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.

  • 3
    $\begingroup$ I don't understand where is the problem here (maybe I have misread I don't know). $\phi^\ast$ and $\partial^\mu \phi$ commute. $\endgroup$ Aug 14 at 11:11
  • $\begingroup$ @JeanbaptisteRoux Thank you! How about in a case that a pauli matrix lie between $\phi^{*}$ and $\partial_{\mu}\phi$, then can we still exchange them? $\endgroup$
    – Daren
    Aug 14 at 13:59
  • $\begingroup$ I haven't read the whole question, but if you are asking whether or not you can exchange the ordering of scalar fields, yes, you can. And this holds even if there is a pauli matrix in between scalars, yes. $\endgroup$
    – schris38
    Aug 14 at 18:21
  • $\begingroup$ @schris38 Thank you very much! $\endgroup$
    – Daren
    Aug 15 at 1:05
  • $\begingroup$ @schris38 However, I find a case. Consider a two complex scalar case, and think the term $(\phi^{*})_i (\sigma^2)_{ij} (\partial_{\mu}\phi)_j$, since $\sigma^2$ have an overall minus sign after transpose, it seems that $(\phi^{*})_i (\sigma^2)_{ij} (\partial_{\mu}\phi)_j = -(\partial_{\mu}\phi)_i (\sigma^2)_{ij} (\phi^{*})_j$. $\endgroup$
    – Daren
    Aug 15 at 2:50

1 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...

  • $\begingroup$ Thank you for your answer! Yes, I understand your answer. But we see the effect that $\partial_{\mu}\Phi_i\sigma^2_{ij}\Phi_j^{*}$ really have sign difference with $\Phi_i^{*}\sigma^2_{ij}\partial_{\mu}\Phi_j$, this just I point out in the end of my post! Since their have a sign difference, you can check my derivation of $j^{\mu a}$, the last equality in $j^{\mu a}$ can not be satisfied! This is my question! $\endgroup$
    – Daren
    Aug 17 at 11:04
  • $\begingroup$ Hi. Sorry, but I do not understand why should the last step hold. I think it shouldn't for the simple reason that $\Phi_j^*$ has a $j$ index! If you want to rename that index, you should rename the $j$ indices for the whole term, i.e. $\partial_{\mu}\Phi_i(\sigma^2_{ij})\Phi^*_j\rightarrow\partial_{\mu}\Phi_j(\sigma^2_{ji})\Phi^*_i=-\partial_{\mu}\Phi_j(\sigma^2_{ij})\Phi^*_i=-\Phi^*_i(\sigma^2_{ij})\partial_{\mu}\Phi_j$. I do not quite fully understand yet what you are discussing in the previous comment to be honest. Could you please ellaborate more? $\endgroup$
    – schris38
    Aug 17 at 13:04
  • $\begingroup$ Thanks for your comment, I just want to get the expression in Peskin's QFT book, which lie in page 34, on chapter 2 problem 2 (d). So could you please check that equation on Peskin's book? In my understanding, the book need the last step of $j^{\mu a}$. But in our discussion, the last step may not true, this is my question. However, here have an overall minus sign difference between my post and book, this don't bother because the sign conversion of Noether charge. $\endgroup$
    – Daren
    Aug 17 at 13:45
  • $\begingroup$ Sorry for not understanding earlier the source of your problem/question. $\endgroup$
    – schris38
    Aug 17 at 16:15
  • 1
    $\begingroup$ Thanks for your edited answer! $\endgroup$
    – Daren
    Aug 18 at 1:48

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