Commutation relations for an interacting scalar field

Click For Summary
SUMMARY

The discussion centers on the application of canonical commutation relations for free scalar fields to interacting fields in quantum field theory (QFT), as presented in Schwartz's "Quantum Field Theory and the Standard Model." Participants clarify that while the Hilbert space for interacting theories may seem different due to varying states, it remains the same as that of free theories, primarily due to the introduction of a different Lagrangian rather than a change in states. However, Haag's theorem indicates that a rigorous description of interacting quantum fields cannot rely on the same Hilbert space as free fields. For further understanding, A. Duncan's book is recommended for insights into perturbative treatments in QFT.

PREREQUISITES
  • Understanding of canonical commutation relations in quantum mechanics
  • Familiarity with Lagrangian mechanics in quantum field theory
  • Knowledge of Haag's theorem and its implications in QFT
  • Basic concepts of quantum electrodynamics (QED)
NEXT STEPS
  • Read "Quantum Field Theory and the Standard Model" by Schwartz for foundational concepts
  • Study Haag's theorem and its implications for interacting quantum fields
  • Explore "The Conceptual Framework of Quantum Field Theory" by A. Duncan for perturbative treatments
  • Investigate the role of Lagrangians in quantum field theories, focusing on interactions
USEFUL FOR

Physicists, particularly those specializing in quantum field theory, graduate students in theoretical physics, and researchers exploring the foundations of quantum mechanics and field interactions.

eoghan
Messages
201
Reaction score
7
TL;DR
Why are the commutation relations for interacting fields assumed to be the same as that of free fields?
Hi there,

In his book "Quantum field theory and the standard model", Schwartz assumes that the canonical commutation relations for a free scalar field also apply to interacting fields (page 79, section 7.1). As a justification he states:

This is a natural assumption, since at any given time the Hilbert space for the interacting theory is the same as that of a free theory.

I do not understand this explanation. Can you please elaborate?

I mean, how can the Hilbert space of the interacting theory be the same as the one of the free theory? In an interacting theory, I expect to have different states than in a non interacting theory, so the Hilbert space should be different, isn't it?
 
  • Like
Likes   Reactions: dextercioby, vanhees71 and PeroK
Physics news on Phys.org
eoghan said:
In an interacting theory, I expect to have different states than in a non interacting theory
Not necessarily, just a different Lagrangian. For example, consider a non-interacting theory of electrons and photons. It would have electron states and photon states, with a Lagrangian that had a kinetic term for electrons, a mass term for electrons, and a kinetic term for photons.

If we now add the electron-photon interaction to this theory, we add the QED coupling term to the Lagrangian, but we don't change any states: there are still electron states and photon states. All we have done is introduced new possibilities for transitions between states, such as scattering processes.

eoghan said:
the Hilbert space should be different
Not for the reason you give; but Haag's theorem and related results do, in fact, show that you cannot describe interacting quantum fields using the same Hilbert space that describes free quantum fields. Most treatments of QFT seem to more or less ignore these results and assume that there is some rigorous mathematical basis for the methods they describe.
 
  • Like
Likes   Reactions: eoghan, vanhees71 and PeroK
If you are interested in, why FAPP one can ignore Haag's theorem in the usual perturbative treatment of QFTs, see

A. Duncan, The conceptual framework of quantum field theory, Oxford University Press, Oxford (2012).
 
  • Like
Likes   Reactions: dextercioby
@PeterDonis Very nice answer, now it is clear to me why the Hilbert space (Haag's theorem apart) is supposed to be the same.

@vanhees71 Nice book. I just had a look at its content on Amazon and it's definitely on my list once I will be done with the Schwartz!
 
  • Like
Likes   Reactions: vanhees71 and PeroK

Similar threads

  • · Replies 1 ·
Replies
1
Views
748
  • · Replies 1 ·
Replies
1
Views
973
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 61 ·
3
Replies
61
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 75 ·
3
Replies
75
Views
10K