Conserved charge commutation relation under $SU(2)$ symmetry in two complex Klein-Gordon fields I'm trying to show that conserved charges of two complex equal-mass Klein-Gordon fields under $SU(2)$ transformation fulfill the following commutation relation:
$$ [Q^j, Q^k]=i\epsilon^{jkl}Q^l .$$
I have the expression for $Q^i$ in terms of fields, their conjugate momenta and Pauli matrices. It looks that I just need to use $[\pi_a(x), \phi_b(y)]$ but I still fail to get the expected result.
The Lagrangian is $$\mathscr L=\partial_\mu\Phi^\dagger\cdot\partial^\mu\Phi-m\Phi^\dagger\cdot\Phi ,$$  where $$\Phi=\begin{pmatrix} \phi_1 \\ \phi_2\end{pmatrix}
$$ 
And the conserved charge is $$Q^j=i\int d^3x(\phi_a^*(\tau^j)_{ab}\pi_b^*-\pi_a(\tau^j)_{ab}\phi_b),$$ where $\tau^j$ is Pauli matrices multiplied by 1/2 factor. 
 A: I am posting this answer to help anyone go through the charge algebra.
Lets consider a general Lagrangian composed by complex scalar fields $\phi_i(x)$. If $\Phi$ is a multiplet containing these fields, using an implicit sum over repeated indexes we can write:
\begin{equation}
\mathscr{L} = \partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - V(\Phi)=\partial_{\mu}\phi^{\dagger}_i(x)\partial^{\mu}\phi_i(x) - V(\Phi)
\end{equation}
The above expression for the Lagrangian is not important for what is to come but, I have written it to make one thing clear: we must keep using the hermitian conjugate $^{\dagger}$ since the fields $\phi_i(x)$ are quantized. We write them as,
\begin{equation}
\phi_i(x) = \int_{}^{} \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{p}}}}(a_{\vec{p}}e^{ip \cdot x} + b^{\dagger}_{\vec{p}}e^{-ip \cdot x}), \quad\quad\quad \phi^{\dagger}_i(x) = \int_{}^{} \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{p}}}}(a^{\dagger}_{\vec{p}}e^{-ip \cdot x} + b_{\vec{p}}e^{ip \cdot x}) \end{equation}
where $b^{\dagger}$ and $a$ are two distintic creation/anihilation operators, each acting on its own space and $p^0=E_{\vec{p}}$. Considering the usual commutation relations for the creation/anihilation operators,
\begin{equation}
[a_{\vec{p}},a_{\vec{y}}]=[a^{\dagger}_{\vec{p}},a^{\dagger}_{\vec{y}}]=0, \quad\quad [a_{\vec{p}},a^{\dagger}_{\vec{y}}] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{y})
\end{equation}
the expression for $\phi_i(x)$ implies relevant results for what is about to come regarding equal time commutation relations:
\begin{equation}[\phi_i(t,\vec{x}),\phi_j(t,\vec{y})] =  [\phi_i(t,\vec{x}),\phi^{\dagger}_j(t,\vec{y})] = [\phi^{\dagger}_i(t,\vec{x}),\phi^{\dagger}_j(t,\vec{y})] = 0 
\end{equation}
(different field indexes, in this case $i$ and $j$, mean a distinct pair of creation/anihilation operators. Since each anihilation/creation operators act in its own space they commute implying that $\phi_i$ and $\phi_j$ commute as well.)
Lets consider a group G with hermitian generators $T_a$ and structure constants $f_{abc}$ such that $[T_a,T_b]=-if_{abc}T_c$. Lets check the non-trivial transformation of $\phi_i$:
\begin{equation}
\phi_i \rightarrow \phi_i´= e^{-i(T_a)_{ij}\theta_a}\phi_j = \phi_i -i(T_a)_{ij}\theta_a\phi_j = \phi_i + \theta_a\delta_a\phi_i, \quad \delta_a\phi_i=-i(T_a)_{ij}\phi_j
\end{equation}
\begin{equation}
\phi_i^{\dagger} \rightarrow \phi^{\dagger}_i´= (e^{-i(T_a)_{ij}\theta_a}\phi_j)^{\dagger} = \phi^{\dagger}_i +i\phi^{\dagger}_j(T_a)_{ji}\theta_a = \phi^{\dagger}_i + \theta_a(\delta_a\phi_i)^{\dagger}, \quad (\delta_a\phi_i)^{\dagger}=+i(T_a)_{ji}\phi^{\dagger}_j
\end{equation}
If G leaves the Lagrangian invariant we can write the Noether currents as:
\begin{equation}
j^{\mu}_a(x) = \frac{\partial\mathscr{L}}{\partial(\partial_{\mu}\phi_i(x))}\delta_a\phi_i(x) 
+ (\delta_a\phi_i(x))^{\dagger}\frac{\partial \mathscr{L}}{\partial(\partial_{\mu}\phi^{\dagger}_i(x))}
\end{equation}
The conserved Noether charges, $Q_a(t)$, are the spatial integral of the $\mu=0$ component. The conjugate momenta, $\pi_i(x)$, to $\phi_i(x)$ is :
\begin{align*}
\pi_i(x) &= \frac{\partial\mathscr{L}}{\partial(\partial_{0}\phi_i(x))} = \partial_{0}\phi^{\dagger}_i(x) = \int_{}^{} \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{E_{\vec{p}}}{2}}(a^{\dagger}_{\vec{p}}e^{-ip \cdot x} - b_{\vec{p}}e^{ip \cdot x})d^3x
\end{align*}
This expression implies another set of relevant equal time commutation relations for what is about to come:
\begin{equation}
[\pi_i(t,\vec{x}),\pi_j(t,\vec{y})]= [\pi_i(t,\vec{x}),\pi^{\dagger}_j(t,\vec{y})]=[\pi^{\dagger}_i(t,\vec{x}),\pi^{\dagger}_j(t,\vec{y})] = 0
\end{equation}
\begin{equation}
[\pi_i(t,\vec{x}),\phi^{\dagger}_j(t,\vec{y})]= [\pi^{\dagger}_i(t,\vec{x}),\phi_j(t,\vec{y})] = 0
\end{equation}
\begin{equation}
[\pi_i(t,\vec{x}),\phi_j(t,\vec{y})]= -i\delta_{ij}\delta^{(3)}(\vec{x}-\vec{y}), \quad [\pi_i(t,\vec{x}),\phi_j(t,\vec{y})]^{\dagger}=[\phi^{\dagger}_j(t,\vec{y}),\pi^{\dagger}_i(t,\vec{x})] = +i\delta_{ij}\delta^{(3)}(\vec{y}-\vec{x}) 
\end{equation}
Writing the charge $Q_a(t)$ as:
\begin{align*} \label{eq:chargea}
Q_a(t) &=\int j^0_a(x)d^3x = \int \frac{\partial\mathscr{L}}{\partial(\partial_{0}\phi_i(x))}\delta_a\phi_i(x)
+ (\delta_a\phi_i(x))^{\dagger}\frac{\partial \mathscr{L}}{\partial(\partial_{0}\phi^{\dagger}_i(x))} d^3x= \\
&\quad  = \int \pi_i(x)\delta_a\phi_i(x) + (\delta_a\phi_i(x))^{\dagger} \pi_i^{\dagger}(x)d^3x= \\
&\quad = \int \pi_i(x)(-i(T_a)_{ij}\phi_j(x))+(i(T_a)_{ji}\phi^{\dagger}_j(x))\pi^{\dagger}_i(x)d^3x = \\
&\quad = \int -i\pi_i(x)(T_a)_{ij}\phi_j(x) + i\phi^{\dagger}_j(x)(T_a)_{ji}\pi^{\dagger}_i(x)d^3x = \\
&\quad = \int -i\pi_i(x)(T_a)_{ij}\phi_j(x) + i\phi^{\dagger}_i(x)(T_a)_{ij}\pi^{\dagger}_j(x) d^3x
\end{align*}
(the last equality is just a re-labeling of indexes).
We can now properly show that $[Q_a(t),Q_b(t)] = if_{abc}Q_c(t)$
\begin{align*}
[Q_a,Q_b] &= \iint d^3xd^3y \Big[-i\pi_i(T_a)_{ij}\phi_j + i\phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j , -i\pi_l(T_b)_{lm}\phi_m + i\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] = \\
&\quad= \iint d^3xd^3y \Big[-i\pi_i(T_a)_{ij}\phi_j,-i\pi_l(T_b)_{lm}\phi_m \Big] + \Big[-i\pi_i(T_a)_{ij}\phi_j , i\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] + \\
&\quad+\Big[i\phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j , -i\pi_l(T_b)_{lm}\phi_m \Big] +\Big[i\phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j, i\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] \quad(1)
\end{align*}
Working out the second commutator we find it to be zero according to the previous commutation relations:
\begin{equation}
\Big[-i\pi_i(T_a)_{ij}\phi_j , i\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] = (T_a)_{ij}(T_b)_{lm}\Big[\pi_i\phi_j , \phi^{\dagger}_l\pi^{\dagger}_m\Big] = 
\end{equation}
\begin{equation}
=(T_a)_{ij}(T_b)_{lm} \Big( \pi_i\Big[\phi_j , \phi^{\dagger}_l\Big]\pi^{\dagger}_m + \Big[\pi_i , \phi^{\dagger}_l\Big]\phi_j\pi^{\dagger}_m + \phi^{\dagger}_l\pi_i\Big[ \phi_j , \pi^{\dagger}_m \Big] + \phi^{\dagger}_l\Big[\pi_i , \pi^{\dagger}_m\Big]\phi_j\Big) = 0
\end{equation}
The same result follows for the third commutator and so we have reduced $[Q_a(t),Q_b(t)]$to the following:
\begin{equation} 
(1) \rightarrow \iint d^3xd^3y \Big[-i\pi_i(T_a)_{ij}\phi_j , -i\pi_l(T_b)_{lm}\phi_m \Big] +  
\Big[i\phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j , i\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] = 
\end{equation}
\begin{equation} 
= \iint d^3xd^3y -\Big[\pi_i(T_a)_{ij}\phi_j , \pi_l(T_b)_{lm}\phi_m \Big]   
-\Big[\phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j , \phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m\Big] =
\end{equation}
\begin{equation} 
= \iint d^3xd^3y \Big[\pi_l(T_b)_{lm}\phi_m,\pi_i(T_a)_{ij}\phi_j \Big] + \iint d^3xd^3y  
\Big[\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m , \phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j \Big] \tag{2}
\end{equation}
Now, lets work out the first commutator in (2):
\begin{equation}
\iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big[\pi_l\phi_m,\pi_i\phi_j \Big] = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_l\phi_m\pi_i\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = 
\end{equation}
\begin{equation}
= \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_l(\pi_i\phi_m - [\pi_i,\phi_m])\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij} \Big(\pi_l(\pi_i\phi_m +i\delta_{im}\delta(\vec{x}-\vec{y}))\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_l\pi_i\phi_m\phi_j + i\pi_l\delta_{im}\delta(\vec{x}-\vec{y})\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_i\pi_l\phi_j\phi_m + i\pi_l\delta_{im}\delta(\vec{x}-\vec{y})\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_i(\phi_j\pi_l - [\phi_j,\pi_l])\phi_m + i\pi_l\delta_{im}\delta(\vec{x}-\vec{y})\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big(\pi_i\phi_j\pi_l\phi_m - i\pi_i\delta_{jl}\delta(\vec{x}-\vec{y})\phi_m + i\pi_l\delta_{im}\delta(\vec{x}-\vec{y})\phi_j - \pi_i\phi_j\pi_l\phi_m \Big) = \iint d^3xd^3y (T_b)_{lm}(T_a)_{ij}\Big( - i\pi_i\delta_{jl}\delta(\vec{x}-\vec{y})\phi_m + i\pi_l\delta_{im}\delta(\vec{x}-\vec{y})\phi_j \Big) = \int d^3x (T_b)_{lm}(T_a)_{ij}\Big( - i\pi_i\delta_{jl}\phi_m + i\pi_l\delta_{im}\phi_j \Big) = \int d^3x \Big( - i\pi_i(T_a)_{ij}(T_b)_{jm}\phi_m + i\pi_l(T_b)_{lm}(T_a)_{mj}\phi_j \Big)\end{equation}
\begin{equation}
= \int d^3x \Big( i\pi_i[(T_b),(T_a)]_{im}\phi_m  \Big) =\int d^3x \Big( i\pi_i(-if_{abc}(T_c)_{im})\phi_m \Big) = if_{abc}\int d^3x -i\pi_i(T_c)_{ij}\phi_j
\end{equation}
Using the above commutation relations we can work out the second commutator in (2) in a similar way. Thus we arrive at:
\begin{equation}
\int d^3xd^3y  
\Big[\phi^{\dagger}_l(T_b)_{lm}\pi^{\dagger}_m , \phi^{\dagger}_i(T_a)_{ij}\pi^{\dagger}_j \Big] = (...) = if_{abc} \int d^3x i\phi^{\dagger}_i(T_c)_{ij}\pi^{\dagger}_j 
\end{equation}
Finally, combining these results we arrive at:
\begin{equation}
if_{abc}\int -i\pi_i(T_c)_{ij}\phi_j d^3x + if_{abc} \int i\phi^{\dagger}_i(T_c)_{ij}\pi^{\dagger}_j d^3x= 
\end{equation}
\begin{equation}
if_{abc} \int -i\pi_i(T_c)_{ij}\phi_j + i\phi^{\dagger}_i(T_c)_{ij}\pi^{\dagger}_j d^3x  = if_{abc} Q_c(t)
\end{equation}
This is called charge algebra.
