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 $\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$)


1 Answer 1


It is just a compact way of saying $$(\partial_{\mu}\phi^{*})\phi - (\partial_{\mu}\phi)\phi^{*}$$

The double arrow is just to remember that the derivative applies on the field and on the conjugate field.

Note: Srednicki defines this with the opposite sign on p.135: $$\phi^* (\partial_\mu \phi) - (\partial_\mu \phi^*) \phi$$

  • $\begingroup$ 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. $\endgroup$ Feb 17 at 21:34

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