Field transformations in Weinberg's QToF

In summary, Weinberg's scalar field is different from the usual definition in that it transforms the fields actively.
  • #1
brodekind
5
0
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
[itex]
\Phi'(x') = \Phi(x) \text{ with } x' = \Lambda x
[/itex]
meaning
[itex]
\Phi'(x) = \Phi(\Lambda^{-1}x)
[/itex]
and the vector field as
[itex]
V'(x') = \Lambda V(x)
[/itex]
meaning
[itex]
V'(x) = \Lambda V(\Lambda^{-1}x)
[/itex]

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

So if I were to interpret [itex]U(\Lambda) \Phi(x) U(\Lambda)^{-1}[/itex] naturally as [itex]\Phi'(x)[/itex], 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.
 
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  • #2
What Weinberg calls x', other call x. So, you can take [itex]\Lambda^{ -1 }[/itex] of Weingerg's to be the [itex]\Lambda[/itex] of the other textbooks.
 
  • #3
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
Thanks for your inputs.
Well, I thought about active versus passive. But it seems to me that he is transforming the states actively. i.e.
[itex]
D(\Lambda) |p\rangle = \text{something}|\Lambda p\rangle,
[/itex]
only the fields passively. Next time in the library I will have a look at that book though.
 

Related to Field transformations in Weinberg's QToF

1. What is the significance of field transformations in Weinberg's QToF?

Field transformations play a crucial role in Weinberg's QToF theory as they allow for the study of how particles interact and evolve in different reference frames.

2. How are field transformations mathematically represented in Weinberg's QToF?

Field transformations in Weinberg's QToF are represented by a set of equations known as Lorentz transformations, which describe how physical quantities, such as energy and momentum, change under different reference frames.

3. Can you give an example of a field transformation in Weinberg's QToF?

One example of a field transformation in Weinberg's QToF is the transformation of a particle's energy and momentum from one reference frame to another. This allows for the study of how particles behave and interact in different frames of reference.

4. How do field transformations affect the behavior of particles in Weinberg's QToF?

Field transformations in Weinberg's QToF affect the behavior of particles by allowing for the study of how they interact and evolve in different reference frames, providing a more comprehensive understanding of their behavior and properties.

5. Are there any limitations to using field transformations in Weinberg's QToF?

While field transformations are a crucial aspect of Weinberg's QToF theory, there are certain limitations to their use. For instance, they are only applicable in the context of special relativity and may not accurately describe the behavior of particles in extreme conditions such as those found in black holes.

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