# Trace in QCD lagrangian

• A
I have a question about the use of trace in QFT in general - more specifically the use of trace in the lagrangian in the effective theory concerning chiral symmetry in QCD. I am slowly trying to get a hang of everything, and most things i am able to calculate, but i still have som very specific areas where i lack a deeper physical intuition/understanding. The biggest hole in my learning is the use of trace.

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!

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

nrqed
Homework Helper
Gold Member
I have a question about the use of trace in QFT in general - more specifically the use of trace in the lagrangian in the effective theory concerning chiral symmetry in QCD. I am slowly trying to get a hang of everything, and most things i am able to calculate, but i still have som very specific areas where i lack a deeper physical intuition/understanding. The biggest hole in my learning is the use of trace.

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!

The key point about using traces is that it gives quantities that are invariant under the chiral transformation. If you have a quantity that transforms as
$U \rightarrow A U B$ (with $A A^\dagger = B B^\dagger =1$ ) then the quantity $D_\mu U D^\mu U^\dagger$ for example transforms as
$$D_\mu U D^\mu U^\dagger \Rightarrow A D_\mu U D^\mu U^\dagger A^\dagger$$
so it is not invariant but the trace of that is invariant.