# Differential operator acting on scalar fields

1. Mar 30, 2017

### shedrick94

1. The problem statement, all variables and given/known data
I really cannot seem to be able to follow the logic of how you would use the product rule when using 4 vector differential operator. ∂μ is the differential operator, Aμ is a scalar field and φ and φ* is it's complex conjugate scalar field. I have the answer, I'd just really like to know how to follow the logic of the answer.

2. Relevant equations
μ(Aμφφ*)=...?

3. The attempt at a solution
Literally have no idea what I'm doing.

2. Mar 31, 2017

### stevendaryl

Staff Emeritus
$A^\mu$ is a vector field, not a scalar field. But what is unclear about the product rule? (Note: I would not use the same index $\mu$ for both $\partial_\mu$ and $A^\mu$. I would let one of them be $\nu$. If you really mean for the indices to be equal, then you can set them equal at the end.)

You have a product $A^\nu \phi \phi^*$ you're taking a derivative with respect to $x^\mu$. So the product rule says:

$\partial_\mu (A^\nu \phi \phi^*) = (\partial_\mu A^\nu) \phi \phi^* + A^\nu (\partial_\mu \phi) \phi^* + A^\nu \phi (\partial_\mu \phi^*)$

3. Mar 31, 2017

### shedrick94

Yeh sorry I did mean to put that A is a scalar field. Should be a bit more explicit about the problem I'm having as I wrote that last night after spending almost all my evening on it and I had given up.

Basically the problem is to do with the klein gordon current with an electromagnetic field and part of the current is the below equation and you have to be able to recognize that the product rule expansion of what I've shown on the LHS is the RHS.

iq(\phi^{*}A_{\mu}\partial^{\mu}\phi + \phi^{*}\partial_{\mu}A^{\mu}\phi + \phi A_{\mu}\partial^{\mu}\phi^{*} + \phi\partial_{\mu}A^{\mu}\phi^{*})=2iq\phi^{*}\phi A_{\mu}

ie that,

\partial_{\mu}(LHS)=\partial_{\mu}(RHS)

I'm particularly struggling to understand when and where you can move the fields around. For example, are you allow to just move the scalar fields φ and φ* around wherever you please?