Explicit Symmetry Breaking in Chiral Lagrangian 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!
 A: 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.
A: 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.
