Lagrangian for quarks and pions

Question

I need to understand how starting from the free Lagrangian $$ \mathscr{L} = \bar{q}(i \not\partial - \hat{m})q $$

and based on the chiral angle associated with the pion field and the quark field rotated by the chiral transformation $$ q'= \exp\left[i\dfrac{\pi \cdot \tau\gamma^5}{2 f_\pi} \right]q $$

arrive at $$ \mathscr{L} = \bar{q}(i \not\partial - m)q - \dfrac{1}{2f_\pi}\partial_\mu \pi\cdot\bar{q}\gamma^\mu\gamma^5 \tau q - i \dfrac{ {m}}{f_\pi} \pi\cdot\bar{q}\gamma^5 \tau q + \dfrac{ {m}}{2 f_\pi^2}\pi^2\bar{q}q $$ where the quark field operators $q'$ are written as $q$.

Answer 1

You have neglected to write down the chiral angle, but that's OK, one may consider the dimensional constant $1/(2f_\pi)$ as the effective angle, since its dimensions are matched by the pions, and simply work to first order in it.

You then have $$ \mathscr{L} = \bar{q}'(i \not\partial - {m})q'= \bar{q}\exp\left[i\dfrac{\pi \cdot \tau\gamma^5}{2 f_\pi} \right](i \not\partial - {m})\left (\exp\left[i\dfrac{\pi \cdot \tau\gamma^5}{2 f_\pi} \right]q \right ) $$ $$ =\bar{q}(i \not\partial -m)q -\bar q \left (\not\partial {\pi\over 2f_\pi}\cdot \tau \gamma^5 \right )q -m\bar q \left ( {\pi\over f _\pi}\cdot \tau \gamma^5 \right )q +O(1/f_\pi^2). $$

Move on to the next order, now. The term proportional to m is self-evident, the second order expansion of the exponential that you have, in your target expression, $$ {m\over 2 f_\pi^2} \bar q \pi\cdot \pi q, $$ but your garbled target expression is incomplete!

You have omitted the term lacking m, involving the gradient of π. For that, you need this, $$ e^{-A}\partial_\mu e^{A}= \partial_\mu A -\tfrac{1} {2} [A,\partial_\mu A]+ O(A^3),\\ A= i\pi\cdot \tau \gamma^5/(2f_\pi). $$ Yielding the first order term, and the second order one, $$ {1\over 4 f_\pi^2} \bar q (\pi\times \not \partial \pi )\cdot \tau q + O(1/f_\pi^3). $$ This cross-product term is the one you missed!