# Field transformations in Weinberg's QToF

1. May 23, 2013

### brodekind

Ahoy,

I was reading parts of Weinberg's QFT book vol. I and was surprised at his definition of a scalar field or Lorentz transformations on fields in general. Usually (e.g. Maggiore, Modern Intro to QFT) I see the scalar field defined as Lorentz transforming via
$\Phi'(x') = \Phi(x) \text{ with } x' = \Lambda x$
meaning
$\Phi'(x) = \Phi(\Lambda^{-1}x)$
and the vector field as
$V'(x') = \Lambda V(x)$
meaning
$V'(x) = \Lambda V(\Lambda^{-1}x)$

Weinberg on the other hand defines a scalar field (eq. (5.1.2)) as
$U(\Lambda) \Phi(x) U(\Lambda)^{-1} = \Phi(\Lambda x)$
and in general in eqs. (5.1.16), (5.1.17)
$U(\Lambda) \Psi(x) U(\Lambda)^{-1} = D(\Lambda^{-1}) \Psi(\Lambda x)$
with $D$ a transformation matrix corresponding to the representation of the Lorentz group $\Psi$ furnishes.

So if I were to interpret $U(\Lambda) \Phi(x) U(\Lambda)^{-1}$ naturally as $\Phi'(x)$, I see that Weinberg transforms contrarily to the usual definition with the primes. Why is this sensible? Is it related to the 'prime' definition acting on classical fields, whereas Weinberg's definition is on operator fields?
I can't wrap my head around it. All books that I know of transform the states the same way, so there is a problem if they transform the field operators contrarily. Having thought about it for some time, I am unable to resolve this (apparent) paradox and would be glad if somebody could shed some light on this issue.

2. May 23, 2013

### samalkhaiat

What Weinberg calls x', other call x. So, you can take $\Lambda^{ -1 }$ of Weingerg's to be the $\Lambda$ of the other textbooks.

3. May 23, 2013

### dextercioby

It's the so-called <active> vs. <passive> point of view when describing symmetries of space-time. Read the relevant discussion in Fonda and Ghirardi's 1970 text on quantum symmetries.

4. May 24, 2013

### brodekind

$D(\Lambda) |p\rangle = \text{something}|\Lambda p\rangle,$