Commutation relation between charges in current algebra Consider the vector current and axial vector current like
$$j^{a\mu} = \bar{\psi}\gamma^{\mu}\frac{\tau^{a}}{2}\psi,$$
$$j^{a\mu}_5 = \bar{\psi}\gamma^{\mu}\gamma_5\frac{\tau^a}{2}\psi,$$
where $\tau^a$ is pauli matrices.
Then the charges defined by
$$Q^a = \int{d^3x j^{a0}(x, t)},\space \space \space Q^a_5 = \int{d^3x j_5^{a0}}.$$
I am having trouble proving the following commutation relations:
$$[Q^a, Q^b] = i\epsilon^{abc}Q^c$$
$$[Q^a, Q_5^b] = i\epsilon^{abc}Q_5^c$$
$$[Q_5^a, Q_5^b] = i\epsilon^{abc}Q^c$$
I tried
\begin{eqnarray}i\epsilon^{abc}Q^c &=& i\epsilon^{abc}\int{d^3x\bar{\psi}\gamma^{0}\frac{\tau^{c}}{2}\psi} = \int{d^3x\bar{\psi}\gamma^0\biggr[\frac{\tau^a}{2}, \frac{\tau^b}{2}\biggl]\psi} = \int{d^3x\bar{\psi}\gamma^0\frac{\tau^a}{2}\frac{\tau^b}{2}\psi} - \int{d^3x\bar{\psi}\gamma^0\frac{\tau^b}{2}\frac{\tau^a}{2}\psi},
\end{eqnarray}
but I think this is not the correct way.
How can I prove these relations?
 A: Let us simplify the charges a little bit
\begin{equation}
Q^a = \int \overline{\psi} \gamma ^0 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x = \int \psi ^{\dagger} \gamma ^0 \gamma ^0 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x = \int \psi ^{\dagger} \frac{\tau ^a}{2} \psi \mathrm{d}^3 x ,
\end{equation}
and similarly $Q_5^a = \int \psi ^{\dagger} \gamma _5 \frac{\tau ^a}{2} \psi \mathrm{d}^3 x$.
In the derivation, we need to use the following trick
\begin{equation}
\left[ A B , C D \right] = A \left\{ B , C \right\} D - C \left\{ D , A \right\} B ,
\end{equation}
when $\left[ A , C \right] = \left[ B , D \right] = 0$ or $\left\{ A , C \right\} = \left\{ B , D \right\} = 0$.
Now knowing that $\left\{ \psi _i \left( x \right) , \psi _j \left( y \right) \right\} = \left\{ \psi _i ^{\dagger} \left( x \right) , \psi _j ^{\dagger} \left( y \right) \right\} = 0$, we have
\begin{equation}
\begin{split}
\left[ Q^a , Q^b \right] & = \iint \left[ \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) , \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \right] \mathrm{d}^3{x} \mathrm{d}^3{y} \\
& = \iint \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \left\{ \psi _j \left( x \right) , \psi _k^{\dagger} \left( y \right) \right\} \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \mathrm{d}^3{x} \mathrm{d}^3{y} - \iint \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \left\{ \psi _l \left( y \right) , \psi _i^{\dagger} \left( x \right) \right\} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \mathrm{d}^3{y} \\
& = \iint \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \delta _{j k} \delta ^{\left( 3 \right)} \left( \vec{x} - \vec{y} \right) \frac{\tau ^b_{k l}}{2} \psi _l \left( y \right) \mathrm{d}^3{x} \mathrm{d}^3{y} - \iint \psi _k^{\dagger} \left( y \right) \frac{\tau ^b_{k l}}{2} \delta _{l i} \delta ^{\left( 3 \right)} \left( \vec{x} - \vec{y} \right) \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \mathrm{d}^3{y} , \\
\end{split}
\end{equation}
apply the $\delta$ functions
\begin{equation}
\begin{split}
\left[ Q^a , Q^b \right] & = \int \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \frac{\tau ^b_{j l}}{2} \psi _l \left( x \right) \mathrm{d}^3{x} - \int \psi _k^{\dagger} \left( x \right) \frac{\tau ^b_{k i}}{2} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} \\
& = \int \left( \psi _i^{\dagger} \left( x \right) \frac{\tau ^a_{i j}}{2} \frac{\tau ^b_{j l}}{2} \psi _l \left( x \right) - \psi _k^{\dagger} \left( x \right) \frac{\tau ^b_{k i}}{2} \frac{\tau ^a_{i j}}{2} \psi _j \left( x \right) \right) \mathrm{d}^3{x} , \\
\end{split}
\end{equation}
rename the dummy indices
\begin{equation}
\begin{split}
\left[ Q^a , Q^b \right] & = \int \psi _i^{\dagger} \left( x \right) \left[ \frac{\tau ^a}{2} , \frac{\tau ^b}{2} \right]_{i j} \psi _j \left( x \right) \mathrm{d}^3{x} \\
& = \int \psi _i^{\dagger} \left( x \right) \imath \epsilon ^{a b c} \frac{\tau ^c_{i j}}{2} \psi _j \left( x \right) \mathrm{d}^3{x} = \imath \epsilon ^{a b c} Q^c . \\
\end{split}
\end{equation}
Same for $Q_5^a$.
