Position Representation in QFT?

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Discussion Overview

The discussion centers around the representation of position in Quantum Field Theory (QFT) compared to Quantum Mechanics (QM). Participants explore how position measurements are modeled in QFT, given that position is not treated as an observable in the same way it is in QM. The scope includes theoretical considerations and conceptual clarifications regarding the nature of position in these frameworks.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that position is not an observable in QFT, leading to questions about how position measurements are represented or modeled.
  • Others propose that a position operator and position representation can be defined in relativistic QFT, similar to QM, referencing specific literature for further exploration.
  • A participant emphasizes the distinction between the 'x' in QM as an operator and the 'x' in QFT as a continuous index, suggesting that this difference is significant for understanding position representation.
  • Some participants argue that the field argument 'x' in QFT does not relate to the position observable and its eigenvalues, proposing that a true position operator should be constructed using the Newton-Wigner recipe.
  • Another viewpoint suggests that it may be possible to reformulate QFT such that 'x' becomes an operator, akin to many-particle QM, referencing specific sections of literature to support this claim.
  • One participant reiterates that position cannot be measured or modeled in QFT, asserting that initial position dependence must be removed from equations before practical application, highlighting the relevance of the S-matrix in QFT.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of position in QFT, with some agreeing that position is not an observable while others propose alternative formulations. The discussion remains unresolved, with multiple competing perspectives on how position can be represented in QFT.

Contextual Notes

There are limitations regarding the assumptions made about the nature of position in QFT and the dependence on specific definitions of observables. The discussion does not resolve the mathematical steps involved in these formulations.

LarryS
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I have read that, in QFT, unlike QM, there is no position probability density function because position is not considered an observable.

Then how is a position measurement represented/modeled in QFT?

As always, thanks in advance.
 
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Position operator and position representation can be defined in relativistic QFT just as well as in ordinary quantum mechanics. You may want to check

E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?", Found. Phys., 32 (2002), 673.

before forming your own opinion.

Eugene.
 
Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
 
tom.stoer said:
Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.

I agree completely, the field argument 'x' in QFT has no relationship to the position observable and its eigenvalues. The true position operator should be built by the Newton-Wigner recipe.

Eugene.
 
tom.stoer said:
Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
Even though it is true in the standard formulation of QFT, I think there is a way to reformulate QFT such that x becomes an operator just as in many-particle QM. See Sec. 3 of
http://xxx.lanl.gov/abs/0904.2287 [Int. J. Mod. Phys. A25:1477-1505, 2010]
 
referframe said:
I have read that, in QFT, unlike QM, there is no position probability density function because position is not considered an observable.

Then how is a position measurement represented/modeled in QFT?

As always, thanks in advance.

As you well say, position is not an observable in QFT and cannot be measured/modeled. As a consequence, any initial position dependence must be eliminated from the equations before going to the lab, and this is the reason for which the only physically relevant QFT object is the S-matrix, which does not depend of position (spatial coordinates are integrated out and eliminated from the matrix).

As Mandl and Shaw note in his classic QFT textbook, this is a fundamental difference between QFT and QM.
 
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