Chiral Perturbation Theory : Some quick questions

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SUMMARY

This discussion focuses on Chiral Perturbation Theory, specifically the expansion of the kinetic term in the Lagrangian involving pion fields and tau matrices. The kinetic term is represented as f2/4 Tr[Dμ Σ Dμ Σ], where Σ is defined as ei πa · τa/f. Participants clarify the correct interpretation of indices and the implications of hermitian conjugation on structure constants. The discussion emphasizes the importance of maintaining proper index notation and the distinction between color and SU(2) flavor indices in traces.

PREREQUISITES
  • Understanding of Chiral Perturbation Theory
  • Familiarity with SU(2) and SU(3) group theory
  • Knowledge of matrix exponentiation and hermitian conjugation
  • Proficiency in tensor notation and index manipulation
NEXT STEPS
  • Study the derivation of the kinetic term in Chiral Perturbation Theory
  • Learn about the properties of SU(2) and SU(3) matrices
  • Explore the implications of hermitian conjugation in quantum field theory
  • Research the role of trace operations in gauge theories
USEFUL FOR

Researchers and students in theoretical physics, particularly those focusing on quantum field theory, Chiral Perturbation Theory, and gauge symmetries. This discussion is beneficial for anyone working with Lagrangians involving pion fields and matrix representations.

Hepth
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I just want to make sure that I am doing some things correctly. I'll be using
http://www.physics.umd.edu/courses/Phys741/xji/chapter5.pdf
from about 5.64 on.

The kinetic term :
<br /> \frac{f^2}{4} Tr[D_{\mu} \Sigma D^{\mu} \Sigma^{\dagger}]<br />
Now if I want to expand this out, as \Sigma =e^{i \vec{\pi}^a \cdot \tau^{a}/f} (5.64)

Now, first, am I missing something. The pion fields have a vector sign, AND an index, AND are dotted into the generators. Now I assume "a" is the index of the color group, and the pion "vector" is really just a 1-D vector, and so the vector sign is redundant.
Then as the tau is a matrix, and the pion now a scalar, the cdot between them is wrong and it should be implied that it is a normal multiplication and not a dot product.

So it should either be \vec{\pi} \cdot \tau or \pi^a \tau^a with a sum implied. Right?

Now on to the next question:
Expanding it out:
Exp[x] = 1 + i x -1/2 x^2

So knowing that there are matrices and vectors involved, how does it expand:
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{2 f^2} (\pi^a \tau^a) (\pi^a \tau^a) <br />
or
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{2 f^2} (\pi^a \tau^a) (\pi^b \tau^b) <br />
or
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{4 f^2} [(\pi^a \tau^a) (\pi^b \tau^b) +(\pi^b \tau^b) (\pi^a \tau^a) ]<br />
etc. Do i need to add new indices each time I expand it, and how do I keep track of order.
The order will matter as when I take the conjugate transpose for the second term in the kinetic part itll reverse the order of the generators.

This is probably what I'm most stuck on. My lagrangian is more complicated than just a kinetic term and so I want to be really careful about the ordering of the generators.

Question 3:
The trace that is implied is a trace over COLOR (the generators)? Or a trace over dirac structure? Or both? The strong interaction term that appears between heavy mesons and the chiral fields includes a similar trace.

Question 4:
When taking the hermitian conjugate of the fine structure constants does (f^{abc})^{\dagger} = f^{cba} = -f^{abc}. Do I change the order of indices thusly, as if it were a matrix?
 
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Hepth said:
So it should either be \vec{\pi} \cdot \tau or \pi^a \tau^a with a sum implied. Right?

Right. The idea is that there are three scalar fields, ##\pi^1, \pi^2, \pi^3##, which if we like we can collect into a "3-vector" (not a spatial 3-vector) denoted ##\vec{\pi}##.

Hepth said:
So knowing that there are matrices and vectors involved, how does it expand:
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{2 f^2} (\pi^a \tau^a) (\pi^a \tau^a) <br />

In general if you have more than two appearances of an index in a factor you're doing Einstein notation wrong. Your last term here is at best ambiguous. You should rename the dummy index from a to something else in the second term in the product.

Hepth said:
or
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{2 f^2} (\pi^a \tau^a) (\pi^b \tau^b) <br />
or
<br /> e^{i \pi^a \tau^a/f} = 1 + \frac{i}{f} \pi^a \tau^a - \frac{1}{4 f^2} [(\pi^a \tau^a) (\pi^b \tau^b) +(\pi^b \tau^b) (\pi^a \tau^a) ]<br />

These two are equivalent and both correct. Keep in mind that any index you sum over is a dummy index which can be renamed at will. So ##\pi^a \tau^a## and ##\pi^b \tau^b## are the same matrix. One thing you might try is defining ##A = \pi^a \tau^a## and expanding out exp(iA/f), which should be clearer, and then reinserting the definition of A, keeping in mind that you should write e.g. ##A^3 = (\pi^a \tau^a)(\pi^b \tau^b)(\pi^c \tau^c)## so that its clear what pairs of indices are summed over.

Hepth said:
The order will matter as when I take the conjugate transpose for the second term in the kinetic part itll reverse the order of the generators.

This is pretty easy once you realize a few things. First, the tau matrices are hermitian, so ##A^\dagger = (\pi^a \tau^a)^\dagger = (\pi^a \tau^a) = A##. And in the higher order terms, say the quadratic one ##(AA)^\dagger = A^\dagger A^\dagger = AA##. So hermitian conjugation is easy.

Hepth said:
The trace that is implied is a trace over COLOR (the generators)? Or a trace over dirac structure? Or both? The strong interaction term that appears between heavy mesons and the chiral fields includes a similar trace.

The trace is over the indices of the tau matrices. This is an SU(2) flavor index--not a color index or a Dirac index (we don't have any color or Dirac indices!).

Hepth said:
When taking the hermitian conjugate of the fine structure constants does (f^{abc})^{\dagger} = f^{cba} = -f^{abc}. Do I change the order of indices thusly, as if it were a matrix?

The structure constants are just numbers, not matrices. And they are real. So hermitian conjugation does nothing to them. In general, if you write out something's indices explicitly, it becomes a number. Matrices are just quantities with suppressed indices. If you write out all the indices, everything becomes a number.
 
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Thank you. You answered all of my questions. I didn't mean color in the 2nd to last, don't know why I said that. I meant SU(3) flavor. uds
 

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