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  Misuse of $\mathbf J^2$ in classifying Poincare reps

+ 4 like - 1 dislike

$SO(1,3)$ has an infinite number of representations, classified by the Casimir invariant $p^2$.

$SO(3)$ also has an infinite number of representations, classified by the Casimir invariant $\mathbf J^2$.

Since representations are diffeomorphic if and only if their Casimir invariants are the same, we are justified in this method of classification.

In the case of $SO(3)$, the physical interpretation is:

  • $\mathbf J$ generates rotations of the particle’s rest frame.

  • $\mathbf J^2$, the total spin of a particle, is the dimension of the vector space in which we have chosen to embed the particle.

Now I am baffled by the fact that we use $\mathbf J^2$, i.e. total spin, to classify $SO(1,3)$. That's the wrong Lie group! How is this not nonsense?

$p^2$ is the correct Casimir invariant - what happened to that?

  • Why isn't $p^2$ sufficient? - it's a Casimir invariant, and so it should give us all the classification information (i.e. tell us if reps are diffeomorphic)!

  • Now, suppose that we do things correctly (i.e. discard $\mathbf J^2$) and use $p^2$ to classify representations.

    • Are there “fermions” or “bosons” corresponding to $m$ taking on half or whole integer values in this case?

    • Finally, the representation $m^2=3$ is not isomorphic to $m^2=\pi$ (because $p^2$ is a Casimir invariant). Same with $m^2=2$ and $m^2=2.00000001$. However, in most field theory textbooks, $m>0$ is treated as one case. It's all a blob to them. What?!!!

This post imported from StackExchange Physics at 2014-11-02 16:29 (UTC), posted by SE-user Dave
asked Oct 31, 2014 in Theoretical Physics by Dave (15 points) [ no revision ]

''representations are diffeomorphic if and only if their Casimir invariants are the same'' only holds in the semisimple case, and only if all Casimirs are taken into account.

SO(1,3) =SO(3) x SO(3) is semisimple but its irreps are classified by two SO(3) Casimirs.

$p^2$ is a Casimir (but not the only one) of the Poincare group, not of SO(1,3).

Actually the identity $S(1,3) = SO(3)\times SO(3)$ does not make sense since the right-hand side is  compact whereas the left-hand side is not. The identity holds true at the level of complex Lie algebras.

2 Answers

+ 5 like - 0 dislike

Actually, the (orthochronous) Poincaré group has two Casimir invariant. One is $P_\mu P^\mu$ the other is $W_\mu W^\mu$, where $W_\mu$ is the Pauli-Lubanski vector. The unitary irreducible representations are fixed by assigning a pair o eigenvalues $(m^2,w)$ of these Casimir operators.   If the first eigenvalue $m^2$ is strictly positive, there is a reference frame where  $P^k=0$ and $P^0= m>0$, in that reference frame $W^0=0$ and $W^k = m J^k$. $J^k$ is the spin of the particle with all standard properties. Notice  that, in the considered case,   $W^\mu W_\mu = -m^2 J^2$ so the eigenvalues of the second Casimir operator are exactly the ones of the total spin $w = -m^2j(j+1)$, up to the value of the mass and the sign.

So, for $m^2>0$ fixed, the representations are classified by the eigenvalues of $J^2$. 

In case $m^2=0$, there is no rest frame with the particle and the picture becomes more complicated and $W^\mu$ carries the information of the helicity of the particle (actually there is a subcase called the "continuous spin" representation, but I never seen any physical interpretation of that case).

answered Nov 3, 2014 by Valter Moretti (2,085 points) [ revision history ]

If $m=0$, the irreps of the Poincare group are not classified by the values $p^2=0$ and $W^2=0$ of the Casimirs. - The massless continuous spin representation cannot occur in local field theories, and hence is presumably unphysical.

Yes, I agree...

@ArnoldNeumaier: Why is continuous spin rep forbidden to have a local field theory?

Please ask this as a main question, as the answer is nontrivial.

+ 2 like - 0 dislike

Why is continuous spin rep forbidden to have a local field theory?

As far as I can see the question has been already stated by Jia Yiyang.

I wait for answers.

Regarding myself,  I am by no means an expert on this issue,  I can just quote what everybody can read on p. 29 of Haag's textbook "local quantum physics" (2nd edition) regarding the representations different from the massive with spin  one $p^2=m^2>0$ with $p^0>0$ and the massless with helicity one $p^2=0$ and $p^0\geq 0$.  The metric has signature $-,+,+,+$ for Haag.

The operator $P^0$, considered in its role as an observable, is interpreted as the total energy of the system and, correspondingly, $P^i$ are the components of the linear momentum. One of the most important principles of  QFT, ensuring stability, demands that energy should have a lower bound, this is not the case int he last three classes ($p^2=-k^2$, $p^2=m^2$ and $p^0<0$, $p^2=0$ and $p^0\leq 0$). In class $p^\mu=0$ (and thus $W^\mu=0$) all states have zero energy-momentum, all states are translation invariant. If they are also Lorentz invariant we have the trivial representation appropriate to the vacuum state. Other possibilities [for this representation] which carry a non-trivial unitary representation of the Lorentz group have only marginal physical interest, though mathematically, the analysis of this class, carried out by Bargmann and by Naimark, is very interesting.

answered Nov 4, 2014 by Valter Moretti (2,085 points) [ revision history ]
edited Nov 4, 2014 by Valter Moretti

Actually Haag uses the mainly minus convention.

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