The free-field KG lagrangian density for complex scalar field is given as $$\hat{\mathcal{L}}_{\text{KG}}=(\partial_\mu\hat{\phi}^\dagger)(\partial^\mu\hat\phi)-m^2\hat{\phi}^\dagger\hat{\phi}$$ By making the minimal substitution $$\partial^\mu\mapsto\hat{D}^\mu=\partial^\mu+iq\hat{A}^\mu$$ we could introduce electromagnetic field into the equation and obtain $$\hat{\mathcal{L}}=\hat{\mathcal{L}}_{\text{KG}}+\hat{\mathcal{L}}_{\text{INT}}$$ The expected results should be $$\hat{\mathcal{L}}_{\text{INT}}=-iq\left(\hat{\phi}^\dagger(\partial^\mu\hat\phi)-(\partial^\mu\hat\phi^\dagger)\hat{\phi}\right)\hat{A}^\mu+q^2\hat{A}^\mu\hat{A}_\mu\hat{\phi}^\dagger\hat{\phi}\tag{1}$$ However, I did the subsitution myself and kept getting $$\hat{\mathcal{L}}_{\text{INT}}=iq\left(\hat{\phi}^\dagger(\partial^\mu\hat\phi)+(\partial^\mu\hat\phi^\dagger)\hat{\phi}\right)\hat{A}^\mu+q^2\hat{A}^\mu\hat{A}_\mu\hat{\phi}^\dagger\hat{\phi}$$ But I could not understand how i got it wrong. Could someone please explain how to get the correct expression equation (1)?

  • $\begingroup$ @FrodCube No they are not. There is a relative sign between the first two terms. That is what he is asking about. $\endgroup$
    – hft
    May 11 at 0:33
  • $\begingroup$ OP, can you provide me with the textbook/lecture notes you've been using? $\endgroup$ May 11 at 18:00

1 Answer 1


The error that you've made is that the factor $(\partial_{\mu} \phi^{\dagger})$ in the Klein-Gordon Lagrangian should be written as $(\partial_{\mu} \phi)^{\dagger}$. This makes a difference, since the hermitian conjugate is applied to the gradient of the field, $\partial_{\mu}\phi$, not $\phi$. Ordinarily, you could get away with your notation, since $(\partial_{\mu}\phi)^{\dagger}=\partial_{\mu}\phi^{\dagger}$. However, since $(D_{\mu}\phi)^{\dagger}=\partial_{\mu}\phi^{\dagger}-iqA_{\mu}\phi^{\dagger} \ne \partial_{\mu}\phi^{\dagger} +iqA_{\mu}\phi^{\dagger} = D_{\mu}\phi^{\dagger}$, you'll get into trouble. In other words, when you make the substitution $\partial_{\mu} \mapsto D_{\mu}$ apply the hermitian conjugate to $D_{\mu}\phi$, not $\phi$.

I hope this clears things up!

  • 2
    $\begingroup$ -1. I don't know where you have seen that $\partial_\mu^\dagger = \partial_\mu$ but it is entirely wrong. Moreover, you completely missed the point of OP, since OP explicitly says that its notation for $+$ is in fact $\ast$, the complex conjugate. Even in the operator formalism, one shouldn't take the hermitian conjugate instead of the complex conjugate. $\endgroup$ May 11 at 7:43
  • $\begingroup$ Hi! Please provide some arguments on why $\partial_{\mu}^{\dagger}=\partial_{\mu}$ is entirely wrong, I would like to learn more about it. $\endgroup$ May 11 at 9:24
  • 1
    $\begingroup$ Hi, first, sorry if I've been rude. Second, our Klein-Gordon field belongs to $L^2(\mathbb{R}^{1,3})$, so we can define $\langle f | g \rangle = \int d^4x f^\ast g$. So $\langle f | \partial_\mu g \rangle=\langle f |\partial_\mu | g \rangle \stackrel{\text{p.p.}}{=}-\langle \partial_\mu f | g \rangle=- \langle f |\partial_\mu^\dagger | g \rangle $, where $\text{p.p.}$ stands for "integration by part". $\endgroup$ May 11 at 9:41
  • $\begingroup$ Does that mean that $\partial_{\mu}=-\partial_{\mu}^{\dagger}?$ I know nothing about $L^p$ spaces... I will probably make a question about this. $\endgroup$ May 11 at 12:31
  • $\begingroup$ @JeanbaptisteRoux In this case $\partial_\mu = \partial_\mu^\dagger$ because this $\dagger$ refers to the action on the Fock space of the theory, while the derivative acts on the field coordinates. $\endgroup$
    – FrodCube
    May 11 at 16:03

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