Noether charge on complex scalar field

Question

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.

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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.

Answer 1

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...