Chiral Symmetry and Charge Algebra I'm following Section 5.1 in Cheng and Li's Particle Physics book and I am having trouble reproducing some of the commutation relations.
The axial current is given by
$$
(J_A^a)^\mu = \bar{\psi}_\alpha\gamma^\mu\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\gamma_5\psi_\beta
$$
where $\lambda^a$ are the Gell-Mann matrices. The associated charge is then
$$
Q^{5a} = \int d^3 x\ \bar{\psi}_{\alpha}(x)\gamma^0\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\gamma_5\psi_\beta(x)
$$
The 'normal' current is given by
$$
(J^a)^\mu = \bar{\psi}_\alpha\gamma^\mu\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\psi_\beta
$$
and has associated charge
$$
Q^{a} = \int d^3 x\ \bar{\psi}_{\alpha}(x)\gamma^0\left(\frac{\lambda^a}{2}\right)_{\alpha\beta}\psi_\beta(x)
$$
I am trying to show that
$$
[Q^a,Q^{5b}] = if^{abc}Q^{5c}
$$
where $f^{abc}$ are the structure constants of $SU(3)$. My work so far is,
\begin{equation}
\begin{split}
[Q^a,Q^{5b}] &= \int d^3x\ d^3 y\ \left[\bar{\psi}_\alpha(x)\gamma^0 (T^a)_{\alpha\beta}\psi_\beta(x),\bar{\psi}_{\gamma}(y)\gamma^0(T^b)_{\gamma\delta}\gamma_5\psi_\delta(y)\right]\\[0.25cm]
&=\int d^3x\ d^3 y\ (T^a)_{\alpha\beta}(T^b)_{\gamma\delta} \left[\psi_\alpha^\dagger(x) \psi_\beta(x),\psi_{\gamma}^\dagger(y)\gamma_5\psi_\delta(y)\right]
\end{split}
\end{equation}
Using,
$$
[A B, C D]=A\left\{B, C\right\}  D - AC\left\{B,D\right\} + \left\{A,C\right\}DB - C\left\{A,D\right\}B
$$
and the normal anti-commutation relations for fermions,
\begin{multline}
\left[\psi_\alpha^\dagger(x) \psi_\beta(x),\psi_{\gamma}^\dagger(y)\gamma_5\psi_\delta(y)\right]
=\\
\psi_\alpha^\dagger(x)\left\{\psi_\beta(x),\psi_\gamma^\dagger(y)\right\}\gamma_5\psi_\delta(y) -\psi_\alpha^\dagger(x)\psi_\gamma^\dagger(y)\left\{\psi_\beta(x),\gamma_5\psi_\delta(y)\right\}
-\psi_\gamma^\dagger(y)\left\{\psi_\alpha^\dagger(x),\gamma_5\psi_\delta(y) \right\}\psi_\beta(x)
\end{multline}
The first term is consistent what with I expect. However, I don't know how to deal with the anti-commutators that contain $\gamma_5$.
Any help would be appreciated!
 A: This relation will be useful for you: 

$$ [\psi^\dagger(t,\textbf x) \Gamma^A \lambda^a \psi(t,\textbf x), \psi^\dagger(t,\textbf y) \Gamma^B \lambda^b \psi(t,\textbf y)] = \delta^{(3)}(\textbf x-\textbf y) \psi^\dagger(t,\textbf x)[\Gamma^A \lambda^a, \Gamma^B \lambda^b]\psi(t,\textbf y) \tag{1} $$

In your case, 
$$ \Gamma^A \to 1,\quad \Gamma^B\to^5,\quad \lambda^a \to T^a,\quad \lambda^b\to T^b \tag{2} $$
Now instead of using the commutator relation you wrote, I'd use  
$$ [A B, C D] = A[B,C]D + [A,C]BD + CA[B,D] + C[A,D]B \tag{3}$$
and we see that $A$ with $D$, as well aas $B$ with $C$ won't contribute, since the commutator of a Dirac matrix with a Gell-Mann matrix is zero. Therefore, 
$$ [\Gamma^A \lambda^a, \Gamma^B \lambda^b] = [\Gamma^A, \Gamma^B]\lambda^a\lambda^b + \Gamma^B\Gamma^A [\lambda^a,\lambda^b] \tag{4}$$
Finally, $[1,\gamma^5]=0$ and $[T^a, T^b]=\text{i} f^{abc}T^c$. 

As mentioned in a comment, this can be proven by writing the product of spinors and Dirac/Gell-Mann matrices in index notation: 
$$ \psi^\dagger \Gamma \lambda \psi = \psi^\dagger_{\alpha, i} \Gamma_{\alpha\beta} \lambda_{ij}\psi_{\beta, j} \tag{5} $$
By doing this, you can pull the $\Gamma$'s and $\lambda$'s out of the anti-commutator, as if they were just scalar numbers (which, in index notation, they are!). 
