- #1

Phis

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Bear with me if you think this is extremely trivial. I have put this question on another forum before, but people seem to think that this is so trivial, that it doesn't deserve an answer. I haven't had QFT courses nor advanced particle physics, so a lot of this is new to me. Everybody seems to think this is trivial, but at the moment it confuses me a lot! I feel that a lot of other smaller things will fall into place after i understand this tiny bit.

To get a context or some form of reference, I am using the article https://arxiv.org/pdf/hep-ph/9806303v1.pdf and it's specifically the jump from the lagrangian in 4.18 to the one in 4.25 on page 40 and 41, that confuses me.

I have a couple of interpretations myself...

Has it something to do with Jacobi’s formula and the fact that ##det(e^{A})=e^{Tr(a)}##

Where we have ## U(\phi) = exp(i \frac{\Phi}{f}) ## where ## \Phi(x) \equiv \lambda \phi = \sqrt{2} \cdot [3x3 matrix] ## and the 3x3 matrix is the one containing the pions, kaons and eta's. Because U is a hermitian unitary matrix belonging to SU(3) we know that ## det(U) = 1## and therefore ## e^{Tr(U)} = 1 ## which is the same as saying that U is traceless and of course the Gell-mann matrices are traceless.

Or is it something much simpler with calculating/summing the eigenvalues of a linear algebra problem where we have a diagonalized matrix and seek our observables.

Or is it maybe that the order of matrices in our lagrangian is put in a way that we have ## [1x4] \cdot [4x4] \cdot \ldots \cdot [4x4] \cdot [4x1]## and we end up with a [1x1] matrix. Because the trace of a scalar is just the scalar it self, we can just as well write our lagrangian as a trace because we then may make use of the cyclic properties of the trace which helps a great deal when making transformations.

Or is it... ?

I can definitely see why we WANT to use it - mostly due to its cyclic properties, and the fact that we can pull all constants outside - but i can not see WHY we are allowed to write our lagrangian as a trace all of a sudden? I get that the lagrangian needs to be a scalar, since it is just ## L=T-V ## but why exactly make it a trace, what is the reason for the only interesting thing happening only in the diagonal, and why a sum, and not say a product or...?

I hope someone can give an in depth and hopefully simple explanation. Thank you very much!