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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 ?

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  • $\begingroup$ 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 $\delta\phi = \phi'(x')-\phi(x)$ $\endgroup$
    – M. M. R.
    Commented Mar 14, 2017 at 16:45
  • $\begingroup$ 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 ? $\endgroup$
    – StarBucK
    Commented Mar 14, 2017 at 21:09
  • $\begingroup$ Oh my god yes, sorry! I forgot to write the $\delta$ in front of $x$! $\endgroup$
    – M. M. R.
    Commented Mar 14, 2017 at 21:12

1 Answer 1

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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.

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  • $\begingroup$ I think what you mean is $\delta \phi = \phi'(x) - \phi(x)$ compares the transformed field to the original field at the same point $x$. The transformed field $\phi'(x)$ is moved. Suppose $\phi(x)$ is a sine wave, let $x = \pi/2$. Let $\phi'(x)$ be the transformed field by moving it to the right by some amount (assuming the new sine wave does not align with the old sine wave), clearly $\phi'(x)$ at $x = \pi/2$ will have a different value compared to $\phi(x)$ at $x = \pi/2$. $\endgroup$
    – mathemania
    Commented Nov 3 at 8:50

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