I'm reading about the classification of relativistic massive particles in Weinberg's "The Quantum Theory of Fields", and I found something that doesn't convince me. In Chapter 2, paragraph 5, having previously dealt with the translations, Weinberg decomposes the action of a proper orthochronous Lorentz trasformation on the Hilbert space of some relativistic particle as

$$ U(\Lambda)\,|p,\sigma\rangle=\frac{N(p)}{N(\Lambda p)}\sum_{\sigma'}D_{\sigma',\sigma}(W(\Lambda,p))\ |\Lambda p,\sigma'\rangle $$

Where the $|p,\sigma\rangle$'s are eigenvectors of the four-momentum operator with some extra degrees of freedom specified by the labels $\sigma$, the $N$'s are normalization coefficients due to the definition of the $|p,\sigma\rangle$'s in terms of the boost of a standard eigenvector $|k,\sigma\rangle$ and $D_{\sigma',\sigma}(W(\Lambda,p))$ is a representation of the element

$$ W(\Lambda,p)=L^{-1}(\Lambda p)\ \Lambda\ L(p) $$

of the "little group", the subgroup of the Lorentz group that fixes $k$ (here $L(p)$ is the boost that brings $k$ to $p$). By decomposing the action of $U(\Lambda)$, Weinberg reduces the question of the classification of relativistic particles to that of finding the irreducible representations of the little group, determined by the standard eigenvector one chooses in order to define the four-momentum eigenvectors. In particular, for massive particles with positive energy, he chooses the four-momentum

$$ k^{\mu}=(m,0,0,0) $$ i.e. the particle-at-rest with mass $m$ four-momentum. He then concludes that the little group for this $k$ is SO(3), as every boost modifies $k$ and no rotation modifies it, so that, given the mass $m$ of the particle (and given that its energy is positive-definite), this last is classified up to the behaviour of the four-momentum eigenstates under three-dimensional rotations. Since the generators of rotations and translations do not commute, while these rotational transformations do commute with the momentum operator, they must act on non-orbital SO(3)-degrees of freedom, i.e. on which one calls the spin degrees of freedom of the particle. Up until here, everything is clear to me (and very cleverly posed, in my opinion). He then passes to the classification of the representations of the little group SO(3). He says that irreducible finite-dimensional representations of SO(3) are labeled by numbers $j$ which can take integer or half-integer values, i.e. spin can be of integer or half-integer value. However, we know that that the half-integer values are really values in the representations of SU(2), not of SO(3). At the Lie algebra level, the representations are the same, as the Lie algebra is the same, but the one-one correspondence between the representations of the Lie algebra and those of the Lie group holds only for simply connected Lie groups, in this case SU(2). So when he classifies the representations of SO(3), shouldn't he be taking $j$ to be exclusively integer? I know that the correct answer is integer AND half-integer, but he doesn't give at all an explanation of why it is possible to use SU(2) instead of SO(3) (at least not to the point where I'm with the reading, I apologize if he does so afterwards). I think I read somewhere else that this has something to do with the fact that the representations one needs to define on the Hilbert space are really projective representations, so that the minus sign that one gets when SU(2)-rotating the state by $2\pi$ really accounts for a phase shift of $e^{i\pi}$. By this line of reasoning, the projective representations of SO(3) do in fact coincide with the non-projective representations of SU(2). Should this be the correct way to see it, I'd like someone to elaborate on this. Thank you in advance.


1 Answer 1


$SO(3)$ symmetry means that the amplitudes $|\langle \psi|\phi\rangle|^2$ is invariant under rotations in the rays. Remember that a ray $\mathcal{R}_{\psi}$ is specified by a family of vectors $e^{i\phi}|\psi\rangle$. This means that the linear or anti-linear operators that describes how the vectors are changed by symmetric transformations furnishes a projective representation, not a unitary one.

Later, in this same chapter, Weinberg start a program to find protective representations of symmetries that cannot be absorbed by a unitary representation in a trivial way, like a redefinition of the unitary operators. Then he shows that there is non-trivial ways to absorbe a projective representation by changing the symmetric group.

The $SO(3)$ have non-trivial projective representations, but this reps could be absorbed by an unitary one of a different group, a large one, the $\mathrm{Spin}(3) \cong SU(2)$.

There is two reasons to have a genuine projective representation, a central extension in the Lie algebra and non-simple connectedness of the Lie group. The first is an algebraic feature (local property of the Lie group) and the second a topological feature (global property).

The case of $SO(3)$ is the topology. There is a closed loop that goes from the identity to the $2\pi$ rotation that is trapped. There is non continuous deformation of this loop that shrink it close to the identity. For each unitary representation with spin $j$ there is a projective representation related to the paths that winding one time on the group. This rep could be identified to a spin $j+1/2$ unitary representation of $SU (2)$. All this is called Covering the Group.


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