# Technical question regarding lagrangian and action variation in QFT

In my textbook the lagrangian density of the system is taken to be $\mathcal{L} = -\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \frac{1}{2}m^2 \phi^2$ where $\phi \equiv \phi(x)$ is a complex scalar field.

We then try to determine the equation of motion by using that the variation of action must vanish. The author then introduce a variation $\delta \phi$ and quickly writes:

$$0 = \delta s = \int d^4x\ \left[-\frac{1}{2} \partial ^{\mu} \delta \phi \partial_{\mu} \phi -\frac{1}{2} \partial ^{\mu} \phi \partial_{\mu} \delta\phi -m^2\phi\delta \phi\right]$$

So my question is simply, what is going on here? How do you obtain the expression in the brackets? How does it relate to $\mathcal{L}$.

For the record, the tensor notation is still new to me, so my confusion might be related to this.

• you must regard $\delta$ as a differential operator (aka, linearity+Leibnitz rule). The justification of this rule lies in rigorous the definition of $\delta$. – AccidentalFourierTransform Nov 30 '16 at 21:05
• I think it is using $\delta(ab) = a\delta (b)+\delta(a) b$. – user130529 Nov 30 '16 at 21:06
• so $\delta(\phi \partial_{\mu} \phi) = \phi \delta \partial_{\mu} \phi + \delta \phi \partial_{\mu} \phi$ ? – Fabric Nov 30 '16 at 21:21
• Yes, if $\delta$ works like a differential operator as mentioned by AccidentalFourierTransform (and I see no reason why it wouldn't). – user130529 Nov 30 '16 at 21:36

Formally you can define your variation as $\delta \mathcal S = \int \mathrm{d}^d x \ \mathcal L [\phi + \delta \phi] - \mathcal L [\phi]$, where you treat $\delta \phi$ as small, i.e. terms which contain higher orders than just linear are to be neglected.
This is equivalent to taking $\delta$ as a differential operator, e.g.
$\delta( \phi^3) = (\phi + \delta \phi)^3 - (\phi)^3 = \phi^3 + 3 \phi^2 \delta \phi + \mathcal{O}(\delta \phi^2) - \phi^3 = 3 \phi^2 \delta \phi$, as expected by simple differentiation of $\phi^3$ wrto $\phi$.