After the answers by joshphysics and user37496, it seems to me that a last remark remains.

The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates **Lie algebras** of symmetric operators with **unitary representations** of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.

In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:

(1) They are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)

and

(2) they enjoy the **commutation relations** of some Lie algebra $\ell$:
$$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$
on a common invariant domain ${\cal D}\subset {\cal H}$.

As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique **simply connected** Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the **universal covering** of all the other Lie groups whose Lie algebra is $\ell$ itself.

All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element.

If (1) and (2) hold, the natural question is:

**Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group $\cal G$ just admitting $\ell$ as its Lie algebra, such that**
$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$

**Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.**

If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint opertors $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are **conserved quantities**.
Otherwise the situation is a bit more complicated, nevertheless one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generator when $\cal G$ is $SL(2,\mathbb C)$).

Well, the fundamental theorem by Nelson has the following statement.

**THEOREM** (Nelson)

*Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator
$$\Delta := \sum_{i=1}^N A_i^2$$
is ***essentially self-adjoint** on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then:

*(a) Every $A_i$ is essentially self-adjoint on $\cal D$,*

*and*

*(b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements:*
$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$
*where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$*.

Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.

The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own Lie algebra.

As another application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected) Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming $X$ and $P$ into the standard operators. Following this way one builds up an alternate proof of Stone-von Neumann's theorem.

As a last comment, I stress that usually ${\cal G}_\ell$ is **not** the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is the appearance of annoying phases as explained by Josh, and overall phases do not affect states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume quantum states that are coherent superpositions of both **integer** and **semi-integer** spin. Otherwise an **internal** phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize **superselection rules**.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti