Graduate Position is no more an operator in QFT

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SUMMARY

In quantum field theory (QFT), there is no universally accepted position operator, particularly for massless particles, due to the limitations imposed by relativistic frameworks. While the Newton-Wigner position operator can be defined for massive particles, it does not transform as a Lorentz vector, which complicates its interpretation. The discussion highlights that the concept of position loses significance in the context of many identical particles, where number density becomes more relevant. The Standard Model's local relativistic QFT does not accommodate a position operator in the traditional sense, emphasizing the need for precise definitions when discussing this topic.

PREREQUISITES
  • Understanding of quantum mechanics and quantum field theory (QFT)
  • Familiarity with the Newton-Wigner position operator
  • Knowledge of Lorentz transformations and their implications in physics
  • Basic grasp of the Standard Model of particle physics
NEXT STEPS
  • Research the implications of the Newton-Wigner position operator in quantum mechanics
  • Study the role of number density in quantum field theory
  • Explore the limitations of massless particle representations in QFT
  • Examine the relationship between position operators and Lorentz invariance
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Physicists, quantum mechanics researchers, and students of quantum field theory seeking to deepen their understanding of the position operator's role and limitations in relativistic frameworks.

Heidi
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In quantum mechanics there is no operator for time (problem with unbounded energy).
position is no more an operator in field theory. was there still a problem in QM?
 
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Say there are many particles of same kind, position lose its meaning and number density takes its palce.
 
In general there is no position operator in relativistic quantum theory, at least not within the only kind of relativistic QT that's successful in describing the real world in terms of the Standard Model, which is local relativistic QFT.

However, for all massive particles you can define a position operator having the usual properties. Since only massive particles have a useful non-relativistic limit, there is no contradiction between having a position operator in non-relativistic quantum theory and local relativistic QFT.

The representations of the Galilei group for massless particles doesn't lead to a physically interpretable quantum theory. See also my comment on this here:

https://www.physicsforums.com/threa...r-the-gravitational-field.997062/post-6433476
 
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Is there a position operator in QFT? The question does not make sense until one defines what exactly one means by "position operator". There is operator that satisfies some properties one would expect from a decent position operator, but not all. In particular, the Newton-Wigner position operator does not transform as a Lorentz vector.
 
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I was referring to the usual definition of position operators, as explained in

https://arnold-neumaier.at/physfaq/topics/position.html

It's of course not part of Lorentz four-vector operator, because otherwise time would be an operator too, but that cannot be by construction, because the energy spectrum is bounded from below.
 
Demystifier said:
Is there a position operator in QFT? The question does not make sense until one defines what exactly one means by "position operator". There is operator that satisfies some properties one would expect from a decent position operator, but not all. In particular, the Newton-Wigner position operator does not transform as a Lorentz vector.
Position in relativistic physics is an interesting thing. It happens that the classical position operator IS a Newton-Wigner operator and, also, does-not transform as a 4-vector (can't give a reference, is still in peer review).

Moreover, I think the Newton-Wigner position function (https://arxiv.org/abs/2004.09723) of Hamiltonian mechanics has the same property, though I'm not sure, I haven't read the article in full details.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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