Question on derivatives of Hermitian conjugate scalar fields

["Don't panic!"]

Hi,

I know this question may seem a little trivial, but is there any real difference between

$\left (\partial_{\mu} \phi \right)^{\dagger}$ and $\partial_{\mu} \phi^{\dagger}$

and if so, could someone provide an explanation?

Many thanks.

(Sorry if this isn't quite in the right section, this is my first ever post).

 

[strangerep]

[...] is there any real difference between $\left (\partial_{\mu} \phi \right)^{\dagger}$ and $\partial_{\mu} \phi^{\dagger}$

Assuming that by $\partial_\mu$, you mean a partial derivative wrt a real parameter, then no.

 

["Don't panic!"]

Yes, sorry I meant wrt $x^{\mu}$, where $x^{\mu}$ is the usual space-time coordinate 4-vector, such that $\partial_{\mu}\equiv\frac{\partial}{\partial x^{\mu}}$

Is it just convention then to, for example, express the Dirac Lagrangian density as

$\cal{L}=i\overline{\Psi}\gamma^{\mu}\partial_{\mu} \Psi - i\overline{\left(\partial_{\mu}\Psi\right)}\gamma^{\mu}\Psi-\overline {\Psi}m\Psi$

(I've seen it expressed like this in several lecture note PDFs and textbooks that I've read)? Is it written this way just so the notation

$\frac{i}{2}\overline{\Psi}\gamma^{\mu}\overleftrightarrow{\partial_{\mu}} \Psi = i\overline{\Psi}\gamma^{\mu}\partial_{\mu} \Psi - i\overline{\left(\partial_{\mu}\Psi\right)}\gamma^{\mu}\Psi$

makes sense?

I seem to remember, maybe incorrectly, that it has to be operated on in this fashion so that it preserves the Hermicity of the Lagrangian for Spinor fields $\Psi$ and $\overline{\Psi}$ ?

Thanks for your time!

 

[strangerep]

[...] Is it written this way just so the notation

$\frac{i}{2}\overline{\Psi}\gamma^{\mu}\overleftrightarrow{\partial_{\mu}} \Psi = i\overline{\Psi}\gamma^{\mu}\partial_{\mu} \Psi - i\overline{\left(\partial_{\mu}\Psi\right)}\gamma^{\mu}\Psi$

makes sense?

It seems to me that one may understand this by just acting with $\partial$ on everything to its left. The constant $\gamma^\mu$ passes through the derivative. Then revert to conventional right-acting derivative notation, inserting parentheses so as to preserve the meaning of the expression. But in this case, one could move the overbar inside the parentheses, afaict.

BTW, note also that the overbar typically means a Dirac adjoint, so you've got a $\gamma^0$ floating around in there.

I seem to remember, maybe incorrectly, that it has to be operated on in this fashion so that it preserves the Hermicity of the Lagrangian for Spinor fields $\Psi$ and $\overline{\Psi}$ ?

I was never a fan of the $\overleftrightarrow{\partial_{\mu}}$ notation, and prefer to exhibit the Hermicity explicitly. But one must be conversant with the notation nevertheless, since many textbooks use it.

 

["Don't panic!"]

Thanks for your help, much appreciated!