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As far as I know, a representation is a homomorphism from the group to a vector space $V$ which preserves the group multiplication, i.e., if $(\pi,V)$ is a representation of the group $G$, then whenever $g_1,g_2\in G$ and $g_1\cdot g_2 = g_3$ then

\begin{equation} \pi(g_1)\pi(g_2) = \pi(g_1\cdot g_2). \end{equation}

In Matthew Schwartz's QFT and the Standard Model, chapter 8 titled Spin 1 and Gauge Invariance, he says (page 110) that fields $\phi(x)$ form a representation of the group of translations because $\phi(x) \to \phi(x+a)$. I did not understand how this works and how one would define the group multiplication law.

After some scrounging and stumbling across Peter Woit's book, I found the following definition of a representation on a functional space (page 7, eq. (1.3)): given a group action of $G$ on a space $M$ (given as a set of points), functions on $M$ come with an action of $G$ by linear transformations $\pi$ given by \begin{equation} (\pi(g))f(x) = f(g^{-1} \cdot x). \end{equation} This made me conclude that it is better to think of the fields as being the 'vector' space on which the linear transformations associated with the group acts. Therefore since this 'vector space' is infinite-dimensional, I would think that it has infinite degrees of freedom. But what Schwartz says the problem is that the degrees of freedom are the indices that the fields carry (i.e. a vector field $A_\mu(x)$ has 4 d.o.f., tensor field $T_{\mu\nu}$ have $4\times 4 = 16$ and so on) and the problem is to embed the proper spin $J$ d.o.f. which are $2J+1$ in number into these fields. I do not understand how the indices are actually the d.o.f. but not the fields themselves. Can someone explain?

Qmechanic
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QFTheorist
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    Is this of any help? https://physics.stackexchange.com/q/349166/226902 "How are quantum fields related to representations of the Lorentz group?" See also https://physics.stackexchange.com/q/127989/226902 – Quillo Jan 31 '24 at 13:11
  • A first problem is that your definition of representation is sloppy, and I don't say this to play smart pants, but wondering whether the confusion you have arises from a misconception. Fields are not representations, but rather form a vector space $V$, over which the Poincaré group $G$ is represented through $\rho: G\to \operatorname{Aut}V$. Fields must live in a representation of the group by physical assumption: in flat spacetime, physics must remain unchanged under Poincaré transformations, so the fields describing this physics must transform under a definite representation of that group. – Albert Jan 31 '24 at 13:41
  • @Albert I said in later in my post that '...fields as being the 'vector' space on which...' – QFTheorist Jan 31 '24 at 13:57
  • My confusion was why are the spacetime indices considered to be the d.o.f. instead of the fields themselves. – QFTheorist Jan 31 '24 at 13:57
  • Fields are not particles. Internal degrees of freedom of the particles, such as spin, are encoded by the addition of indices to the fields. These fields with indices also must live in a representation of the Poincare group, so these indices are acted upon by a separate finite-dimensional representation of the group, whose action can therefore be represented as a matrix acting on these "vectors of fields". For example, in the vector representation, we'd have $\phi^{\mu} (x)\stackrel{g}{\mapsto}\pi (g)^{\mu}_{\ \ \ \nu} \phi^{\nu}( g^{-1} x)$. – Albert Jan 31 '24 at 14:30
  • That makes it so much clearer @Albert thanks. – QFTheorist Jan 31 '24 at 15:12
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    You are welcome, @QFTheorist. It is important to remember that fields are operators acting on the Fock space where the (multi-)particle states live; ideally, and put simply, one tries to conceive a QFT in such a way that free particle states are created by the action of the field operators on the vacuum state. – Albert Jan 31 '24 at 15:24

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Fields are reducible representations (by which I mean, fields are elements of the vector space on which a reducible representation of the Poincar'e group acts). On the other hand, particles are irreducible representations.

Schwartz is trying to explain how $e.g.$ the $3$ irreducible d.o.f. of a spin-1 particle can be embedded into the 4 reducible d.o.f. of a spin-1 field or $e.g.$ the 1 irreducible d.o.f. of a spin-0 particle can be embedded into the 1 reducible d.o.f. of a spin-0 field.

Prahar
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  • Hi @Prahar. Thanks for your answer. What was not clear to me actually is that why are we considering the spacetime indices of fields to be the d.o.f. Maybe this is a naive question but I never really understood it. – QFTheorist Jan 31 '24 at 13:53
  • We are not doing that. The dofs of the particles are the physical ones. However, the physical dofs have a complicated transformation law under Lorentz transformations. The field dofs on the other hand, transform "nicely" under Lorentz transformations. We prefer to work with the latter and to do that, we need to figure out how the physical particle dofs should be embedded into the field dofs. – Prahar Jan 31 '24 at 13:58
  • Thanks @Prahar. Why aren't the field dofs infinite? Because it is a function of spacetime, so shouldn't the field dof actually be infinite? My question is why is the spacetime index the dof of the field rather than its values in spacetime itself. – QFTheorist Jan 31 '24 at 14:08
  • As I said, we are NOT taking the spacetime index to be the dofs. The dofs of a field (and particle) ARE infinite for the precise reason you mentioned. – Prahar Jan 31 '24 at 14:10
  • @QFTheorist - The spacetime indices are the so-called "internal dofs". By abuse of notation, these are often simply referred to as "dofs" since the spacetime dependence of fields is understood to always be present. – Prahar Jan 31 '24 at 14:30
  • that's what I was confused about. Thanks! – QFTheorist Jan 31 '24 at 15:13