Noether charge on complex scalar field For complex scalar field, we write the Lagrangian as:
$$
\mathcal{L}=\partial_{\mu}\phi^{*}\partial^{\mu}\phi-m^2 \phi^{*}\phi
$$
with the $U(1)$ symmetry, and under infinitesimal transformation:
$$
\phi \rightarrow \phi +\alpha (i \phi) \\
\phi^{*} \rightarrow \phi^{*} +\alpha (-i \phi^{*})
$$
The Noether current:
$$
\begin{aligned}
j^{\mu}&=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\Delta \phi +\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi^{*})}\Delta \phi^{*} \\
&=i[(\partial^{\mu}\phi^{*})\phi-(\partial^{\mu}\phi)\phi^{*}]
\end{aligned}
$$
However, in many cases, we see the Noether current is given in this formula: (such as P & S's QFT book on page 18)
$$
j^{\mu}=i[(\partial^{\mu}\phi^{*})\phi-\phi^{*}(\partial^{\mu}\phi)]
$$
So why the second term is switched between $(\partial^{\mu}\phi)$ and $\phi^{*}$?
If you have any comment or answer, I am really appreciate it.

New editing from comment:
Now if we consider the case of two complex Klein-Gordon field with the same mass: (as in P & S's QFT book on problem 2 (d) on page 34)
Now we can write the Lagrangian as
$$
\mathcal{L}=\partial_{\mu}\Phi^{*}_i\partial^{\mu}\Phi_i-m^2 \Phi^{*}_i\Phi_i
$$
where $i=1,2$
$$
\begin{align}
    \Phi &= \begin{bmatrix}
           \Phi_1 \\
           \Phi_2
         \end{bmatrix}
  \end{align}
$$
$\Phi_1$ and $\Phi_2$ are two independent Klein-Gorden fields. Now we know that this Lagrangian have $U(2)$ symmetry, and due to
$$
U(2)\simeq \frac{SU(2)\times U(1)}{\mathbb{Z}_2}
$$
$\Phi$ under a infinitesimal transformation:
$$
\Phi \rightarrow \Phi + i(\alpha\ + \vec{\theta}\cdot \vec{\sigma}^a/2)\Phi
$$
where $\vec{\sigma}$ is the Pauli matrix.
Now according to the Noether current (related to parameter $\vec{\theta}$ only):
$$
\begin{aligned}
j^{\mu a}&=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi_i)}\Delta \Phi_i +\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi^{*}_i)}\Delta \Phi^{*}_i \\
&=\frac{i}{2}[(\partial^{\mu}\Phi^{*}_i)(\sigma^a_{ij})\Phi_j-(\partial^{\mu}\Phi_i)(\sigma^a_{ij})\Phi_j^{*}] \\
&=\frac{i}{2}[(\partial^{\mu}\Phi^{*}_i)(\sigma^a_{ij})\Phi_j-\Phi_i^{*}(\sigma^a_{ij})(\partial^{\mu}\Phi_j)] \\
\end{aligned}
$$
$\textbf{Now my question is that I am troubled for the last step.}$ I already know that $\Phi_i^{*}$ and $\partial^{\mu}\Phi_j$ commutate
Consider the case of $\sigma^2$:
$$
(\partial^{\mu}\Phi_i)(\sigma^2_{ij})\Phi_j^{*}=-i(\partial^{\mu}\Phi_1)(\Phi_2^{*})+i(\partial^{\mu}\Phi_2)(\Phi_1^{*})
$$
while
$$
\Phi_i^{*}(\sigma^2_{ij})(\partial^{\mu}\Phi_j)=i(\partial^{\mu}\Phi_1)(\Phi_2^{*})-i(\partial^{\mu}\Phi_2)(\Phi_1^{*})
$$
So their have a minus sign difference here, in this situation, $\textbf{the last equality of $j^{\mu a}$ can not be satisfied}$. While Peskin's QFT book require the last expression, so how to explain this.
 A: Well, basically you ignore the anti-symmetry of the Pauli matrices. For example, since $i$ and $j$ are dummy indices, I can perform the substitution $i\leftrightarrow j$ and hence starting from the first term
$$\partial_{\mu}\Phi_i\sigma_{ij}^2\Phi_j^*=
\Phi_j^*\sigma_{ij}^2\partial_{\mu}\Phi_i=
\Phi_i^*\sigma_{ji}^2\partial_{\mu}\Phi_j=-
\Phi_i^*\sigma_{ij}^2\partial_{\mu}\Phi_j$$
which in turn is equal to (upon expanding by plugging in the values of $i$ and $j$)
$$-\Phi_i^*\sigma_{ij}^2\partial_{\mu}\Phi_j=-
\Phi_1^*(-i)\partial_{\mu}\Phi_2-\Phi_2^*(+i)\partial_{\mu}\Phi_1=-
\partial_{\mu}\Phi_2(-i)\Phi_1^*-\partial_{\mu}\Phi_1(+i)\Phi_2^*$$
The order does not count, since $\sigma_{ij}^{a},\ a=1,2,3$ are simply numbers/elements of the Pauli matrices.
EDIT:
I think I realise where the error may be found. Let's take things from the beginning. We have two complex Klein Gordon fields $\phi_a(x),\ a=1,2$. Before going any further, we should note that [and this justifies why you do not have terms that are formed by the product of $\pi_a(x)$ with $\phi_a(x)$ and terms that are formed by the product of $\pi_a^*(x)$ with $\phi_a^*(x)]$ the conjugate momenta of the two scalar fields is given by
$$\pi_a(x)=\frac{\delta\mathcal{L}}{\delta\dot{\phi}_a(x)}=\dot{\phi}^*_a(x)
\\
\pi^*_a(x)=\frac{\delta\mathcal{L}}{\delta\dot{\phi}^*_a(x)}=\dot{\phi}_a(x)$$
Now, one must think about the transformation that is associated with the current you are referring to: if I am to follow your steps (with the only difference being a minus sign-there I think your mistake may lie) and say that the transformation is one such that
$$\delta\phi^i_a(x)=-\frac{i}{2}\sigma^i_{ab}\phi_b(x)$$
Then, the complex conjugate for the latter shall be
$$\delta\phi^{*i}_a(x)=+\frac{i}{2}\phi^*_b(x)(\sigma^i_{ba})^*=
+\frac{i}{2}\phi^*_b(x)\sigma^i_{ab}$$
The expression for the current yields
$$j^{i\mu}=\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi_a(x)}
\delta\phi^i_a(x)+
\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi^*_a(x)}
\delta\phi^{*i}_a(x)$$
which yields upon substituting
$$j^{i\mu}=\frac{i}{2}\Big[-
\partial^{\mu}\phi^*_a(x)
\sigma^i_{ab}\phi_b(x)+
\partial^{\mu}\phi_a(x)
\sigma^i_{ab}\phi^*_b(x)\Big]$$
Now, choosing $\mu=0$ yields
$$Q^i=\int d^3x j^{i0}=
\int d^3x\frac{i}{2}\Big[-
\dot{\phi}^*_a(x)
\sigma^i_{ab}\phi_b(x)+
\dot{\phi}_a(x)
\sigma^i_{ab}\phi^*_b(x)\Big]=
\int d^3x\frac{i}{2}\Big[-
\pi_a(x)
\sigma^i_{ab}\phi_b(x)+
\pi^*_a(x)
\sigma^i_{ab}\phi^*_b(x)\Big]$$
The only difference with the book can be found into the ordering of the scalar operators (and I think I have now came to the part where your question is!). I have struggled with that for some time and my only reasonal explanation is that the $\sigma$ matrix should be starred. I lay my arguments below
$$\pi_a^*(x)\sigma^i_{ab}\phi^*_b(x)=
[\phi_b(x)(\sigma^i_{ab})^*\pi_a(x)]^*=
[\phi_a(x)(\sigma^i_{ba})^*\pi_b(x)]^*=
[\phi_a(x)\sigma^i_{ab}\pi_b(x)]^*=
\phi^*_a(x)(\sigma^i_{ab})^*\pi^*_b(x)$$
I am always reluctant on assuming that my points of disagreement with a book are actually errors, but I do not see some other way around it. You can think about it yourself too. I think I haven't "cheated" somewhere in the considerations above because the only things we have used is the fact that the field is scalar and the hermitian property of the Pauli matrices...
