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I am struggling to see the connection between the symmetries of Lagrangians and particle multiplets. If I had three quark fields arranged in a vector $\Psi(x) = (u(x),d(x),s(x))$, these three fields have a combined Lagrangian

$$ \mathcal{L}= \bar{\Psi}(x)(i\gamma^\mu\partial_\mu-m)\Psi(x) $$

I can see that this Lagrangian has an $SU(3)$ symmetry. When I quantise my theory, I promote my fields $u,d,s$ to quantum operators on an infinite dimensional Fock space. The interpretation of the operators $\hat{u},\hat{d},\hat{s}$ is that they are particle creation operators that create up, down and strange quarks respectively. However, I have also read that quarks live in the fundamental representation of the Lie algebra of $SU(3)$.

My question:

From my studies of quantum field theory, I thought particles live in infinite dimensional Fock spaces, not finite dimensional representation spaces. How do I arrive at the particle multiplets from the symmetry of the Lagrangian?

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In general, a lagrangian can be invariant under more than one transformation so there can be several symmetries and representations related. Specifically, the one particle states furnish representations of the Poincare Group and therefore live in an infinite dimensional space. On the other hand flavor (or color) symmetry does not have anything to do with Poincare, the states live in another space, which is finite dimensional. There will be, for instance, many spaces related to a given quark: The infinite dimensional Fock space associated to momentum and spin, the three dimensional space associated to color and the two (left) and one (right) dimensional space associated to electroweak interaction gives you an explicit example.

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  • $\begingroup$ That makes sense, thank you. Should I really view the field operators as $ \phi(x) \otimes u $ etc. Where $\phi(x)$ acts on the Fock space and tells us where to create the particle, while $u$ tells us what type of particle to create and that acts on the internal 3D Hilbert space? $\endgroup$ – Matt0410 Jun 4 '18 at 14:33
  • $\begingroup$ @Matt0410 Yes. But remember that what you are decomposing is the representation. The same field has different representations. $\endgroup$ – Diracology Jun 4 '18 at 18:52

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