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.


2 Answers 2


It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on here which I will briefly outline.

The first thing I want to briefly discus is the relevence of representations of the LORENTZ group vs. representations of the POINCARE group.

Lorentz group: Representations are finite dimensional, given by two half integers $(j_1, j_2)$, and not unitary. Spacetime translation is not included. Representations of this group are representations of local field operators $\hat \phi_i(x)$. Under a Lorentz transformation, the indices of $\hat \phi_i(0$) will get mixed together while the spacetime point $0$ will be fixed. Two simple examples include the spin $0$ representation, where $i$ is just $0$, and the $(1/2, 1/2)$ representation, where we instead use the symbols $\hat A_\mu$ and $\mu = 0... 3$. The fact that these finite dimensional representations (which just juggle around the field indices) are not unitary is totally uninteresting because there isn't a natural inner product of local operators, like there is with states, so who cares that they're not unitary!?

Poincare group: Representations are infinite dimensional, unitary, given (by Wigner's theorem) by either a mass $m^2$ and a half integer spin, or in the $m = 0$ case by a half integer helicity. These correspond to particle states. The fact they they are unitary is important because you want the group action to preserve the natural inner product on states. The fact they are infinite dimensional is not really unexpected, the same thing happens in non relativistic QM! For instance, in the spin $0$ case, single particle states are given by a momentum $| p \rangle$. Under a Poincare transformation,

$$ U(\Lambda, a) | p \rangle = e^{- i p a} | \Lambda p \rangle. $$

The fact that you need to include a state $| \Lambda p \rangle$ for each Lorentz transformation $\Lambda$, this representation is infinite dimensional because your basis of single particle states is spanned by an infinite number of $p$'s. But this is also the case in non relativistic QM, that you have a basis spanned by an infinite number of momenta, and isn't special to quantum field theory, so it's not really that big a deal. It's true that the Poincare group is non compact because of boosts, but its also non compact because of translations, which is why QM is infinite dimensional (there's an infinite number of positions).

Relationship between the two: You can create single particle states by acting on the vacuum with field operators. In the spin $0$ case a single particle state is given by

$$ | p \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \hat \phi(\vec x) | 0 \rangle $$

and in the spin $1$ case by

$$ | p, \varepsilon \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \varepsilon^\mu \hat A_\mu(\vec x) | 0 \rangle $$

(modulo maybe some integration measure I forgot.) Here $\varepsilon^\mu$ is a polarization vector satisfying $\varepsilon^\mu p_\mu = 0$. Because single particle states are super positions of states like, say, $\hat A_\mu | 0 \rangle$, the spins of the physical particle states can be found by restricting oneself to the rotation part of the Lorentz group, $U(R)$, and just using the standard angular momentum addition formula, where the irreducible representations of the $j_1 \otimes j_2$ representation is $j_1 +j_2 \oplus j_1 + j_2 - 1 \oplus \ldots \oplus |j_1 - j_2|$. So in the $(1/2, 1/2)$ representation of the Lorentz group, we have $1/2 \otimes 1/2 = 1 \oplus 0$. However, the non physical $0$ representation is projected out by the condition $\varepsilon^\mu p_\mu$, getting rid of the negative norm states, yadda yadda yadda. The point is that $j_1 \otimes j_2$ give the possible particle spins, but some of them will be non physical.

The main point, though, is that representations of the Lorentz group are used for talking about field operators, while representations of the Poincare group are used for talking about particle states. Once you realize this, it's completely reasonable that representations of the Lorentz group are not unitary (why should they be?) and that representations of the Poincare group are not finite dimensional (why should they be?).

  • $\begingroup$ Thanks for the answer! But there is actually something I do not get. In many books, fields are presented as objects living in infinite-dimensional (not necessarily unitary) representations of the Poincaré group. Indeed the Lagrangian is constructed as a Poincaré invariant. Why are you not considering translations in your answer when talking about field operators and just the Lorentzian part? $\endgroup$
    – Gravitino
    Commented Sep 16, 2021 at 14:03

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}

  • $\begingroup$ 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?) $\endgroup$
    – balu
    Commented Feb 20, 2018 at 20:48
  • $\begingroup$ @balu I'm a bit late, but the reason you can't embed the photon in a single (scalar) field is that there simply aren't enough degrees of freedom to allow for the polarisation states $\endgroup$ Commented Dec 20, 2020 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.