I'm trying to expand $(D_\mu \phi)^\dagger (D^\mu \phi)$ in a Lagrangian. Where:

$$D=\partial +ieA \tag{1}$$

$A$ is a photon field.

I'm uncertain of how to expand the first half (I wrote this question as a comment in https://physics.stackexchange.com/questions/516077/prove-d-mu-phi-dagger-d-mu-phi-dagger).

Is my take on this correct? Should $A$ get a complex conjugate?

My take on it:

$$(D_\mu \phi)^\dagger = \phi^\dagger D_\mu ^\dagger = \phi^\dagger(\partial_\mu +ieA^\dagger_\mu) \tag{2}$$

and hence

$$\phi^\dagger D_\mu ^\dagger = \phi^\dagger(\partial_\mu +ieA^\dagger_\mu) (\partial^\mu +ieA^\mu ) \phi\\=\phi^\dagger \partial^2\phi + \phi^\dagger \partial_\mu(ieA^\mu\phi)+ \phi^\dagger ieA^\dagger_\mu \phi - \phi^\dagger e^2 A^\dagger_\mu A^\mu \phi $$

Can this be further expanded? Is it correct?


1 Answer 1


The rewrite $(D_\mu \phi)^\dagger = \phi^\dagger D_\mu^\dagger$ is not valid since this $D_\mu$ acts on this occurrence of $\phi,$ not on the other one.

Also, $A$ is Hermitian (real) since the photon has no electrical charge.

Thus, $$ (D_\mu \phi)^\dagger (D^\mu \phi) = (\partial_\mu\phi + ieA_\mu\phi)^\dagger (\partial^\mu\phi + ieA^\mu\phi) = (\partial_\mu\phi^\dagger - ieA_\mu\phi^\dagger) (\partial^\mu\phi + ieA^\mu\phi) \\ = \partial_\mu\phi^\dagger \, \partial^\mu\phi + ieA^\mu \, \partial_\mu\phi^\dagger \, \phi - ieA_\mu \, \phi^\dagger \, \partial^\mu\phi + e^2 A_\mu A^\mu \, \phi^\dagger\phi , $$ where it also has been used that $A$ commutes with $\phi$ and $\phi^\dagger.$

  • $\begingroup$ Oh right, it makes sense that $D_\mu$ acts on a specific a $\phi$. So if A depicted an electron then it would have a hermitian conjugate? Thank you for your help. $\endgroup$ Jun 6, 2020 at 13:40
  • 1
    $\begingroup$ Yes, the electron field is represented by a complex spinor. $\endgroup$
    – md2perpe
    Jun 6, 2020 at 13:49

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