Decomposed isospin

Isospin is (approximately) conserved in the strong interaction. Thus you can decompose the total isospin of the two-pion system in irreducible parts. Pions are pseudoscalar mesons with isospin 1 (forming a triplet, ##\pi^+##, ##\pi^-##, and ##\pi^0##). From the theory of angular momentum, i.e., the representation theory of the Lie algebra of su(2), i.e., the commutation rules for angular momenta you know that the direct product of two spin-1 representations splits into the direct sum of the trivial, the ##s=1##, and ##s=2## representation: ##1 \otimes 1=0 \oplus 1 \oplus 2##.

In this case it's simpler to work with a real representation of the pion fields ##\vec{\pi} \in \mathbb{R}^3##. Then the isospin rotations are simply described by real ##\mathrm{SO}(3)## matrices. The split into irreducible parts of the tensor product is then straight forward.

The ##s=0## part of the product of two pion fields is found by looking for a scalar built as a bilinear form of ##\vec{\pi}##. That's uniquely given by the scalar product ##\vec{\pi} \cdot \vec{\pi}=\delta_{ab} \pi^a \pi^b##. To find the corresponding amplitude for this channel thus you simply trace over the index pair ##(a,b)##
$$\delta_{ab} T^{abcd}=(3A+B+C) \delta^{cd}.$$
This gives you the first equation in (1).

The ##s=1## piece is given by the vector product ##\epsilon_{abe} \pi^a \pi^b##. Thus you have to contract the amplitude with ##\epsilon_{abe}## leading to
$$\epsilon_{abd} T^{abcd}=B \epsilon_{cde}+C \epsilon_{dce}=(B-C) \epsilon_{cde},$$
giving you the 2nd line of Eq. (1).

Finally the ##s=2## part is represented by the traceless symmetric part of the tensor ##\vec{\pi} \otimes \vec{\pi}##, i.e., ##3 \pi^a \pi^b -\vec{\pi}^2 \delta^{ab}##. So you just have to symmetrize your amplitude wrt. the index pair ##(a,b)## and subtract the trace, leading to
$$T^{(ab)cd}=2A \delta^{ab} \delta^{cd}+ (C+D) \delta^{ac} \delta^{bd}$$
The trace is
$$\delta_{ab} T^{(ab)cd}=(6A+C+D)\delta_{cd}$$
and thus you get
$$T^{(ab)cd}-\frac{1}{3} (6A+C+D) \delta^{ab} \delta^{cd} = \frac{2}{3} (C+D) \delta^{ac} \delta^{bd},$$
which gives you the corresponding amplitude in the 3rd line of Eq. (1).

 

How to get that [itex] 3 \pi^a \pi^b - \delta^{ab} \pi^2 [/itex] is seen by the fact that this combination is the only traceless : (a=b and sum) [itex] 3 \pi^a \pi^a - \delta^{aa} \pi^2 = 3 \pi^2 - 3 \pi^2 =0 [/itex] and symmetric: (interchanging a,b you get the same result)...
In general these type of combinations can be easier obtained (I think) from Young Tableaux...

For the other question, you just have:
[itex] (C+D) \delta^{ac} \delta^{bd} - \frac{1}{3} (C+D) \delta^{ab} \delta^{cd} [/itex]
[itex] \frac{1}{3}(C+D) [3\delta^{ac} \delta^{bd} - \delta^{ab} \delta^{cd} ] =\frac{1}{3}(C+D) [2\delta^{ac} \delta^{bd} + \delta^{ac} \delta^{bd} - \delta^{ab} \delta^{cd} ] [/itex]

The last two terms in the brackets cancel out....

 

I didn't understand the first questions.....
the cancelation of deltas you give are a general feature and doesn't apply for a given representation...
A fast way is to use mathematica to show you that the combination is zero for all a,b,c,d which was the way I used to give a fast answer... As for how to see that, well you could play with symmetries? or try to put some given values on a,b,c,d.... ?

 

How to derive these "projections" is not such a simple thing, because you need representation theory of the rotation group. A great book about this is Lipkin's "Lie groups for pedestrians".

 

Sure, you can as well treat the entire business also with the usual isospin states. Then you represent your pions with ##\pi^{\pm}## and ##\pi^0## fields rather than in the real SO(3) representation of isospin. Of course, this is entirely equivalent. The relation between the fields is
##\pi^{\pm}=\frac{1}{\sqrt{2}} (\pi^1 +\pm \mathrm{i} \pi^2), \quad \pi^{0}=\pi^3.##