# Why do we look at the representations of $SO(3)$ in QM?

I have a bit of an understanding issue why the representations of $SO(3)$ are so important for Quantum Mechanics. When looking at its Irreps one gets the Spin and Angular Momentum operators and thus their physical significance is fairly obvious.

What I don't get is WHY these Irreps give what they give. When introducing the angular momentum operator by canonical quantization $\vec{L} = \vec{r} \times \vec{p}$ where $\vec{r}$ and $\vec{p}$ are operators one arrives at an operator whose entries satisfy the commutation relationships that the images of the Irreps of $SO(3)$ satisfy as well (i.e. $[L_1, L_2] = iL_3$ (up to some $\hbar$s)).

Wigner's theorem then tells us that every single vector operator acting on states that are eigenstates of a rotationally invariant operator (if I understood the theorem correctly).

My main question now is: Is the declaration of $SO(3)$ as a symmetry group a postulate? Is it "obvious?" If yes, why? When we talked about symmetries in classical mechanics it was either a postulate (Noethers Theorem gives us a conserved quantity for the symmetry postulate that the laws of nature are the same everywhere and constant in time) or one could explicitly calculate it (i.e. that a Hamiltonian or Lagrangian is invariant under the act of certain operations like rotations). Which (if any) of that is it in the case of $SO(3)$ and Quantum Mechanics?

I hope someone can shed some light on this for me.

You already received several answers. However the fundamental physical reason is elementary: In classical, quantum and relativistic physics the physical laws describing an isolated physical system in an inertial reference frame are the same (are invariant) if you rotate (with an element of $SO(3)$) the system (there are many other symmetries depending on the theory, but the point is that rotations are symmetries). This is a fundamental postulate of all physics, valid at small (noncosmological) scales at least. So, for instance a curve describing the evolution of the system in the space of states remains a curve describing (another) evolution of the system if we apply the same rotation to the former curve at every time. There are many other examples. You understand that, unless the space of states be the physical space isomorphic to $\mathbb R^3$, the action of $SO(3)$ on the states cannot be implemented directly using the matrices $R\in SO(3)$. Instead you should faithfully "represent" the elements group $SO(3)$ in terms of natural transformations of the space of states. In quantum mechanics, for many theoretical reasons (already mentioned Wigner and Kadison theorems) these natural transformations are given by unitary (and anti unitary) operators, as theoretical consequence of the fact that the most elementary expected preservation property of these symmetries is that of probabilty transitions of pairs of pure states. (Actually one should use projective unitary representations, but I do not think it is the case to enter into the details here.) Another intersting fact of $SO(3)$ (actually $SU(2)$) representations is that, due to the compactness of the group, all possible unitary representations are constructed out of the irreducible ones via the standard procedure of direct sum (this is not a trivial fact, because usually one should deal with the much more complicated tool of the direct integral).

• So really it is along the lines of what I said above: Similarly as to where we demand the isotropy of space we get momentum conservation through Noether's theorem, we get the conservation of spin from the demand of QM systems being invariant under rotations of the system. – user17574 Aug 3 '14 at 20:33
• Yes I think you are right. However two remarks are worth. (1) Conservation of physical quantities as consequence of existence of symmetries is not a basic direct fact as, e.g., the validity of isotropy, and it concerns dynamical symmetries. (2) Conservation of spin is not automatic this way. What is conserved in view of isotropy of isolated physical systems and various versions of Noether's theorem is the just the total angular momentum. – Valter Moretti Aug 4 '14 at 7:58

One can make sense of the introduction of $\mathrm{SO}(3)$ into quantum mechanics as follows:

1. Consider a physical system in three spatial dimensions which we'll think of as residing in a box sitting on a table, like a table-top experiment.

2. Suppose that we prepare the system in a particular way, so that the system is measured to be in a (pure) state $|\psi\rangle$.

3. Suppose that we rotate the box relative to our measurement apparatus. This spatial rotation will be mathematically described by an element $R\in\mathrm{SO}(3)$.

4. We now ask ourselves "what will the state of the system be now that it's been rotated relative to our measurement apparatus?"

5. To answer this question, we realize that we could have rotated the system according to any rotation $R\in\mathrm{SO}(3)$, so we are really looking for a mapping $\alpha:\mathrm{SO}(3)\to\mathscr{F}(\mathcal H)$ that assigns to each rotation $R$, a function $\alpha(R)$ which tells us how a given state $|\psi\rangle$ changes when the system is rotated. In other words, $\alpha(R)(|\psi\rangle)$ will represent the state of the rotated system.

6. Next we think about what properties this mapping should have. Firstly, we demand that each mapping $\alpha(R)$ be surjective (just like rotations themselves are) and that it preserves transition probabilities (since rotations of the system shouldn't change the physical predictions).

7. This is where Wigner's Theorem on symmetries in quantum mechanics then comes into play. It tells us that such a mapping will be represented by a unitary or antiunitary mapping on the Hilbert space up to phase. For $\mathrm{SO}(3)$, it can then be shown that each $\alpha(R)$ must be unitary because rotations can always be written as squares.

8. It follows (after a bit more argumentation) that the action of rotations on the states of a quantum mechanical system consists of a projective, unitary representation of $\mathrm{SO}(3)$ acting on the Hilbert space of the system. More info on this step here: Idea of Covering Group

9. The angular momentum operator then falls naturally out of this as it is defined as the infinitesimal generator of this representation.

The very important group in physics is the Lorentz group $SO(3,1)$. It is built from the $SO(3)$ group and Lorentz boosts. The algebra of the Lorentz group generators (boosts $L_{i}$ and 3-rotations $R_{i}$) doesn't have the separation on $SO(3)$ and boosts' parts. But by introducing the "new" generators $J^{i}_{\pm} = \frac{1}{2}(R_{i} \pm iL_{i})$ we may represent the Lorentz group as the direct product of two $SO(3)$ or $SU(2)$ groups algebra (the last is homomorphic to SO(3)). These representations has dimension $2j_{\pm} + 1$. The sum of eigenvalues of $J^{3}_{+} + J^{3}_{-}$ refers to spin. The spin plays very important role in classification of irrep of the Lorentz group (and so in classification of one-particle states). But, of course, the Lorentz group works better rather with $SU(2)$ reps, because they may describe both fermions and bosons.

Also the SO(3) group plays important role as the little group for the massive states of the Poincare group. As it can be shown in consequence, the square of one of the Casimir operator of this group (the square of Pauli-Lubanski operator) gives the square of $SO(3)$ group operator.