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How to use the standard techniques of projection operators to obtain the equation (1) by the first formula? Thanks
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....
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".
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.... ?
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 isThank you for your recommendation. By the way , [itex]T^{I=2}(s,t)=<I=2,I_3=0|T^{abcd}|I=2,I_3=0> [/itex]? I and I3 are isospin and component of isospin for initial final state,respectively