How can i derive conserved charge of $SO(3)$ internal symmetry? I'm studying on QFT for gifted amateur written by Tom Lancaster chapter 13.1 and i'm not fully understanding about derivation of conserved charge of $SO(3)$ internal symmetry.
So if there's three fields $\phi^1, \phi^2, \phi^3$ and they transform into each other by 3 dimensional rotation matrix $SO(3)$ and the lagrangian $$\mathcal{L} = \frac{1}{2}[(\partial_\mu\phi^1)^2-m^2(\phi^1)^2 +(\partial_\mu\phi^2)^2-m^2(\phi^2)^2 +(\partial_\mu\phi^3)^2-m^2(\phi^3)^2]$$
then this lagrangian is invariant under $SO(3)$ matrices so that there must exist some kind of conserved quantities by Noether's theorem.
The book says

but as i calculated, my answer couldn't reach to the eq (13.15).
If conserved current is given as eq (13.14), then the charge would be spatial integral of zeroth component of the current so i replaced the field and time derivative of the field by mode expansion of it.
$$J^{3,0}_N = \dot{\phi}^1\phi^2 - \dot{\phi}^2\phi^1$$
$$\hat{\phi}^{\alpha = 1,2,3} = \int \frac{d^3p}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_p}}(\hat{a}_{p \alpha} e^{-i\vec{p}\cdot\vec{x}} + \hat{a}^{\dagger}_{p \alpha}e^{i\vec{p}\cdot\vec{x}})$$
$$\partial_0 \phi^\alpha = \dot{\phi}^\alpha = \int \frac{d^3p}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_p}}(-iE_p\hat{a}_{p \alpha} e^{-i\vec{p}\cdot\vec{x}} + iE_p\hat{a}^{\dagger}_{p \alpha}e^{i\vec{p}\cdot\vec{x}}) \\
= \int \frac{d^3p}{(2\pi)^{3/2}}\cdot-i\frac{\sqrt{E_p}}{\sqrt{2}}(\hat{a}_{p \alpha} e^{-i\vec{p}\cdot\vec{x}} - \hat{a}^{\dagger}_{p \alpha}e^{i\vec{p}\cdot\vec{x}})$$
Then the current would be
$$J^{3,0} = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{2}\bigg[-i(\hat{a}_{p1} e^{-i\vec{p}\cdot\vec{x}} - \hat{a}^{\dagger}_{p1}e^{i\vec{p}\cdot\vec{x}})(\hat{a}_{p2} e^{-i\vec{p}\cdot\vec{x}} + \hat{a}^{\dagger}_{p2}e^{i\vec{p}\cdot\vec{x}}) + i(\hat{a}_{p2} e^{-i\vec{p}\cdot\vec{x}} - \hat{a}^{\dagger}_{p2}e^{i\vec{p}\cdot\vec{x}})(\hat{a}_{p1} e^{-i\vec{p}\cdot\vec{x}} + \hat{a}^{\dagger}_{p1}e^{i\vec{p}\cdot\vec{x}})\bigg]$$
The terms with $e^{-2i\vec{p}\cdot\vec{x}}$ disappear since $[\hat{a}_{p \alpha},\hat{a}_{q \beta}] = 0 $
The current equation can be sorted into
$$J^{3,0} = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{2}\bigg[-i(\hat{a}_{p1}\hat{a}^\dagger_{p2} - \hat{a}^\dagger_{p1}\hat{a}_{p2} - \hat{a}_{p2}\hat{a}^\dagger_{p1}+\hat{a}^\dagger_{p2}\hat{a}_{p1}\bigg]$$
And this is the part where i'm stuck.
$$ Q^{3} = \int\int d^3 x\frac{d^3p}{(2\pi)^{3}}\frac{1}{2}\bigg[-i(\hat{a}_{p1}\hat{a}^\dagger_{p2} - \hat{a}^\dagger_{p1}\hat{a}_{p2} - \hat{a}_{p2}\hat{a}^\dagger_{p1}+\hat{a}^\dagger_{p2}\hat{a}_{p1}\bigg]$$
I have no idea how to deal with integral $d^3x$ and the term $\frac{1}{(2\pi)^3}$
I guess my entire calculation might have been wrong.
Could you please let me know how to reach the eq (13.15) in text book?
 A: When you define your current you have quantities that multiply like $\dot{\phi}^1 \phi^2$ and a series expansion of those has different momenta $p^1, p^2$. In fact it is the very integral on spatial coordinate that simplify things because
$$ \int \text{d}^3{x} e^{\mathrm{i} p \cdot x} = (2\pi)^3 \delta(p) $$
and you use this result to cancel out one of the two momenta integrals that appear. You just missed this part.
Edit:
For example consider the first term $\dot{\phi}^1 \phi^2$ then
$$
\dot{\phi}^1 \phi^2 = - \mathrm{i} \int \frac{\text{d}^3 p}{(2\pi)^{\frac{3}{2}}} \frac{\sqrt{E_p}}{\sqrt{2}} \left( \hat{a}_{p1} e^{-\mathrm{i}p\cdot x} + \hat{a}_{p1}^\dagger e^{\mathrm{i}p\cdot x} \right)
\int \frac{\text{d}^3 \tilde{p}}{(2\pi)^{\frac{3}{2}}} \frac{1}{\sqrt{2 E_{\tilde{p}}}} \left( \hat{a}_{\tilde{p}2} e^{-\mathrm{i}\tilde{p}\cdot x} + \hat{a}_{\tilde{p}2}^\dagger e^{\mathrm{i}\tilde{p}\cdot x} \right)
$$
and integrating in $x$
$$
\int \text{d}^3 x \,\dot{\phi}^1 \phi^2 = \frac{- \mathrm{i}}{2} (2\pi)^{\frac{3}{2}} \int \frac{\text{d}^3 p}{(2\pi)^{\frac{3}{2}}}
\int \frac{\text{d}^3 \tilde{p}}{(2\pi)^{\frac{3}{2}}} \left( \hat{a}_{p1} \hat{a}_{\tilde{p}2} \delta(p+\tilde{p}) + \hat{a}_{p1} \hat{a}_{\tilde{p}2}^\dagger \delta(p-\tilde{p}) + \hat{a}_{p1}^\dagger \hat{a}_{\tilde{p}2} \delta(p-\tilde{p}) + \hat{a}_{p1}^\dagger \hat{a}_{\tilde{p}2}^\dagger \delta(p+\tilde{p})\right)
$$
