# How do the infinite dimensional representations of the Poincaré group work?

I'm familiar with representations of the Lorentz group, that take the form:

\begin{equation} e^{-\frac{i}{2}\omega_{\mu \nu}J^{\mu \nu}} \end{equation}

where $J^{ij}=\epsilon^{ijk}R^k$ for $i,j=1,2,3$ are the generators of rotations in the i-j plane (or in the k axis) and $J^{0i}=K^{i}$ for $i=1,2,3$ are the generators of boosts in the i direction. By combining this two sets of generators into $J_{\pm}=R \pm iK$ one can make two invariant subgroups spanned by the new generators, both satisfying su(2) algebra. Then finding every representation is just finding every $(2j_++1,2j_-+1)$ representation of each su(2) algebra.

However, because boosts are non-compact, this transformations are not unitary so are not fit fot quantum mechanics. This is when I get confused, how does the Poincaré group solve this? if Lorentz group finite representations acts on finite dimensional objects such as spinors $\psi$ or vectors $A_{\mu}$... what does the infinite dimensional representations of the Poincaré group act on? How are the generators written?

I get the idea that in infinite dimensional representations the transfomations acts on fields of vectors rather than on vectors, but since transformations are rigid (i.e. don't depend on space-time coordinates) rotating or boosting a vector field seems to be just rotating each vector in a different point in space in the same way, so I don't see the difference.

I've try reading introductory books on QFT such as Schwartz, Ryder, Peskin or Maggiore but I don't fully understand the implications of infinite dimensional representations of the Poincaré group.

I'm sorry for any errors this might have, I'm just starting to study group theory and particle physics and it's overwhelming sometimes.

Extending the Lorentz group to the Poincare group doesn't solve this problem - the solution is found in the way the Lorentz and Poincare groups act on the Hilbert space $\mathcal{H}$ of the theory, which is infinite dimensional. If $\mathcal{H}$ were finite dimensional, it would admit no (nontrivial) unitary representations. But $\mathcal{H}$ isn't finite dimensional.
If we denote a unitary transformation associated with some restricted Lorentz transformation $\Lambda$ by $\mathcal{U}(\Lambda)$, then state vectors transform as $|\psi\rangle\mapsto \mathcal{U}(\Lambda)|\psi\rangle$. Now suppose we have some collection $(\phi^a(x))$ of operator valued fields. Using the same logic as one uses to change back and forth between the Heisenberg and Schrodinger picture, we can transform the fields instead of the states using $\phi^a(x)\mapsto \mathcal{U}(\Lambda)^{-1}\phi^a (x)\mathcal{U}(\Lambda)$. The fields will, in general, mix with each other under this tranformation: \begin{equation} \mathcal{U}(\Lambda)^{-1}\phi^a(x)\mathcal{U}(\Lambda)=D_b^a(\Lambda)\phi^b(\Lambda^{-1}x). \end{equation} It's not too hard to see that $\Lambda\mapsto D_b^a(\Lambda)$ is a representation of $SO^+(1,3)$. We usually only consider finitely many fields, so we can break this representation up into finitely many finite dimensional irreducible representations of $SO^+(1,3)$. These representations need not be unitary since they're not the representation acting on the Hilbert space $\mathcal{H}$ - they act on the fields.
Here's an example - take the case of a vector field $V^\mu(x)$ transforming as $V^\mu(x)\mapsto \Lambda_\nu^\mu V^\nu(\Lambda^{-1}x)$. Then we have generators $\mathcal{M}^{\mu\nu}$ which are operators on $\mathcal{H}$ that satisfy to $\mathcal{O}(\omega^2)$: \begin{equation} \mathcal{U}\left(1-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\right)=1-\frac{i}{2}\omega_{\mu\nu}\mathcal{M}^{\mu\nu},\end{equation}so that: \begin{equation} e^{+\frac{i}{2}\omega_{\mu\nu}\mathcal{M}^{\mu\nu}}V^\alpha(x)e^{-\frac{i}{2}\omega_{\mu\nu}\mathcal{M}^{\mu\nu}}=\left(e^{-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}}\right)^\alpha_\beta \cdot V^\beta(\Lambda^{-1}x). \end{equation}
• Could you elaborate on the point "The fields will, in general, mix with each other under this tranformation"? What exactly is the reason for this? I mean I know that e.g. in QED we're trying to find the quantum analogue of the 4-vector potential $A^\mu$, so we introduce 4 operator-valued fields $A^\mu (\mu = 1, 2, 3, 4)$ and expect a similar transformation behavior as in electrodynamics and also a similar Lagrangian. But is there any there any other (deeper) justification for needing multiple fields? (I.e. naively put, why can't we have a single field $\phi$ representing the photon field?) – balu Feb 20 '18 at 20:48