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Chiral Perturbation Theory : Some quick questions

  1. Oct 16, 2012 #1


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    I just want to make sure that I am doing some things correctly. I'll be using
    from about 5.64 on.

    The kinetic term :
    \frac{f^2}{4} Tr[D_{\mu} \Sigma D^{\mu} \Sigma^{\dagger}]
    Now if I want to expand this out, as [itex] \Sigma =e^{i \vec{\pi}^a \cdot \tau^{a}/f}[/itex] (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 [itex]\vec{\pi} \cdot \tau[/itex] or [itex]\pi^a \tau^a[/itex] 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:
    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)
    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)
    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) ]
    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 [itex](f^{abc})^{\dagger} = f^{cba} = -f^{abc}[/itex]. Do I change the order of indices thusly, as if it were a matrix?
    Last edited: Oct 16, 2012
  2. jcsd
  3. Oct 16, 2012 #2
    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}##.

    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.

    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.

    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.

    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!).

    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.
    Last edited: Oct 16, 2012
  4. Oct 16, 2012 #3


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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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