Explicit Symmetry Breaking in Chiral Lagrangian

Question

Currently I am going over explicit symmetry breaking in chiral Lagrangians. In particular, consider a term \begin{equation} \mathcal{L}_{\text{mass}} = \bar{Q}MQ \end{equation} where $Q = (u, d)^T$ and \begin{equation} M = \begin{pmatrix} m_u & 0\\ 0 & m_d \end{pmatrix} . \end{equation} Following some course notes, we write \begin{equation} Q = e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}}\tilde{Q}. \end{equation} Then, \begin{equation} \bar{Q}MQ = \bar{\tilde{Q}}e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} M e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} \tilde{Q} . \end{equation} We then replace the quark bilinear with its vacuum expectation value, \begin{equation} \langle 0| \bar{\tilde{Q}_i}\tilde{Q}_j |0\rangle = -v^3\delta_{ij}. \end{equation} which gives \begin{equation} \begin{split} \bar{Q}MQ &= -v^3\text{tr} \left[ e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} M e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} \right]\\[0.25cm] &= -v^3 \text{tr}\left[ e^{-\frac{i}{F}\gamma_5\vec{\pi}\cdot\vec{\sigma} }M\right]\\[0.25cm] &= -v^3\left[ \text{tr}(U^\dagger M) + \text{tr}(MU) \right] \end{split} \end{equation} where, $$ U = \exp\left(\frac{i}{F}\vec{\pi}\cdot{\sigma}\right). $$

I am unsure about the equivalence between the second to last line and the last line. Any help would be appreciated!

Answer 1

Well, there should be several corrections. First of all this symmetry acts on the flavor space of $2 \times 2 $ matrices. So the transformation doesn't work with the $\gamma_5$ matrix, but corresponds to transformation : $$ \chi \rightarrow L \chi \qquad \xi \rightarrow R^{*} \xi \qquad \Psi = \begin{pmatrix} \chi \\ \xi \end{pmatrix} $$ Where $\chi$ and $\xi$ are the components of Dirac spinor, and $L, R$ are independent so far unitary matrices. For the reference see chapter 83 of Srednicki. The axial symmetry corresponds to the choice $R = L^{\dagger}$.
$$ -v^3 \ \text{tr} \left[e^{-\frac{i}{2 F} \vec{\pi} \cdot \sigma} M e^{-\frac{i}{2 F} \vec{\pi} \cdot \sigma} \right] = -v^3 \ \text{tr} \left[e^{-\frac{i}{F} \vec{\pi} \cdot \sigma} M \right] $$ This identity follows from the trace cyclic property: $\text{tr} A B C = \text{tr} C A B$.

Next use $ \text{tr} A B = \text{tr} B^{\dagger} A^{\dagger}$, so one can replace $U^{\dagger} M^{\dagger}$ by $\frac{1}{2} (U^{\dagger} M^{\dagger} + M U)$ which, with the given choice of $M$ gives the resulting expression. However, there are some $1/2$ missing.

Answer 2

This is a storm in a teacup. It is fueled by the conflation of two traces, which the experienced consider as self-explanatory, but confuses the novices. The exponent on your axial rotation on fermions lives in $$ \gamma_5 \otimes \vec \sigma = \begin{pmatrix} 1& 0 \\ 0 &-1\end{pmatrix} \otimes \vec \sigma, $$ in the Weyl basis, so it rotates R and L fermions in opposite directions in SU(2).

By contrast, the pion chiral field is just an SU(2) group element, $$ U = \exp\left(\frac{i}{F}\vec{\pi}\cdot{\sigma}\right). $$

You start with a trace summing over your large chiral component space and SU(2), for which I use the term Tr, but after you do the R/L components sum, you are left with a plain SU(2) matrix sum, for which I use tr, \begin{equation} \begin{split} \bar{Q}MQ &= -v^3\text{Tr} \left[ e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} M e^{-\frac{i}{2F}\gamma_5\vec{\pi}\cdot\vec{\sigma}} \right]\\[0.25cm] &= -v^3 \text{Tr}\left[ e^{-\frac{i}{F}\gamma_5\vec{\pi}\cdot\vec{\sigma} }M\right]\\[0.25cm] &= -v^3\text{tr}\left [ (e^{-\frac{i}{F} \vec{\pi}\cdot\vec{\sigma}}+e^{\frac{i}{F} \vec{\pi}\cdot\vec{\sigma} } )M\right] =-v^3 \text{tr}\left [(U^\dagger +U)M \right ], \end{split} \end{equation} ready for your Gell-Mann—Oakes—Renner application.

Note that, even if M were m times the identity matrix, so, preserving vector isospin, the axial violation would still force your trace to involve a pion bilinear (mass term) proportional to it! So your pion mass squared will be proportional to m.