Field transformations in Weinberg's QToF

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Discussion Overview

The discussion revolves around the definitions and transformations of scalar and vector fields as presented in Weinberg's Quantum Field Theory (QFT) compared to other texts. Participants explore the implications of these definitions on the understanding of Lorentz transformations in the context of quantum field operators versus classical fields.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant notes a discrepancy between Weinberg's definition of scalar fields and those found in other texts, highlighting the transformation of fields as operator fields rather than classical fields.
  • Another participant suggests that the difference in notation may stem from the distinction between active and passive viewpoints in describing symmetries of space-time.
  • A further contribution questions whether Weinberg's approach actively transforms states while passively transforming fields, indicating a potential inconsistency in the treatment of transformations.

Areas of Agreement / Disagreement

Participants express differing interpretations of Weinberg's definitions and their implications, indicating that multiple competing views remain without a clear consensus on the matter.

Contextual Notes

Participants reference the need to consider the definitions and contexts in which transformations are applied, suggesting that assumptions about the nature of fields (operator vs. classical) may influence interpretations.

brodekind
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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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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.
 
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.
 
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.
 

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