# What does $\overset\leftrightarrow{\partial_{\mu}}$ means?

I have a scalar complex field: $$\phi(x) = \phi_{1} + i \phi_{2}\;$$ so $$\;\phi^{*}(x) = \phi_{1} - i \phi_{2}$$ where $$\phi_{1}, \; \phi_{2}$$ are real scalar fields.

Then I have something like $$\;\phi^{*}\overset\leftrightarrow{\partial_{\mu}}\phi \;$$. What does this $$\;\overset\leftrightarrow{\partial_{\mu}}$$ means?

(PS: I Know that $$\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$$)

It is just a compact way of saying $$(\partial_{\mu}\phi^{*})\phi - (\partial_{\mu}\phi)\phi^{*}$$
Note: Srednicki defines this with the opposite sign on p.135: $$\phi^* (\partial_\mu \phi) - (\partial_\mu \phi^*) \phi$$
• Note: The convention is not universal, outside of QFT-contexts it is also often used to just mean that the derivative acts in both directions: $a \overset{\leftrightarrow}{\partial} b = a \overset{\leftarrow}{\partial} b + a \overset{\rightarrow}{\partial} b$ – this may be especially convenient when handling vector identities (with $\nabla$ in place of $\partial$) and non-commutative operations (e.g. as in $\vec a \times \overset{\leftarrow}{\nabla}$). It is also useful in caculations done in Einstein notation. Feb 17 at 21:34