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The Conformal group contains the Poincare group. Typically, if you take a representation of a group and then look at it as a representation of a subgroup, the representation will be reducible. I often hear that CFT's cannot have particles, and I have some understanding of this since $P_\mu P^\mu $ is not a Casimir of the conformal algebra. However, I would think reps of the conformal group should still be reducible representations of the Poincare group, and thus have some particle content.

Is it known how to decompose representations of the conformal group into reps of Poincare? Can we understand it as some sort of integral over masses that removes the scales of the theory? Are there any significant differences between reps of Poincare appearing in reps of the conformal group and the usual representations we are familiar with from QFT?

I'd appreciate any information or a reference that treats this thoroughly from a physics perspective.

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Possible duplicate: physics.stackexchange.com/q/27598/2451 Related: physics.stackexchange.com/q/78552/2451 –  Qmechanic Sep 26 '13 at 16:14
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You can have a look at the discussion here physics.stackexchange.com/q/77222. The summary is that a generic irrep of conformal group decomposes in an integral of Poincare irreps. There are exceptions, free scalar/fermion in any dimension form an irrep of Poincare and conformal algebra. Maxwell, i.e. massless spin-one is conformal in 4d –  John Sep 26 '13 at 16:31

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Reps of the conformal group are still reps of the Lorentz group, that's correct. The operator content of a CFT contains scalars, currents, tensors, spinors etc. However, just because you have a local operator $\phi(x)$ or $\psi_\alpha(x)$ doesn't mean you have bosonic (resp. fermionic) particles in the theory.

Let's recall what we mean by "particle". It's an excitation of the vacuum. In a free theory, for example, we write a bosonic field $\phi$ as

$$\phi(x) \sim \int e^{ipx} a^{\dagger}(p) + c.c.$$ so acting on the vacuum you get a linear superposition of modes $$|p\rangle = a^{\dagger}(p) |0\rangle.$$ By acting several times with $\phi$, we can construct multi-particle states $|p,q,\ldots\rangle.$

There is a more general way of developing the particle picture, and that's through the spectral representation, http://en.wikipedia.org/wiki/K%C3%A4ll%C3%A9n%E2%80%93Lehmann_spectral_representation or sec. 7.1 in Peskin-Schroeder. This means that you write the propagator as $$ < \phi(x) \phi(0) > \; = \int_0^\infty \rho(\mu^2) \Delta(x,\mu^2)$$ where $$\Delta(p,\mu^2) = \frac{1}{p^2 - \mu^2 + i0}$$ in $d=4$ dimensions. $\rho(\mu^2)$ is called the spectral density and encodes which states contribute to the two-point function. It can be calculated as (see Wiki or Peskin-Schroeder) $$\rho(p^2) = \sum_{\text{states } s} \delta^d(p - p_s) |\langle s | \phi \rangle|^2$$ with the sum running over all states in the theory and $p_s^\mu$ is the 4-momentum of state $s$. If you have a free boson with mass $m$, you can derive that $$\rho(p^2) \sim \delta^d(p^2 - m^2).$$ Peskin-Schroeder plot a typical graph of $\rho$ for an interacting theory in Fig. 7.2. There, you will have a bunch of other terms for $p^2 > m^2$ as well, when $\phi$ overlaps with heavier states.

Anyway, back to CFT. In a free CFT (d = 4), the two-point function is given by $$<\phi(x)\phi(0)> = \frac{1}{x^2}$$ and we have the spectral representation $$ \rho(p^2) \sim \delta^4(p^2)$$ which we interpret as there being a single massless excitation. But a generic scalar operator in CFT has some anomalous dimension $\delta$, so the two-point function will be $$\frac{1}{x^{2+2\delta}}, \quad \delta > 0.$$ In that case, you get $$ \rho(p^2) \sim (p^2)^{\delta-1}$$ which is continuous. You asked if "we [could] understand it as some sort of integral over masses that removes the scales of the theory?" - and you see that it's indeed what happens, in a quantitative way. There is no scale, only a parameter $\delta$ controlling the power-law scaling of the contribution of each state. As a consequence, there is no discrete set of states with a well-defined energy: $\phi$ just creates a big, scale-invariant blob of states.

That's essentially why people make a difference between "fields" and "operators". Superficially, both look the same: they are reps of the Lorentz group and generate the Hilbert space of the theory. However, in a normal QFT, fields give you particles and you can do scattering. Generically, in a CFT the operator don't have particle-like excitations, so we need to use a different term - "operator". A field in QFT is of course a local operator, but a local operator in CFT isn't necessarily a field.

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