The following is loosely related to this question:
[…], the most general renormalisable Lagrangian that is invariant under both Lorentz transformations and gauge transformations is
$$\mathcal{L}=-\frac14F_{\mu\nu}F^{\mu\nu}+\left(D_\mu\phi\right)^\ast D^\mu\phi-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2\tag{1}$$In a more explicit form, this Lagrangian can be written as
$$\mathcal{L}=\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)
+\partial_\mu\phi^\ast\partial^\mu\phi$$
$$+ie\color{red}{A^\mu\left(\phi\partial_\mu\phi^\ast – \phi^\ast\partial_\mu\phi\right)}
+e^2A_\mu A^\mu\phi^\ast\phi
-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2\tag{2}$$
Where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and $D_\mu\phi\equiv \partial_\mu\phi + ieA_\mu\phi$
My question is regarding how the term in red was deduced in eqn. $(2)$ of the authors' extract.
As an attempt to try to get this red term I write out each stage slowly:
$$\mathcal{L}=-\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\left(\partial_\mu\phi^\ast – ieA_\mu\phi^\ast\right)\left(\partial^\mu\phi + ieA^\mu\phi\right)-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2$$
$$=\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\partial_\mu\phi^\ast\partial^\mu\phi+\color{blue}{ie\partial_\mu\phi^\ast A^\mu\phi- ieA_\mu\phi^\ast\partial^\mu\phi}+e^2A_\mu A^\mu\phi^\ast\phi
-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2$$
From now on focussing on the term in blue, the most simplified way I can write this is as $$ie\left(\partial_\mu\phi^\ast A^\mu\phi- A_\mu\phi^\ast\partial^\mu\phi\right)$$
but in order for this to be consistent with the red term in $(2)$ we must have
$$\partial_\mu\phi^\ast A^\mu\phi- A_\mu\phi^\ast\partial^\mu\phi=A^\mu\phi\partial_\mu\phi^\ast – A^\mu\phi^\ast\partial_\mu\phi\tag{?}$$
I have two questions regarding this $(\mathrm{?})$ equality:
- Looking at the first terms on the left and right hand sides of $(\mathrm{?})$, what allowed the author to commute the $A^\mu$ to the front of the derivative, in other words why is $\partial_\mu\phi^\ast A^\mu\phi\stackrel{?}=A^\mu\phi\partial_\mu\phi^\ast$ true? The reason I question this step is because the derivative is respect to the spacetime coordinate $x^\mu$, and the four-vector potential, $A^\mu$, depends on this $\mu$ also.
- Nextly, looking at the last terms on the left and right hand sides of $(\mathrm{?})$, for equality to make sense we must have $A_\mu\phi^\ast\partial^\mu\phi\stackrel{?}=A^\mu\phi^\ast\partial_\mu\phi$. But what logic allowed the index for $A_\mu$ to be raised and the index for $\partial^\mu$ to be lowered? This is the manipulation I thought was being invoked: $A^\mu\partial_\mu\phi=\eta^{\alpha\mu}A_\alpha\eta_{\beta\mu}\partial^\beta\phi=\eta^{\alpha\mu}\eta_{\beta\mu}A_\alpha\partial^\beta\phi=\delta^\alpha_\beta A_\alpha\partial^\beta\phi=A_\alpha\partial^\alpha\phi=A_\mu\partial^\mu\phi$ where in the last step I simply relabelled $\alpha\to\mu$ since it is a dummy index. I'm doubtful of this manipulation as the four-vector potential is ambiguous as its' contravariant version, $A^\mu=\left(\varphi, \vec A\right)$ and the covariant version is $A_\mu=\left(\varphi, -\vec A\right)$ or $A^\mu=\left(\varphi, \vec A\right)$ and $A_\mu=\left(-\varphi, \vec A\right)$ and there are of course other variants where the negative sign is present in the contravariant four-vector potential, $A^\mu$. So I think that actually there should be a minus sign present, in other words, $A^\mu\partial_\mu\phi=-A_\mu\partial^\mu\phi$. The convention followed for the Minkowski metric throughout the above is $\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$.
Best Answer
No reason why it should not be allowed to commute the three functions $\partial_\mu \phi^\ast(x)$, $A^\mu(x)$ and $\phi(x)$. At this point, these are still classical, commuting fields and not yet field operators: $$\partial_\mu \phi^\ast A^\mu \phi\equiv \frac{\partial \phi^\ast}{\partial x^\mu} A^\mu \phi= A^\mu \frac{\partial \phi^\ast}{\partial x^\mu}\phi=A^\mu \phi\frac{\partial \phi^\ast}{\partial x^\mu}=\frac{\partial \phi^\ast}{\partial x^\mu}\phi A^\mu=\ldots$$
$A_\mu B^\mu= A^\mu B_\mu$ holds for arbitrary 4-vectors $A^\mu$, $B^\mu$ (in this case, vector fields): $$A_\mu B^\mu = \eta_{\mu \nu} A^\nu B^\mu = A^\nu \eta_{\mu \nu}B^\mu =A^\nu \eta_{\nu \mu}B^\mu=A^\nu B_\nu=A^\mu B_\mu,$$ so $A_\mu \partial^\mu \phi^\ast=A^\mu\partial_\mu \phi^\ast$ is correct.
With $A^\mu =(\varphi, \vec{A})$ and the convention $\eta= {\rm diag}(1,-1,-1,-1)$, you have $A_\mu =(\varphi, -\vec{A})$ and not $(-\varphi, \vec{A})$.