# $SO(3)$, $SU(2)$ and symmetries in quantum mechanics [duplicate]

A rotation in the vector space $\mathbb{R}^3$ is represented by the known 3x3-matrices.

But at this point I'm really confused how to get from there to Quantum Mechanics. The group of $\mathrm{SO}(3)$ contains all this matrices, but the representation of the rotation operator is a $(2j+1)\times(2j+1)$ matrix. Could someone say a few words to this?

And I'm also confused what's the difference between $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$. (in groups the difference is clear, but they both apply rotations to our kets?)

• – Qmechanic Jan 30 '15 at 16:13

It's not that hard to see how a rotation can end up being represented by a matrix of dimension $(2j+1)\times(2j+1)$. The key concept is that this matrix acts on a subspace $V$ of the Hilbert space $\mathcal H$; that is, $V$ contains state vectors (kets). Generally, $V$ is required to be an invariant subspace in the sense that if $v\in V$, then under a rotation $v$ will in general go to some different vector $v'$ but it will nevertheless stay in $V$.

The easiest way to see this is by way of example, so let me show how this works for $j=2$. There are in general many possible realizations of $V$, but the cleanest realization is as the vector space of functions $f:\mathbb R^3\to \mathbb C$ which are homogeneous polynomials of degree 2, and which are 'traceless' in the sense that $$⟨f⟩=\int_{S^2} f(\hat{\mathbf{r}})\,\mathrm d \Omega=0.\tag1$$ This vector space is best analysed in a convenient basis, and the cleanest one is $$B=\{x^2+y^2-2z^2, xz, yz, xy, x^2-y^2\}.$$ It is fairly easy to see that $V$ is closed under rotations, because each vector component will go into a linear combination of $x,y$ and $z$, and multiplying any two such combinations will again give a homogeneous polynomial. Rotations will also not affect the tracelessness condition (1).

To calculate the effect of a rotation $R\in\mathrm{SO}(3)$, you simply take a given $f\in V$ to the function $G(R)f\in V$ which is given by $$(G(R)f)(\mathbf r)=f(R^{-1}\mathbf r).$$ (The reason for the inverse is so that the operators $G(R)$ have the nice property that $G(R_1\circ R_2)=G(R_1)\circ G(R_2)$, so that $G$ itself is a homomorphism between $\mathrm{SO}(3)$ and the group of unitary transformations on $V$, $\mathrm{U}(V)$.)

For any given $R$, $G(R)$ is a geometrical transformation but it is also, at a simpler level, a linear transformation in a finite-dimensional vector space $V$ with basis $B$, so you can simply represent it by its matrix with respect to this basis. Thus, for example, a rotation by 90° about the $+x$ axis would be represented by the matrix $$\begin{pmatrix} -\tfrac12&0&0&0&\tfrac12\\ 0&0&0&-1&0\\ 0&0&-1&0&0\\ 0&1&0&0&0\\ \tfrac32&0&0&0&\tfrac12\\ \end{pmatrix}.$$ (Work it out!)

The others have given more detail on how this works mathematically - the function $G$ being a representation of the group $\mathrm{SO}(3)$ - but I think that examples of this sort help a lot in visualizing what's going on.

The representation of the abstract Lie group $\mathrm{SO}(3)$ on the usual space $\mathbb{R}^3$ is known as the fundamental representation of the group. No other representations usually occur in classical mechanics because we mostly just have $\mathbb{R}^{3N}$ as the space of spatial coordinates for $N$ objects on which the rotations act. The rotations must act upon your "space of classical states", but it is clear that ordinary spatial coordinates will always transform in this fundamental representation.

In quantum mechanics, transformations/symmetries like rotations have to be implemented (i.e. represented) upon the space of states of the theory, which is essentially the projective Hilbert space associated to a quantum system by taking the Hilbert space spanned by all independent states and projectivizing it.

Thus, any quantum mechanical Hilbert space must carry a unitary projective representation of $\mathrm{SO}(3)$ for us to be able to "measure" angular momentum, since angular momentum generates the rotations - the operators of angular momentum lie in the Lie algebra $\mathfrak{so}(3) \cong \mathfrak{su}(2)$, and every linear representation of the Lie algebra induces a linear representation of the universal cover of the Lie group, which are in bijection to projective representations of the group(s) covered by it.

Therefore, to find all possible projective representation of the rotations, we seek all linear representations of their universal cover, which is the (double) cover $\mathrm{SU}(2)$. Classifying all possible representations of this kind, one finds (e.g. by looking at Verma modules) that all these unitary representations are already fully described by giving the expectation value of the Casimir operator $\vec L^2$, commonly written $l(l+1), l \in \mathbb{Z} \vee l \in \mathbb{Z}+\frac{1}{2}$, and the associated representation space has dimension $2l+1$.1

Of these, only the ones with integer $l$ are proper linear representations of $\mathrm{SO}(3)$, while the half-integer ones map a "rotation" by $2\pi$ to a reflection, but since that is just an overall phase, all of them are projective representations of the rotations. $l$ is also the total angular momentum of a state in such a representation.

1These are only the irreducible representations, but every other representation can be built out of them. The commonly presented basis for these spaces are eigenvectors of $L_z$ with eigenvalues in $\{-l,-l+1,\dots,l-1,l\}$

$SO(3)$ is 2-connected and it turns out that $SU(2)$ is its universal (simply connected) covering group. Since there is then a covering homomorphism $\gamma:SU(2)\to SO(3)$, which is a local isomorphism, one can then consider (as a consequence of Peter-Weyl) theorem, the irreducible representations of $SU(2)$. These can be parametrised by the points in the spectrum of the Casimir operator, which is generally interpreted (up to renormalisation) with the spin $j$ (actually $j(j+1)$, but it doesn't really make a difference since it is the possible to recover $j$). It is then observed that elements in the representation spaces transform according to some well-known transformation rules: for $j=0$ everything is left invariant, so elements in this vector space (trivially $\mathbb C$) behave like scalars. Elements from the representation space of $j=1/2$ behave like spinors (they don't get back to themselves after a rotation of $2\pi$ but they take a sign). For $j=1$ you recover the matrices of $SO(3)$, so the elements are interpreted as vectors, and so on (cf. Wigner-Eckart theorem).