U*, defined on the same Hilbert space, if for any eigenvalue <sup>λ</sup> of* <sup>O</sup>*, the corresponding eigenspace <sup>E</sup>λ*(O) *is an invariant subspace, i.e.,* <sup>∀</sup> **<sup>v</sup>** <sup>∈</sup> *<sup>E</sup>λ*(O)*, also* <sup>U</sup>**<sup>v</sup>** <sup>∈</sup> *<sup>E</sup>λ*(O)*.*

**Definition 2** (**Commuting operators**)**.** *Given two linear operators* O1*,* O<sup>2</sup> *defined on a Hilbert space, they are commuting if the commutator is null, that is:*

$$[\mathbb{G}\_1, \mathbb{G}\_2] = \mathbb{G}\_1 \mathbb{G}\_2 - \mathbb{G}\_2 \mathbb{G}\_1 = 0.1$$

Since the operator L is self-adjoint, it is easy to prove an invariance result, which holds for all linear and self-adjoint operators admitting a complete set of eigenvectors generating the Hilbert space.

**Theorem 1** (**Invariance theorem**)**.** *The linear operator* O *is invariant under a linear transformation* U *if and only if* [O, U] = 0*.*

**Proof.** If the commutator is zero, we have that, for any eigenvector **v** of the operator O:

$$\mathbb{UU} = \mathbb{UU} \qquad \Longleftrightarrow \qquad \mathbb{U}(\mathbb{UU}) = \mathbb{O}(\mathbb{U}\mathbf{v})\_{\prime}$$

hence O(U**v**) = *λ*U**v**, meaning that O is invariant. Conversely, if O is invariant, this means that, by linearity:

$$\mathbb{U}(\mathbb{U}\mathbf{v}) = \lambda(\mathbb{U}\mathbf{v}) = \mathbb{U}(\lambda\mathbf{v}) = \mathbb{U}(\mathbb{O}\mathbf{v})\_{\prime}.$$

implying that [O, U] = 0, because the eigenvectors generate the whole Hilbert space, by assumption.

It is straightforward to note that the operator L is invariant under the action of three different linear operators, i.e.:

$$\mathbb{M}\_1 = i \left( z \frac{\partial}{\partial y} - y \frac{\partial}{\partial z} \right), \ \mathbb{M}\_2 = i \left( x \frac{\partial}{\partial z} - z \frac{\partial}{\partial x} \right), \ \mathbb{M}\_3 = i \left( y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right), \tag{3}$$

where the imaginary unit *i* is necessary to guarantee that the operators are self-adjoint (we recall that in the framework of quantum mechanics, the operators M1, M2, and M<sup>3</sup> are the components of the pseudo-vector "angular momentum", and the invariance of an operator under the action of all three is called *rotational invariance* or *invariance under the rotation group SO*(3)).

Given its rotational invariance, it is more convenient to study the eigenvalue problem by employing the spherical polar coordinates, which depend on the original variables through the following relations:

$$r = \sqrt{x^2 + y^2 + z^2}, \qquad \theta = \arccos\left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right), \qquad \phi = \arctan\left(\frac{y}{x}\right),$$

where (*r*, *θ*, *φ*) ∈ [0, +∞) × [0, *π*] × [0, 2*π*]. By using the standard formulas, we also reformulate the Laplacian operator in terms of partial derivatives with respect to the spherical polar coordinates, i.e.:

$$
\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \left( \frac{\partial^2}{\partial \theta^2} + \cot \theta \frac{\partial}{\partial \theta} + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \right), \tag{4}
$$

whereas the operator <sup>M</sup><sup>3</sup> becomes <sup>M</sup><sup>1</sup> <sup>3</sup> <sup>=</sup> <sup>−</sup>*<sup>i</sup> <sup>∂</sup> ∂φ*.

Plugging the expression (4) into L yields the following form for the operator:

$$\mathcal{L} = -\frac{1}{2} \left( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right) + \frac{A(\theta, \phi)}{2r^2} + \frac{r^2}{2},\tag{5}$$

and consequently, the eigenvalue problem becomes:

$$
\left[\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} - \frac{A(\theta,\phi)}{r^2} - r^2 + 2\lambda\right] \psi(r,\theta,\phi) = 0,\tag{6}
$$

where *A*(*θ*, *φ*) is the following self-adjoint operator (An alternative formulation of the problem (6) takes place when *A* is a constant, i.e., *A* := *l*(*l* + *q* − 2), where *q* is the dimension of the space and *l* is an integer number. This problem is usually solved numerically. Another kind of degeneracy would occur, and although a deep analysis of such a case deserves future research, it is beyond the scope of our paper.).

$$A(\theta,\phi) = -\frac{\partial^2}{\partial\theta^2} - \cot\theta\frac{\partial}{\partial\theta} - \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}.\tag{7}$$

Based on the change of variables, it is necessary to modify the Hilbert space of the solutions as well:

$$\tilde{\mathcal{H}} = \left\{ \psi : \mathbb{R}^3 \longrightarrow \mathbb{R} \mid \psi \in \mathcal{C}^2(\mathbb{R}^3) \cap L^2(\mathbb{R}^3),$$

$$\psi(r, \theta, \phi + 2\pi) = \psi(r, \theta, \phi), \lim\_{r \to \infty} \psi(r, \theta, \phi) = 0 \right\}.\tag{8}$$

The next theorem is very relevant for the subsequent analysis of degeneracy.

**Theorem 2.** *The following relation holds in spherical polar coordinates:*

$$A(\theta, \phi) = \mathbb{M}\_1^2 + \mathbb{M}\_2^2 + \mathbb{M}\_{3\prime}^2$$

*where the operators* M*i, for i* = 1, 2, 3*, are defined by (3).*

**Proof.** The sum of the squares of the operators defined in (3) reads as:

$$\mathcal{M}\_1^2 + \mathcal{M}\_2^2 + \mathcal{M}\_3^2 = -\left(z\frac{\partial}{\partial y} - y\frac{\partial}{\partial z}\right)^2 - \left(x\frac{\partial}{\partial z} - z\frac{\partial}{\partial x}\right)^2 - \left(y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y}\right)^2$$

$$= \dots = -x^2 \left(\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) - y^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial z^2}\right) - z^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)$$

$$+2\left(xy\frac{\partial}{\partial x}\frac{\partial}{\partial y} + xz\frac{\partial}{\partial x}\frac{\partial}{\partial z} + yz\frac{\partial}{\partial y}\frac{\partial}{\partial z} + x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} + z\frac{\partial}{\partial z}\right).$$

Now, we recall the well-known identities among partial derivatives:

$$\frac{\partial}{\partial x} = \sin\theta\cos\phi \frac{\partial}{\partial r} + \frac{\cos\theta\cos\phi}{r} \frac{\partial}{\partial\theta} - \frac{\sin\phi}{r\sin\theta} \frac{\partial}{\partial\phi'},$$

$$\frac{\partial}{\partial y} = \sin\theta\sin\phi \frac{\partial}{\partial r} + \frac{\cos\theta\sin\phi}{r} \frac{\partial}{\partial\theta} - \frac{\cos\phi}{r\sin\theta} \frac{\partial}{\partial\phi'},$$

$$\frac{\partial}{\partial z} = \cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r} \frac{\partial}{\partial\theta}.$$

Applying the above formulas to the latest expression we obtained for the sum of squares yields:

$$\mathbb{M}\_1^2 + \mathbb{M}\_2^2 + \mathbb{M}\_3^2 = \dots = -\frac{\partial^2}{\partial \theta^2} - \cot \theta \frac{\partial}{\partial \theta} - \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} = A(\theta, \phi).$$

Back to the identification of the solution of (6), we can proceed by the separation of the variables. The eigenvalues of <sup>L</sup><sup>1</sup> are countable; more precisely for all *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>, they have the form:

$$
\lambda\_n = n + \frac{3}{2}
$$

.

The associated eigenfunctions read as:

$$\psi(\cdot) = \psi\_{nlm}(r,\theta,\phi) = R\_{n\ell}(r)Y\_{\ell m}(\theta,\varphi) = \left[N\_1 \, r^{\ell} \, e^{-r^2/2} \mathfrak{L}\_{n/2-\ell/2}^{\,(\ell+1/2)}(u)\Big|\_{u=r^2}\right] [N\_2 \mathcal{Y}\_{\ell m}(\theta,\varphi)],\tag{9}$$

where the terms:

$$R(r) \doteq N\_1 r^{\ell} e^{-r^2/2} \mathfrak{L}\_{n/2-\ell/2}^{(\ell+1/2)}(\mathfrak{u})\Big|\_{\mathfrak{u}=r^2}$$

and

$$\mathcal{Y}\_{\ell m}(\theta,\varphi) = \mathcal{N}\_2 \mathcal{Y}\_{\ell m}(\theta,\varphi).$$

are respectively called the *radial part* and the *angular part*. In the expression (9), we have that:

• the coefficient *N*1, *N*<sup>2</sup> are normalization constants, as in every eigenvalue problem, with respect to the norm of the Hilbert space, that is:

$$\int\_0^\infty r^2 |R(r)|^2 \, dr = 1 \qquad \text{and} \qquad \int\_0^{2\pi} d\varphi \int\_0^\pi |Y\_{\ell m}(\theta, \varphi)|^2 \sin \theta \, d\theta = 1;$$


$$
\mathfrak{L}\_j^{(h)}(u) = \frac{e^u}{u^h} \frac{d^j}{du^j} \left( u^{j+h} e^{-u} \right);
$$

• The functions Y*<sup>m</sup>*(*θ*, *ϕ*) are the *spherical harmonics*:

$$\mathcal{Y}\_{\ell,\mathfrak{m}}(\theta,\varphi) = \epsilon^{im\mathfrak{q}} \sin^m \theta \left(\frac{1}{\sin \theta} \,\frac{d}{d\theta}\right)^{\ell+m} \sin^{2\ell} \theta \,\varphi$$

for *m* > 0, whereas Y<sup>∗</sup> -,*m*(*θ*, *<sup>ϕ</sup>*)=(−1)*m*Y*<sup>m</sup>*(*θ*, *ϕ*) for *m* < 0. They are simultaneous eigenfunctions of the operator *<sup>A</sup>*(*θ*, *<sup>ϕ</sup>*) and of the operator <sup>M</sup><sup>1</sup> 3, in compliance with the following equations:

$$A(\theta,\varphi)\mathcal{Y}\_{\ell m}(\theta,\varphi) = \ell(\ell+1)\mathcal{Y}\_{\ell m}(\theta,\varphi),\tag{10a}$$

$$
\widetilde{\mathcal{M}}\_3 \mathcal{Y}\_{\ell m}(\theta, \varphi) = m \mathcal{Y}\_{\ell m}(\theta, \varphi). \tag{10b}
$$

It is well known that the "degeneracy" of an eigenvalue *λ* of a linear operator is the property for which the eigenspace corresponding to *λ* has dimension greater than one. Under such a circumstance, we can state that the eigenvalue *λ* is *degenerate* as well. When the spectrum of a linear operator has a *degeneracy*, a problem usually arises: given a degenerate eigenvalue *λ*, it is not possible to guarantee that a related eigenvector **v** is selected unambiguously.

From the secular Equations (10a) and (10b), the three simultaneous secular equations:

$$
\mathcal{L}\psi\_{n\ell m}(r,\theta,\varphi) = \lambda\_n \psi\_{n\ell m}(r,\theta,\varphi),
\tag{11a}
$$

$$A(\theta,\varphi)\psi\_{n\ell m}(r,\theta,\varphi) = \ell(\ell+1)\psi\_{n\ell m}(r,\theta,\varphi),\tag{11b}$$

$$
\tilde{\mathcal{M}}\_3 \psi\_{n\ell m}(r, \theta, \varphi) = m \psi\_{n\ell m}(r, \theta, \varphi) \tag{11c}
$$

follow, and the degeneracy of the spectrum of the operator L<sup>1</sup> given in (5) is, in other words, due to the dependence of the eigenfunctions on the given parameters. Namely, the eigenfunctions *ψn<sup>m</sup>*(*r*, *θ*, *ϕ*) depend on the three parameters *n*, -, *m*, whereas the eigenvalues *λ<sup>n</sup>* of the operator L<sup>1</sup> in (11a) depend on *<sup>n</sup>* only, being independent of the other two parameters -, *m*. The next result, whose proof is rather straightforward, describes the *commutation property* of the operators.

**Theorem 3** (**Commutation theorem**)**.** *The linear operators* O1*,* O2*,* ... *,* O*<sup>n</sup> acting on the same Hilbert space are pairwise commuting if and only if there exists a basis of the Hilbert space formed by all simultaneous eigenfunctions of* O1*,* O2*,* ... *,* O*n.*

Theorem (3) provides an important connection with the degeneracy of the spectrum of an operator, as the next theorem shows.

**Theorem 4** (**Degeneracy theorem**)**.** *If a linear operator* O*, acting on a Hilbert space, is invariant under at least two linear transformations* U1, U2*, acting on the same Hilbert space, which are not pairwise commuting, then the spectrum of the operator* O *has a degeneracy.*

**Proof.** By *reductio ad absurdum*, suppose that the spectrum of the operator O has no degeneracy. Since O is invariant under the linear transformation U1, we can apply Theorem 1, from which the commutation relation [O, U1] = 0 follows. Hence, there exists a basis of the Hilbert space formed by all simultaneous eigenfunctions {*y* (1) *<sup>i</sup>* } of <sup>O</sup> and <sup>U</sup>1. For the same reason, there exists a basis of the Hilbert space formed by all simultaneous eigenfunctions {*y* (2) *<sup>j</sup>* } of <sup>O</sup> and <sup>U</sup>2. Since the spectrum of the operator <sup>O</sup> has no degeneracy, it follows that the two sets of eigenfunctions {*y* (1) *<sup>i</sup>* } and {*y* (2) *<sup>j</sup>* } are the same set, but this conclusion is absurd because the operators U1, U<sup>2</sup> do not commute with each other, and then, there cannot exist a basis of the Hilbert space formed by all simultaneous eigenfunctions of the non-commuting operators U1, U2.

At the present stage, based on Theorem 4, we can state that the degeneracy of the spectrum of the operator <sup>L</sup>˜ is not surprising, in that this operator is invariant under the action of the three operators M1, M2, M3, which fail to be pairwise commuting.

**Definition 3** (**Complete set of operators**)**.** *If the linear operators of the set* {O1*,* <sup>O</sup>2*,* ... *,* <sup>O</sup>*n*} *are all pairwise commuting and there exists no other linear operator commuting with them, except the trivial operators, then the set is called the complete set of operators.*

Each complete set of operators is endowed with the following key property. Provided that an operator has a degenerate spectrum, that is the knowledge of an eigenvalue does not allow selecting its eigenfunction unambiguously in the corresponding eigenspace, such a degeneracy can be removed. Basically, if a certain eigenfunction is also an eigenfunction of all the operators in the complete set with respect to a fixed eigenvalue for every operator simultaneously, then the operator's degeneracy is eliminated. The notion of *ladder operators* is very helpful to outline our procedure.
