Can the finite dimensional irreducible $(j_+,j_-)$ representations of the Lorentz group $SO(3,1)$ be unitary?

Question

Since the Lorentz group $SO(3,1)$ is non-compact, it doesn't have any finite dimensional unitary irreducible representation. Is this theorem really valid?

One can take complex linear combinations of hermitian angular momentum generator $J_i^\dagger=J_i$ boost generator $K_i^\dagger=-K_i$ to construct two hermitian generators $N_i^{\pm}=J_i\pm iK_i$. Then, it can be easily shown that the complexified Lie algebra of $SO(3,1)$ is isomorphic to that of $SU(2)\times SU(2)$. Since, the generators are now hermitian, the exponentiation of $\{iN_i^+,iN_i^-\}$ with real coefficients should produce finite dimensional unitary irreducible representations. The finite dimensional representations labeled by $(j_+,j_-)$ are therefore unitary.

$\bullet$ Does it mean we have achieved finite dimensional unitary representations of $SO(3,1)$?

$\bullet$ If the $(j_+,j_-)$ representations, are for some reason are non-unitary (why I do not understand), what is the need for considering such representations?

$\bullet$ Even if they are not unitary (for a reason I don't understand yet), they tell how classical fields transform such as Weyl fields, Dirac fields etc. So what is the problem even if they are non-unitary?

Valter Moretti · Answer 1 · 2016-11-19T11:37:22.257

$SO(1,3)$ is a real Lie group so its Lie algebra is real too, you are not allowed to combine generators with complex coefficients if you are looking for representation of that group. Referring to the (non-unitary) fundamental representation made of $4 \times 4$ real matrices you consider, $N_i^\pm$ does not belong to the real Lie algebra of $SO(1,3)$. Exponentiating real linear combinations of $N_i^\pm$ you, in fact, obtain a unitary representation of a group. Unfortunately the group is not $SO(1,3)$.

The use of complex extensions of the Lie algebra of $SO(1,3)$ is however helpful when classifying the representations of the proper real Lie algebra of $SO(1,3)$, since classifying all complex representations includes a classification of the real representations too, and the complex Lie algebra of $SO(1,3)$ is isomorphic to a direct sum of a pair of Lie algebras of $SU(2)$ whose theory is relatively simple.

Unitarity is necessary in Quantum Theory due to Wigner's theorem which establishes that, picturing the states of a quantum system in a Hilbert space, all symmetries are represented by unitary or anti unitary operators.

Actually the problem is more complicated due to the appearance of phases (pure states are defined as unit vectors up to phases) which may destroy the composition law of the Poincaré group (a central extension is necessary). However a theorem by Bargmann proves that the Poincaré group is not affected by this problem.

@ Valter Moretti- Let me try to understand. 1. The representations considered by taking complex combinations of generators are unitary but they are not representations of $SO(3,1)$. Is it correct? Are they representations of $SL(2,\mathbb{C})$? 2. You said, exponentiating real linear combinations of $N_i^{\pm}$ one can obtain unitary representations. Which particular combinations are you referring to? — SRS, Nov 19 '16 at 14:12

score 5 · Accepted Answer · edited Apr 13 '17 at 12:39

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The statement "Non-compact groups don't have finite-dimensional unitary representations" is a heuristic, not a fact. $(\mathbb{R},+)$ is a non-compact Lie group that has non-trivial finite-dimensional unitary representations. However, the Poincaré group and the Lorentz group really don't have any finite-dimensional unitary representations.

Your construction fails because the complexification $\mathfrak{so}(1,3)_\mathbb{C}$ is only isomorphic to the complexification $(\mathfrak{su}(2)\oplus\mathfrak{su}(2))_\mathbb{C}$, not to the real Lie group. You found a unitary representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ itself, but this doesn't give you a unitary representation of the complexification, nor of $\mathfrak{so}(1,3)$.

We care about those finite-dimensional representations of $\mathrm{SO}(1,3)$ even if they are not unitary because these are the representations on the target spaces of fields. The representation that needs to be unitary is the representation on the quantum space of states, but not on the target space of fields. Clearly, a vector field transforms in the "standard" representation of $\mathrm{SO}(1,3)$ and doesn't care that it's not unitary because the target space is $\mathbb{R}^{1,3}$ which isn't even a complex vector space to begin with! There is no "problem" with these representations, they just are not the representations we need on the Hilbert space of states, which are projective representations of $\mathrm{SO}(1,3)$, which are equivalent to unitary linear representations of $\mathrm{SL}(2,\mathbb{C})$, its universal cover. For more on the necessity of projective representation, see this Q&A of mine.

edited Apr 13 '17 at 12:39

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answered Nov 19 '16 at 16:51

ACuriousMind

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@ ACuriousMind- The $(j_+,j_-)$ representations that I find, are they anyhow related to the representations of $SL(2,\mathbb{C})$? – SRS Nov 19 '16 at 18:21
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@SRS Sure, since they are reps of $\mathrm{SO}(1,3)$, they are also reps of its universal cover. – ACuriousMind Nov 19 '16 at 18:24
Although the representation is labeled as $(j_+,j_-)$, while constructing representations of $SO(3,1)$, I have to go back to $J_i$ and $K_i$ from $N_i^{\pm}$. Exponentiate them and they are non-unitary. Direct exponentiation of $N_i^{\pm}$ doesn't give the representation of $SO(3,1)$. Going from $J_i,K_i\rightarrow N_i^{\pm}$ is merely a trick to solve for generators easily. Is that correct? – SRS Nov 19 '16 at 18:36
@SRS I don't know what you mean by "to solve for generators easily". Going to the $N_i$ means presenting $\mathfrak{so}(1,3)\mathbb{C}$ as $(\mathfrak{su}(2)\oplus\mathfrak{su}(2))\mathbb{C}$, which we do because we already know the representation theory of $\mathfrak{su}(2)$ from non-relativistic rotations! – ACuriousMind Nov 19 '16 at 18:39
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Exponentiating $\textbf{N}^{\pm}$ would give representations of $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$ which we do not want. For Weyl spinor (1/2,0), after finding that $\textbf{N}^{+}=0$ and $\textbf{N}^-=\sigma/2$, we go back to $\textbf{J}=\frac{\sigma}{2}$ and $\textbf{K}=i \frac{\sigma}{2}$. That is solve for $\textbf{J},\textbf{K}$ and exponentiate them with real coefficients. That would give the representation of $SO(3,1)$ and not the direct exponentiation of $\textbf{N}^{\pm}$. Is it correct now at least in parts? – SRS Nov 19 '16 at 18:55
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@SRS Yes, that's correct. – ACuriousMind Nov 19 '16 at 18:56

Can the finite dimensional irreducible $(j_+,j_-)$ representations of the Lorentz group $SO(3,1)$ be unitary?

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