# Peskin's treatment of Pions as Goldstone Bosons

After restoring the mass terms in the Lagrangian \begin{align} \mathcal{L}=\bar{u} i \not D u+\bar{d i} \not D d-m_{u} \bar{u} u-m_{d} \bar{d} d, \end{align} one obtains equations of motion for the quark field \begin{align} Q=\left(\begin{array}{l}{u} \\ {d}\end{array}\right) \end{align} given by \begin{align} i \not D Q=\mathbf{m} Q, \text{ and} \qquad -i D_{\mu} \overline{Q} \gamma^{\mu}=\overline{Q} \mathbf{m} \end{align} where \begin{align} \mathbf{m}:=\left(\begin{array}{cc}{m_{u}} & {0} \\ {0} & {m_{d}}\end{array}\right). \end{align} This leads to \begin{align} \partial_{\mu} j^{\mu 5 a}=i \overline{Q}\left\{\mathbf{m}, \tau^{a}\right\} Q, \end{align} where $$\tau^{a}=\sigma^{a} / 2$$ are the generators of SU(2). Using this equation to rewrite the matrix elements $$\left\langle 0\left|j^{\mu 5 a}(x)\right| \pi^{b}(p)\right\rangle=- i p^{\mu} f_{\pi} \delta^{a b} e^{-i p \cdot x},$$ where $$f_\pi$$ is the pion decay constant, \begin{align} \left\langle 0\left|\partial_{\mu} j^{\mu 5 a}(0)\right| \pi^{b}(p)\right\rangle&=- p^{2} f_{\pi} \delta^{a b}\\ %% %% &= \left\langle 0\left|i \overline{Q}\left\{\mathbf{m}, \tau^{a}\right\} \gamma^{5} Q\right| \pi^{b}(p)\right\rangle. \end{align} Peskin claims that the final line is an invariant quantity multiplied by \begin{align} \operatorname{tr}\left[\left\{\mathbf{m}, \tau^{a}\right\} \tau^{b}\right]=\frac{1}{2} \delta^{a b}\left(m_{u}+m_{d}\right). \end{align} This claim is quite non-transparent to me. I can't seem to identify where the trace might arise from, and what the invariant quantity might be.

He concludes that the quark mass terms which break the axial vector symmetry give rise to pions with masses of the form \begin{align} m_{\pi}^{2}=\left(m_{u}+m_{d}\right) \frac{M^{2}}{f_{\pi}}, \end{align} where we group the remaining invariant quantity into a new parameter $$M$$... This is shortly after a paragraph in which he argued that an on shell-pion must be massless ($$p^2=0$$), as expected from Goldstone's theorem. How is this final equation consistent with this claim?

• page and Eqs in P&S? – Oбжорoв Sep 21 at 8:57
• All appears on page 670. – user4580791 Sep 21 at 9:01

The generator $$\tau^a$$ comes from the creation of the pion state. The calculation then decouples into the standard trace over all the color factors, that can be taken outside the expectation value, and an expectation value. The expectation value is an invariant for which they introduce the symbol $$M$$ ($$M^2/2$$ to be specific). They don't calculate it, just mention it can be measured experimentally.