Mass term in chiral lagrangian and chiral lagrangian

Question

Mesons in chiral lagrangian are described by the chiral action $U = e^{2i\pi^a T^a/f_\pi}$, \begin{equation} \mathcal{L}_{chiral} = \frac{f_\pi^2}{4} \text{Tr} \left( \partial^\mu U^\dagger \partial_\mu U \right) + \frac{\sigma}{2} \text{Tr} \left(MU + U^\dagger M^\dagger \right) \end{equation}

The mass term linear in quark masses, $M = \operatorname{diag} (m_u, m_d, m_s)$ and proportional to the chiral condensate: \begin{equation} \langle \bar{\psi}_{-i} \psi_{+j} \rangle \approx - \sigma U_{ij}. \end{equation}

This leads to the dispersion laws, \begin{equation} m_\pi^2 = \frac{2\sigma}{f_\pi^2} (m_u + m_d), \;\;\;\;\; m_\eta^2 = \frac{2\sigma}{f_\pi^2} (m_u + m_d + 4m_s)/3 \end{equation}

How to explain that the coefficient of the mass term is $\sigma$??

Answer 1

Even simpler: The QCD partition function satisfies $$ \frac{\partial\log Z}{\partial m_q} = \langle \bar{q}q\rangle $$ which follows directly from the form of the mass term in the QCD Lagragian. Now compute the same object in chiral perturbation theory. The ground state is $U=1$, and $$ \frac{\partial\log Z}{\partial m_q} = -\sigma $$

Answer 2

I strongly suspect you could write the perfect answer to your question after looking at Georgi's Weak interaction book, or Sec 28.2.2 of M Schwartz's book, or 5.5 of TP Cheng & LF Li, or 4.1.2 of this classic review of chiral perturbation theory. I'll give you a trail map, which means all normalizations below will be suspect...

All of these references, obsessed with "telling the truth" do not emphasize what all pros have in the back of their mind: the basic correspondence of the SSBroken axial currents in their fundamental QCD quark reincarnation versus their pseudoscalar meson chiral lagrangian reincarnation, $$ J_5^{a~~\mu}= \bar q \gamma^\mu \gamma_5 T^a q ~~~~\leftrightarrow ~~~~~f_\pi \partial^\mu \pi^a, $$ normally connected through the less transparent PCAC bridge, $$ \langle 0| J_5^{a~\mu}(x)|\pi^b(p)\rangle=ip^\mu f_\pi e^{-ipx}\delta_{ab}. $$ The divergence of this SSB current fails to vanish only because of the explicit breaking due to the quark masses, $\partial_\mu J_5^{a~\mu}= f_\pi m_\pi^2 \pi^a$, (~$m_q \bar q \gamma_5 \lambda^aq$). The crucial link is that both reincarnations transform identically under the L and R transformations of the chiral SU(3)s.

Dashen's theorem, or, more conventionally, the Gell-Mann—Oakes—Renner relations connect the pseudoscalar masses of the above to the chiral condensation parameter of QCD, $\sigma=-\langle \bar q q\rangle$ for the three light quarks, the cube of a quarter of a GeV, $$ f_\pi^2 m_\pi^2= \sigma (m_u+m_d),\\ f_\pi^2 m_\eta^2= \sigma (m_u+m_d+4m_s)/3. $$ (Your funny $U_{ij}=\delta_{ij}$ in chiral condensation in QCD.$^\sharp$)

When you introduce the explicit breaking term in the chiral lagrangian, The P of PCAC, $$ \frac{\sigma'}{2} \text{Tr} \left(MU + U^\dagger M^\dagger \right), $$ you check the breaking term transforms like the corresponding quark mass terms in the fundamental QCD lagrangian, but, as yet, you don't know what the mystery arbitrary parameter $\sigma '$ might be. Compute the quadratic terms of the pseudoscalars in the above (the linear ones vanish from the tracelessness of the Gell-Mann matrices, $T^a=\lambda^a/2$).

The coefficient of the $\pi_3^2/f_\pi^2$, for example, is $-\sigma' (\operatorname{Tr}M\lambda_3^2)/2= -\sigma'(m_u+m_d)$, and of the $\eta^2/f_\pi^2$, is $-\sigma' (\operatorname{Tr}M\lambda_8^2)/2= -\sigma'(m_u+m_d+4m_s)/3$.

Comparing with the GOR masses, up to the flakey sign (hiding in M?), you may identify $\sigma= \sigma'$.

Having done that, one confirms that the vacuum energies of the two reincarnations, QCD and chiral lagrangian, also match, $$m_u\langle \bar u u\rangle + m_d\langle \bar d d\rangle + m_s\langle \bar s s\rangle= \sigma (m_u+m_d+ m_s)= \sigma \operatorname{Tr}M,$$ but this is not quite compelled by their symmetry structure.

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$^\sharp$ _{Veery loosely, just a trailmap, eqn. (5.223) of Cheng & Li, $$ m_\pi^2 f_\pi^2 \propto \langle \pi^a| [Q_5^a,[Q^b_5,H(0)]|\pi^b\rangle \propto m_q \langle \bar q q\rangle, $$ no summation over flavor indices.}