Derivation related to gauge theory of quantum spin antiferromagnets I was reading Chapter 5 of the book by N. Nagaosa on Quantum Field Theory in Strongly Correlated Electronic Systems. In the first section, he introduced the complex-field ($z_{\alpha}$) representation of "classical" spin fields:
$$\boldsymbol{\Omega}(x)=\sum_{\alpha, \beta}z^*_{\alpha}(x) \boldsymbol{\sigma}_{\alpha,\beta}z_{\beta}(x)$$
where $\boldsymbol{\sigma}$ are Pauli matrices and $\alpha = \uparrow, \downarrow$, and the constrain on spin field leads to :
$$|z_{\uparrow}(x)|^2+|z_{\downarrow}(x)|^2=1$$
and when he try to use $z_{\alpha}(x)$ fields to represent the non-linear sigma model (which originates from the 1d quantum antiferromagnetic Heisenberg model), he got in equation (5.1.6):
$$\frac{1}{g}\int d^2x \left( \partial_{\mu} \boldsymbol{\Omega}(x)\right)^2=\frac{1}{g}\int d^2x \left( \partial_{\mu}z^*_{\alpha}\partial_{\mu}z_{\alpha}+\left(z^*_{\alpha} \partial_{\mu} z_{\alpha}\right) \left(z^*_{\beta} \partial_{\mu} z_{\beta}\right) \right)$$
also the Berry phase term in equation (5.1.11):
$$\frac{1}{2}\boldsymbol{\Omega}\cdot\left( \partial_{\tau}\boldsymbol{\Omega} \times \partial_{x}\boldsymbol{\Omega} \right)=\partial_{\tau}\left( z^*_{\alpha}\partial_xz_{\alpha} \right)-\partial_x\left( z^*_{\alpha}\partial_{\tau}z_{\alpha} \right)$$
my question is how to derive these two equations? I have tried to do a brute force calculation but finally only got very messy results, I don't know if there is any tricks during the derivation or if there is easier way (instead of expanding all the derivatives) to do the calculation?
 A: After discussing with some friends, I realised that in order to easily prove the two equations above, we need to know the following identities for Pauli matrices:
$$\boldsymbol{\sigma}_{\alpha,\beta}\cdot \boldsymbol{\sigma}_{\gamma,\delta}=2\delta_{\alpha,\delta} \cdot \delta_{\beta,\gamma}-\delta_{\alpha,\beta} \cdot \delta_{\gamma,\delta}$$
for the poof of the first equation and
$$\epsilon_{i,j,k}\sigma^i_{\alpha,\beta} \sigma^j_{\gamma,\delta} \sigma^k_{\mu,\nu}=2i \left( \delta_{\alpha,\nu} \delta_{\beta,\gamma} \delta_{\delta,\mu}-\delta_{\alpha,\delta}\delta_{\beta,\mu}\delta_{\gamma,\nu} \right)$$
for the proof of second equation, while for the second equality above can be checked numerically (for example, in Mathematica).
Bear these in mind, come back to the first equation (non-linear sigma part):
$$\begin{align}
\left( \partial_{\mu}\boldsymbol{\Omega}(x) \right)^2 &=\sum_{\mu}\sum_{i}\sum_{\alpha,\beta,\gamma,\delta}\partial_{\mu}\left( z^*_{\alpha}(x) z_{\beta}(x) \right) \partial_{\mu}\left( z^*_{\gamma}(x) z_{\delta}(x) \right) \sigma^i_{\alpha,\beta} \sigma^i_{\gamma,\delta} \\
&=\sum_{\mu}\sum_{\alpha,\beta,\gamma,\delta} \left( 2\delta_{\alpha,\delta} \cdot \delta_{\beta,\gamma}-\delta_{\alpha,\beta} \cdot \delta_{\gamma,\delta} \right) \partial_{\mu}\left( z^*_{\alpha}(x) z_{\beta}(x) \right) \partial_{\mu}\left( z^*_{\gamma}(x) z_{\delta}(x) \right) \\
&=\sum_{\mu}\sum_{\alpha,\beta}2\partial_{\mu}\left( z^*_{\alpha}z_{\beta}\right) \partial_{\mu}\left( z^*_{\beta}z_{\alpha} \right)-\partial_{\mu}\left( z^*_{\alpha}z_{\alpha} \right)\partial_{\mu}\left( z^*_{\beta}z_{\beta} \right)
\end{align}$$
remember that
$$
\begin{align}\sum_{\alpha}\partial_{\mu}\left(z^*_{\alpha}a_{\alpha}\right)&=0 \\
\rightarrow \sum_{\alpha}\left( \partial_{\mu}z^*_{\alpha} \right) z_{\alpha}&=-\sum_{\alpha}z^*_{\alpha}\partial_{\mu}z_{\alpha}
\end{align}$$
one can finally arrive at
$$\left( \partial_{\mu}\boldsymbol{\Omega}(x) \right)^2 =
4\times \big\{ \partial_{\mu}z^*_{\alpha}\partial_{\mu}z_{\alpha}+
\left(z^*_{\alpha}\partial_{\mu}z_{\alpha}\right) \left(z^*_{\beta}\partial_{\mu}z_{\beta} \right) \big\}$$
and the factor $4$ can be absorbed into the coupling constant $1/g$.
As for the second equation, after using the second identity above, one would get:
$$\begin{align}
\frac{1}{2}\boldsymbol{\Omega}\cdot\left(\partial_{\tau}\boldsymbol{\Omega}\times \partial_{x}\boldsymbol{\Omega}\right)&=i\sum_{\alpha,\beta,\gamma,\delta,\mu,nu} \left( \delta_{\alpha,\nu} \delta_{\beta,\gamma} \delta_{\delta,\mu}-\delta_{\alpha,\delta}\delta_{\beta,\mu}\delta_{\gamma,\nu} \right) z^*_{\alpha}z_{\beta} \partial_{\tau}\left( z^*_{\gamma}z_{\delta} \right) \partial_x\left( z^*_{\mu}z_{\nu}\right) \\
&=i \sum_{\alpha,\beta,\delta}z^*_{\alpha}z_{\beta} \partial_{\tau}\left( z^*_{\beta}z_{\delta} \right) \partial_x\left( z^*_{\delta}z_{\alpha}\right)-i\sum_{\alpha,\beta,\gamma}z^*_{\alpha}z_{\beta} \partial_{\tau}\left( z^*_{\gamma}z_{\alpha} \right) \partial_x\left( z^*_{\beta}z_{\gamma}\right) \\
&=i\sum_{\alpha,\beta,\delta}z^*_{\alpha}z_{\beta} \partial_{\tau}\left( z^*_{\beta}z_{\delta} \right) \partial_x\left( z^*_{\delta}z_{\alpha}\right)-z^*_{\alpha}z_{\beta} \partial_{\tau}\left( z^*_{\delta}z_{\alpha} \right) \partial_x\left( z^*_{\beta}z_{\delta}\right) 
\end{align}$$
from now on, one just need to expand all the derivatives and will find that most term cancels with each other, finally we will get:
$$\begin{align}
\frac{1}{2}\boldsymbol{\Omega}\cdot\left(\partial_{\tau}\boldsymbol{\Omega}\times \partial_{x}\boldsymbol{\Omega}\right)&=-i\sum_{\alpha}\partial_{\tau} z^*_{\alpha}\partial_x z_{\alpha}-\partial_{x} z^*_{\alpha}\partial_{\tau} z_{\alpha} \\
&=-i \sum_{\alpha} \partial_{\tau}\left( z^*_{\alpha} \partial_x z_{\alpha}\right)-\partial_x \left( z^*_{\alpha} \partial_{\tau}z_{\alpha} \right)
\end{align}$$
there is a factor $(-i)$ difference and I think it should be a typo in the textbook.
