0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\phi\to\phi^\prime$? $\endgroup$ – Orient Nov 27 '17 at 8:53
  • $\begingroup$ 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. $\endgroup$ – Ari Nov 28 '17 at 4:06
-1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.