10
$\begingroup$

I read several textbooks of QFT and find that there are two ways to classify the particles or fields. The first one is to study the irreducible representation of Lorentz group (or exactly the universal covering group $SL(2,C)$). Then we find irreducible but not unitary representation $(i,j)$ which is finite dimensional and use them to represent different kinds of field. The second one is to study the unitary representation of Poincare group and we can classify particles by mass and spin.

Then my question is:

  1. Why do we not study the finite dimensional irreducible representation of Poincare group, like Lorentz group? Some people will say that the useful representation in Quantum Mechanics is unitary representation and Poincare group which is not compact do not have finite dimensional unitary rep. However this argument is not convincing, because it cannot explain why we still study the finite rep of Lorentz group.

  2. Except the "trivial" rep., does there exist any other finite dimensional irreducible rep. of Poincare group? Here "trivial" means the rep. that we can get from enlarging the original rep. of Lorentz group by letting translation act trivially on original representational space.

For example, we have a faithful rep. of Poincare group, $\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}$, where $\Lambda$ is Lorentz transformation and $x$ is translation. This is a reducible but indecomposable representation. We can always define an irreducible rep. of Poincare group by

$$f:\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}\rightarrow D_{(i,j)}(\Lambda)$$ where $D_{(i,j)}(\Lambda)$ is the irreducible rep. of Lorentz group. So is there other finite dimensional irreducible rep. of Poincare group?

  1. It seems that we use Lorentz group's rep. to classify the fields and use Poincare group's rep. to classify the particles. Because the isometry of Minkovski spacetime is Poincare group, why do we only use Lorentz group's rep. to classify the fields and don't take the whole Poincare group into consideration?
Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ All unitary irreducible representations of Lorentz group are infinite dimensional. This is the reason. In fact, unitary reps. of the Poincaré group are studied in general not only those of Lorentz group when defining the notion of elementary particle in the sense of Wigner. Dealing with fields the translational part acts trivially, for this reason is usually disregarded when viewing fields as section on some vector bundle based on the spacetime. $\endgroup$
    – Valter Moretti
    Commented Apr 7, 2016 at 17:46
  • $\begingroup$ @ValterMoretti Thanks. ACuriousMind and your answer have solved the question 1,3. Do you have any idea of question 2? $\endgroup$
    – 346699
    Commented Apr 8, 2016 at 5:08
  • $\begingroup$ Actually not, did you try to have a look at Barut Raczac's textbook on representations? $\endgroup$
    – Valter Moretti
    Commented Apr 8, 2016 at 5:30

3 Answers 3

Reset to default
7
$\begingroup$

Let $\phi : \mathbb{R}^4\to V$ be a field with (complex) target vector space $V$, transforming in a finite-dimensional projective representation $\rho_\text{fin} : \mathrm{SO}(1,3)\to\mathrm{U}(V)$. As it is a field, the representation of the translations $\mathbb{R}^4$ on $V$ is the trivial one, since the field transforms as $\phi(x)\overset{x\mapsto x+a}{\mapsto} \phi(x+a)$. Hence, the field transforms in a finite-dimensional representation $\sigma_\text{fin}$ of the Poincaré group, but the non-trivial, i.e. interesting, part is the representation of the Lorentz group. Hence, your premise that we "only study finite-dimensional representations of the Lorentz group" is wrong, it's just that the finite-dimensional translations are always represented by their trivial representation.

In the quantum field theory, the field now becomes operator-valued, acting upon some Hilbert space $\mathcal{H}$. Since the quantum field theory shall have Poincare symmetry, there must be a projective unitary representation $\sigma_\text{U} : \mathrm{SO}(1,3)\ltimes\mathbb{R}^4\to\mathrm{U}(\mathcal{H})$ upon this space of state. By one of the Wightman axioms, we have that $$ \sigma_\text{fin}(\Lambda,a)\phi(\Lambda^{-1} x-a) = \sigma_\text{U}(\Lambda,a)^\dagger \phi(x)\sigma_\text{U}(\Lambda,a)\quad \forall \Lambda\in\mathrm{SO}(1,3),a\in\mathbb{R}^4$$ where on the l.h.s., $\sigma_\text{fin}$ is a finite-dimensional matrix acting upon the vector $(\phi^1,\dots,\phi^{\dim(V)})$, and on the r.h.s., the $\sigma_\text{U}$ are operators on $\mathcal{H}$ are multiplied with each component operator $\phi^i$.

We study the finite-dimensional representations because of this relationship - we have to know the finite-dimensional representations to be able to give the "classical" field, and we have to know the infinite-dimensional unitary representation to know how the Poincaré symmetry acts on states, and because the irreducible unitary representations correspond to particles by Wigner's classification. Since the Poincaré group is just as non-compact as the Lorentz group, these are also all infinite-dimensional.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks. According to your answer, I have understood my question 1,3. Do you have any idea about question 2? $\endgroup$
    – 346699
    Commented Apr 8, 2016 at 5:06
  • $\begingroup$ @user34669: I have no idea, except that they are probably physically irrelevant. We don't need any non-trivial finite-dimensional representations of translations since the fields always have to transform trivially. $\endgroup$
    – ACuriousMind ♦
    Commented Apr 8, 2016 at 11:52
  • $\begingroup$ Yes. we don't need to consider this case in physics. The question 2 is only a mathematical interest. $\endgroup$
    – 346699
    Commented Apr 8, 2016 at 13:03
  • $\begingroup$ @ACuriousMind Once again, interesting and does point to the crucial fact that requiring unitarity has to do with quantum theory. But again, the representation $\sigma_{\text{fin}}$ of translation on the field is obviously infinite dimensional!! (Unless I'm missing an argument saying the the space of fields is finite dimensional...) $\endgroup$
    – Noix07
    Commented Jan 30, 2020 at 16:11
  • 1
    $\begingroup$ "Hence the field transforms in a finite dimensional representation $\sigma_{\text{fin}}$": what is transformed? the field. In what space? the space of fields, i.e. of functions. So, as it does not been explicitly mentionned, the reason one is interested in unitary representations of the Poincaré group has to do with the axiom of quantum mechanics that says that states are described by rays. Then by Wigner's theorem, projective representations on ray may be lifted to true representation, either unitary or antiunitary. And finally, for the connected subgroup of Poincaré containing the identity.. $\endgroup$
    – Noix07
    Commented Jan 31, 2020 at 13:37
2
$\begingroup$

The matrix reps of the Poincare group can be found by solving the commutation relations. See arXiv:math-ph/0401002v3 2 Jul 2007 A Derivation of Vector and Momentum Matrices.

Improve this answer
$\endgroup$
0
1
$\begingroup$

About point 2, This paper shows some non-trivial finite-dimensional representations of the Poincaré algebra. One interesting thing that happens is that, since Poincaré group is not compact and continuous, we cannot guarantee the representations are either irreducible or completely reducible (or decomposible). Indeed, the paper deals with the indecomposable ones.

Improve this answer
$\endgroup$

Your Answer

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