Active and passive transformation in field theory

Question

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
\begin{equation} x^\nu\to x^\nu-\epsilon^\nu\quad\Rightarrow\quad\phi(x)+\epsilon^\nu\partial_\nu\phi(x) \end{equation}(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?

Answer 1

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

Whereas if the transformation was passive the fields would be written as :

$\phi(x') = \phi(x) - \epsilon^\mu \partial_\mu \phi(x)$.

The only difference is the position of $'$ in both the cases. I hope this answers your questions.

Answer 2

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.