# Active and passive transformation in field theory

From last paragraph of page 14 to first line of page 15 of this pdf (http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf) as below:

Consider the infinitesimal translation
$$x^\nu\to x^\nu-\epsilon^\nu\quad\Rightarrow\quad\phi(x)+\epsilon^\nu\partial_\nu\phi(x)$$(where the sign in the field transformation is plus, instead of minus, because we’re doing an active, as opposed to passive, transformation).

So $x$ is changed to $x^\nu-\epsilon^\nu$ and field configuration itself isn't changed ($\phi\to\phi$).
Active transformation is when we move a particle (field, this case, i think) instead of move coordinates which is passive, so that this is passive transformation, i think.

Why is this transformation active?

Consider $x^{'\mu} = x^\mu - \epsilon^\mu$. The active transformation of fields would require the change of $\phi(x)$ to $\phi'(x)$. The new field at x is given by $\phi'(x) = \phi(x) + \epsilon^\mu \partial_\mu \phi(x)$.
$\phi(x') = \phi(x) - \epsilon^\mu \partial_\mu \phi(x)$.
The only difference is the position of $'$ in both the cases. I hope this answers your questions.
• What is meaning of $x\to x^\prime=x-\epsilon$? Is it that translate all of some kine of "object" at $x$ to $x^\prime$ or that translate all of "coordinate" $x$ to $x^\prime$? And what is meaning of the change of $\phi\to\phi^\prime$? I think I understand the case of $\phi(x)\to\phi(x^\prime)$. It just change of $x\to x^\prime$ and field isn't changed. But what is difference between this and the case of $\phi\to\phi^\prime$? – Orient Nov 27 '17 at 8:53
• The difference between the cases comes when you take into account internal transformations(e.g. gauge transformations). There $\phi \rightarrow \phi'$ but x remains unchanged. – Ari Nov 28 '17 at 4:06
The way that we have to define $$\mathcal{L}_L\psi(x)=\psi(L^{-1}x)$$ is to guarantee the isomorphism of Lorentz group L, where $$\psi(x)$$ bears a representation of that group.