Why can't we interprete /x> in relativistic QFT as position eigenfunc?

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SUMMARY

The discussion centers on the interpretation of the state |x⟩ in relativistic Quantum Field Theory (QFT) and its distinction from the state |1x⟩ = Φ(field operator)(x)|0⟩. In relativistic QFT, the normalization of the field operator Φ is adjusted to ensure Lorentz invariance, resulting in |1x⟩ being fundamentally different from |x⟩. This difference is highlighted by the fact that the function described in equation (59) of the referenced paper is Lorentz invariant but does not conform to a δ-function representation.

PREREQUISITES
  • Understanding of Quantum Field Theory (QFT)
  • Familiarity with Lorentz invariance in physics
  • Knowledge of field operators in quantum mechanics
  • Basic concepts of eigenstates and eigenfunctions
NEXT STEPS
  • Study the implications of Lorentz invariance in Quantum Field Theory
  • Examine the differences between non-relativistic and relativistic QFT
  • Read the paper referenced (http://de.arxiv.org/abs/hep-th/0202204) focusing on equations (55)-(60)
  • Explore the mathematical properties of δ-functions in quantum mechanics
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Physicists, graduate students in theoretical physics, and researchers interested in the foundations of Quantum Field Theory and relativistic quantum mechanics.

fxdung
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Why can't we interprete /x> in relativistic QFT as position eigenstate?And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?
 
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fxdung said:
And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?
In non-relativistic QFT it's the same. In relativistic QFT one changes the normalization of ##\phi## so that it becomes Lorentz invariant, the consequence of which is that ##|1x\rangle## differs from ##|x\rangle##. See e.g. my
http://de.arxiv.org/abs/hep-th/0202204
Eqs. (55)-(60). In particular, the function (59) is Lorentz invariant, but is not a ##\delta##-function.
 

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