From: Maxime Bagnoud Subject: Particles & representations (old: longitudinal mass) Date: 16 Aug 2000 22:07:42 GMT Newsgroups: sci.physics.research Summary: [missing] I remember someone asked me to be more explicit about my claim that "particles in QFT are irreducible representations of the Poincare group". Unfortunately, the thread disappeared from my news server while I was 4 days in the mountains. Though I don't remember whom I have to answer, I'll try to do my best anyway... Let me first recall what is an irreducible representation: A representation of the group G is a group homomorphism from G to the group of vector space endomorphisms (IOW linear transformations) of a representation space V, i.e. R:G------>End(V) g------>M(g) s.t.: - M(g)*M(h) = M(g*h) (the group multiplication rule is compatible with the matrix product in End(V) - M(1) = 1 (the image of the identity of the group G is the identity matrix from V to V) - M(g^{-1}) = M^{-1} (g) (the image of the inverse of a group element is the matrix inverse of the image of the element) A representation is called irreducible if it cannot be split into a direct sum of two or more other representations (in matrix words, this means (more or less): if there are no basis in V such that M(g) is block-diagonal for every g in G, i.e. if there are no invariant subspaces under the action of the M(g)'s). By a slight abuse of terminology, we also call representation of G the representation space V (and that's what I will call a representation in everything which follows). Now, let me come back to physics. Relativistic quantum field theories all have the property of being covariant under any Poincare transformations, which are basically changes in the space-time coordinate system that leave the metric invariant. As a result, you want your concept of a particle to be something which is also invariant under any Poincare transformation (e.g.: you don't want to be able to change an electron into a photon by changing the frame of reference in your space-time). That's certainly the case for scalars. But nature is more subtle, allowing different spins than 0. For example, you can take vectors, as well, and Poincare transformations will mix all their components, making it meaningless to call (for example) A_1 and A_2 (first & second component of a vector potential) by different names. When you tensor two vectors to get a second-rank tensor (e.g. A_i x B_j = E_ij), the game becomes more interesting: You get three combinations of the components which are not mixed between them by any Poincare transformation. The first is the traceless symmetric combination 1/2 (E_ij + E_ji - 1/2 Tr(E) g_ij), which is the real spin-2 part, which can be used to describe quantum fluctuations of the metric. The second is the antisymmetric part 1/2 (E_ij - E_ji), which has in fact spin one, and is called Kalb-Ramond field in supergravity and superstrings theories. The third is the trace part (which is well-known from linear algebra to be an invariant) and is then a scalar. As the components of these three different objects are never mixed between each other by Poincare transformations (and are as such irreducible representations of the Poincare group, while the tensor E_ij itself is not) and have different spins, they describe different particles. I hope this was a relatively convincing way to make you understand that "particles in QFT are irreducible representations of the Poincare group". The second question was something like: "Are spin (or helicity for massless) and mass really enough to classify all particles?" The answer is: Yes and no... Yes, when talking about relativistic quantum field theories in general. Then, mass and spin are the only features that make us able to distinguish between them, because these are the only features which always survive a Poincare transformation. Why? Because they commute with all generators of the Lie algebra and thus with the whole group of transformation and this is exactly the definition of the Casimir operators. As the Poincare group has only two Casimir's, P^2 and W^2, which set mass and spin but nothing else, you cannot classify further. On the other, this is only half the truth for special cases, because you will say: photons and gluons are both massles and have both spin one, BUT they don't follow the same physics. This is because, even if the kind of allowed couplings is severely limited by renormalizability, dimensionality and so on, you can always give different interactions to particles of the same group-theoretical kind. For example, you can make a QED with two different photons and give them the same kind of interactions with electrons, but with different coupling constants, say e and 2e. Then, they will have different physics, while being the same particle. You will then ask: How am I cheating in case two? This leads us to a more complete answer. Spin and mass are enough to classify all possible *free* relativistic quantum fields. Different interactions can then make them have a different physics, but their (tree-level) propagator will always be the same! I took quite some time to get there, so I hope this was clear enough. If not, don't hesitate to ask me further. Greetings, Maxime.