# Infinitesimal transformation of fields

In my QFT course, we are doing some infinitesimal transformations of scalar fields.

We do the following :

$$\phi'(x')=\phi'(x+\delta x) =\phi'(x)+\delta x^\mu \partial_\mu \phi(x)$$

But i don't get why it is $\partial_\mu \phi(x)$ and not $\partial_\mu \phi'(x)$ ?

Why would the derivative of $\phi'$ be the same as the derivative of $\phi$ ?

Is it because $\phi'=\phi+\delta \phi$ and we only keep the first order terms ? So $\delta x^\mu \partial_\mu \phi'(x)=\delta x^\mu \partial_\mu (\phi+\delta \phi)(x)=\delta x^\mu \partial_\mu \phi(x)$ at first order ?

• Yes you're right. If you call $\delta_0\phi(x) = \phi'(x)-\phi(x)$, than $\delta x^\mu\partial_\mu\phi'(x) = x^\mu\partial_\mu(\phi(x)+\delta_0\phi(x)) = x^\mu\partial_\mu\phi(x)$ at first order. In our course we defined $\delta_0\phi$ as form variation, in order to distinguish it from the value variation $\delta\phi = \phi'(x')-\phi(x)$ – M. M. R. Mar 14 '17 at 16:45
• You mean $\delta x^\mu \partial_\mu \phi'(x)=\delta x^\mu \partial_\mu (\phi(x)+\delta_0 \phi(x))=\delta x^\mu \partial_\mu \phi(x)$ instead of $\delta x^\mu \partial_\mu \phi'(x)=x^\mu \partial_\mu (\phi(x)+\delta_0 \phi(x))=x^\mu \partial_\mu \phi(x)$ right ? – StarBucK Mar 14 '17 at 21:09
• Oh my god yes, sorry! I forgot to write the $\delta$ in front of $x$! – M. M. R. Mar 14 '17 at 21:12

Your conclusion is correct: $$\delta \phi = \phi'(x) - \phi(x),$$ involves a variation in which we compare the field at two distinct points relative to the same coordinate system. To first order, or for an infinitesimal transformation, these are the same and we get the desired equality.